Chapter VI Various Types of Eulerian Trails

VI. 1

Chapter VI

VARIOUS TYPES OF EULERIAN TRAILS

Let us consider the proof of Theorem IV.l where we applied the Splitting Lemma to reach the conclusion that a connected eulerian graph G has an eulerian trail T . If we apply the Splitting Lemma systematically, then we can produce such T (see the chapter on algorithms for finding eulerian trails, but also Corollary V.13). However, the Splitting Lemma tells us more than that: namely, that we are not too restricted at any given stage in producing an eulerian trail, whence we conclude that G must have many eulerian trails, and that one might impose restrictions (apart from the nuturd restriction expressed by the Splitting Lemma) on the course of some eulerian trail of G. This chapter deals precisely with the question as to which type of restrictions can be imposed.

VI. 1. Eulerian Trails Avoiding Certain

Transitions

In this section we deal first with a restriction which acts locally and is purely set theoretical, so to say. Consider the set E,' for an arbitrary vertex Y of an eulerian connected graph G. How many transitions {e(v),f(v)} can be chosen as forbidden transitions at 'u where e and f are arbitrary elements of E, ? In other words: what is a necessary and sufficient condition for the existence of an eulerian trail T of G such that T avoids any of the prescribed transitions ? Kotzig, [KOTZ68a, Theorem 11, answered this question in the following theorem for which we recall Definition 111.41.

Theorem VI.l. Let G be a connected eulerian graph, and for every v E V(G),let P ( v ) be a partition of E,*.The following statements are equivalent.

VI. Various Types of Eulerian Trails

VI.2

1) G has a P(G)-compatible eulerian trail.

I I<

2) For every v E V ( G ) and every class C E P ( v ) C P ( G ) , C $dG(V).

Proof. 1) implies 2). Let T be a P(G)-compatible eulerian trail of G. Consider for arbitrary v E V ( G )and any C E P ( v ) all sections e , u,f of T for which C n {e(v), f(v)} # 0. Since T is P(G)-compatible we conclude e ( u ) E C if and only if f ( u ) $Z C ; i.e., I C n {e(v), f(v)} I= 1. Since every edge of G appears precisely once in T we conclude I C I< $ d G ( v ) . 2) implies 1). We proceed by induction on a(G) and observe that the implication is true for a(G) = 0. For, a(G) = 0 implies I C I= 1 for every C E P(G), whence we conclude that a run through the cycle G is necessarily P(G)-compatible. Hence suppose a(G) > 0.

Now choose E V ( G ) with dG(v) > 2, and choose C, E P ( v ) such that I C, I= maz{l C I / C E P(v)}. Take any el(v) E C,, and let e,(v),e,(v) be any two distinct elements of E: - C,; they exist because of the choice of v and I C, I< idG(”). Applying the Splitting Lemma we conclude w.1.o.g. that G,,, is connected (see conclusion 2) of the proof of that lemma). Let C2 E P ( v ) denote the class containing e2(u) and let P - ( v ) := P ( v ) - {C,, C 2 } .Now we define

P‘(w) := P ( w ) P’(u1,2)

for every

:= {{e,(v>h

w E V(G,,,) - {v, v , , ~ }

{e2(4H >1

W 1 , 2 )

=

u

>IC2 I= 1 =IC2 I= 1

P‘(4

~ E V ( G,2I 1

-

We observe the following facts regarding G and P ( G ) ;they follow from statement 2) and the choice of v,el, e2.

a) For every w E V(G,,,) - {v,v,,,} and every C’ E P ( w ) c P(G) we haveIC’I< $ d G ( w ) = $dG,,,(w).

b) I C” I > I C, - {el (v)} I for some C” E PI(.) implies C” E P(u) and I cz151ct’ c,I
VI. 1. Eulerian Trails Avoiding Certain Transitions

VI.3

We conclude from a) and b) that statement 2 ) is fulfilled regarding Gl,2 and P(G1,2).Furthermore, a(G,,,) < a(G). The theorem now follows by induction.

Remark. Kotzig says in [KOTZ68a] that his Theorem 1 (i.e. Theorem VI.l) is the answer to a question posed by Nash-Williams at the Graph Theory Colloquium held at Tihany in 1966. However, the proceedings of that Colloquium do not contain any reference to this or a similar problem posed by Nash-Williams (who, in a private communication, told the author of this book that he does not recall ever having posed this problem). But in this case, it is irrelevant who posed the problem; what counts are the various directions into which this theorem points, if one makes special assumptions concerning G and P(G). The following result [ItOTZ56a] is partly a consequence of Theorem VI.l and partly a generalization of Petersen’s method to show that a 4-regular graph has a 2-factorization7 [PETESla]. It was originally proved without the use of Theorem VI.l (which was published only years later), just as the original proof of Theorem VI.l does not make use of the Splitting Lemma.

Corollary VI.2. Let G be a connected eulerian graph with an even number of edges. Then the following is true: 1) If T is an eulerian trail of G, then there are two edge-disjoint spanning subgraphs of G, G, and G, say, such that

and the edges of G, and G, alternate in T . 2) If G is the edge-disjoint union of two subgraphs G, and G, with dG,(u) = d G b ( u )for every u E V ( G ) ,then G has an eulerian trail in

which edges of G, and G, alternate.

Proof. A) Coloring the edges of T alternatingly with T (red) and b (blue)

as one runs through T,yields the subgraphs G, and G, as described in 1) (for u 0 , the initial and end vertex of T , we have dGr(u0) = dGa(uO)

precisely because I E ( G )I is even).

B) Part 2) of the corollary follows from Theorem VI.l by observing that every P ( u ) consists of two classes of equal size.

VI. Various Types of Eulerian Trails

VI.4

From part A) of the proof of Corollary VI.2 we deduce directly an important result contained in Petersen’s famous paper [PETESla].

Corollary VI.3. Every 2k-regular graph having an even number of edges can be decomposed into two k-factors. Consequently, every 4-regular graph can be decomposed into two 2-factors. Theorem VI.l can be generalized in a way similar to the generalization of Theorem IV.l expressed in Corollary V.2. This has been done in [KOTZSSa] as well.

Corollary VI.4. Let G be a graph such that every component of G contains at least one odd vertex, and let P ( T Jbe ) a partition of E: for every TJ E V(G). The following statements are equivalent. 1) G has a P(G)-compatible decomposition into p open trails (where 2p is the number of odd vertices of G).

2) For every

ICl5

TJ

E V(G) and every class

C

E P(TJ) c P(G),

+dG(?J)).

Proof. Since every component of G contains odd vertices by hypothesis, it suffices to prove Corollary VI.4 for connected graphs with odd vertices.

As in the proof of Corollary V.2, form an eulerian graph Go from G by = joining pairs of odd vertices by an additional edge each. Define Po(v) P ( v ) for even vertices TJ of G; and for odd v , define P,(v)= P ( v ) U {{eo(v)}} where eo(u)corresponds to the edge eo E E(Go)-E(G) joining TJ and some other odd vertex of G. W O ) = U V E V ( G )P0 (TJ)satisfies statement 2) of Theorem VI.l if and only if P ( G ) satisfies statement 2 ) of Corollary VI.4. On the other hand, a P(Go)-compatible eulerian trail T of Go corresponds to a P(G)compatible decomposition S of G into p open trails, and vice versa. These equivalences and the validity of Theorem VI.l imply the validity of Corollary VI.4. Unfortunately, the parallelisms between Theorem IV.l and Theorem VI.l, and between Corollary V.2 and Corollary VI.4, cannot be extended to a statement on P(G)-compatible path/cycle decompositions following Corollary V.3. This is the case, even if P(G) = X ( G ) is a system of transitions. Later on in the considerations on compatibility (see the third volume of this monograph), we shall discuss certain examples.

VI. 1. Eulerian Trails Avoiding Certain Transitions

VI.5

Next we specify P(G)for the connected eulerian graph G without 2valent vertices, by assuming that P(G) is a system of transitions X ( G ) (see Definition 111.41). Thus we have the next corollary as a trivial consequence of Theorem VI.1; it can be proved, however, by simply using the Splitting Lemma (i.e., without relying on Theorem VI.l). Corollary VI.5. Let G be a connected eulerian graph without 2-valeat vertices, and let X = X ( G ) be a system of transitions of G. Then an eulerian trail T exists which is compatible with X . In particular, T exists if X = X,, is the system of transitions induced by an eulerian trail T’ of G. Consequently, we can say that if S(G) > 2, then G has at least two mutually compatible eulerian trails. But what is the largest number of mutually compatible eulerian trails in a connected eulerian graph G with S(G) > 2 ? We shall deal with this question in the next section of this chapter. We further specialize by assuming that G is a 4-regular graph embedded in some surface F ,and that X ( v ) c X ( G ) is a pair of intersecting transitions for every v E V ( G )(see Definition III.42a). Corollary VI.6.If G is a connected 4-regular graph embedded in some surface 3,then G has an A-trail.

Proof. Choose X ( G ) to be a system of intersecting transitions (see above), and apply Corollary VI.5. The edge-sequence 1,2,. . ., 1 2 in Figure V.l gives an example of an Atrail of the octahedron. However, not only can Corollary VI.6 be proved directly by applying the Splitting Lemma (Exercise VI.3), but it is also a consequence of the following result which can also be obtained by applying the Splitting Lemma (Exercise VI.4). Lemma VI.7. Let G be a connected eulerian graph embedded in some surface 3. Then G has a non-intersecting eulerian trail.

It follows from the definitions of an A-trail, respectively of a non-

intersecting eulerian trail (see Definition III.42a) that these two concepts coincide in the 4-regular case (or, more generally, if G is an eulerian graph with A ( G ) < 6). However, as we shall see in the section on A-trails, there exist 2-connected plane graphs having 4- and 6-valent vertices only, without A-trails (but they do have non-intersecting eulerian trails by Lemma VI. 7).

VI.6

VI. Various Types of Eulerian Trails

We observe that Corollary VI.6 and Lemma VI.7 are folklore results for the planar case (see, for example, [TUTT4la], [KOTZ68c, Theorem lo], and [HARA69a, Exercise 11.21). For graphs embedded in some arbitrary surface F,Belyj [BELY83a] proves a result which does not rely on surface embeddings as such but rather on a predefined cyclic order of E,* for every v E V ( G )(this is implicit in Belyj’s paper, although he does not mention it - therefore his definitions lack clarity to some extent). However, both Corollary VI.6 and Lemma VI.7 are trivial consequences of the Splitting Lemma, as is the following lemma (from which the above two results follow immediately) which is the essence of Belyj’s paper, (see [BELY83a: Theorem 11); we leave its proof to the reader (see Exercise VI.4).

Lemma V I A . Let G be a connected eulerian graph with a prescribed order O+(E,*)of the half-edges incident with 21, for every 21 E V ( G ) .Then G has a non-intersecting eulerian trail. Remark. As a matter of historical record, we note that Kotzig, apparently unaware of [TUTT4la], proves Corollary VI.6 for 3 being the plane (calling a-line what is called here A-trail) by applying Corollary VI.5 restricted to 4-regular graphs (see [KOTZ68c, Theorem 31). On the other hand, the paper [TUTT4la] starts with the statement “Consider a plane network v at all of whose nodes exactly 4 lines meet. By Euler’s theorem, it is possible to find unicursal paths in such a network which do not cross themselves ...” (network = graph, unicursal path = eulerian trail, H.F.). This statement, while correct (see Corollary VI.6), does not automatically follow from Theorem IV.l (see also Euler’s original paper). Lemma VI.8 has its generalizations for arbitrary graphs as well; we leave their proofs as exercises (their statements in [BELY83a] are less general than the ones given here, but it is apparent that the author had in mind the more general statements - see his Definitions 6 and 7).

Corollary VI.9. Let G be a connected graph having precisely two odd vertices, and let O+(E,*)be given as in Lemma VI.8. Then G has a non-intersecting open covering trail. Corollary VI.10. Let G be a connected graph, and let O+(E:) be given as in Lemma VI.8. Then G has a decomposition into n mutually nonintersecting open non-intersecting trails (where 2 n is the number of odd vertices of G).

Belyj’s definitions are a little bit foggy as he does not use any O+(E:) but simply speaks of the vertices q,. . . ,21, incident with v in G (in

VI.l.l. P(D)-Compatible Eulerian Trails in Digraphs

VI.7

his Definition 1, he replaces v with the n-gon v1v2 - . - v " and lets vi be adjacent to v' - which is problematical, the more so as he allows G to have loops and multiple edges). Consequently, his Theorem 2 is false in the non-planar case: trying to establish an analogue to Lemma VI.8 for eulerian digraphs he claims that such graphs have a non-intersecting eulerian trail i f and only i f at every vertex v, arcs incident to v alternate with arcs incident f r o m v. Figure VI.1 shows an eulerian orientation D2,4 of the K2 which, if embedded on the torus or on the projective plane, yields a non-intersecting eulerian trail contradicting Belyj's statement. I

Y

Figure VI.l. An A-trail of 0, where in- and out-going arcs are not alternating in the vertices x and y. I

VI.l.l. P(D)-Compatible Eulerian Trails in Digraphs We now try to rephrase some of the preceding results for digraphs. The digraph D,, of Figure VI.l shows that Theorem VI.l has no simple analogue fo; digraphs: for, if we define p(D2,4)

= x(D2,4)

= {{l,8), {4,5 ) , {2, 7)> {3, 6))

>

then there is no eulerian trail of D2,4 compatible with X ( D 2 , 4 ) .This should not be surprising, though, since - generally speaking - for every arc e incident to some vertex v of the connected eulerian digraph D ,there

VI.8

VI. Various Types of Eulerian Trails

are id(v) - 1 = i d ( u ) - 1 arcs which cannot be (immediately) consecutive to e in any eulerian trail of D (namely, all arcs incident to u and different frorn e ) . That is, the fact that D is a digraph and not a graph, in itself imposes a certain partition P(u) of At for every u E V ( D ) - namely the one consisting of two classes one of which is ( A t ) + ,the other being (At)-. For this ‘natural’ partition P ( v ) and the corresponding P ( D ) , a P(D)-compatible eulerian trail is just any eulerian trail of D.

On the other hand, for an arbitrary P ( D ) , the existence of a P(D)compatible eulerian trail implies statement 2) of Theorem VI.1 (with D replacing G in that statement). Figure VI.2 below shows, however, that this statement is not sufficient for the existence of a P(D)-compatible eulerian trail.

So, let us consider an arbitrary partition system P ( D ) of D. Because of the Splitting Lemma (applied to D such that D,,, and Dl,3are eulerian where e , is an arc incident to u - hence e 2 , e3 are arcs incident from u),one may suspect that if for every u E V ( G ) and any half-arc e- E C- E P ( u ) C P ( D ) (where e - is incident to u),there are two half-arcs e r , e l E C+ E P(u) (where ef ,e l are incident from TJ), and C+ # C - , then D has a P(D)-compatible eulerian trail. In other words, if P ( D ) leaves, at every vertex u,for every e- E (At)- (at least) two choices to define a transition { e - , e t } , i = 1,2, is this sufficient for the esistence of a P(D)-compatible eulerian trail of D ? The answer is negative as one can see from the digraph of Figure VI.2 by trial and error. However, the following result for digraphs (which seems to be new) can be considered an analogue to Theorem VI.l.

Theorem VI.ll. Let D be a connected eulerian digraph, and for every u E V ( D ) ,let P ( u ) be a partition of A:. If D has a P(D)-compatible eulerian trail, then 1)for every u E V ( D )and every C E P(u) c P ( D ) ,lCl< $d,(v). If statement 1) holds true and if P ( D ) satisfies 2)for every u E V ( D )with d,(u) > 2, IP(u)I> id,(u) then D has a P(D)-compatible eulerian trail.

+ 2,

Proof. It follows from the discussion preceding Figure VI.2 that it suffices to prove that the statements 1) and 2 ) imply the existence of a P(D)-compatible eulerian trail in D.

VI.l.l. P(D)-Compatible Eulerian Trails in Digraphs

VI. 9

Y

X

Figure VI.2. A 3-regular digraph together with a system of transitions X (marked by little arcs) having no eulerian trail compatible with X. Consider an arbitrary v E V ( D ) with d,(v) > 2. Statement 2) implies even for the largest class C, E P(v) that I C, I< $d,(v) (hence statement 1) is relevant only for the 2-valent vertices). Consequently, if C, is not affected by the splitting procedure yielding D1,2,then we still have for Cl E PI(.) c P ( D 1 , 2 )I,C, 1 $ d D ( v ) - 1 = gdD,,,(v) (compare this notation with the one used in the proof of Theorem VI.l). Furthermore, statement 2) implies that for such v E V ( D ) ,I P ( v ) 12 4 holds. Consequently, we deduce the following property: 3) I f d D ( v )2 4 for some a) f o r every

21

E V ( D ) ,then either

C E P(zJ), C f l (A:)+ # 0 if and onZy if C n (A:)- = 0

or

b) C,,,C’,C’’ E P ( v ) ezist with C, n ( A ; ) + # 0 # C, n ( A ; ) - and C’i l (A;)+ = C” n (A:)+ = 0 (note: I ( A ; ) + I= id,(v) and I P ( 4 I> $,(v) + 2). Now we assume the theorem to be false and choose a counterexample D with minimal a ( D ) 2. 0. The case a ( D ) = 0 being treated the same way as in the proof of Theorem VI.l (just replace the letter D with the letter G), we conclude from the resulting contradiction that some v E V ( D )

VI. 10

VI. Various Types of Eulerian Trails

is not 2-valent. Choose such a v and consider for P ( v ) the two cases corresponding to 3)a) and 3)b) above.

In the first case we simply apply repeatedly the Splitting Lemma until v has been replaced with k = id,(v) 2-valent vertices, and the digraph D’ obtained is connected and eulerian. We choose the notation in such a way that we have for these k 2-valent vertices P’(vii) = { { u ; ( v ) } , {u:(v)}} with u i ( v ) E (A:)- and u f ( v ) E (A;)+ for i = 1,- - - ,k. Since a(D’) < a ( D ) we conclude that for P(D’) = ( P ( D )- P ( v ) )U lJf.-l P’(vii), a P(D’)-compatible eulerian trail of D’ exists which corresponds to a P(D)compatible eulerian trail of D because of the construction of P(D’). In the second case let ul(v) E Con (At)+,u 2 ( v ) E C’ n (At)-, u3(v) E C” n (At)- be arbitrarily chosen. By the proof of the Splitting Lemma, D1,2or Dl,3 is connected (both are eulerian anyway); w.1.o.g. Dl,2 is connected. Let P’(w)for w # v,v1,2,P’(v1,2), P’(v) and P(Dl 2 ) be defined the same way as we did in the proof of Theorem VI.l. From the discussion preceding 3) we deduce that P(D1,2)satisfies statement 1), and the choice of Co implies

i.e., P(D,,,) satisfies statement 2 ) as well (observe that 3)b) implies d D ( v ) 2 6). From this and c ( D , , ~ )< a ( D ) we deduce the existence of a P(Dl 2)-compatible eulerian trail in D,,, which, by construction, corresponds to a P(D)-compatible eulerian trail of D. In both cases 3)a) and 3)b) we obtained a P(D)-compatible eulerian trail of D , which contradicts the choice of D. Theorem VI.ll now follows.

The lower bound in the inequality of statement 2) of Theorem VI.ll is best possible; this is demonstrated by the digraph Do of Figure VI.3 and its reduction to Dh. Looking at any 4-valent vertex v of Do, it is clear that for I P ( v ) I= 3 the splitting of v satisfying P(D,)-compatibility is uniquely determined. And this is, basically, the reason why one cannot improve the lower bound in statement 2 ) without further assumptions concerning the structure of some arbitrary eulerian digraph D and concerning P ( D ) . For, by applying sequences of splitting procedures according to the Splitting Lemma, one may arrive at a connected eulerian digraph with just two 4-valent

VI.l. 1. P(D)-Compatible Eulerian Trails in Digraphs

VI. 11

vertices and all the other vertices being 2-valent, and a system of transitions analogous to X ( D 2 , , ) as defined in the discussion following Figure VI.l. It follows from the above that replacing (4,5) with (4)) (5) and (3,6} with {3), (6) in X ( D 2 , 4 )does not alter the problem. And if we alter P ( x ) in Figure VI.3 by letting Ci = C, UC,, C; = C2n ( A : ) - , Ci = C, n (A:)+, Ci = C,, and P’(x) = {Ci,C;, C;, C;} (where Ciis represented in Figure VI.3 by the arcs marked i), then for

P’(D) = ( P ( D )- P ( z ) )u P y x )

,

there is no P‘(D)-compatible eulerian trail of D either.

Figure VI.3. Do and P(D,) (where the different classes of P(v) are represented by different numbers for every v E V ( D o ) )and its unique transformation to Dh and P(D6) following P(Do)-compatibility. Dt, has no P(D6)-compatible eulerian trail; therefore, Do has no P(Do)-compatible eulerian trail.

So, let us specialize concerning the structure of D and the structure of P(D)* Definition VI.12. For a digraph D with given partition system P ( D ) , we say P ( D ) is almost transitional if and only if for every v E V(D)

VI. Various Types of Eulerian Trails

VI. 12

and every C E P ( T Jc) P ( D ) , the inequality I C 15 2 holds. In this case we write X * ( D )instead of P ( D ) and X * ( v )instead of P(TJ). Corollary VI.13. If we replace in Theorem VI.ll P ( D ) with X * ( D ) (where X * ( D ) is an almost transitional partition system), and 2) with 2’) every

id,(.)

TJ

E V ( D )with d,(v) > 2 satisfies d,(v) 2 8 and I X * ( v )12

+ 1,

then we obtain a true statement. The proof of Corollary VI.13 can be reduced to an application of Theorem VI.ll (if one properly applies the Splitting Lemma at every TJ with d,(v) > 2), and is therefore left to the reader (Exercise VI.7). Moreover, by the same argument used in Exercise VI.7, one can lower the lower bound in 2’) to ?jd,(v) if every vertex TJ satisfies d,(v) > 8 unless it is 2-valent. Thus we obtain another corollary. Corollary VI.14 If we replace in Corollary VI.13 d,(v) 2 8 with d,(v) > 8 and I X * ( v ) 12 $ d D ( v ) 1 with I X * ( v ) 12 ;d,(v), then we obtain a true statement.

+

By a further specialization concerning d,(v) and P ( D ) we obtain another result. Corollary VI.15. Let D be a connected eulerian digraph, and let a system of transitions X ( D ) be given. If S(D) > 6, then D has an eulerian trail compatible with X ( D ) . Proof. By Corollary VI.14, it suffices to replace every 8-valent vertex in an appropriate manner with four 2-valent vertices. The application of Corollary VI.14 to the digraph thus obtained then finishes the proof of Corollary VI.15. If D has no 8-valent vertices, then D and X ( D ) satisfy the hypothesis of Corollary VI.14, and Corollary VI.15 follows. If however, D has an 8-valent vertex 21 indeed, then let a: E (At)+ and a; E (At)- such and are that w.1.o.g. X ( v ) = { { u i , a ~ }i, = 1,2,3,4}, and connected. If ( D 1 , 3 ) 2 , 4 and (D1,3)4,2 are both disconnected, then by the Double Splitting Lemma (Lemma 111.27), at least one of ( D 1 , 4 ) 2 , 3 and ( D 1 , 2 ) 4 , 3 is connected; say, ( D 1 , 4 ) 2 , 3 is connected. Moreover, by the Splitting Lemma either ( ( 0 1 , 4 ) 2 , 3 ) 3 , 1 or ((D1,4)2,3)3,2 is connected; say, H = ( ( 0 1 , 4 ) 2 , 3 ) 3 , 1 is connected. But then H is a connected eulerian digraph obtained from D by replacing 21 with four 2-valent vertices, and

VI.l.l. P( D)-Compatible Eulerian Trails in Digraphs

VI. 13

such that every ( X ( D )-X(u))-compatible eulerian trail of H corresponds to an X(D)-compatible eulerian trail of D. If, however, (D1,3)2,4 01' (D1,3)4,2 is connected then a further application of the Splitting Lemma yields the same conclusion (note that eulerian digraphs are bridgeless in any case by Corollary IV.9). Applying this construction to every 8 - d e n t vertex of D we obtain an eulerian digraph to which Corollary VI.14 can be applied. Corollary VI.13 now follows. Figure VI.2 shows that Corollary VI.15 is best possible concerning the restriction on the minimum valency esceeding 2; and there are infinitely many examples demonstrating this fact. To see this, take any 3-regular digraph D and subdivide any two arcs ul, u2. Then join the two subdivision vertices s,, s2 by two arcs of the form (sl, s2) and two arcs of the form ( s 2 ,sl), and define X(sl) and X ( s 2 )as indicated by Figure VI.4.

a

I

s1

2

s2

D

D,

a

Figure VI.4. The 3-regular digraph D, constructed from the 3-regular digraph D.

We leave it as an exercise to check that D, has an eulerian trail compatible with X ( D , ) = X ( D ) U X ( s l ) U X ( s 2 ) if and only if D has an eulerian trail compatible with a given X ( D ) . Consequently, if D is the digraph of Figure VI.2 or any 3-regular digraph obtained from this digraph by repeated application of the extension indicated by Figure VI.4,

VI.14

VI. Various Types of Eulerian Trails

and if X(D) is defined correspondingly, then D has no X(D)-compatible eulerian trail. However, if X(D) = X, is the transition system of an eulerian trail T of D,then we obtain a somewhat stronger result which can be viewed as the analogue to Corollary VI.5 for digraphs. Corollary VI.16. If D is a connected eulerian digraph with S(D) > 4 and T is an eulerian trail of D, then a T-compatible eulerian trail T‘ of D exists.

Proof. In view of Corollary VI.15 it suffices to construct from D a connected eulerian digraph D‘ without 4- and 6-valent vertices with X(D’) being induced by X, such that no transition of T at a 6-valent vertex of D defines A: where x is 2-valent in D’. In fact we can do better by constructing D’ in such a way that X(D’) defines an eulerian trail of D‘. For this purpose we assume D to have a 6-valent vertex v (otherwise, by Corollary VI.15, nothing has to be proved). Write T in the form

+ ,- - - , e- 3,v,e+3 , . . . , x T = x , . . . , e l , v,e t , - - . , e-2,v,e2 where X(v) = { { e i ( v ) - ,ei(v)+}li= 1,2,3} ei(v)+ E (At)+, i = 1,2,3. Then

X, with e j ( v ) - E ( A : ) - ,

is an eulerian trail of D which follows T everywhere except in where it behaves like a T-compatible eulerian trail of D. Doing this step by step for every 6-valent vertex of D we finally obtain an eulerian trail T’ which coincides with T in all but the 6-valent vertices; and in these vertices it behaves like a T-compatible eulerian trail of D. Denoting by D’ the digraph obtained from D by replacing every 6-valent vertex with three 2-valent vertices such that T‘ corresponds to an eulerian trail of D‘, we have arrived at an eulerian digraph without 4- and 6-valent vertices with a system of transitions X ( 0 ’ ) induced by T and defining an eulerian trail of D’. By the construction of D‘, X(0’) respectively, any eulerian trail of D’ resulting from the application of Corollary VI.15 to D’, corresponds to a T-compatible eulerian trail of D. Corollary VI.16 now follows. It is clear that Corollary VI.16 cannot be extended to include eulerian digraphs with 4-valent vertices (see the discussion preceding Definition

VI.1.1. P(D)-Compatible Eulerian Trails in Digraphs

VI.15

VI.12). It is not clear which 2-regular digraphs have an eulerian trail compatible with a given eulerian trail; a related problem will be considered in subsection VI.1.2. We continue our search for results on digraphs analogous to the results previously established for graphs. As for Corollary VI.2, it is clear that if we proceed as in the proof of that Corollary, then we can write D = D,U D, (with D, and Db being arc-disjoint), while the degree condition is being transformed into

1’) idDr(.) = odDb(v), odD,(v) = idDb(.) for every

‘u

E

V(D)

(any ‘red’ arc incident to ‘u is matched by a ‘blue’ arc incident from ’u, and vice versa). However, statement 2 ) of Corollary VI.2 is no longer valid for eulerian digraphs: to see this, consider any connected eulerian digraph D with an even number of arcs and a 4-valent cut vertex x (such digraphs exist ! - see Exercise VI.9). Then we can write

D = D,U D,,D,nD, = x, and Di is connected for i = 1,2. Thus idDl(.) = idD,(x) = o d D l ( x ) = odD,(x) = 1. Consider any bluered coloring of the arcs in accordance with an eulerian trail of D,and interchange the color classes on D,. Now assume D has been chosen s O(mod2), i = 1,2, and w.1.o.g. the arc in such a way that I A( D i) 1 incident to x in D, has been colored red. Then 1’)(see above) is still fulfilled for this new blue-red coloring, but we have

that is, every eulerian trail of D arriving at x via the red arc in D, must continue from x along the red arc in D,. Generally speaking, we are faced with the (somewhat more general) problem of having to define P ( v ) for any v E V ( D ) ,where P ( v ) consists of two classes: the blue half-arcs incident to and from v forming the first, the corresponding red half-arcs the second; moreover, P(v) satisfies 1’). Let P ( D ) = UP(,). Question: when does D have a P(D)-compatible eulerian trail ? The answer to this question is given in the following theorem for which we need some more terminology.

VI. 16

VI. Various Types of Eulerian Trails

Let D be an eulerian digraph, and for every v E V ( D ) ,let P ( v ) be a partition of A: into two classes,

Moreover, for i = 1,2, let Pi(.) be written as

Pi(,)= Pi,l(v) U Pi,2(v) with Pi,l(v) n Pi,2(2)) =0 where Pi,l(v) contains precisely the half-arcs of Pi(.) which are incident to v (hence Pi,,contains only half-arcs incident from TI). L,et P ( D ) = U V E V ( D )P ( v ) , and let a new digraph D , be derived from D in the following way: every 21 E V ( D )is replaced by two vertices v1,2 and t ~ , ,and ~ , the incidences in D, are defined by

Theorem VI.17. Let D be a connected eulerian digraph and let P ( D ) be defined as above. Then the following statements are equivalent. 1) D has a P(D)-compatible eulerian trail. 2) D , has an eulerian trail.

Proof. If D has a P(D)-compatible eulerian trail T , then, by definition of P ( D ) ,if T reaches an arbitrary v E V ( D )via an half-arc belonging to Pi(u), then this half-arc belongs to Pj,l(~), and T must leave 21 via an halfarc belonging to Pj+l,2(u) where we put i 1 = 1 if i = 2. Consequently, because of the definition of the incidences in D,, T defines a unique eulerian trail of D,.

+

Conversely, if D , has an eulerian trail T p , then T p defines a unique eulerian trail T of D by definition of D,. Moreover, by this definition it follows that the half-arcs defining a transition of T p in 211,2 or 212!1 belong to different classes of P ( v ) for any v E V ( D ) . That is, T is P(D)-compatible. However, Corollary VI.3 has an analogue for digraphs. Corollary VI.18. Let D be a k-regular digraph having an even number of arcs. Then D can be decomposed into two arc-disjoint subdigraphs D, and D, such that d D i ( v ) = k for every v E V ( D ) and i = 1 , 2 . In particular, if k = 2, then D is the arc-disjoint union of two 1-factors.

VI.1.2. Aneulerian Trails, Bieulerian Digraphs and Orientations VI. l i

Proof. The first part of the corollary is trivial - just use the bluered coloring of the arcs of D as described above. As for the second part, split every vertex v into two 2-valent vertices v- and v+ such that odDo(zl-) = idDo(.+) = 0 for the digraph Do thus obtained. By this definition, Dois bipartite since every vertex of Do is either a source or a sink. Consequently, each component of Do has an even number of arcs. Hence the arcs can be 2-colored with blue and red such that no vertex of Do is incident with two arcs of the same color. Considering this arccoloring in D we have at every vertex z1 E V ( D ) idGb(v)= 0dGb(w)= idG,(v) = 0dG,(u) = 1. That is, each of G, and G, is a 1-factor of D. We note in passing that a generalization of the idea just used will be employed in the next subsection. Corollaries VI.6 - VI.8 can be reformulated for digraphs as well provided the orientation of the arcs alternates in @ ( A ; ) . We state the most general result only, leaving its proof as an exercise (Exercise VI.10).

Corollary VI.19. Let D be a connected eulerian digraph with a prescribed order @(At) of the half-arcs incident with w , for every w E V ( D ) , such that two consecutive elements in O+(A:) are not both incident to (from, respectively) w. Then D has a non-intersecting eulerian trail. Note. It may be an interesting research problem to reformulate some of the preceding results for mixed graphs. Also, on the basis of the results obtained so far, it should not be too difficult to reformulate for digraphs what has been said concerning open trails, trail decompositions and path/cycle decompositions in graphs. We did not go into details here because we want to concentrate on graphs.

VI.1.2. Aneulerian Trails in Bieulerian Digraphs and Bieulerian Orientations of Graphs Definition VI.20. Let D be a connected digraph whose valencies are all even, and let TG be an eulerian trail of the graph G underlying D. The sequence To in D corresponding to TG is called an antidirected eulerian trail, in short an aneulerian trail, if for every section a i , vi+l, ai+l of T D , a; and are both either incident to zli+l or incident from v ; + ~ . D is called aneulerian if it has an aneulerian trail. If D is eulerian and has an aneulerian trail, then we call D a bieulerian digraph.

VI. Various Types of Eulerian Trails

VI. 18

The following is an immediate consequence of Definition VI.20.

Corollary VI.21. If D is a bieulerian digraph, then for every o E V ( D ) , id,(v) = odD(o) O(mod2) and, consequently, I A ( D ) 10 (mod2).

=

Note that if D is just aneulerian, then the first part of the conclusion of Corollary VI.21 is false in general, while the second part, I A ( D ) 10 (mod 2), remains valid.

As a generalization of the construction of Do in the proof of Corollary VI.18, we define for an arbitrary digraph D the digraph 07 as obtained from D by splitting every o E V ( D ) into o-,o+ E V ( D 7 ) and by letting o - , vs respectively, be incident with the elements of (At)-, (A:)+ respectively.

Theorem VI.22. Let D be an eulerian digraph. The following statements are equivalent. 1) d,(v)

3

O(mod4) for every o E V ( D ) ,and DT is connected.

2) D is bieulerian. Proof. If d,(v) 0 (mod4), then id,(v) = odD(v) E 0 ( m o d 2 ) , hence GT , the graph underlying D+ , is eulerian by definition of DT . Since D+ is connected by hypothesis, so is GT. An eulerian trail of GT corresponds to an aneulerian trail of D+ and to an aneulerian trail of D by definition of D+. Hence D is a connected eulerian digraph; hence it has an eulerian trail. That is, it has an aneulerian trail as well as an eulerian trail; i.e., D is bieulerian.

Conversely, if D is bieulerian, then d,(v) 5 O(mod4) for every o E V ( D )by Corollary VI.21. By definition of DT, an aneulerian trail of D corresponds to an aneulerian trail of DT; hence 07 is connected. This finishes the proof of the theorem. The two-regular digraph on two vertices shows that the condition for D+ to be connected, cannot be deleted in the statement of Theorem VI.22. In fact, Theorem VI.22 is only a special case of the next result.

Theorem VI.23. Let D be an eulerian digraph with d,(v) 0 (mod4) for every o E V ( D ) . Then D has a unique decomposition S,, into maximal aneulerian subdigraphs of D , and I s,, /=c(D7).

Proof. Consider 0 7 , and let its components be denoted by C,,. . . ,C,, k 2 1. For any i E { 1,.. .,k}, let Gi be the graph underlying Ci.

VI.1.2.

Aneulerian Trails, Bieulerian Digraphs and Orientations VI. 19

Since D is eulerian and d D ( v ) G 0 (mod4) for every v E V ( D ) ,we have d(v+) = d(v-) O(rnod2) in 07 for every v+,v- E V ( D 7 ) . Consequently, Gi is eulerian, and an eulerian trail of Gi corresponds to an aneulerian trail of Ciby definition of DT. Consequently, if D, . . . D, are the subdigraphs of D corresponding to C, , . . . ,C, respectively, then S ,, = {D,, . . . ,D,} is a decomposition of D into aneulerian subdigraphs (note that there is a 1 - 1 correspondence between aneulerian trails of Ciand aneulerian trails of Di, 1 _< i 5 k). Moreover, precisely because of the definition of 0 7 , and because C,,. . . ,C, are its components, S,, is uniquely determined and Diis a maximal aneulerian subdigraph of D for i = 1,.. .,k. Consequently) I S ,, I= c ( D 7 ) . This finishes the proof of the theorem. )

)

We observe that if D is 2-regular, then the aneulerian subdigraphs of D are necessarily the maximal aneulerian subdigraphs of D simply because the vertices of 07 are 2-valent. Which 4-regular graphs have a bieulerian orientation)i.e. an eulerian orientation admitting an aneulerian trail ? The answer we offer is not a good one from an algorithmic point of view; but it relates 4-regular graphs and 3-regular graphs in a way which will play an important role in the context of the Compatibility Problem (We note that the above question was originally stated in [BERM78b]).

Theorem VI.24. A 4-regular graph G, has a bieulerian orientation if and only if there is a hamiltonian, bipartite, 3-regular graph G, with hamiltonian cycle H such that G, N G,/L where L = E(G,) - E ( H ) . Proof. If G, has a bieulerian orientation) then fix such an orientation) call it 0,) and construct the bipartite digraph DT as before. The aneulerian trail of 0, guarantees that D+ is connected. Therefore) for H being the cycle of the underlying graph Go,G, = Go U {v+v-/v E V ( D , ) } is a 3-regular graph with a hamiltonian cycle H ; by the construction of D+ and G,, the latter is bipartite as well, and G, N G J L where L = {v+v-/v E V(D,)} = E(G3) - E ( H ) . Conversely) if G, N G,/L, and if H = G, - E(L) is a hamiltonian cycle of the 3-regular graph G,, then orient the edges of H such that H contains no path of length 2; i.e., the digraph Do thus obtained from H has only sources and sinks. The bipartition of V ( D o )defined by the sources, sinks respectively, coincides with the bipartition of V(G,) since Do is connected. Thus, every e E L joins a source with a sink since G,

VI.20

VI. Various Types of Eulerian Trails

is bipartite. Hence, D, obtained from Do by identifying the same pairs of vertices as in forming G,/L from G,, is 2-regular, and a run along the aneulerian trail of Do corresponds to an aneulerian trail of D,. Consequently, since G, is underlying 0, by hypothesis and by construction, G, has a bieulerian orientation. The theorem now follows. Thus, G , being bipartite is an essential property in order that G, has a bieulerian orientation. However, G, must not be bipartite if it has such an orientation. This fact is substantiated by the next result (see also [BERM79d]).

Proposition VI.25. If G is a bipartite graph with dG(u) for every TJ E V ( G ) ,then it has no bieulerian orientation.

0 (rnod4)

Proof. Suppose an orientation D of G has an aneulerian trail To starting w.1.o.g. at v E V, along the arc (v,w)where w E V,, and {V,, V,} is the bipartition of V ( G ) = V ( D ) . Then the next arc in To is of the form (z,w), with z E V,. Then the arc following (5, w)in To is of the form (z,y) with y E V,, a.s.0. Consequently, the first component of an arc of D always belongs to V,, while the second component lies in V,. That is, D is not an eulerian orientation of G , and therefore, G cannot have a bieulerian orientation. On the other hand, we have for the class of connected eulerian graphs G where each cycle is a block of G, the following strong result. Although this is a rather narrow class of graphs, it will play an important role in the next two sections of this chapter.

Theorem VI.26. If G is a connected graph with d G ( v ) 0 (mod4) for every v E V ( G ) ,and whose cycles are the blocks of G , then every eulerian orientation of G is bieulerian.

Proof. We proceed by induction on bn(G),the number of blocks of G. In any case, bn(G) 2 2. If bn(G) = 2, then, by hypothesis, G has just one vertex incident with just two loops. Orient each of them arbitrarily in one of the two possible ways. This gives an eulerian orientation D of G; passing through one loop according to its orientation and through the other against its orientation, defines an aneulerian trail of D. Hence the theorem holds for bn(G) = 2. For bn(G) > 2 consider an end-block B, of G and its only vertex v (G cannot have 2-valent vertices by hypothesis, so every end-block of G is a loop). Depending on the degree of v we distinguish between two cases.

VI. 1.2. Aneulerian Trails, Bieulerian Digraphs and Orientations VI.21

1) dG(v) = 4. Then we form G, obtained from G - B, by suppressing in G - B, the 2-valent vertes v (see Figure VI.5).

Y

X

X

Y

G Figure VI.5. G , obtained from G by deleting the loop B , and suppressing the 2-valent vertex v of G - B,; possibly 2

= y.

G I satisfies the hypothesis of the theorem and has fewer blocks than G. Let D be an arbitrary eulerian orientation of G , and let D, be the eulerian orientation of G , induced by D. By induction, D,is bieulerian. Assuming ) A ( D ) , we have (v,y) E A ( D ) and (x,y) E A (D,). w.1.o.g. ( 2 , ~ E Considering now any aneulerian trail T, of D,, we see that T, can be extended to an aneulerian trail T of D by replacing the arc (z,y) with (2, v), v, A ( B , ) - ,v,(v, y), where A ( B , ) - means that we pass through B, against its orientation (w.1.o.g. T, passes (2,y) from x to y. Observe that for an aneulerian trail, the inverse sequence is also an aneulerian trail).

> 4.

Because of 1) handled already, we assume w.1.o.g. that no 4-valent vertex of G is incident with a loop of G. Consequently, if we look a t the block-cut vertex graph bc(G), then no end-vertes of bc(G)is adjacent to a 2-valent vertex of bc(G). Consequently, since bc(G) is a tree, there is a cut vertex z of G such that all blocks containing z are loops, except possibly one of them. W.1.o.g. z = v. Because 8 5 d,(v) 0 (rnodd), v is incident with two loops, B, and B, say. Now form G , = G - ( B , U B2) . 2 ) d,(v)

VI.22

VI. Various Types of Eulerian Trails

G, satisfies the hypothesis of the theorem because 4 5 dG(v) - 4 = dG,(v). Let D be an eulerian orientation of G as in case 1)with ( 2 , ~E) A ( D ) . The orientation D, of G, induced by D is eulerian; by induction, D, is even bieulerian. Replace in any aneulerian trail T, of D, the arc (x,TJ)with (2,TJ), v,A ( B , ) - , TJ, A(B,), thus obtaining an aneulerian trail of D (A(&)- means, as above, that we pass through B, against its orientation, while A(B,) stands for passing through B, according to its ) 2 to orientation; also here we assume w.1.o.g that T, passes ( 2 , ~ )from u ) . This finishes the proof of the theorem.

Proposition VI.25 and Theorem VI.26 name classes of graphs representing the extremal cases concerning the theme of bieulerian orientations of 4regular graphs (no bieulerian orientations at all, every eulerian orientation is bieulerian). Proposition VI.25 does not account for all 4-regular graphs without bieulerian orientation. For example, if G is obtained from two 4-regular graphs Gjby subdividing an edge ei with one vertex ZI;, i = 1 , 2 , and then identifying the two 2-valent vertices TJ,and v,, then G is a 4regular graph with cut vertex TJ arising from identifying v, and v,. It is straightforward to see that G has a bieulerian orientation if and only if G, and G, have such orientations. In fact, one can prove the validity of the following statement. Lemma VI.27. A connected 4-regular graph G has a bieulerian orientation if and only if for each block B c G which is homeomorphic to a 4-regular graph B*, B* has a bieulerian orientation.

One proceeds similarly to show the validity of the next lemma. Lemma VI.28. For a connected 4-regular graph G, every eulerian orientation is bieulerian if and only if for every block B of G which is homeomorphic to a 4-regular graph B*, B* has the property that every eulerian orientation is bieulerian.

The proofs of Lemmas VI.27 and VI.28 are left as exercises (Exercise VI. 11). Note. Observe for the statements of the preceding two lemmas that G may contain cycles which are blocks of G; for those there is no homeomorphic 4-regular graph.

Because of the Lemmas VI.27 and VI.28, it suffices to deal with 2connected graphs if one wants to determine all 4-regular graphs which have bieulerian orientations, and/or the property that every eulerian

VI. 1.2. Aneulerian Trails, Bieulerian Digraphs a n d Orientations VI.23

orientation is bieulerian. The example above which has no bieulerian orientation at all, has, however, a cut vertex; but in [BERM79d], a 2connected, 4-regular graph without bieulerian orientation is given. To see that also Theorem VI.26 does not cover all graphs having the property that an eulerian orientation is necessarily bieulerian, we loolc at an arbitrary eulerian orientation D, of K5.It is a routine matter to check that D5has a hamiltonian cycle (in fact, as we shall see later, every eulerian orientation of a complete graph on an odd number of vertices, is hamiltonian). Thus if we draw such a hamiltonian cycle as the outer pentagon, then there are, formally, only two choices for orienting the inner pentagram (see Figure VI.6). These two eulerian orientations are isomorphic, though, and the vertex-sequence 1,3,2,4,3,5,4,1,5,2,1determines the aneulerian trail corresponding to these orientations. Thus the K5 has an eulerian and thus a bieulerian orientation which is unique up to isomorphisms. However, if we label the vertices of the K5beforehand, then the I(, has several eulerian orientations which are, however, bieulerian as well. In fact, the number of bieulerian orientations of any 4-regular graph containing a prescribed arc (2,y) is even, [BERM79d, Theorem 6.21.

Figure VI.6. For a given cyclic orientation of the outer pentagon of the K5,the two possible choices of orienting the inner pentagram yield isomorphic eulerian orientations of the K5,with the isomorphism indicated by the vertex labeling .

VI.24

VI. Various Types of Eulerian Trails

As we have seen in Corollary VI.18, every 2-regular digraph can be decomposed into two l-factors. In general, there will be more than one such decomposition. Hence the question: what is a necessary and sufficient condition for a 2-regular digraph to have a unique l-factorization ? This question is answered as a consequence of the following proposition. Proposition VI.29. Let D be a 2-regular digraph. Then there is a 1-1correspondence between the l-factors of D and the maximal independent arc sets of 0;.

Proof. If F is a l-factor of D , then we have odF(v) = i d F ( v ) = 1 for every v E V ( D ) . That is, one of the arcs incident with v in F is incident from v+, the other to v- (if a loop is a component of F , then these two arcs are identical). Hence F corresponds in 0 3 to an independent arc set covering all vertices of D+, i.e. to a maximal independent arc set, of DT. This arc set is uniquely determined by the construction of DT , Conversely, let M be a maximal independent arc set of DT. We have IA(C)I = O(mod2) for every cycle C of G because DT is bipartite D;

by construction. Moreover, since D is 2-regular, the vertices of 0; are 2-valent. It follows that M covers all vertices of 0;. Consequently, for every v E V ( D ) there is precisely one arc of M incident from v+ and precisely one arc of M incident to v-. That is, M corresponds to a uniquely determined subgraph F of D satisfying i d F ( v ) = o d F ( v )= 1; i.e., F is a l-factor of D.Proposition VI.29 now follows.

For a 2-regular digraph D with l-factor F,D - F is a l-factor F’ of D as well; and M‘ = A(DT) - M is a maximal independent arc set of 0; if M is such an arc set; and if M corresponds to F , then M’ corresponds to F’. This follows from the definition of D 7 . Moreover, if DT has at least two components, one of which is C,say, then we have for the above M and M’ that

M“ = ( M n A @ ) ) u (M’ n (DT - C)) is a maximal independent arc set of DT different from M and M’. These considerations together with Proposition VI.29 yield the following relation between the concept of a l-factorization and that of being bieulerian.

Corollary VI.30. A 2-regular digraph has a unique l-factorization if and only if it is bieulerian.

VI. 1.3. Do-Favoring Eulerian Trails in Digraphs

VI.25

We note in passing that Proposition VI.29 is only a different way of stating and proving [BERM78b, Theorem 2.21; Corollary VI.30 appears as a statement in the introduction of [BERM79d], but it also follows from [BERM78b, Theorems 2.3 and 2.41. These are summarized as follows.

Corollary VI.31. Let D be a 2-regular digraph. Then the number of 1-factors of D equals 2c(D;) and, consequently, the number of 1factorizations of D equals 2c(DT)-1. Corollary VI.31 follows easily from the proof of Proposition VI.29 and the subsequent considerations; its proof is therefore left as an exercise. The paper [BERM78b] finishes with the question: is it true that a 4valent graph with a n odd circuit can be directed so as t o be aneulerian ? The question has a trivial positive answer. For, if G is any connected eulerian graph on an even number of edges, then let T be any eulerian trail of G and orient the edges of G alternatingly following the orientation induced by T or in the opposite direction. This yields an aneulerian orientation of G. Apparently, in the above question the word aneulerian should be replaced with bieulerian. But then the answer is negative as the following example shows: take three copies of any 4-regular graph G which has no bieulerian orientation (e.g., by Proposition VI.25 G can be chosen to be bipartite). Subdivide in each of the three copies an edge. Thus we have three graphs G,, G,, G, with just one 2-valent vertex a, b, c respectively, and all the other vertices are 4-valent. Consequently, G+ = G, U G, U G, U {ab, bc, cn} is a 4-valent graph having a triangle (i.e. an odd cycle); but G+ cannot have a bieulerian orientation because G, has no bieulerian orientation (see Lemma VI.27).

VI.1.3. D,-Favoring Eulerian Trails in Digraphs We finish this section with another type of restriction on eulerian trails in digraphs (see [BERK78a]).

Definition VI.32. Let D be a connected eulerian digraph, and let Do be a subdigraph of D. An eulerian trail T of D is called Do-favoringif and only if for every ZI E V ( D ) ,T traverses every arc of Do incident from 21 before it traverses any arc of D, := D - Do incident from 21.

Of course, every connected eulerian digraph D can be written as D = DoUD, with A ( D , ) n A ( D , )= 0; just take D, = D , V ( D o )= V ( D )

VI.26

VI. Various Types of Eulerian Trails

and A(Do)= 8. In this case any eulerian trail of D is Do-favoring.') Also, Definition VI.32 indicates that the existence of a D,-favoring eulerian trail T may depend on the choice of an initial vertex for T . We prove two theorems on the existence of Do-favoring eulerian trails depending on the structure of D, := D - Do (see Proposition 111.24).

Theorem VI.33. Let D be a connected eulerian digraph, and for given v E V ( D ) let Do C D be chosen such that D, = D - Do is a spanning in-tree of D with root v. Then a D,-favoring eulerian trail starting (and ending) at v exists. Conversely, if T is an eulerian trail of D starting (and ending) at 21, and if we mark at every w E V ( D ) ,w # v, the last arc of T incident from w,then D,, the subdigraph of D induced by the marked arcs, is a spanning in-tree with root v (and hence T is a ( D - D,)-favoring eulerian trail of 0). Proof. Let Do C D be chosen such that D, = D - Do is a spanning in-tree of D with root v. Construct T by starting at vertex v with any arc (v,x), choose any arc of Do incident from x,if such arc exists; choose the arc of D , incident from x, otherwise. Continue this way until this procedure terminates at some y E V ( D ) . Then y = v; otherwise, T contains more arcs incident to y than it contains arcs incident from y, contradicting D being eulerian. Suppose T does not contain all arcs of D. Then let z be a vertex incident with arcs not contained in T . Since D is eulerian and T is a closed trail, idD-*(z) = o ~ ~ - ~#( 0.z Moreover, ) z # v by the very construction of T. By definition of D,, there is a path P ( z , v ) C_ D,joining z to v. Write

p ( z > = *, ( z 7

u l ) ,u l ,

-

* * 7

uk:,(uk, v),

;

possibly z = uk and u1 = v (i.e., P ( z , v ) may contain just one arc). By the construction of T it follows that ( z , u l ) is not contained in T ; therefore, also (u1,u2)is not contained in T (note that (u1,u2)can be contained in T only if all arcs incident to u1 are contained in T), a.s.0. In particular, (uk,v) is not contained in T , contradicting the fact that i d T ( v ) = odT(v>= i d D ( v ) = odD(v). Thus, T contains all arcs of D.

1' This observation is contained in [BERK78a] where the author attributes Theorem IV.8 to I.J. Good, an error apparently inherited from [I
VI. 1.3. Do-Favoring Eulerian Trails in Digraphs

VI.27

This and the construction of T imply that T is a Do-favoring eulerian trail of D. Next suppose T to be an eulerian trail of D starting at some vertex u. For every vertex w # v,mark the last arc of T incident from w and denote by D, the subdigraph induced by these marked arcs. By definition of D,, T is a Do-favoring eulerian trail of D for Do = D - D,. All we have to show is that D, is a spanning in-tree rooted at u.

In any case, D, has the following property: odDl(w) = 1 for every

20

E V ( D )- { u } ,

OdDl ( u ) = 0

.

(*)

Hence it suffices to show that D, is connected (see Theorem 111.31). Suppose this is not so. Then there is a component B, of D, which does not contain v, and by (*), odg, (w) = odDl(w) = 1 for every 20 E V(B,). By the finiteness of D, it follows that B, contains a cycle and, therefore, B, contains a non-trivial strongly connected component. By Lemma 111.24.5), B, even contains a non-trivial strongly connected component C,such that B, (and therefore 0,)has no arc incident from C,. Now, if T is the last vertex of C, in T such that T , ( T , s), s is a section of T where ( T , s) E A ( D , ) ,then it follows from the choice of C, that s E V(C,); and by the choice of T , T terminates in s. Since T is an eulerian trail, s = u which contradicts the choice of B, C D, - { u } . Hence D, is connected implying that D, is a spanning in-tree rooted at u. The theorem now follows. Theorem VI.33 will play an essential role in establishing what has become known as the BEST-Theorem which gives a formula for the number of eulerian trails in an eulerian digraph ([AARD5la]). The first part of the proof of Theorem VI.33 presented here follows along the lines of the proof of [KAST67a, p.76-771, '[AARD5la, Theorem 5b] respectively; the second part is partly extracted from the proof of [BERK78a, Theorem 11. In fact, this second part is more general and somewhat longer than the corresponding (and basically identical) proofs in [AARD5la, KAST67a1, but it permits a shortening of the proof of the next theorem. Nevertheless, I believe that a sketch of the proof as presented in those papers is appropriate because of the relevance of the BEST-Theorem. One proceeds as follows: After establishing (*) it suffices to show that D, is acyclic. Assume the edges of D labeled by 1,2,. ..,IA(D)I as one runs through T. It follows

VI.28

VI. Various Types of Eulerian Trails

that if ( T , s ) and (s, t ) are arcs of D,, then the label of ( T , s ) is smaller than that of ( s , t ) ( ( s , t ) is the last arc of T incident from s and hence can be passed only after all arcs incident to s have been passed, one of which is ( T , s)). Consequently, there cannot be any cycle in D,; otherwise, there would be arcs (r’,s’) and (s‘, t’) in such a cycle with ( T ’ , s’) having a larger label than (s’, t’).

We can generalize Theorem VI.33 concerning the choice of Do. Theorem VI.34. Let D be a connected eulerian digraph. Let D, D be chosen such that odD,(v) 2 1 for every ZI E V ( D , ) C V ( D ) ,and let Do = D - D,. Then the following statements 1) and 2) are equivalent. 1 ) D has a D,-favoring eulerian trail. 2) D, has precisely one (non-trivial) strongly connected component with the property that no arc of D, is incident from C,.

C,

3) Moreover, every Do-favoring eulerian trail of D must start at some vertex of C,,and for any vertex of C,there is a Do-favoring eulerian trail of D starting at that vertex.

Proof. By hypothesis and by the finiteness of D , D, has at least one cycle; we conclude in a manner similar to the second part of the proof of Theorem VI.33 (see the arguments following (*)) that any Do-favoring eulerian trail T of D starts in some vertex of C, where C, is a strongly connected component of D, with no arc of D, incident from C,. By repeating this argument we conclude that D, has precisely one such strongly connected component C,. Hence we conclude not only that 1)implies 2), but also the validity of the first part of 3). Conversely, assume the validity of statement 2), and let v be any vertes of C,. We want to show that D has a D,-favoring eulerian trail T starting at ZI. In order to make use of the first part of the construction of T in the proof of Theorem VI.33, we observe that because of the structure of D, there is a spanning in-tree 0: of D, rooted at v (see Proposition 111.24.6)). Now write

A ( D ) = K,U K, U K3 with Ki n K j = 0 for i # j , i, j E { 1 , 2 , 3 } where

VI. 1.3. Do-Favoring Eulerian Trails in Digraphs

and set

Ko=K,UK3

VI.29

.

Clearly, ( K 3 )= Do;denote ( K O = ) DA.

Of course, Di may not be a spanning in-tree of D. If there is some x E V ( D )- V(Di), then all arcs of D incident with x belong to K3 KO since V ( D i ) = V ( D , ) . Using the Splitting Lemma repeatedly we can replace each such x with 2-valent vertices, and by suppressing these 2valent vertices we arrive at a connected eulerian digraph D’with I(, u K2 A(D’) and V ( D i ) = V(D’); i.e., Di is a spanning in-tree of D’ rooted at 2). If V ( D )= V ( D i ) ,then define D’:= D. By Theorem VI.33, D‘ has a (D’ - Di)-favoring eulerian trail T’ which corresponds in a natural way to a DA-favoring eulerian trail T of D. However, T may not be D,-favoring, but the construction of T’ as performed in the proof of Theorem VI.33 gives us a high degree of freedom concerning the choice of traversing the arcs of D‘ - K,. Namely: for any vertex w E V ( D ’), having reached w in the construction of T‘ one can choose any a E A(D’)- K, incident from w and not yet traversed as the next arc to be traversed by TI. That is, we can impose the additional restriction that at w, T‘ traverses first the arcs of D’- D,incident from w and then traverses the arcs of K, incident from w, before it traverses any arc of K,. Consequently, if T’ has been constructed this way, then T is a D,-favoring eulerian trail of D. The theorem now follows. Theorem VI.34 is nothing but a slight generalization of [BERK78a, Theorem 11 where V( D ,) = V ( D ) is part of the hypothesis (unnecessarily so, as we have seen). Moreover, the proof of Theorem VI.34 basically follows along the lines of the proof of Berkowitz’ result, except that in the proof presented here, certain technicalities do not have to be explained in detail because of the.possibility of using parts of the proof of Theorem VI.33 and this theorem itself. On the other hand, on comparing the proofs of Theorems VI.33 and VI.34 it becomes apparent that Theorem VI.33 could be derived from Theorem VI.34 in much the same way as we derived Theorem VI.34 from Theorem VI.33. What is the most general structure that a subdigraph D, of a connected eulerian digraph D can have in order to imply the existence of a (D-D,)favoring eulerian trail T ? An examination of the proofs of the preceding two theorems shows that D,must not contain more than one non-trivial strongly connected component C, with the property that no arc of D,

VI.30

VI. Various Types of Eulerian Trails

is incident from C,. But this condition is not sufficient even if D , is connected; this can be seen from the digraph D* of Figure VI.7. On the other hand, if we let Db = (Do - { ( e , d ) } ) U { ( d , e)}, then D* has a DI,-favoring eulerian trail T*starting at either a or b. It should be noted, however, that in constructing T* one no longer has the freedom to choose any arc of DI, incident from d. In fact, T*is necessarily of the form

T* = . . . c, (c, d ) , d, ( d , e ) , e, ( e , d),d, ( d , c ) . . .

Figure VI.7. An eulerian digraph D* having no Dofavoring eulerian trail (the arcs of D j are marked with i, i = 0 , l ) .

So, what is the reason for D* having a Db-favoring but no Do-favoring eulerian trail ? A second look at the proofs of the Theorems VI.33 and

V1.34 reveals that reason: in those proofs we first produced a closed trail T starting at v, and proceeding indirectly, we had a vertex z and a path P ( z ,v) C D , with no arc of P ( z ,v) belonging to T ; we thus obtained a contradiction. Moreover, in the proof of Theorem VI.34 we first looked at a subgraph 0; rather than at the whole of D,. So, what if we go the other way around ? That is, generally speaking, for a given : 2 D with connected eulerian digraph D and D, E D , can we find 0 D, D r such that D has a ( D - Dr)-favoring eulerian trail T+ which induces a ( D - D1)-favoring eulerian trail T ? Applying this reasoning to the above D* and D,, 0;= D* - 0; respectively, we see that (Di)+ := 0; U { ( d , c ) } satisfies the hypothesis and statement 2) of Theorem VI.34. A (D* - (Di)+)-favoring eulerian trail T+ of D* (rooted at a or b) exists and induces necessarily T* as described above. As for the above D, C D*, D:+ := D, U { ( e , d ) , ( d , c ) } also implies the esistence of a (D* - Dt+)-favoring eulerian trail of D*; there is, however,

s

VI.1.3. Do-Favoring Eulerian Trails in Digraphs

VI.31

Df := D, U { ( e ,d ) } with D,

C Df C Of+ such that Df has two nontrivial strongly connected components with no arc of Df incident from any of them. This implies that there is no ( D - Df)-favoring eulerian trail of D* (see the argument preceding Figure VI.7); consequently, there cannot be any D,-favoring eulerian trail of D* because the transition from 0: = D - Df to Dodoes not affect an arc incident from d.

The preceding discussion and Theorem VI.34 itself lead to the following theorem which answers our original question.

Theorem VI.35. Let D be a connected eulerian digraph, and let D, be an arbitrary subdigraph of D. Any two of the following statements are equivalent. 1) D has a ( D - D,)-favoring eulerian trail. 2) There exists a digraph Df with D, & Dlf C D such that for every v E V(D)

a)

OdD+(V) 1

= o d ~(V) ] if and only if

OdDl

(v) # 0;

b) otherwise, odD+(v) = 1 and Df has precisely one non-trivial strongly connected component C, with no arc of Dlf incident from C,. 3) There exists a digraph Dlf with D, C Dlf C D such that a) D has a ( D - Df)-favoring eulerian trail;

b) for every Di with D, & 0; & D f , if (z,y) E A(Di - D,), then OdDl (X) = 0.

4)

D, contains a spanning in-forest D l such that

a) for some 2ro E V ( D , ) and for every x E V ( D , ) - {vo}, odD- (x)= 0 if and only if odD,(x)= 0, and odDF(v,)= 0;

b) D has a spanning in-tree B with root vo and D l C B.

Proof. 1) implies 2). Let T be a ( D - D,)-favoring eulerian trail starting at w. If odD,(v) 1 for every 2r E V(D),then choose Df = D,; otherwise, mark the last arc of T which is incident from v, and let D: consist of D, plus the marked arcs. In any case, D, C Df and D r satisfies the degree conditions of 2a),b). Moreover, T is a ( D - Dlf)-favoring eulerian trail because of the choice of the elements of A ( D r ) - A ( D , ) . By Theorem VI.34, 0: has precisely one non-trivial strongly connected

>

VI. Various Types of Eulerian Trails

VI.32

component follows.

C,with no arc of 0: incident from C,. The implication now

2) implies 3). Take 0: as defined by 2)a), b). By Theorem VI.34, this D: satisfies 3)a). Now consider any 0; with D, C Di C D:, and suppose A ( D i -0,) # 8; let (x,y) E A(D{ - 0,).By definition of D: in 2)a),b), an arc of D: - D, is necessarily incident from a vertex z with odDl( z ) = 0. Hence (2,y) E A ( D i - 0,) implies odDl(x) = 0; thus 3)b) holds as well. 3) impZies4). Start with D r as described in 3), and consider a (D-D:)favoring eulerian trail T+ of D. If there is w E V ( D ) different from the initial vertex vo of T+ such that the last arc of T+ incident from w is not in O f , then mark this arc. Note that in this case none of the arcs incident from w lies in 0:.

We define

0:'

= 0:

if no such w exists

;

otherwise, = (A(D:) U {a E A;/od,l+(w) = 0 anda has been marked})

.

In any case, by definition of D:+, T+ is even a ( D - Dr+)-favoring eulerian trail of D , and 0:' satisfies 3)b) as well. Moreover, V(D:+) =

V(D). Marking for every v # vo the last arc of T+ incident from 21 yields a spanning in-tree B of D rooted at vo (Theorem VI.33), and B follows from the very definition of D:+. Define D l by V(D,) = V ( D , ) and A ( D T ) = A ( B ) n A(D,); thus D l is a spanning in-forest of D,. Let (x,y) be any arc of B not in 0;; then x # vo. If (x,y) $ A(@), then it follows from the definition of 0:' and 0:' 2 D, that odDl+(z)= 0 =

If (x,y) E A(D:), then (x,y) $ A(D,) by definition of 0;; and by 3)b) with D{ = D:, o d D l ( z )= 0 follows. We summarize: D l is a spanning in-forest of D, , and if z # vo for some vo E V(D,) (which is the root of B indeed) satisfies odD;(x) = 0 then o d D l ( x )= 0 (for, z not being the root of B implies (z,y)E A ( B - A(D,)) for some y). Since odDl(x)= 0 implies od D l (x) = 0 anyway and od,; (vo) = odB(uo)= 0 , and because D l E B with V ( B )= V ( D ) ,the proof of the implication is finished. OdDl (x).

VI. 1.3. Do-Favoring Eulerian Trails in Digraphs

VI.33

4)impZies 1). Let DT C D, be chosen as described in 4 ) a ) and let B be a spanning in-tree of D with root vo and Dl C B. Applying Theorem VI.33 we obtain an eulerian trail T of D starting at vo and such that for every v # vo, the last arc of T incident from v belongs to B. Because of the freedom to choose the order in which the arcs of A$ - A ( B ) appear in T for every z1 E V ( D ) ,and which we utilized in the proof of Theorem VI.34, we are able to construct T in such a way that the arcs of A$ n ( D - 0,) appear in T before any of the arcs of A$ n D, are used. This is true even in the case where an arc (x,g)E B does not belong to D,; for, in this case odD; (z) = odD,(x) = 0 by 4 ) a ) , i.e. A t n A ( D , ) = 8, i.e., A: C D - D,. In the case of u 0 , if A:o n A ( D , ) # 8, we then proceed with the construction of T by starting along an arc of AZo n A ( D - D,), and each time we arrive at v0 we continue along an arc of A$o n.4(D - 0,) not traversed before, as long as there is such an arc. Consequently, T is a ( D - D,)-favoring eulerian trail of D. This finishes the proof of the implication. Theorem VI.36 now follows. We observe that for the digraph D* of Figure VI.7 and its subdigraphs D,, D{ = D* - DA, statement 4 ) a ) holds for both D, and D { , while

statement 4)b)holds for 0; only. We leave it as an exercise to check the details (Exercise VI.12).

Of course, Definition VI.32 could have been formulated more generally to include graphs and mixed graphs as well. In the case of graphs, for example, one would then be faced with a question such as: does G have a n eulerian orientation D such that a spanning tree of G corresponds t o a spanning in-tree of D with root v for some v E V ( G )= V ( D ) ? I haven't found any work in this direction. However, at least one can say that the answer to this question can be given in polynomial time. Namely, choose in a spanning tree B c G an arbitrary vertex v as root and orient B so as to become an in-tree D, with root v. It is then Theorem IV.ll by which one decides the existence of an eulerian orientation DG of G such that D , c DG: just take the unique mixed graph H whose underlying graph is G and whose arcs are the elements of A(D,). In fact, it is precisely the proof of Theorem IV.ll as presented in [FORD62a, Theorem 7.1, p.601 which tells us that the existence of D, can be decided in polynomial time. Repeating this procedure for every 21 E V(B)= V ( G ) it follows that the above question has a polynomial time answer.,) '1 This has been pointed out to me by Prof. Cai Maocheng, Beijing, during

VI.34

VI. Various Types of Eulerian Trails

Another interesting question (although not related to the preceding problem) is the following. Let G be a connected eulerian graph, and let an ordered set (el, e 2 , .. .,e m } C E(G) be given. Is there some eulerian trail T which can be written in the form

T = . . . ,el, . . .,e 2 , .. . ,e 3 , .. . , e m , . . .

?

1)

This question has a positive answer if X(G) 2 m - 1; and if X(G) 2 2 m then one can even prescribe the direction in which these m edges are traversed by T . These and more general results are contained in [CAIM89a]; they are derived by using [JACKMd, Theorem 11.

VI.2. Pairwise Compatible Eulerian Trails As we have seen in Corollary VI.5, if G is an arbitrary connected eulerian graph without 2-valent vertices and X any system of transitions of G, then there exists an eulerian trail T of G compatible with X. This is true in particular if X = X,, is the system of transitions induced by an eulerian trail T' of G. Whence the following question arises: what is the m a x i m u m number of pairwise compatible eulerian trails in a connected eulerian graph G ? This question was first asked by A.J.W. Hilton and B. Jackson with the latter proving that if S(G) = 21; > 2 , then there are at least m a x ( 2 , k - 1) pairwise compatible eulerian trails, [JACK87b]. In the same paper, B. Jackson put forward the following conjecture which is the starting point for the results of this section and their discussion. Conjecture VI.36. If G is a connected eulerian graph with S(G) = 2k > 2 , then G contains at least 21; - 2 pairwise compatible eulerian trails.

This conjecture has been verified in [FLEISSa] for the class of eulerian graphs G where each block of G is a cycle. Such graphs demonstrate that the bound 21; - 2 as quoted in the above conjecture, is best possible in general: for, if the edge e is incident with a cut vertex v and d ( v ) = 2k, and i f f is the only other edge incident with v and belonging to the same block as e, then {e(v), f(v)} is a separating transition and can thus not be my visit t o China in the summer of 1987. This question was asked by Prof. Li Qiao, Hefei (Anhui Province), during the same visit. Recently, G. Sabidussi asked me the same question.

VI.2. Pairwise Compatible Eulerian Trails

VI.35

contained in X , where T is an arbitrary eulerian trail of G. Consequently, in this case there are a t most 2 1 - 2 possible choices for e(u) to form a transition with other half-edges incident with u. But of course there are instances where a graph contains 21; - 1 pairwise compatible eulerian trails. This is exhibited in Figure VI.8 for k = 2.

Figure VI.8. G with 6(G) = 4 and three pairwise compatible eulerian trails TI = u,1, w, 2, u,3 , w , 4, u;T2 = U , 1,W, 3, U , 4, W, 2, U ; T3 = U , 1,W, 4, U , 2, W, 3, U .

In fact, one might conjecture that if one strengthens the hypothesis of Conjecture VI.36 by additionally assuming G to be 2-connected, then one can even find 21; - 1 pairwise compatible eulerian trails. Figure VI.8 points into this direction. In the case where G = 1<2k+lsuch a conjecture has already been formulated by Kotzig, [I
VJ. Various Types of Eulerian Trails

VI.36

on the study of what are called isotropic systems which were introduced by A. Bouchet some years ago (see e.g. [BOUC87a]). Let us consider Conjecture VI.36 for the case where G consists of just one vertex v and m 2 2 loops e l , . . . ,em. As we shall see, although this graph is quite trivial in many respects, it takes a non-trivial effort to solve Conjecture VI.36 even in this 'trivial' case. For the sake of simplicity we consider G' = S ( G ) instead of G; thus we need not speak of half-edges, and the following discussion will be simpler. We note, however, that the concept of compatibility has to be restricted to v, the only vertex having a valency exceeding 2. In G' we choose the notation of the edges in such a way that e: and ey correspond to ei in G, i = 1,. . . ,m. Together with G' we consider its edge graph L(G') which is isomorphic to K d ,d = d(v). This situation is exemplified in Figure VI.9 for m = 3.

e;

e'3

G'

L(G'>

Figure VI.9. The eulerian graph G' and its edge graph

L(G'). The 'impossible' transitions {e:(v), ey(v)}, i = 1 , 2 , 3 , of G' define a 1-factor L = { e ~ e ~ , e ~ e ~ ,ofe ~ e ~ } L(G') 11 K 6 .

Of course, whatever the shape of an eulerian trail T of G' might be, none of the transitions {e:(v),ey(v)} can define a section e:,v,ey of T , i = 1,. . . ,rn, simply because each of these transitions separates G'. These are

VI.2. Pairwise Compatible Eulerian Trails

VI.37

the 'impossible' transitions; in L(G') they define a 1-factor L = {eley/i = 1,. . . ,m}. Of course, any other transition at 21 in G' defines a section of some eulerian trail of G'. In L(G') a transition other than the 'impossible' ones defines an edge not belonging to L. We can say even more:

An eulerian trail of G' corresponds t o a hamiltonian cycle H of L(G') with E ( H ) 3 L, and vice versa. (*I

For, by the above considerations, an eulerian trail of G' (written as an edge sequence) must be of the form

where we assume T to start at 21 and with the notation chosen in such 6j Sj+l a way that {eij ,eij } = {eij,e:'.> for j = 1,.. .,m, and {il,. . .,i,,,} = (1,. . . ,m}. This sequence, re-interpreted as a sequence of vertices of L(G') defines nothing but a hamiltonian cycle H of L(G') with L C E ( H ) . By the same token, a hamiltonian cycle H of L(G') with L C E ( H ) defines an eulerian trail of G'.

Now, if Tland T' are two compatible eulerian trails of G', then the corresponding hamiltonian cycles Hl and H, of L(G') necessarily satisfy

E(H,) n E(H,) = L

.

(**I

This follows from the preceding considerations and the fact that X, fl X , = 8 by the very definition of compatibility. Observing now that if H is a hamiltonian cycle of L(G') corresponding to an eulerian trail of G', then (*) implies that E ( H ) - L is a 1-factor of L(G'), and considering (**), we are led to the following theorem which is therefore equivalent to the validity of Conjecture VI.36 in the case where G consists of just one vertex and m 2 2 loops, [FLEI86a].

Theorem VI.37. Let n = 2m > 2 be a positive even integer, and let L be a 1-factor of the complete graph K,. Then there exists a 1-factorization L = {Ll,. ..,Ln-l} of K, such that L = Ll and L U L, is a hamiltonian cycle of K,, for i = 2, . . .,n - 1.

Proof. To obtain a 1-factorization as stated in the theorem we consider K , obtained from K , by contracting the edges of the 1-factor L = (el,. . .,em} C K,. Let the notation be chosen in such a way that e j = eie;, i = 1,.. . ,m, thus obtaining V(I(,) = {e{,ey/i = 1,. . ., m } ;

VI. Various Types of Eulerian Trails

VI.38

and for the sake of simplicity denote V ( K m )= {el,. . . ,em} (Observe that this notation is in accordance with the discussion preceding the statement of Theorem VI.37). Let e = eilei2 be a fixed edge of Km. By the above choice of notation we have in A', the identities ei, = e!11 e!'2 1 , eiz - e!I? e'!1 2 and eil ,ei2 E L. Thus, e corresponds to a set He of four edges in Ii , He = {e:le:2, eyl e::, eil e::, e:: ei2} (see Figure VI.10); and this correspondence between e E E ( K m ) and He C E(K,) defines a bijection between the edges e of Kmand the subsets He of I(, such that for e # f,H e n H , = 8. Depending on the parity of m we distinguish between two cases.

el

e

ei

*e.

ei E L

1

1

el.' '2

e'! 'I

Kn

Figure VI.10. e E E(I',)

1) m

G

corresponding to He C E ( K n ) .

1( m o d 2 ) . Then I(, has a decomposition 7 i = {H,,/r = 1,. . .

into hamiltonian cycles H I ,. . .,Hsider an arbitrary but fixed H € 7-f. The bijection e

'

2

(see Theorem 111.50). Con-

He enables us to describe the 5-regular subgraph (where UeEE(H) He is the extension of the above bijection to E ( H ) ) as follows (see Figure VI.11): f-)

% = ( U e E E ( H ) He U1;) C I<,

Take the m-sided prism whose side edges are the elements of L , and add the diagonals of the side faces. We define the notation in Pg in such a way that { j j / j = 1,.. . ,m } = { e i / i = 1,. . .,m} is the vertexset of the boundary of thc 'top' face, while = 1,. . .,m} = { e ; / i = 1,.. . ,m} is the corresponding set of the 'bottom' face, and L = { j j j y / j = 1,.. . ,m}.

{fy/j

VI.2. Pairwise Compatible Euleriaii Trails

/

VI.39

\

Figure V I . l l . The 5-regular graph P;i defining an m-sided prism together with the diagonals of the side faces. We want to show that Pz has a cycle cover with four hamiltonian cycles C,,C,,C,,C, such that Cin Cj= L for i # j, 1 5 i, j 5 4. To achieve this we first decompose E(P;i)- L into four classes:

Ml = {f!f! 3 3 + 1 /3' = 1, ..., m} M, = { f; fi>l/j= 1,.. . ,m} M3 = { fify+l/j = 1,.. . , m} M4 = { fy fi+l/j= 1,.. . ,m} with the subscripts reading modm. The next step is to define for i = 1 , 2 a partition of M iinto two classes,

Mi= with

u

VI.40

VI. Various Types of Eulerian Trails

In any case, it follows from the very definition of Mi and C:, i = 1,2,3:4, that { C ~ , C ~ , Cis~ a, 1-factorization C~} of E ( P i ) - L. To see that Ci= C{UL is a hamiltonian cycle of PE; for i = 1,2,3,4, we observe that Ciis a 2-factor of P i for i = 1,2,3,4, which, for i = 1 , 2 , cannot have more than one component precisely because of m G 1( m o d 2 ) , while for i = 3,4, Ciis a cycle independent of m 1(mod2). Therefore, Ciis a hamiltonian cycle for i = 1 , 2 , 3 , 4 . These four hamiltonian cycles are described in Figure VI.12. Precisely because of the above bijection e c-f He satisfying He n H j = 0 for e # f,and because 7-1 is a cycle decomposition of K , into hamiltonian cycles, therefore, the hamiltonian cycles Ci,Cj*,1 5 i, j 5 4, corresponding to H , H* € 'H also satisfy Ci n C; = L. Consequently, if we redefine C:(H) := C: for i = 1 , 2 , 3 , 4 and arbitrary H E 'H, then we have for

L' = {C:(H)/i= 1,2,3,4; H E 3-1) that

m-1 ILC'1=4.13-11=4--=2m-2=n-2 ; 2 hence C = L' U {L}is a 1-factorization of K , as required. This finishes the case m = 1( m o d 2 ) .

2) m = 0 (mod 2). Observing that the unique 1-factorization of the K4 satisfies the theorem, we may assume m 2 4.

By Theorem 111.49, I<, has a 1-factorization L* = {Ll, . . . ,Lm-l} such thatLiULjisahamiltoniancycleofI(,forj=i+l,i+2, 15 i 5 m-1, where we put L, = L,, L,+l = L2. Let

h!*=

(H, := L2k u L2,+1/2

m-2

5k52 L2k, L2k+1 7

L*}

*

As in the case m = 1(mod2) we consider for a fixed H E 3-1' the correHe U L) c I<,; let the four sponding 5-regular subgraph P& = (UeEE(H)

VI.2. Pairwise Compatible Eulerian Trails

vi.41

Figure VI.12. The hamiltonian cycles C,,C2,C,,C, covering P;i and satisfying Cin Cj = L for i # j , 1 5 i , j 5 4. classes Mi, i = 1,2,3,4, and the partition M i= Mj,o U Mj,l for j = 1 , 2 be defined as above. Since m.is even, we have, however, a simpler definition of Cl, i = 1,2,3,4; namely:

c;= 4 , 1 "M2,o c;= M1,o " M2,1 c; = M3 c; = M,

Again, it follows from the very definitions of M iand Ci, i = 1,2,3,4, that {Ci,C;,Ci, Ci} is a 1-factorization of Pi - L , and it follows by an

VI.42

VI. Various Types of Eulerian Trails

even simpler argument that Ci = Cf U L is a hamiltonian cycle of P;i: for i = 1,2, the pattern top edge, L-edge, bottom edge, L-edge yields a hamiltonian cycle precisely because m is even (compare this with Figure VI.12), while for i = 3,4, Ciis a hamiltonian cycle regardless of the parity of m (compare this with the construction of C, in the case m 1(mod 2)). By the construction just described we again obtain a set L' of pairwise disjoint 1-factors C:(H) of I<, (with C:(H) defined as in the case m E 1(mod 2)), C' = { C : ( H ) / i= 1,2,3,4,H E If*} , but now

m-4 I C' I= 4- 1 31* I= 4 =n -8 . 2 *

To obtain a 1-factorization L of K , as required it remains to be shown that the 7-regular subgraph G* of K , with L c E(G*)and corresponding to L,UL,UL, c E(Km),has a 1-factorization L'' = { L ,L;, . . .,Lg} such that Lj* U L is a hamiltonian cycle of I<, for j = 1,. . .,6. L = C" U L' will then be a 1-factorization of I{, as required; this follows from the bijection e cf He with He Hf = 8 for e # f , already used in the case m f 1 (mod 2), and the fact that 7 f * U { L , U L, U L,} is a decomposition ofI<,;and

ILl=n-8+7=n-1.

Let H be the hamiltonian cycle of K , induced by L , and L,, and consider the 5-regular subgraph P;i 3 L of Kn as before (see Figure VI.ll). G* is then obtained from P i by adding for every edge e = eiej E L, the elements of He = {ele;, eye;, e:ey, eye;}. Assuming w.1.o.g. e1e2 E L, we classify the edges of G*- L as follows: For i = 1,2,3, if ere, E L i , then

By definition, {Li,Ly, N i / i = 1,2,3} is a partition of E ( G * )- L. This partition is the basis for constructing the 1-factors L;, 1 5 j 5 6, as required. We distinguish between two cases.

a) rn = 0 (mod4). We consider the hamiltonian cycle H' of I<, induced by L, and L, and the corresponding 5-regular subgraph P i f of I<, with L C E(P&). Again, P i , can be viewed as obtained from the m-sided prism (where the boundary edges of the top (bottom) face are precisely the edges of LL U L$(L;U Ly), while the side edges of the prism are the

VI.2. Pairwise Compatible Eulerian Trails

VI.43

edges of L ) by adding the diagonals of the side faces. These added edges are precisely the edges of N2 U N 3 . Now we partition N2 U N3 into two 1-factors C;,C; of I<, such that C: U L and C; U L are hamiltonian cycles of K n (see the construction of C4 = C A U L with C;= hf4 in the preceding cases; e.g., see Figure VI.12. Moreover, observe that N2 U N3 is the disjoint union of two even cycles). Having assumed e1e2 E L , we may, w.l.o.g., further specify the notation in such a way that

L - e2i-1e2i/i = 1,.. . ,? ,

} { L2= { e2;e2;+,/i= 1,.. . , F, puttingem+l = e l } . 1-

With this notation at hand, we next define a partition { N i ,N r } of ATl by

N;= {eii-3eyi-2/i = I , . . . , m }

Ni'=ivl- N i ,

u { ell. el . / i = 1,.. . , y , 41-1 42 }

and a partition {L:, L:} of Lh U Lg by

L: ={eii-2eii-l/i = I,. . ., y } U { e ~ i e ~ i + = l / i1 , . . ., T m , putting e$+l = el,l}

LZ =(LL u Lg) - L!

.

A straightforward argument now shows that because of m z 0 (mod 4),

are 1-factors of I<, such that Cj* U L is a hamiltonian cycle of for j = 3,4, each corresponding to the hamiltonian cycle induced by L , U L, in K , (see also Figure VI.13). Since

3

G*-L= U ( L : U L ~ U N ~ ) i= 1

and i= 1

i= 1

VI. Various Types of Eulerian Trails

VI.44

Figure VI.13. The 1-factors C; and Ci of G* having the property that C; U L and Cz U L are hamiltonian cycles of K, (the edges of L appear as the side edges of the prism). the only edges of

G*- L not yet considered are those of

Viewing these edges as the boundary edges of the top and bottom faces of the rn-sided prism and proceeding much the same way as in the construction of Cf and C; at the beginning of this case 2), we conclude as before that C;=L',ULy and C:=Ll,/ULL are 1-factors of I(, such that C; U L and C$U L are hamiltonian cycles of I(,. Thus, for L3 = Cj*, 1 5 j 5 6, we have found a 1-factorization L" = { L, L;, . . .,Lg} of G* as required, provided m 0 (mod 4).

=

b) m E 2 (mod4). We have m 2 6 because m 2 4 in any case. Looking at the construction of Cj*,1 5 j 5 6, in case a) i t follows for j = 1 , 2 that C; is a 1-factor of I<,, independent of the parity of m, while for j = 5 , 6 , Cj* is a 1-factor of I(, provided rn 0 (rnod2). But C;,C; are no C'; longer 1-factors of I<, if rn $ 0 (rnod4). In fact, if one constructs c;, step by step starting with e i e y , eye; respectively, as in case a), then one

VI.2. Pairwise Compatible Eulerian Trails

VI.45

necessarily obtains dc;(ei) = dc:(ey) = 2 and dc;(ey) = dc:(ei) = 0. However, by a local transformation concerning C;, 1 5 j 5 6, as constructed in case a) and involving the 12 edges adjacent to eiey in G*, we shall also succeed in this case. Thus, in what follows let Cj*, 1 5 j 5 6, be the sets of edges as above, subject to the following modification concerning the upper bounds for the indices i in the definitions of AT{ and L!: namely, in the first term of each of Ni and L:, replace T with while in the second term of each of these sets replace with (putting, of course, el,+l = ei in this modified definition of L:).

y,

Considering the edge eqem E ,?3(Kn)we assume w.1.o.g. that the edge e”e’ belongs to C; (observe that because of the notation chosen in case a),mtl eqem I E N3 c C; U C;). This assumption implies e:el,, eye:. E C; where emel E L2 and ele, E L,; consequently, eGek,ekel,l,el,er E C,*. Moreover, we still have eie; E C;, and due to the above modification of the upper bounds of the indices i we also have e A - l e k , eAe\ E C;. This implies eye;, ek-,el,, eke? E Ci. As for C; and Cl, we have eie;, eyer E C; and eye;, eie; E Cl. Leaving C; and C; unchanged we first define

c;*= c;,c;*= c; To obtain the remaining four 1-factors Ct* of G*, i = 1,3,4,6, we exchange certain edges incident with ei ,ey respectively; namely:

Cy*= (C;- { e k e : , eye’,}) u {eke:, eie’,} C;*= (C; - { e i e l } ) u {eye;} I1 It C:* = (Ci - {erne,}) u{eke;} C,** = (C: - {eye;, eie’,}) u { e i e t , eye’,} . Having observed that in this case {Ct/l 5 i 5 6) is a partition but not a 1-factorization of E(G*)- L , we now conclude from the very definition i 6, and m 2(mod4) that for LT = Ct*, 1 5 i 5 of C:*, 1 6, Ll1 = { L ,L;, . . . ,Lg} is a 1-factorization of G*. Moreover, precisely because L U C: is a hamiltonian cycle of G* if m G 0 ( m o d 4 ) , it follows from the definition of CT* that L U C:* is a hamiltonian cycle of G* if m E 2 (mod 4). This final observation settles case b).

< <

Hence in both cases rn 3 0 (mod4) and m = 2 (mod 4) we have found a 1-factorization of G* such that L = Lt U Ll1 is a 1-factorization of Kn

VI. Various Types of Eulerian Trails

VI.46

as required (as for C', see the discussion of case 2) preceding case a)). This settles the case rn G 0 (mod 2); and because the case m G 1(mod 2) has been settled before, Theorem VI.37 now follows.

Theorem VI.37 now enables us to verify Conjecture VI.36 for eulerian graphs G whose cycles are the blocks of G (see the discussion following the statement of Conjecture VI.36 and preceding Theorem VI.37). Theorem VI.38. Let G be a connected eulerian graph with S(G) > 2. Suppose every cycle of G is a block of G. Then G contains b(G) - 2 pairwise compatible eulerian trails, and among any S(G) - 1 eulerian trails of G there are at least two which are not compatible with each other.

Proof. The second part of the conclusion of the theorem follows from the discussion performed immediately after the statement of Conjecture VI.36; hence it suffices to show that under the given hypothesis, G contains b(G) - 2 pairwise compatible eulerian trails. We proceed by induction on p =I V ( G )I. Suppose p = 1. Since S(G) > 2, the only vertex v of G is incident with m 2 2 loops. Thus we have the situation discussed after Figure VI.8 and preceding the statement of Theorem VI.37. There it was shown that the case p = 1 is equivalent to Theorem VI.37. Hence Theorem VI.38 is valid for p = 1. For p > 1 we proceed algorithmically by repeatedly applying Theorem VI.37. We start with an eulerian trail T of the eulerian graph G with V ( G ) p > 1 and b(G) > 2, and consider the cycle C, with E(C,) = E(G) whose traversal in one of the two possible directions corresponds .. . , u p } . to To = T . Denote V(G)= {q,

I

I=

Let G, be obtained from C, by identifying the vertices v,,,,. . . ,' u , , ~ ~of C, where 2k, = d(v,) 2 S(G),and where these k, vertices correspond to the replacement of u1 E V(G) by k, 2-valent vertices following To. Hence G, has precisely one vertex of vdency exceeding 2, namely u,. We note that the very structure of G (every cycle of G is a block of G) implies that for i = 1 (with G,,, = G,) two half-edges incident with ui belong to the same block of G,,, if and only i f these half-edges belong t o the same block of G. (*> We rephrase statement (*) in a more descriptive way by saying that G,,, and G have the same block distribution in vi.

VI.2. Pairwise Compatible Eulerian Trails

VI.47

By the discussion preceding Theorem VI.37 and by this theorem itself, G, has eulerian trails Tl,l,T1,2,. . .,T1,2k1 -2 which necessarily define the same transitions in V ( G )- {v,} (since these vertices correspond to sets of 2-valent vertices in G I ) ,whereas in v1 they behave compatibly (i.e., a pair of half-edges adjacent in v1 defines a transition for at most one Tl,j, 1 5 j 5 2k, - 2). For n = b(G), let T,,,,. . .,Tl,n-2be the first n - 2 eulerian trails constructed above (if 2k1 = n, then take all of them), and for j = 1 , . . . ,n - 2, let C,,j be the cycle corresponding to For each j E (1,. . .,n - 2) we construct G2,j from Cl,j by identifying the vertices v ~ ,. .~.,,v 2 , k 2 where 2k2 = d ( v 2 ) 2 6(G) = n (see the construction of G , from Co above). We conclude that statement (*) holds for i = 2 as well; i.e., G2,jand G have the same block distribution in v2, and therefore, all these n - 2 graphs G2,jhave the same block distribution in v2 (observe that the cycles of G containing vl correspond to one cycle in each G 2 , j ) . Hence, if we look at then the pairs of halfedges incident with v2 and defining the blocks of G2,j are the same for every j E { 1, .. .,n - 2 ) and therefore define a unique 1-factor L of li;kz. Consequently, the application of Theorem VI.37 plus the discussion preceding this theorem yield n - 2 5 2k2 - 2 = d(v2) - 2 eulerian trails T2,P - * - 9 T 2 p - 2 of G which define the same transitions in V ( G )- {v, ,v2} (namely, the transitions of To),but behave compatibly in {vl, v2}.') Next we consider the cycles C2,j corresponding to T2,jand construct correspondingly the graphs G3,j from C 2 , jj, = 1,.. .,n - 2. Now we conclude that statement (*) holds for i = 3, i.e., the graphs G3,j. all have the same block distribution in v3. Consequently, we obtain eulerian trails T3,1, - * 7 T3,n-2 which behave like To in V ( G )- {v,, v2, v3} but behave compatibly in {v1 ,v2,v3}. If p = 2 or p = 3, then the theorem follows from the above. For p > 3, we obtain, by repeated application of the above argument, eulerian trails T p - l , , , - * * 7 T p - L n - 2 of G which behave compatibly precisely at V ( G ){ u p } while in v p they still behave like To. Constructing G,,j from the cycle CP-lJ . corresponding to Tp-l,j,j = 1 , . . .,n - 2 , and observing that these To be more precise one could first view Tz,jas an eulerian trail of G 2 , j , with Tz,i chosen from the 2k2 2 painvise compatible eulerian trails of G2,j, which in turn have been obtained by the application of Theorem VI.37 with respect to G2,j, j = 1,. . .,n - 2. However, one has to start with a fixed 1-factorization of KzkZsatisfying Theorem VI.37.

-

VI.48

VI. Various Types of Eulerian Trails

n - 2 graphs have the same block distribution in up, we apply Theorem VI.37 to the corresponding where 2kp = d(vp) 2 S(G) = n to obtain eulerian trails TP,,,.. . ,Tp,n-2of G which now behave compatibly on all of V ( G ) .That is, T P , , ,... ,Tp,n-2are 6(G)- 2 eulerian trails of G which are pairwise compatible. This finishes the proof of the theorem. It should be noted, however, that a long-standing conjecture of Kotzig, [KOTZ64a,KOTZ64b], and others says that K2m has a 1-factorization {L,, . . . ,L 2 m - l } such that L , U L j is a hamiltonian cycle for i # j, 1 5 i, j 5 2m - 1. It follows immediately that the validity of this conjecture implies Theorem VI.37; for, because of the symmetry of it is no real restriction concerning the 1-factorization to contain a prescribed 1-factor as expressed in the statement of Theorem VI.37. The above conjecture, however, has only been proved so far for some special cases: for those cases where 2m - 1 is a prime, m is a prime, [KOTZ64a], or 2m = 16,28,36,50,244 or 344 (for detailed references on this matter, see [MEND85a, SEAH87a, STIN87al; for 2m = 12 all non-isomorphic solutions of this problem can be found in [PETR80a])l). On the other hand, the proof of Theorem VI.38 as presented here shows that Theorem VI.37 and Theorem VI.38 are equivalent (see also the statement preceding Theorem VI.37). We now turn to the question whether it is possible to use the approach employed in the proof of Theorem VI.38, in order to prove the whole of Conjecture VI.36. Take an eulerian trail To of the connected eulerian graph G without 2-valent vertices and consider C,,the cycle corresponding to To;produce G, as in the proof of Theorem VI.38, apply Theorem VI.38 to G,, a.s.0. The problem we are faced with, however, is that for some i 2 2 the n - 2 graphs G i j may no longer have the same block distribution in v i (see the proof of Theorem VI.38). This problem can be illustrated by the graph of Figure VI.8 and its eulerian trails Tl and T2: starting with the eulerian trail Tl which assumes the role played by To in the proof of Theorem VI.38, and with Co being the cycle corresponding one obtains from C, the graph G, with v, = w as its sole vertex of to TI, valency exceeding 2. GI has two eulerian trails which behave compatibly in w, namely = '1,4, 1,W,2,212,3,3,W,4,211,4

'1 A. Rosa informed me recently that various authors solved Kotzig's conjecture for 2m = 126,170,730,1332,1370,1850,2198,3126,6860.

VI.2. Pairwise Compatible Eulerian Trails

VI.49

and

T1,, = '1,4, 1,w,3, 212,3, 2, w ,4, v , , ~ ( v , , ~and v2,3 denote the 2-valent vertices corresponding to the splitting of 21 following TI). From T,,' and TI,, we obtain the graphs G,,, and G,,? whose sole vertex with valency exceeding 2 is 21. But in G,,, the blocks are defined by the edge sets {1,2} and {3,4}, while in G,,, the blocks are defined by the edge sets {1,3} and {2,4}. Hence G,,, and GI,, have different block distributions in v. However, the problem just discussed leads us to the following conjecture (first put forward in [FLEI86a]) whose validity would prove Conjecture VI.36 and which can be viewed as a generalization of Theorem VI.37.')

Conjecture VI.39. Let n = 2m > 2 be a positive even integer, and let Lo,L', . . . ,L("-,) be 1-factors of K , such that Lo U L(i) is a hamiltonian cycle of I(, for i = 1,.. . ,n - 2. Then there is a 1-factorization { L l , . . . , L n - l } of Kn such that Li U L ( i ) is a hamiltonian cycle for i = 1,..., n - 2 . We observe that in the statement of Conjecture VI.39, the 1-factors L(;),1 5 i 5 n-2, need not all be different, nor does L(') # L ( j ) , 1 _< i < j 5 n - 2, imply L(i)n L ( j )= 0. This takes account of the fact that in a proof of Conjecture VI.36 analogous to that of Theorem VI.38, for some i >, 2, two graphs Gi,jl and Gt,j2may or may not have the same block distribution; and if they have hfferent block distributions, they still may contain some identical block-defining pair of half-edges incident with vi.

If we have L' = . . , = L(n-2),then Conjecture VI.39 reduces to Theorem VI.37 with L = L' = L,-l. Observe that in this case the 1-factor Lo becomes irrelevant: for it does not play any role in the conclusion of Conjecture VI.39; thus, for Lo to form a hamiltonian cycle together with L = L ' = Ln-l is tantamount to saying "let L be any 1-factor of I<,". To see that the validity of Conjecture VI.39 implies the validity of Conjecture VI.36 we go back to the discussion preceding the statement of Conjecture VI.39 using the terminology of the proof of Theorem VI.38: observing that for i = 1, statement (*) in the proof of that theorem remains valid we obtain eulerian trails T,,,, . ..,Tl,n-2behaving compatibly '1 Several other conjectures similar to Conjecture VI.39 have been put

forward in [FLEI86a].

VI.50

VI. Various

Types of Eulerian Trails

precisely in vl; and from the corresponding cycles Cl,l,.. .,Cl,n-2we obtain G2,1,. . .,G2,n--2.Let i 2 2 be the smallest integer for which the n - 2 graphs Gi,,, j = 1,. . . ,n - 2 , do not satisfy (*), i.e., these n - 2 graphs do not have the same block distribution. If no such i 5 p exists, Conjecture VI.36 is valid for this G by a p-fold application of Theorem VI.37. If, however, such i 5 p exists, we observe the following: the transitions of To in 2ri are the same for every G,,,,j = 1,. . .,n -2 (see the definitions of GI and Gi,j for i 2 2 ) , and these transitions correspond to a 1-factor Lo of Kniwhere n 5 ni = d( ui). Similarly, for every j = 1, . . . ,n - 2 , the blocks of Gi,j define a 1-factor L(j) of I<,; such that Lo U L(j) is a hamiltonian cycle of I<, (see statement (*) preceding Theorem VI.37). That is, the hypothesis of Conjecture VI.39 is fulfilled (if n < n,, then L ( j ) = I,(”-’) for n - 2 5 j 5 ni- 2 ) . Assuming the validity of this conjecture we have a 1-factorization C = { L l , .. . ,L,i-l} of Kni such that L j U L ( j )is a hamiltonian cycle of Kni for j = 1,.. . ,ni- 2. That is, L j defines an eulerian trail Ti,, of Gi,j for j = 1,.. . ,n - 2; and since L is a 1-factorization, behave compatibly precisely in q ,. . . ,vi. these n - 2 eulerian trails Ti,j For i < p, we obtain from these Ti,, the corresponding cycles C,,, and, consequently, the graphs Gi+l,j, j = 1,. . . ,n - 2. We then repeat the above argument with i 1 in place of i, a.s.0. That is, the validity of Conjecture VI.39 implies the validity of Conjecture VI.36. The above discussion shows on the other hand, that the validity of Conjecture VI.36 implies something concerning 1-factorizations of K,, n 0 (mod 2). It is not clear, however, whether this ‘something’ is as strong as Conjecture VI.39; i.e, whether these two conjectures are equivalent. For it is not clear yet whether in proving Conjecture VI.36 one really has to deal with all possible systems of block distributions as expressed (in a translated form) in the hypothesis of Conjecture VI.39. We note that [JACK88a, Theorem 11 solves Conjecture V1.36 for the case k = 3 by relying on isotropic systems rather than on Conjecture V1.39.

+

=

So, since we are not able at this point to prove any of the two conjectures in question, one is tempted to replace in Conjecture VI.36 the value 2k - 2 by the function f ( k ) and, knowing that 2k - 2 is an upper bound for f(k) (see Theorem VI.38), ask for a ‘decent’ lower bound. In fact, B.Jackson’s result (see the discussion preceding Conjecture VI.36) asserts that f ( k ) 2 maz(2, k - 1). This result is a direct consequence of a generalization of Corollary VI.43 (see Theorem VI.42 and Corollary VI.43 below). But before stating these results we need some insight into another relation between systems of transitions in a connected eulerian graph and

VI.2. Pairwise Compatible Eulerian Trails

VI.51

hamiltonian cycles containing a given 1-factor of a related graph. Let G be a connected eulerian graph with 6(G) = 2k > 2, and let r systems of transitions of G, X,, , . . , X,,be given. Consider an arbitrary vertex v E V ( G ) ,and let X,(v)be any system of transitions at v if v is not a cut vertex; otherwise, let the system of transitions at v , X,(v), be defined in such a way that it induces a system of transitions at v in every block containing v (since G is eulerian, every block containing v has an even number of half-edges incident with v; hence X,(v)can be defined in the above manner). Let X, = U,Ev(c,X,(v), and let Xi(.) C Xi,i = 1, . ., r , be the corresponding systems of transitions at 21.

Now let e i , . . .,e&,d = d(v) G O(rnod2), be the half-edges incident with v, and consider the complete graph 1C-d with v(I(,) = { e i / l = 1,.. . ,d } . Analogous to what has been said in the discussion preceding Theorem VI.37, we conclude that a system of transitions at v corresponds to a 1-factor L ( v ) of K d . We mark in K d the edges of Lj(v), the 1-factor corresponding to Xj(v), j = 0,. . .,r . We know that constructing an eulerian trail T of G is tantamount to replacing step by step, every vertex v of G with t = ;d(v) 2-valent vertices vl,. . . , vt such that, at each step, the resulting graph G, is connected. Now, if T is compatible with X i , then X,(v) f l X i ( v ) = 0, i = 1,.. .,r . Consequently,

x,(v) corresponds to a 1-factor &(v) L T ( v )n Li(v)= 0, i = 1,.. . ,r

.

in

with (0)

This is a necessary condition for the existence of T compatible with X i , i = 1,.. . ,r . But it is not sufficient because it does not take care of the fact that G, has to be connected (observe that the elements of X,(v) are precisely the sets EGj, j = 1,.. . ,t ) . On the other hand, a consideration of the statement (*) preceding Theorem VI.37, plus the observation that X,(v) is uniquely defined if and only if every block of G containing contains precisely two half-edges incident with v, yields the following conclusion:

If L T ( v )U L,(v) is a hamiltonian cycle of I
(00)

Summarizing (0) and (oo), we seem to be faced with the following question: does K d (with 4 5 d = d(v) E 0 (mod 2)) contain a 1-factor L such that Li(v) n L = 0, i = 0,. . . , T , and L,(v) U L is a hamiltonian

VI. Various Types of Eulerian Trails

VI.52

cycle ? Well, let us consider the case r = 1. If d = 4 then K4 has a unique decomposition into l-factors, any two of which yield a hamiltonian cycle of K4; consequently, regardless of L,(v) n Lo(v)= 8 or # 8, Ii4 contains a l-factor L with L,(v) n L = 8 such that Lo(v)U L is a hamiltonian cycle of K4 (note that in this case L,(v) n Lo(v) # 8 yields L,(v) = Lo(v)). If d > 4, then d 2 6, and by Theorem VI.37, Kd has a l-factorization { L , ) . . ., Ld-1) such that Lo(v)= L , and Lo(v) U Liis a hamiltonian cycle for i = 2 , . . . ,d - 1. d 2 6 implies a d < d - 2; i.e., the l-factor L,(v) (which has $ d edges) satisfies L,(v) n Li= 8 for at least one i E ( 2 , . . . ,d - 1). Whence we conclude that the above question has a positive answer if T = 1 (we note that the same conclusion could have been reached without using Theorem VI.37, by employing a) compatibility results of section VI.l, b) the Splitting Lemma, and c) the structure of Lo(v))* So, the problem we are really faced with is this (we put Li = Li(v)):

+

Let T 1 l-factors Lo,L,, . ..,L, of the K d be given, where d > 2 is a n even integer. How large can r be such that Kd has a 1 -factor L with L n L, = 0, i = 0,. . . ,r , and L U Lo is a hamiltonian cycle ? (0 o 0) The answer to (0 o 0) follows from the next result (see [HAGG79a]). Observe that a graph of even order satisfying Ore’s Theorem (Theorem 111.75) contains a l-factor because it has a hamiltonian cycle (see [BERM83a] for a generalization of the next theorem). Theorem VI.40. Let H be a simple graph of even order d 2 4 such that d H ( z ) d,(y) 2 d 1 for every pair of nonadjacent vertices of H , and let L be an arbitrary l-factor of H . Then L is contained in some hamiltonian cycle of H .

+

+

Proof. We proceed indirectly. Since K , satisfies vacuously the hypothesis of the theorem and since any prescribed 1-factor L C E(Ii,) is contained in some hamiltonian cycle of Kd by Theorem VI.37, therefore we may choose a graph H having the following properties: a) H is a proper subgraph of K , satisfying the hypothesis of the

theorem,

b) H has a l-factor L not contained in any hamiltonian cycle of H , c)

1 E ( H )I is as large as possible.

Consequently, if x and y are any two nonadjacent vertices of H , then H U (xy) has a hamiltonian cycle C containing L. xy E E(C), otherwise

VI.2. Pairwise Compatible Eulerian Trails

VI.53

H is not a counterexample to the theorem. Writing C as a sequence of vertices with x = x,, y = xd,

we conclude that

By the same argument used in the proof of Ore's Theorem we conclude that d(x) d ( y ) 2 d+ 1 implies the existence of vertices x i and xi+, such that x1xi+,,Xdxi€ E ( H ) , 2 5 i 5 d - 2. If xixi+, 4 L , then the cycle C,with

+

(see the proof of Ore's Theorem) is a hamiltonian cycle of H containing L , contradicting the choice of H . Thus

xixi+, E L , and i is odd .

(4

To obtain a contradiction concerning the choice of H satisfying (*), we decompose H into two subgraphs H , and H2 where H , is bipartite with L C E ( H , ) and, subject to this property, IE ( H , ) I is as large as possible, and H2 = H - H,. Consequently, V ( H , ) = V ( H ) . Furthermore, let H i = H , - L. Observe that H , is connected because of the hypothesis of the theorem; hence the vertex bipartition {A,, B,} of V ( H , ) is uniquely determined, and I A, I=I B, I= $ d because of L C E(H,). W.1.o.g. x1 E A,. We deduce some properties of H,. Consider an arbitrary e = xjzCj+, E L ; w.1.o.g. xj E A, (otherwise, replace the index j 1 with j - 1). Now define a new bipartite graph H;* with bipartition V ( H ; * = ) A; U B; as follows:

+

Observe that

VI. Various Types of Eulerian Trails

VI. 54

Using the fact that L c E ( H ; ) by the definition of H;, we conclude from the choice of H , that IE ( H ; ) Is]E ( H l ) I. This and the above equation yield d H ; ( e ) 2 d H z( e ) for arbitrary e E L . (**I Consequently, if e = zjxj+,, f = zkzk+l E L , then dH;

+ d H ; (f)2 d H 2 (). + d H z (f)

+

*

This inequality together with d H ( e ) = d H ,( e ) dH2( e ) yields 2 d H ; ( e ) 2 d H i (f) 2 d H (e) d H (f)- 4

+

+

+ 2 for any e E L -

Now, since e = zjzcJ+l, f = z k z k + l and w.1.o.g. x j , x k E A , , we obtain from the preceding inequality and because of the choice of H , dH; (e)

+ d H ; (f)2 d - 1 if

z j Z k + l > zkxj+l

6E(H1)

*

(* * *)

(note: the maximality of I E ( H , ) I then implies x.zk+,,zkzj+, E ( H ) whence d H ( x j ) d H ( x k + l ) 2 d 1 by hypothesisj.

+

+

We construct from H , a digraph D as follows: replace every edge xy E E ( H , ) with the arc (y,x) where x E A,; from the resulting digraph DHl produce D := D H 1 / L(where L stands for the arc set in DH1 corresponding to L in H,). For practical reasons we label the vertices of D with the labels of the elements of L C E(H,) in accordance with the contraction procedure. This gives dD(e)

= d H ; (e>

Since e , f E V ( D )are not adjacent for the above e , f E L if and only if E ( H , ) (remember that H , is bipartite !), and because d = 2 I V ( D )1, we obtain from (* * *) the inequality zjzk+,, xkxj+l $!

d D ( e ) + d D ( f ) 2 2 I V ( D )I -1 if

e , f E V ( D )are nonadjacent.

(*** *)

However, since we do not know at this point whether D is strongly connected or not, Meyniel’s Theorem cannot be applied to D (note that D has no multiple arcs because of the definition of DH1).But for the application of Corollary 111.78 strong connectedness is not required. Whence we conclude that D contains a hamiltonian path Po which we write as a vertex sequence, Po = eil ,e i 2 ,. . . ,e j m

VI.2. Pairwise Compatible Eulerian Trails

VI.55

where m = i d =I V ( D )I and {ejl, . . . ,ei,} = V ( D ) = L. Because of the orientation DH1of H , it follows that Po corresponds to an antidirected hamiltonian path in D H l ; therefore, Po corresponds to a hamiltonian path P of H , containing L. If the end-vertices of P are joined by an edge e in H , then PU { e } is a hamiltonian cycle of H containing L , contradicting the choice of H . W.1.o.g. we can relabel the vertices of P such that E ( P ) = E(C)-{z,x,} and thus Po = e l , e 2 , .. .,em for ej = z 2 i - 1 z 2ii ,= 1 , . . ., m. It follows from the hypothesis (see the arguments preceding (*)) that zlzj+l, xdxi E E ( H ) ; and by (*), i is odd. Also, since P C H , and z, E A , by assumption, we have A , = {z2j-1, j = 1,..., $ d } , B, = {x2j/j = l , . . . , i d } . This yields zlzj+l,zizdE E ( H l ) since i is odd. Thus, (zi+l,zl),(zd,zj) E A ( D H l )by definition of D H 1 ,and these two arcs correspond to the arcs (ej, el), ( e , , e j ) E A ( D ) , where e j = zizi+, and j = whence we conclude that Po U {(ej, el), (em,e j ) } is a strongly connected spanning subdigraph of D ; i.e. D is strongly connected. By Meyniel's Theorem, D has a hamiltonian cycle C,. We conclude as above concerning the relation between Po and P that C, corresponds to a hamiltonian cycle Co of Hl with L C E(Co). Since V ( H , ) = V ( H ) ,we obtain the final contradiction to the choice of H . The theorem now follows.

q;

We are now in a position to answer the question (0 o 0) raised before stating Theorem VI.40. This question is equivalent to asking whether the graph H = (ICd L i ) U Lo has a hamiltonian cycle containing Lo. By definition of H and because L j n L j # 0 may hold for some i # j , 0 5 i , j 5 T , we have d H ( u ) 2 ( d - 1 ) - ( T 1) 1 = d - T - 1 for arbitrary v E V ( H ) . Therefore,

u:='=o

+ +

+

d H ( z ) d,(y)

2 2d - 2r - 2 if

2,y

E V ( H ) and z y @ E ( H )

.

+

Consequently, if 2d - 27- - 2 2 d 1 , then Theorem VI.40 guarantees the existence of a hamiltonian cycle C in H with Lo c E(C).That is, since 2d-2r-22d+1

ifandonlyif

1 2

~<-(d-3)

,

and since T is an integer, a first answer to question (0 o 0) is: T can be as large as i d - 2. Moreover, the discussion preceding (0o 0) shows that one can assume T 2 1 in any case. We summarize these arguments in the following result which is therefore implied by Theorem VI.40.

VI.56

VI. Various Types of Eulerian Trails

Corollary VI.41. Let d be an even integer exceeding 2, and let r be a positive integer with r 5 rnaz(1, f d - 2). Let L,;L,, . . . ,L , be 1factors of the complete graph Kd. Then H := (Kd Li)U Lo has a hamiltonian cycle C containing Lo.

Ui=,

Let C and Lo be as in Corollary VI.41, and let LT = C - Lo;L , is a 1-factor of H . That is, we have in I
Let G be a connected eulerian graph with A(G) 2 4, and let v be any vertex of G with d(v) > 2. Suppose the positive integer r satisfies r 5 maz{l,k, - 2) where 2k, = d(v). Let X i ( v ) , 1 5 i 5 r , be arbitrary systems of transitions at v, while X,(v)is a system of transitions at v none of whose elements consists of two half-edges belonging to different blocks of G. Then there is a system of transitions at v , X T ( v ) , such that X T ( v ) n X i ( v ) = 0 for i = 0,. . . , r , and G, is connected, where the incidences at the 2-valent vertices of V(G,) - V(G) are defined by the elements of X,(V). (1) Now let G be a connected eulerian graph without 2-valent vertices. Define k = +6(G), i.e., k = min{k,/v E V(G)), and let r 5 maz{l, k - 2). Denote the vertices of G by v l ,. . . , u p ,p =I V(G) 1, and define Go := G, G j = (Gj-l)vj, j = 1,.. . , p . For the given systems of transitions X iof G, 1 5 i 5 r , which induce the systems of transitions Xi(.) at v = vj, j = 1,.. . , p (see (I)), we have Xi= U;=, X , ( v j ) . Substituting in (I) step by step G = Gj-, and v = vj, and defining X,(vj) for v j E V(Gj-l), j = 1,.. . , p , we arrive by a pfold application of (I) at the connected eulerian graph G, which has 2-valent vertices only; i.e. G, is a cycle. Thus, X T = U;=, X T ( v j )is a system of transitions defining an eulerian trail 2'; and X , n Xi= 0, i.e., T is compatible with Xi, i = 1,. . .,r (Observe that X , = UT=, X,(vj)plays, so to say, a supporting role only since it cannot be defined beforehand). Thus we have obtained the main result of [JACK87b] which is expressed by the next theorem. Theorem VI.42. Let G be a connected eulerian graph with S(G) = 2k > 2 , and let r 5 maz{l, k - 2) be a positive integer. Let there be given systems of transitions X,, . . .,X,. Then there exists in G an eulerian trail compatible with Xi, i = 1,.. .,r.

VI.2. Pairwise Compatible Eulerian Trails

VI.57

Corollary VI.43. If G is a connected eulerian graph with 6(G) = 2k > 2, then G has at least max(2, k - 1) pairwise compatible eulerian trails. The proof of Corollary VI.43 can be derived from Theorem VI.42; therefore it is left as an exercise. We observe that Theorem VI.42 has been improved for even k by B. Jackson and N. Wormald, [JACK88c 1. They show that r 5 k - 1 is sufficient. Moreover , they show that G has k pairwise compatible eulerian trails if 6(G) = 2k > 2, The following considerations show that for the latter result one cannot arbitrarily choose pairwise compatible eulerian trails. The inequality r 5 (1, k - 2) in Theorem VI.42 is best possible, as can be seen from the following example (see also Figure VI.14): Let H be a 2-connected 6-regular graph. Consider two compatible eulerian trails Ti,Ti in H and take three copies of H . Subdivide in each of them an edge and identify the subdivision vertices thus obtaining the graph G with cut vertex v (see Figure VI.14). With Ti,Ti at hand in each of the three blocks of G, these eulerian trails can be extended to two compatible eulerian trails TI and T2 as described in Figure VI.14. G contains no eulerian trail T compatible with Tland T2;otherwise the initial section of T starting with the run through el towards v (which is no loss of generality), must be of the form el ,21, e3 or el, v,e5. In the first case, the next section containing must then be e2, v,e4 (in order not to violate compatibility), in the second case this next section must be e4,w,e2. But then the third section containing v in T must be either of the form e 5 , v , e 6 (in the first case) or of the form e 3 , v , e 6 (in the second case). Hence X, n X, # 0 in the first case, X , n X , # 0 in the second case. Hence G does not contain an eulerian trail compatible with TI and T2. This example also illustrates that a solution of Conjecture VI.36 using a ‘greedy’ method is impossible. This fact, however, has been demonstrated already by the equivalence of Theorems VI.37 and VI.38. There is another approach to solving Conjecture VI.36. Originally put forward by A.J.W. Hilton, it is contained in [FLEI86a] (although in a form differing from the following presentation). We call it the top-down approach as opposed to the bottom-up approach described in the discussion leading to the formulation of Conjecture VI.39. The top-down approach starts with the connected eulerian graph G with 6(G) > 2 where vl, . . .,vt, t 2 1, are the vertices of valency exceeding 2.

VI.58

VI. Various Types of Eulerian Trails

G Figure VI.14. A 6-regular graph G with two compstible eulerian trails Tl = el, v,e2,. . . , e 3 ,21, e4,. . . ,e 5 , 21, e 6 , .. . and T2 = el, v,e4,. . . , e 5 , v,e2,. . . ,e3,21, e 6 , .. .. There is no eulerian trail of G compatible with Tl and T’.

Now produce a set G1 of d(v,) - 2 connected eulerian graphs by splittilig 2 1 ~into 2-valent vertices in d(v,) - 2 ways such that no pair of edges incident with v1 in G appears as a pair of adjacent edges in more than one element of G1.Theorem VI.37 guarantees that GI exists: for if we pair the edges incident with 21, in G in such a way that this pairing induces a corresponding pairing in every block containing vl, then the construction with L C of G1 is precisely the application of Theorem VI.37 to I
9

VI.2.1. Pairwise Compatible Eulerian Trails in Digraphs

VI.59

then proceed as above with G2 in place of GI and v3 in place of v,. But does G2 exist ? Maybe one has to be more careful in the choice of the 2k-2 graphs when one tries to form 6,; maybe one has to consider all possible choices for forming G1 (i.e., all possible l-factorizations resulting from the application of Theorem VI.37. to I(dcvl,- L ) . This is not known at this point. But one is led to the following question: suppose G I ,. . . ,G2k-2 are graphs on the same vertex set V such that Gi - V,(G,) N- G j - V,(Gj) for 1 5 i , j 5 2k - 2, and 2k 5 min{d(v)/v E V ( G )- V2(G)}. Is it then possible to produce for each i, 1 5 i 5 2k - 2, a splitting of the same v E V(G) - V,(G) in each Gi in such a way that for any e , f incident with v in Gi, e and f are adjacent in at most one of the 21; - 2 graphs thus obtained ? A positive answer to this question, however, is tantamount to proving Conjecture VI.39. - We observe that if we drop produced above, are connected the requirement that the graphs G, in order to be a candidate for belonging t o G,, then 6, exists. This is guaranteed by Theorem VI.37. Thus both the top-down and the bottomup approach yield, together with the application of Theorem VI.37, the following result (see [FLEI86a, Theorem 21; there, two proofs of Theorem 2 representing the respective approaches, are given): if G is an eulerian graph with 6(G) 2 4, then there exist b(G) - 1 pairwise compatible trail decompositions. Although this result is basically nothing but a translation of Theorem VI.37 into the language of compatibility in eulerian graphs, it raises the interesting question whether one obtains a true statement if one replaces in this result ‘trail’ by ‘cycle’. It should be noted that in the case of G = K 2 m + l ,m 2 2 , the problem has been considered by Kotzig who asks whether K2m+lhas 2 m - 1 pairwise compatible decompositions into hamiltonian cycles, [KOTZ79a, Definition 2 and Problem 41 (Kotzig uses the term ‘perfect’ instead of ‘compatible’).

VI.2.1. Pairwise Compatible Eulerian Trails in

Digraphs Let us consider the problem of determining the maximum number of pairwise compatible eulerian trails in digraphs, and suppose first that every block of the connected eulerian digraph D is a cycle, and that b(D) > 2. Consider a vertex v. As in the discussion leading to the statement of Theorem VI.37 we construct a new graph whose vertices are the half-arcs incident with v, where, as before, the pairs of half-arcs

VI.60

VI. Various Types of Eulerian Trails

belonging to the same block of D (the ‘impossible’ transitions) define the 1-factor L of the new graph, and every other edge of the new graph joins a half-arc incident to v with one incident from v. Hence this new graph is the K d d where d = i d ( v ) = o d ( v ) , and we have, as before, that a n eulerian &ail T of D corresponds t o a hamiltonian cycle H of I
ci

The theoretical reason behind this problem (which arises from the approach chosen, as we shall see later) can partly be found by studying the proof of Theorem VI.37. Let { L l , . . . ,Ld-,} be a 1-factorization of Kd suchthatLiULjisahamiltoniancycleofKd, 1 < i , j < d , 1 < I i - j 1 < 2 , putting L d = L,; Li U Li+l, i = 4,.. . , d - 2, i E 0 (mod2), defines P i , as above and yields again two hamiltonian cycles Hi,Hi+, of Ii-d,d with Hi n Hi+1 = L. But if we look at the 4-regular graph G, 3 L defined by G, = ( L , U L, U L 3 ) , then the proof of Theorem VI.37 indicates that in general G, will not be decomposable into four 1factors L = CA, Ci, C;, Ci such that L U Ci is a hamiltonian cycle of Kd,d,i = 1 , 2 , 3 (despite the fact that Li U L j is a hamiltonian cycle of K d , i # j , 1 5 i , j 3).

<

Now suppose G, has such a decomposition. We orient the hamiltonian cycles Hi= LUG‘:, i = 1,2,3, in such a way that the edges of L are always passed in the same direction, from top to bottom, say. This is possible because G, is bipartite (see Figure VI.16). Thus G, is transformed into

VI.2.1. Pairwise Compatible Eulerian Trails in Digraphs

I'

2'

3'

4'

1

2

3

4

Figure VI.15.

VI.61

K4 cannot be decomposed into l-factors

Lo,L,, L,, L, such that Lo = L = {ii'/i = 1 , 2 , 3 , 4 } and L U L j is a hamiltonian cycle, j = 1 , 2 , 3 . a bipartite digraph D,.

Precisely because of the definition of G,, D4 defines a 3-regular digraph D6 which can be viewed as obtained from G, by replacing every edge a b E E(G,) by two arcs ( a , b), (b, a ) E A(D,); and D,has a decomposition into three oriented hamiltonian cycles H;, H;, H i corresponding to the respective original hamiltonian cycles L , U L,, L , U L,, L, U L, of G, (see Figure VI.16). Next we define an eulerian trail T of D6 as follows: starting at u E V(D6) = V(G,), for example, we pass the whole of H;; after arriving at u again we continue along H;; and after completing the run through H i we pass the arcs of H;. Consequently, T corresponds to a closed covering walk W of G, in which one starts in u E V(G,) and passes first through the elements of L, U L , according to the run through H; in T ;then one continues along the elements of L, UL,,and finally one passes the elements of L,U L, in accordance with the run through H;, H,* respectively. That is, W has the following properties: W . passes each edge of G, precisely once in each direction, and n o edge of G, is passed the second time immediately after it has been passed f o r the first time. As we shall see in Chapter VIII (see Theorem V111.4)) a closed covering walk of G, with this property implies I V(G,)I= 2 (rnod4).') We claim that '1 [AUBE82h, Theorem] follows by the same argument. However, no use

VI.62

VI. Various Types of Eulerian Trails

X

G4 C K d , d

V

D4

Figure VI.16. The hamiltonian cycles L,UL,, L,UL,, L,U L, of G, (with i marking elements of L J correspond to the respective hamiltonian cycles H,, H,, H3 of G, (whose edges

not belonging to L are marked with 1 2 , 1 3 , 2 3 respectively) which in turn correspond to oriented hamiltonian cycles of D,. The latter define a decomposition of D6 into three oriented hamiltonian cycles. The edges (arcs) of L in G, (0,) are marked with 0. The subscripts t and b stand for 'top' and 'bottom'.

G, is bipartite. To see this, consider D6 and delete the following arcs: if a b E L l , delete the arc of H; joining a and b; if ab E L,, delete the corresponding arc of H;; and if ab E L,, delete the corresponding arc of H; (in Figure VI.16, this corresponds to deleting in D6 the arcs incident from u ) . The subdigraph D, C D, thus obtained is an orientation of G, of Theorem VIII.4 is made in that paper.

VI.2.1. Pairwise Compatible Eulerian Trails in Digraphs

VI.63

such that every vertex is either a source or a sink: i.e. 0,is bipartite, and so is G, (observe that under the given assumption, if one of the arcs incident with u E v(D6)is known to belong to Hi*, i E {1,2,3}, it is known for each of the remaining five arcs to which Hf it belongs; see Figure VI. 16).

On the other hand, one obtains 0, from D, by replacing every u E V(D,)by vt and ub, introducing the arc (ut,vb), and by letting the arcs incident to 21 (from u ) be incident to vt (from vb). Moreover, if D6has a

decomposition into three hamiltonian cycles, one concludes via D, that G, has a 1-factorization {CA,Ci, Ci, C;} such that for L = C;,Hi= L U C;l, i = 1,2,3, is a hamiltonian cycle of G,.

Moreover, since the hamiltonian cycles H;, H;, H i of D, induce an orientation of the hamiltonian cycles L , U L,, L , U L,, L, U L, of G, such that every edge of G, is passed exactly once in each direction by the latter hamiltonian cycles, we conclude that the arrows in Figure VI.16 can be reversed as well. In other words, we have the following result.

Proposition VI.44. Let G, and G, be given as above where G, has a 1-factorization {L,, L,, L,} such that L , U L,, L, U L,, L, U L, are hamiltonian cycles of G,. Then G, has a 1-factorization { L ,Ci, Ci, Ci} such that L U C;l is a hamiltonian cycle of G,, i = 1,2,3, if and only if the above hamiltonian cycles of G, have a cyclic orientation such that each edge of G, is passed once in each direction by the two hamiltonian cycles to whichit belongs. If G, is a bipartite 3-regular graph, it has a 1-factorization {L,, L,, L,} (see Theorem 111.48). If {V,, V,} is a bipartition of V(G,) (which is uniquely determined if and only if G, is connected), and if we replace every edge v1v2 E L , U L, by the arc ( u , , ~ , ) where v, E V, and 21, E V,, while such edge of L, is replaced by the arc (u,, v,), the 2factors L , UL,, L, U L, appear as 1-factors F,, F3 of the digraph 0, thus obtained. Consequently, if I? denotes the inverse orientation of F,, and if F, denotes the cyclic orientation of L, U L, induced by the arcs of D, corresponding to L,, it follows that F, ,F t , F3 define cyclic orientations of L , U L,, L, U L,, and& U L, respectively, such that every edge e of G, is passed precisely once in each direction by the two 2-factors containing e . Together with what has been said before, we conclude the validity of the next result.

Proposition VI.45. Let G, be a 3-regular graph having a 1factorization {L,, L,,.L3} such that Hi= Li U Li+,, i = 1,2,3 (where

VI. Various Types of Eulerian Trails

VI.64

we put L, = L,) is a hamiltonian cycle of G,. The following statements are equivalent.

2 ( m o d 4 ) and G is bipartite. 1) IV(G,) 2 ) Hi, i = 1,2,3, can be assigned one of its two cyclic orientations such that every edge is passed once in each direction by the corresponding two Hi.

In view of Propositions VI.44 and VI.45 and the discussion preceding them, our original approach concerning a 1-factorization {L,, . . . ,Ld} of Kd,d with L , U L,, i = 2 , . . . , d , being a hamiltonian cycle of I 2 with d G 2 ( m o d 4) for which I - , has a l-factorization {L,, . . . ,L&1} with the following properties: a) Li U Li+,, i = 1,.. . ,d - 2,and L , U L, are hamiltonian cycles of ;

b) ( L , U L, U L,) is bipartite. A straightforward argument shows that d 2 10 must hold: for, in the case of K6 the graph ( L , U L, UL,) with the above properties is uniquely determined up to isomorphism and implies that K, - { L , U L, U L,} is a disconnected graph consisting of two disjoint triangles which, therefore, cannot be decomposed into two l-factors. However, the positive outcome and direct consequence of our discussion so far is the next theorem. Theorem VI.46. Let a l-factor L C Kd,d,d 2 2 , be given. Then there is a I-factorization {L,L,, . . .,Ld} of Kd,, with the following properties.

a) L u L , is a hamiltonian cycle of K d d for i = 2 , . . . ,d - 1;

b) if d is odd, then L U Ld is a hamiltonian cycle as well; c ) if d

G

0 ( m o d 4 ) or d = 6 , then LuLd is not a hamiltonian cycle.

Theorem VI.46 tells us something concerning the maximum number NT of painuise compatible eulerian trails of D , namely: d - 2 5 NT 5 d - 1 for d = $ 6 ( 0 ) > 2,; and NT = d - 1 for odd d , while NT = d - 2 for d = 4 (see Figure VI.15). But we have not obtained the best possible result for NT simply because we lost some information by utilizing the same approach as in the undirected case.

VI.2.1. Pairwise Compatible Eulerian Trails in Digraphs

VI.65

So, let us return to the consideration of the original 1-factorization problem of Kd,d. Let L = {viv:/i = 1,...,d} denote a I-factor of Kd d , where V := { v J i = 1,.. .,d}, V' := {t(/i = 1,.. . , d } and {V,V ' } is the vertex bipartition of Kd,d.Now orient the edges of Kd d such that the elements of L are oriented from V' to V ,while the eleme'nts of E(lid,d)- L are oriented from V to V'; let KZ,d denote the digraph thus obtained. Now every hamiltonian cycle of Kd,dcontaining L appears as a hamiltonian cycle of I(d0 d (and vice versa), and if we contract in I<& the arcs (v;, vi), i = 1,.. . ,(i,then we obtain the complete symmetric digraph I<: whose hamiltonian cycles are in 1 - 1-correspondence with those of l i j l , d (and thus with those of Kd d containing L as well). so, the original 1factorization problem of Kd,d(stated at the beginning of this subsection) is equivalent to the question whether K: can be decomposed into hamiltonian cycles. By Theorem 111.51 this is possible whenever d > 1 and d # 4,6. Thus we arrive at the following result. Theorem VI.46.a. Let L be a 1-factor of ICd,d, where d 2 2. Then there is a 1-factorization { L ,L,, . . .,Ld} such that LUL, is a hamiltonian cycle of I
Theorem VI.47. Let D be a connected eulerian digraph such that every cycle of D is a block of D. Then, for d = $6(D) 2 2, NT satisfies:

if d # 4 , 6 NT=d-l N ~ = d - 2 if d = 4 , 6

, .

(Observe that Theorems VI.46 and VI.46.a yield for d' > d that I
To obtain an analogue to Theorem VI.42, Corollary VI.43 respectively, for eulerian digraphs, we are naturally led to formulate and prove a result on bipartite graphs analogous to Theorem VI.40 (see also [BERMHb, Theorem 3.11). Corollary VI.48. Suppose the bipartite simple graph H on 2d > 2 vertices with vertex bipartition ( A , B } has a 1-factor. Suppose that d H ( a ) d H ( b ) 2 d 2 for every a E A and b E B , where ab # E ( H ) . Then every 1-factor of H is contained in a hamiltonian cycle of H .

+

+

VI. 66

VI. Various Types of Eulerian Trails

Proof. We follow the proof of Theorem VI.40. In any case, I A I=! B I= d since H has a 1-factor. Observing that Corollary VI.48 is true for the K,,, by Theorem VI.46, but assuming Corollary VI.48 to be false in general, we choose a counterexample H with I E ( H )I as large as possible, subject to the condition that I V ( H )I is as small as possible. Whence we conclude that d > 2 and that H U { a b } has a hamiltonian cycle C containing an arbitrarily prescribed 1-factor L , for arbitrary nonadjacent a E A , b E B. Thus we can write

where a = a l , b = b,, and L = {a,bi/i = 1,.. . , d } . Since H is bipartite and a, b, $Z E ( H ) the degree condition

implies the existence of some i with 2

5 i _< d - 1 such that

(note that C - {ab} contains d edges of the form e j := a j b j ) . Moreover, for e j = a j b j , e k = a k b k E L Satisfying a j b k , akbj # E ( H ) we obtain

Now let D , and D := D H / L be defined as in the proof of Theorem VI.40 (with H in place of H I ) . Then the preceding inequality yields for every pair e j ,ek of nonadjacent vertices of D

(observe that d H ( e j )= d,(ej)

+ 2 ) . Moreover,

A , := { ( e i ,ei+,)/i = 1,.. . ,d - 1) c A ( D ) since (bi,ai+,) E A ( D H )for i = 1,.. . , d - 1. Hence Po := ( A p ) is a hamiltonian path of D. Furthermore, a,bi,aibd E E ( H ) implies (bi, a,), (b,, a;) E DH whence we conclude ( e i ,e l ) , ( e d ,e ; ) E A ( D ) . Consequently, Po U { ( e i ,e , ) , (e,, e ; ) } is a strongly connected spanning subdigraph of D. That is, D is strongly connected. This and the last inequality

VI.2.1. Pairwise Compatible Eulerian Trails in Digraphs

VI.67

permit the application of Meyniel’s Theorem. By the very definition of D and DH, a hamiltonian cycle of D corresponds to a hamiltonian cycle of H containing L. This contradiction implies the validity of Corollary

VI.48.

Corollary VI.49. Consider I 3, and let L o ,L,, . . .,L , be Then ( K dd Li)U Lo has a l-factors of K d d , where r 5 hamiltonian cycle containing Lo.

uG0

9.

Proof. Denote L = Lo and H = (I
uG0Li)U Lo. We have

2 d - (.+ 1) + 1 = d - r

&(?I)

.

Consequently,

+

d H ( a ) d H ( b ) 2 2d - 2r 2 d + 2 if

ab $ E ( H ) .

Corollary V1.49 now follows from Corollary VI.48. We note in passing that if we replace in Corollary VI.49 the restrictions on d and r with d 2 3 and r 5 rnaz(1, then the conclusion is false 1. If, precisely for d = 3. One only needs to choose L , with Lo n L, however, Lon L , = 8, these weaker restrictions admit the same conclusion as in Corollary VI.49.

y},

I

I=

On the grounds of the preceding discussion we deduce our next result from Corollary VI.49 in the s a n e way we reached Theorem VI.42 by applying Corollary VI.41.

Theorem VI.50. Let D be a connected eulerian digraph with S(D) > 6, and let XI, . . .,X , be any r systems of transitions, where r 5 .46( 0 ) - 4 Then D has an eulerian trail compatible with Xi, i = 1, . . . ,r . This theorem together with the remark following Corollary VI.49 yields the analogue to Corollary VI.43.

Corollary VI.51. Let D be a connected eulerian digraph with S(D) 2 6. Then D has at least maz{ 2, pairwise compatible eulerian trails.

y}

Again, this lower bound for the maximum number of pairwise compatible eulerian trails is, in general, best possible (see Exercise VI.4). However, in view of Theorem VI.47 we put forward the following conjecture (we do not bother, however, to formulate the conjecture on l-factorizations

VI.68

VI. Various Types of Eulerian Trails

in the K d d corresponding to Conjecture VI.39 with the former’s validity implying the validity of Conjecture VI.52). Conjecture VI.52. The maximum number of pairwise compatible eulerian trails in a connected eulerian digraph D with S(D) > 4, is at least 16(D) 2 - 2.

We note in passing that most of the above results appear in [FLEISOa].

Of course, one can try to apply the discussion performed in this section so far to mixed graphs. However, in the case where every block is a cycle, one is no longer faced with seeking special types of 1-factorizations of the 1<2d or I<,,,; one rather has to find the maximum number T of pairwise disjoint 1-factors L,, . . .,L,. in a graph H on 2d vertices such that LiUL is a hamiltonian cycle and L is a prescribed 1-factor of H , i = 1,.. ., T. For, H is in general no longer a regular graph; it satisfies, however, the I<2d. This inequality is strict if and only if inequality K d , d E H the mixed graph has at least one block containing an arc and at least one block containing edges only; for, if a block contains an arc, this arc forces the orientation of the edges of this block because of the cut condition of Theorem IV.ll. In the case of arbitrary eulerian mixed graphs, the situation is more complicated: for in the case analogous to Corollaries VI.41 and VI.49 a hamiltonian cycle containing Lo (which corresponds to the replacement of a 2d-valent vertex with d 2-valent vertices) yields the transformation of a connected eulerian mixed graph M into a connected mixed graph M,. In M , the cut condition is still fulfilled for every vertex, but it may no longer be fulfilled globally. That is, one does not know automatically whether M, has an eulerian trail (compare this with statement (I) following Corollary VI.41). In fact, this problem can arise even if every block of M is a cycle. So, it is not clear yet how to handle the problem of determining the maximum number of pairwise compatible eulerian trails in the case of mixed graphs. We finish this section by remarking that problems on special types of eulerian trails have been reduced to problems on special types of hamiltonian cycles (with some of the latter problems solved implying solutions of some of the former problems). This is the most general way of describing the approach chosen in this section. In the next section we shall go the other way around by discussing a problem related to a special type of eulerian trail in plane graphs, and showing how the eulerian problem is related to a hamiltonian problem.

VI.3. A-Trails in Plane Graphs

VI.G9

VI.3. A-Trails in Plane Graphs As we have seen in section VI.1, the concept of a non-intersecting eulerian trail and that of a n A-trail coincide for connected 4-regular graphs (as well as for connected eulerian graphs with A(G) 5 4). However, while it is true that every connected eulerian graph embedded in some surface F has a non-intersecting eulerian trail (Lemma V1.7), such a statement concerning A-trails is false even if 3 is the plane (or, equivalently, the sphere). This fact is illustrated by the following figure which also shows that the existence of an A-trail may depend on the actual embedding of the planar graph.

Figure VI.17. Two embeddings GI and G, of the same planar graph: G, has no A-trail, G, has one; but both G, and G, have non-interesting eulerian trails.

In Figure VI.17, the (unique) A-trail of G, is defined by the transitions { l‘, 2’}, {3’, 4’}, {5’, 6’) (marked in the figure with little arcs), where {i’/i = 1 , . . . ,6) = Ez. The same transitions define a non-intersecting eulerian trail T,of G,. But T,is not an A-trail of G,. In fact, if one starts in the search for an A-trail of G, by passing edge 1 towards TJ, then the transition { l‘, 2‘) is the necessary consequence in this search (for the

VI. Various Types of Eulerian Trails

VI. 70

only other possible choice for a suitable transition, {l’,6’}, d‘isconnects the graph). Consequently, {3’, 4’) must be the next transition at v (otherwise, {3’, 6‘) disconnects the ‘inner’ triangle from the rest of the graph); thus X ( v ) = { { l’,2’}, {3’, 4’},{5’, 6’}}. Since 5’ and 6’ are not neighbors in the cyclic ordering of the edges incident with v, X ( v ) does not define an A-trail but the non-intersecting eulerian trail T, (which comes, so to say, as close as possible to an A-trail). Hence, G, has no A-trail at all. Figure VI.17 also indicates how one can avoid the existence of an A-trail in a connected planar eulerian graph G with vertex v and three or more blocks containing v: let B,, B,, B, be three such blocks of G, and embed G in the plane in such a way that

E(B,)nbd(F,)

# 8 # E(B,)nbd(F,)

while E(B,)nbd(F,)

,

= {v}

and there is a face boundary bd(F) such that

bd(F) n E p , )

# 0 # b d (F ) n E(B,)

and

E, n b d ( ~n) b d ( ~ , ) # 0 .

Then an argument analogous to the one used in the discussion of Figure VI.17 shows that the plane embedding of G thus obtained does not admit an A-trail. However, the ensuing discussion relating A-trails to special types of vertex splittings will give the theoretical answer concerning the non-existence of A-trails in G, of Figure VI.17 and the plane embedding of G just discussed.

In what follows let G be a connected eulerian plane graph. Since G is plane, a counterclockwise cyclic ordering of the edges incident with v is given by the embedding of G on the plane for every v E V ( G ) ,which we denote by O+(v) = (ei, e l , . . .,e&)where d = d(v). Lemma VI.53. Suppose the plane eulerian graph G with A(G) > 2 has an A-trail 5“. Let v E V(G)with d = d(v) > 2 be arbitrarily chosen and suppose {ei,el} is the first transition defined by T in v. W.1.o.g. T = . . . ,e l , v, e 2 , .. .. Then

T = . . . , e l , z l , e 2 , .. . , e 3 , v , e 4 ,.. .,ed-,,v,ed,. . . where O+(v) = (ei, e;,..

. ,e&-,,e&)

or

O+(v) = (e&,e&-,,. .. ,e;, ei)

;

VI.3. A-Trails in Plane Graphs

VI. 71

and e2i = e2i+l if and only if e2i is a loop, i E (1,. . .,i d } (where we put ed+l = 4Proof. Suppose T contains an extended segment T' of the form

where e2i-1, e2i,f , g are the only edges of T' incident with v, and f # e2i+l, f # el respectively if 2i = d. By rotating, if necessary, the indices in O+(v), we may assume w.1.o.g. that i = 1; hence f = e j where j E (4,. . .,d } . Then T' contains a subsequence C which is a cycle starting and ending in v and containing e2 and ej. Consequently, if s(ek) is a subdivision point of ek, k = 1,3, s(el) lies in the exterior of C if and only if s(e3) lies in the interior of C. Now, because of the assumption i = 1 and because of the possibility of cyclically rotating the initial vertex in the sequence T , we may, w.l.o.g., express T in the form

where T" is the subsequence of T' starting and ending in v and not containing e, and g, while T"' is the remainder of T , i.e. the subsequence of T having g as its first and el as its last edge. By construction, e3 E T"'. Hence, viewing T"' as a closed curve starting and ending in v we conclude from the above that TIt' contains an open curve To,whose ends are s(el) and s(eg). We conclude from the Jordan Curve Theorem (Theorem III.61), that there is a vertex w on C such that at least one section f i , w,fi+l of T is a section of T"', and the subdivision points s ( f i ) ,s(fi+,) lie on different sides of C. But then fi and fi+, do not belong to the same face boundary of G (see Corollary III.6la). This contradiction implies the validity of Lemma VI.53. Lemma VI.53 implies that if we know one transition of an A-trail T at v E V ( G ) ,we know the other transitions of T at v as well; i.e. if {ei ,e;} E X,(v) and O+(v) = (ei, ek, . ..,e&)or O-(v) = (e;, eh, . . .,e&), X,(v) = {{ei, eh}, {e;, ei), . . ., {ed-,, I e&}}. Consequently, we can say more generally:

If T is a n A-trail of the plane eulerian graph G with 6(G) > 2, and if O+(v) = (e;, e;, .. .,e&-, ,e&)for a n arbitrarily chosen E V ( G ) ,

VI.72

VI. Various Types of Eulerian Trails

Observe that the validity of ( I ) rests (implicitly) on the Jordan Curve Theorem which does not hold for surfaces of genus other than 0. However, the concept of an A-trail can be defined for graphs G embedded on arbitrary surfaces. In fact, an A-trail in G yields the same local information expressed by (I). On the other hand, ( I )is a weaker statement than Lemma VI.53; for, the conclusion of Lemma VI.53 says that the A-trail T ‘picks up’ the edges incident with v either in their clockwise or in their counterclockwise cyclic ordering induced by the embedding of G. Which of these two ‘pick up’ procedures is followed by T , depends, of course, on the direction in which one passes through T. In any case, once an orientation of T has been chosen by fixing an initial vertex and an initial edge, T defines a partition of V = V ( G ) ,V = V +U V - , where V+ ( V - ) contains precisely those vertices v at which T ‘picks up’ the edges incident with v according to O+(v) (O-(v)). If T is passed in the opposite direction, these two sets V +and V - naturally interchange their roles. That is, the above partition { V + , V - } of V ( G ) is uniquely determined by T with the meaning of V+,V- respectively, depending on the direction in which T is passed. So, the question arises whether it is possible to obtain this partition of V ( G )independent of an orientation of T and without using any O+(v) and/or O-(v). To see that this is possible, we make use of Theorem 111.68. Consider for the plane eulerian graph G with A(G) > 2 a 2-face-coloring (with 1 and 2 denoting the two colors), and let T be an A-trail of G. Choose an arbitrary v E V ( G ) with d = d(v) > 2 and suppose the half-edges incident with v are labeled in such a way as e:, ek, . . . ,e: that O+(v) = ( e i ,e;, . . . ,e&).We further assume that the face whose boundary contains el and e2,is colored with color 1. Then the faces F2i-l ( F Z i ) with {e2i-l,e2i} C E(bd(F2i-I)) ({e2i,eZj+1} C E(bd(F2;))) are ~01ored 1 (2), i = 1,. . . ,+d. Now, if we split 21 into k = i d 2-valent vertices ~ ( ‘ 1 , . . . ,dk) following X,(v) (i.e., {E:(;)/i= 1,.. . , k} = X,(v)) such that the resulting graph G, is plane, then by ( I ) , the faces F2i-2+6,i = 1 , . . . ,k, appear (homotopically) unchanged in G, while the faces F2i+l-6,i = 1,.. . , k, become one face in G,, whereby 5 = 1 or 6 = 2 depending on the pattern of X,(v) (see Figure VI.18). Thus we are led to the following definition (recall that a 1-face, 2-face respectively, is a face colored 1, respectively 2, in a 2-face-coloring).

Definition VI.54. Let G be a plane, eulerian, 2-face-colored graph

VI.3. A-Trails in Plane Graphs

VI.73

with A(G) > 2, and let v C V(G) with 2k = d(v) > 2 be arbitrarily chosen. Suppose that the notation is chosen in such a way that O+(v) = (ei, eb, . . . ,e&) and ei, eb belong to the boundary of a 1face. For S E {1,2}, define X,(V) = ( ( e ~ i - 2 + 6 , e ~ i - l + , }=/ i 1,. . . ,k} (putting ekk+, = ei) and form the plane graph G, such that Ec(!) = {eLi-,+,, eki-1+6}, i = 1,. . . ,I;, and bd(F) is a face boundary both in G and G,, for every 6-face F of G. Then we say that G, is the result of a (6 1)-splitting applied to E V(G), where we put 6 1 = 1 if 6 = 2 (see Figure VI.18) '1.

+

+

In view of Definition VI.54 and the paragraph preceding it we can say that an A-trail T of the plane eulerian graph G without 2-valent vertices induces a partition {V,,V,} of V(G) such that v E V, if and only if X,(V) = X,+,(V) (putting 6 1 = 1 for 6 = 2). That is, V, contains precisely those vertices at which T induces a &splitting, 6 = 1,2. Of course, V, = 0 or V, = 0 may hold; the same can be said concerning the above partition {V+,V - } (see, e.g., G, in Figure VI.17).

+

Observe that in Definition VI.54, O+(V)was used for practical purposes only. As Figure VI.18 indicates, a &splitting can be defined solely on the grounds of G being a plane eulerian graph and of a fixed 2-face-coloring of G (see [FLEI74a, Definition 2 I).

Corollary VI.55. Let G be a plane eulerian 2-face-colored graph without 2-valent vertices which has an A-trail T , and let {V+,V - } , {Vl, V,} be the two partitions of V(G) induced by T (see above). Then {Vl, V,} = {V+,

v-}.

Proof. Choose for T an initial vertex x and an initial edge e = xy, '1 Combining Definition VI.54 with the discussion preceding it we can say that if v is a cut vertex of G, ~ c I ( F ~ ~ - , +can , ) be a cycle in G, while it is not a cycle in G, but there is a 1-1-correspondence between the walks in G, and the corresponding walks through the components of bd(F2i-,+6) = ~ G , ( b d ( f ' ~ ~ - , + , ) ) . Similarly, if v is a cut in G; i.e., ~G(bd(F,~-~+b)) vertex of G, F2i+l-a might be the same face in G for i = 1,.. . ,k, and so this face becomes one face in G, as well; but in this case CG(bd(F2i+14)) < ~ ~ ~ ( b d ( F ~That ~ + ~is,-taking ~ ) )a.topological point of view one would say that the transition from G t o G, does not alter the homotopical properties of F2i-,+6, i = 1,. . ,k, while it does alter these properties if F2i+l-c -

.

FZj+,-, for some j # i, 1 5 i , j <_ needed only if v is a cut vertex of G.

k. However, such considerations are

VI. Various Types of Eulerian Trails

VI.74

1

a> 1-splitting

e

3

b) 2-splitting

Figure VI.18. G, obtained from G by applying a &splitting to TJ E V ( G ) ,6 = 1,2.

thus inducing an orientation on 2'. As one walks on e from x to y, the face to the right of e has w.1.o.g. color 1. With this choice concerning the orientation of the A-trail T we deduce immediately the following properties (see also Lemma VI.53 and Figure VI.19): 1) for every ' o + ( u ) = (ei,eL, .. . , e & ) , d = d(v), e: is being trsis being traversed away from versed towards TJ if and only if TJ (we put e&+l= ei). 2 ) for every

f = uw E E ( G ) ,i f f is being traversed from

u

to w,

VI.3. A-Trails in Plane Graphs

\

- - - - - - - - - --

VI. 7.5

/

Figure VI.19. Orienting the A-trail T ; the face to the right of an arc always has color 1. the face to the right off has color 1. Consequently, T induces in 21 a 1-splitting if and only if TJ E V - (and hence induces in 21 a 2-splitting if and only if 21 E V + ) .The corollary now follows.

Definition VI.56. Let G be a connected, plane, eulerian, 2-face-colored graph with 6(G) > 2. A partition {V’”’’} of V ( G ) is called an Apartition if and only if G has an A-trail T such that the vertex partition {Vl, V2}induced by T satisfies the equation {V‘,V”} = {Vl, V 2 } . Now consider a partition { V V ” } of V ( G ) where G is a graph as described in Definition VI.56. How can one recognize whether {V’,V”} is an A-partition ? Put differently, we want to characterize the A-partitions of G. For this purpose we introduce some notation. Let TJ E V ( G ) be arbitrarily chosen. We write Gv,6for the graph G, obtained from G by applying the &splitting to 21 (see Figure VI.lS), 6 E {1,2}. For V, = {vl,*--,TJk} V(G), define GV0,6 = (... ((Gul,,5)uz,6). - . ) u k , 6 1 E {1,2}. If V, = 8, then we put Gv0,6= G. With this notation we obtain, on the grounds of Corollaries V.10 and V.13, immediately the following result.

c

Corollary VI.57. Let G be as in Definition VI.56, and let {V’,V”} be a partition of V ( G ) . The following statements are equivalent. 1) {V’,V”} is an A-partition.

VI. 76

VI. Various Types of Eulerian Trails 2 ) At least one of the graphs (Gv,,l)v,,,2 and (G,,,,),,,,,

is a cycle.

Of course, (Gvr,,),,,,, may be a cycle although (GV,,2)V,,,1is not a cycle. This is exemplified by the graph of the octahedron as described in Figure VI.20.a). However, we call an A-partition {V’”’‘} perfect if both (Gvr,l)v,,,2and (GV,,2)V,,,lare cycles. An example of a perfect A-partition is given in Figure VI.20.b). We note in passing that an A-partition in [FLEI74a] is what we call a perfect A-partition here.

V’

V”

Figure VI.20. a) The graph of the octahedron with a nonperfect A-partition. b) The same graph with a perfect A-partition (the elements of V’ (V”)are marked with v’ (v”)).

In fact, in checking whether a given partition {V’”’’} of V(G) is an A-partition (respectively, a perfect A-partition), one does not have to go so far as to decide whether (G,,,,),,,,, or (Gv,,2)v,,,1is connected :respectively, whether both are connected); for, both graphs are 2-regular anyway. To see this we first state a lemma whose proof is left as an exercise (Exercise VI. 15).

Lemma VI.58. If G is a plane 2-face-colored embedding of a connected graph each of whose blocks is a cycle having an edge in common with the

VI.3. A-Trails in Plane Graphs

VI. T i

boundary of the unbounded face of G , G has an A-trail T . Moreover, if the outer face of G has color 1 (2), T induces a 2-splitting (1-splitting) on every 21 E V ( G )- V,(G) implying that T is uniquely determined. With the help of Lemma VI.58 we obtain the following characterization of plane graphs having an A-trail.

Theorem VI.59. The connected, plane, 2-face-colored, eulerian graph G has an A-trail if and only if for some V, V ( G ) and 6 = 1 or 6 = 2 , G,,, is connected and outerplane such that every block of G,,, is a face boundary.,) Proof. Suppose G has an A-trail T . Then let {V,, V,} be the partition of V ( G )- V,(G) induced by T where V, contains precisely those vertices at which T induces a &splitting, 5 = 1 , 2 (see the discussion following Definition VI.54). Assume w.1.o.g. that the outer (unbounded) face of G is a 1-face, and consider H := Gv1,,. Since for every 21 E V, the 1splitting applied to 21 is induced by T , therefore the A-trail T corresponds to an A-trail T H of H (in fact, if T and T H are written as edge sequences, then T = T H ) .Thus H is connected. If H is not outerplane, or if it has a block which is not a face boundary, it has a 1-face Fl other than its outer face. Since T H induces a 2-splitting in every w E V ( H )-V,(H), TH corresponds to an A-trail T, of H, = HVI2,,,where V,, = V(bd(F,))nV,. On the other hand, since 2-splittings do not alter the boundaries (viewed as edge sequences) of 1-faces (see Figure VI.18.b)), F, is a 1-face of H , other than the outer face; and bdHl(F,) is a cycle with dHl(z) = 2 for every x E V ( b d H l (F,)). That is, bdHl (F,) is a non-trivial component of H,, i.e. H , is disconnected, contradicting the fact that T, is an eulerian trail of H,. Whence H is outerplane and each of its blocks is a face boundary indeed, and for V, = V, and 6 = 1, G,,, has the properties as required. Conversely, suppose for some V, V ( G ) and 6 E {1,2} that Gv0,, is connected, outerplane and each of its blocks is a face boundary . By Lemma VI.58, GV0,,.has an A-trail To which corresponds t o an eulerian trail T of G which induces a &splitting in each of the elements of V,. Hence T induces a 1- or a 2-splitting in each of the vertices of G other

G , ,6 is a special type of a Husimi tree which in turn is generally defined as a connected graph in which every edge belongs to at most one cycle (see, e.g. [HARA53a] which attributes the consideration of such graphs to Husimi,

[HUSI50al).

VI.78

VI. Various Types of Eulerian Trails

than the 2-valent vertices. Consequently, T is an A-trail of G. Theorem VI.59 now follows. We cannot conclude from the second part of the proof that the A-trail T of G induces a l-splitting for the elements of V , if the outer face of Gvo,6 has color 1. For it can be that G itself satisfies the hypothesis of Lemma VI.58, and then T induces a 2-splitting for all of V ( G )- V,(G). Figure VI.21 shows that there are connected outerplane eulerian graphs without A- trails.

H Figure ‘1.21. A connected, outerp,sne, simple eulerian graph without any A-trail (observe that H has only 2-, 4and 6-valent vertices).

In fact, if we assume for H of Figure VI.21 that its outer face is a 1face, then the triangle ( a , b , c ) is the boundary of a l-face Fl as well. A l-splitting in any one of a, b, c renders a disconnected graph with one component consisting of a triangle; and H{a,b,cl,2contains E(bd(F,)) as the edge set of a non-trivial component. Hence H cannot have an A-trail.

VI.3. A-Trails in Plane Graphs

VI. 79

We leave it as an exercise to show - as a generalization of the idea used in Figure VI.21 - that if G is a graph which satisfies the hypothesis of Lemma VI.58, then a connected, outerplane, simple eulerian graph H exists which has no A-trail and contains a subgraph homeomorphic (also in the topological sense) to G (Exercise VI.16). It is no surprise, however, that in the discussion centering around Figure VI.21 the graphs considered had cut vertices since it was shown in [REGN76a] that every %connected, outerplane, simple eulerian graph has an A-trail. In what follows we call the elements of E(G)- E(bd(F,)) inner edges of the outerplane graph G whose outer face is F,. Moreover, we shall make use of the following lemmas. For the sake of brevity, we extend the concept of S-splitting t o 2-valent vertices by saying that T induces in such a vertex a &splitting where 6 = 1 or 6 = 2; and in this case we choose 6 E { 1,2} as we find fit. Lemma VI.60. Let H be a plane eulerian 2-face-colored graph which is a (non-trivial) block-chain whose outer face F, is a l-face, and let B,, . . .,Br be the blocks of H denoted in such a way that B, n Bj # 8 if and only if I i - j I< 1. Suppose further that E(bd(F,)) f l E(B,) # 8 for i = 1,., . ,T . Then H has an A-trail if and only if B j has an A-trail which induces a 2-splitting in every cut vertex of H contained in B,. = B, n the cut vertex of H for which Proof. Denote by Bj+l, i = 1,. . .,T - 1. Furthermore, let e,, f,, g,, h j E E(B,) denote those edges which satisfy

and such that w.1.o.g. O+(Z~-,,~)= (e{,g{-l ,..., hi-l, fi ,...), i = 2 , . . . ,T . Whence we conclude that gi-l and e, on the one hand, and f, on the other hand are consecutive in a walk through bd(F,). Now suppose H has an A-trail T. By hypothesis and the notation chosen above, and since w.1.o.g. T starts in 21,2 along the edge gl,T is necessarily of the form

VI. Various Types of Eulerian Trails

VI.80

We note that since T is an A-trail in a plane graph, T therefore traverses all edges of B, before passing an edge of Bi, 2 5 i 5 r , and likewise traverses (after passing f,) all edges of B, before passing any other edges of B,, 2 5 i 5 r - 1; and for i = 2,. . . ,T - 1, T passes the edges of B, in two subtrails, namely f,,. . . ,h,, and g,, .. . ,ei. Consequently, T induces A-trails Ti in B,, i = 1,.. . ,r , as follows:

Thus, Tiinduces a 2-splitting in zi-l,i,zi,i+l E V ( B , ) (in the case dBi(zi-,,i) = 2 or dBi(zi,,+,) = 2 we can also regard the behavior of Tias inducing a 2-splitting). Conversely, if Tiis an A-trail of B, inducing a 2-splitting in the cut vertices of H belonging to B,, i = 1, . . .,r , then w.1.o.g. it can be expressed as above, and T as above can be constructed by decomposing Ti, i = 2,. . . ,r - 1 into the above two subtrails; and T is an A-trail of H . The lemma now follows. Now we investigate certain properties of outerplane graphs.

Lemma VI.61. Let G be a 2-connected outerplane eulerian simple graph which is not a cycle and whose outer face F, is a l-face. Define H := G,,, for some v E V ( G ) with d(v) > 2. Then H has the following properties: 1) If F, # F, is a l-face of G with v E V(bd(F,)), then every w E V(bd(F,))- {v} is a cut vertex of H . 2 ) The cut vertices of H are precisely the vertices defined in 1).

3) H is a non-trivial block-chain.

Proof. To prove property 1) we observe that the outerplanarity of G yields v, w E V(bd(F,))nV(bd(F,)). That is, there exist two simple open curves C,, C, in the plane such that a) C, n G = {v, w},i = 0 , l ;

b) C, - {v, w} lies in the interior of the face F,, i = 0 , l ; and, consequently, c)

C := C, U C, is a simple closed curve in the plane satisfying CnG={v,w}.

VI.3. A-Trails in Plane Graphs

VI.81

By construction and the Jordan Curve Theorem, C divides F,-C, i = 0,1, into two parts which lie on different sides of C. From this and the fact that F, and Fl are 1-faces, we can conclude that each of the two sides of C contains a positive even number of edges incident with v and w , respectively. Since G is a simple graph, each of the two sides of C must contain at least one vertex of G. That is, v and w separate G. More precisely, we can express G as

Consequently, G' and GI' lie on different sides of C (except for v and w,of course, which lie on C). Hence, if we form the plane graph H , by splitting v into two vertices v' and v" in such a way that E,, = E, n E(G') and E,,, = E, n E(G")and such that v' (21") lies on the same side of C as G' (G"), then we can write

H , = H i U H:,

where

H i n H: = {w}

and H I , H r and Hl are all connected and eulerian by (*). Moreover, w is the only element of H I which lies on C (hence HI and Hi' lie on different sides of C except for w); and by construction, w is a cut vertex of H,. Also, by the above construction,

and H can be written as

H = HI U H" , where H' n HI' = {w} and

H' = (Hi),/,l, HI' = (H;)v/t,l

*

That is, H' and HI' lie on different sides of C except for w. This classifies w as a cut vertex of H as well. Now we turn to the proof of property 2). First of all, the 2-valent vertices d'),i = 1,. . .,k, k = i d G ( v ) , arising from the 1-splitting of v, cannot be cut vertices of H ; otherwise, EUc;,is a set of two bridges of H, thus contradicting the fact that H is eulerian.

VI. 82

VI. Various Types of Eulerian Trails

Now consider the path

Po := bd(F') - v C H n G

,

and let x E V(Po)be chosen subject to the condition that x g' V(bd(F,)) if Fl # Fo is a l-face of G and 21 E V(bd(Fl)).By the above and 1) it suffices to show that x is not a cut vertex of H . In any case, the choice of x implies that vx @ E, - {el, e d } , where {el, e d } = E, n E(bd(Fo)). Let the paths Pl and P2 be defined by

p0 = p1UP,, p1n ~ =, IC (possibly E(P,) = 0 or E(P,) = 0). We can consider two cases. a) Suppose there exists x i E V(P,)- {x}, i = 1,2, such that either x1z2 E E ( H ) ,or for some j E (1,. . .,k}, x1&), x 2 v ( j ) E E ( H ) . Then let P3 = P3(x1,x,) be the path joining x1 and x 2 in Po;x E V ( P 3 ) {zl, 2,) follows. Define the cycle Cjof H by

Cj= (E(P3)U {x1x2}) if x1x2 E E ( H ) , Cj = ( E ( P 3 )U E u ( j ) )

otherwise

(note that {xldj),x2dj)} = E,,j, if x1x2 @ E ( H ) ) . In any case, H ( j ) := (V(Cj)) is a %-connected subgraph of H . In fact, Cj c bd(F,), where F, is the unbounded (outer) face of H . Since H and H ( j ) c H are also outerplane with Cj being the boundary of the outer face of H ( j ) ,we can conclude that E, C E ( H ( j ) )for every u E V(Cj) - {xl,x 2 , v(j)} (observe that the elements of E,nE(P,) are consecutive in O+(u)). Consequently, u cannot be a cut vertex of H since H ( j ) is a 2-connected subgraph of H and thus a subgraph of a block of H . Observing that

V(Cj) - {XI,

52, ~ ( j ) = } ~ ( ~ 3 1{XI, - ~ 2 )

we can conclude that x is one of these vertices u and, therefore, x is not a cut vertex of H .

b) Assuming now that xi E V(Pi) - {x}, i = 1,2, with the properties stated in a) does not exist, we immediately deduce the validity of the following equations: E(G)=E1UE2,

E,nE2=0,

E j = E((V(P,)u {v})) , i = 1,2 .l)

VI.3. A-Trails in Plane Graphs

VI.83

Consequently, assuming first that E ( P l ) # 0 # E(P2),it follows that v and x separate G. Moreover, these two equations imply even that in G there exists a face F # Fo with v , x E V(bd(F)).2)By the choice of x, F cannot be a 1-face; hence it is a 2-face both in G and H . Thus, d j ) E V ( b d ( F ) )n V ( H ) for some j E (1,. . .,k}. Similar to case a) let P3 = P3(x1,x2) C Po denote the path containing x and defined by {v(~)x1,21(j)z2} = E,cj,. Using the same symbols, define Cj and H ( j ) as in case a); again Cj C bd(F,), and Cj is a cycle both in G and H . Continuing the argument as in case a), we can conclude that x is not a cut vertex of H if x = u for some u E V(P3)- {z1,x2}. By this and because vx $ E, - {el, ed} by the choice of x, we are led to the following conclusions: x E {x,,x2}, hence vx E E, C E ( G ) which implies vx E {el,ed}. So, either E ( P l )= 0 or E (P 2 )= 0. W.1.o.g. vx = el. Let x’ denote the element of V ( G )- {v} incident with e2 which is defined by O+(v) = (ei, eh, .. . ,e&),and let in this case P3 = P3(x,x’) C Po be the path joining x and x‘ in Po. Again, by using the same symbols as before and by observing that in this case j = 1, we define the same way as before the cycle C, which is the boundary of the outer face of H ( l ) where, as before, H ( l ) := ( V ( C l ) )is a 2-connected outerplane subgraph of H . But also in this case we can conclude E, C E ( H ( l ) ) ;for we can express C, as

C,= Pi u {e2} for Pi

= P3 u {v,vx}

.

This gives the situation of case a) with v and x‘ assuming the roles played by x1 and x2. That is, x is not an end vertex of Pi; so, as before, E, c E(H(’)) follows, and therefore, x is not a cut vertex of H . This finishes the proof of property 2 ) . To see that H is a non-trivial block-chain once again consider Po as defined at the beginning of the proof of property 2), and let xl,x,x 2 be three different vertices lying in this order on Po. We claim that x is not a cut vertex of H if x1 and x 2 belong to the same block B c H .

B is 2-connected since H is eulerian and simple. Hence there is a cycle C C B with z 1 , x 2E V ( C ) .For the path P3= P3(z1,z2)E Po,form the Observe that in the definition of PI it is not said whether P, is the path in Po ‘to the left’ or ‘to the right’ of 2;hence it is no loss of generality, if we write the same index i on both sides of the last equation. 2, This implication is true, however, for every separating pair of vertices in a 2-connected plane graph.

VI. Various Types of Eulerian Trails

VI.84

2-connected subgraph B, of H defined by

B,

:=

c u P3

(possibly B, = C ; B, # B if B, # C). In any case, B, C B because C C B and K ( B , )2 2. Also, x E V ( B , ) since x E V ( P 3 ) ;hence x E V ( B ) . Moreover, P3 c B implies that P3 C bd(F*), where F* is the unbounded (outer) face of the outerplane graph B. Now, we observe as before that the elements of E , n E(P3)are consecutive in O+(z); hence E, C E ( B ) , i.e., x is not a cut vertex of H . That is, since dj), j = 1,.. . , k, is not a cut vertex of H and since V(P,) = V ( H )- { d j ) / j = 1,.. . , k}, a block of H contains at most two cut vertices of H . Consequently, since H is connected but not 2-connected and since H is a simple graph, it follows that a cut vertex of H belongs to two blocks of H ; that is bc(H) is a non-trivial path. This implies the validity of property 3) and finishes the proof of Lemma VI.61. Finally, we prove a property of 2-connected outerplane graphs. Lemma VI.62. Let B be a 2-connected outerplane simple graph whose outer face is F,, and let B- := B - e for some edge e = xy E E(bd(F,)). Denote by F2 # F, the other face of B with e E E(bd(F2)).Then Bhas the following properties: 1) V(bd(F,)) - {x,y} is the set of cut vertices of B-. 2)

B- is a (non-trivial) block-chain with z and y belonging to different end-blocks of B-, and they are not cut vertices of B - .

Proof. The lemma is trivially true if B is a cycle, i.e., if B- is a path. Hence suppose that B is not a cycle. Any w E V(bd(F,))- {z, y} is a cut vertex of B-: this follows from the fact that w E V(bd(F,))nV(bd(F,)) and zy E E(bd(F2))n E(bd(F,)) which implies that there is a plane simple closed curve C containing nothing in B except w and precisely one point of the edge z y (with z y viewed as a topological image of the open unit interval (0,l)). Thus z and y lie on different sides of C; hence, every path from z to y in B- passes through w; i.e., w is a cut vertex of B - . Observe that V(bd(F,)) - {x,y} # 8 because G is a simple graph. Now let z be any cut vertex of B-. Since B is 2-connected while B- is not 2-connected7it follows from Theorem 111.32 that B- is a non-trivial blockchain with z and y belonging to different end-blocks of B-, and they are not cut-vertices of B-. Hence every path P(z, y) C B- contains z , and

VI.3. A-Trails in Plane Graphs

VI.85

so every cycle of B containing z y also contains t. Since bd(F,) is such a cycle, and since z and y are not cut vertices of B - , t E V(bd(F,))-{x, y} follows. This proves the lemma. Now we turn to Regner’s result on A-trails in outerplane eulerian graphs.

Theorem VI.63. Let G be a 2-connected outerplane simple eulerian graph, and consider a 2-face-coloring of G such that the outer face Fo is a 1-face. Let v E V ( b d ( F o ) )with d(v) > 2 be arbitrarily chosen (if such v exists). Then G has an A-trail inducing a 1-splitting in v. Proof. If G has only 2-valent vertices it is a cycle; a run through this cycle represents an A-trail, whence we can assume that G has at least one vertex whose valency exceeds 2. We observe that G being a 2-connected simple graph implies that in this case G has at least three vertices with a valency exceeding 2. Consequently, the smallest graph Go other than a cycle satisfying the hypothesis of the theorem, has precisely three 4valent vertices and three 2-valent vertices; it can be interpreted as being obtained from a plane embedding of the octahedron graph by deleting the edges of the outer face boundary (see Figure VI.22). A 1-splitting in u renders the graph GO,,,having precisely two 4-valent vertices x,y which are necessarily cut vertices of GO,,,. In fact, Gt,l satisfies the hypothesis of Lemma VI.58, whence we can conclude that Go has an A-trail To inducing a 1-splitting in v and a 2-splitting in each of x and y. Using the same argument, we can conclude that if G is an outerplane graph homeomorphic to GO, G has an A-trail T inducing a 1-splitting in the chosen vertex v and a 2-splitting in the other two 4-valent vertices. Now we can proceed by induction. Consider a graph G satisfying the hypothesis of the theorem and having n > 3 vertices of valency exceeding 2, and assume the validity of the theorem for all corresponding graphs H having at most n - 1 such vertices. Consider in G the chosen vertex v and O+(v) = (ei,e;, .. .,e&) where d = d(v) > 2, and w.1.o.g. assume that e l , e d E E(bd(Fo))where Fo is the outer face of G. Now form H = Gv,l and denote Po = bd(Fo)- v. In any case, H is connected, eulerian and outerplane. If we denote the cut vertices of H by xl,.. .,x, according to the order in which they appear in Po,it follows from the argument used in proving property 3) of Lemma VI.61, that xi and xj belong to the same block of H , if and only if I i - j I= 1. Hence we can denote the blocks of H by

VI. Various Types of Eulerian Trails

VI.86

V

Figure VI.22. The 2-connected outerplane eulerian graph Go having an A-trail Towhich induces a 1-splitting in v. B,, . . . ,B,+l such that z1E V ( B , ) ,x, E V(B,+,) and xi-l,xiE V ( B i ) for 2 5 i 5 r. As for the distribution of the j = 1,.. . ,k, d(v) = 2k, in these blocks B,, i = 1,.. .,T 1, we have

+

I {J1), t A k ) } n V ( B J I=]{&I, Jk)}n v(B,+,) I= 1 since G is 2-connected and because of the definition of dj), j = 1,.. . ,k. W.l.o.g., dl)E V ( B l ) ,dk)E V(B,+,) (observe that the notation for Bi,i = 1, . . . ,r 1, can be chosen either “from the left to the right” or “from the right to the left”). For j = 2 , . . . ,k - 1, let ij denote the index such that v(j) E V ( B i j ) .Of course, there may be blocks of H not containing any &I,j = 1,.. .,k; but for j # rn we have ij # ,i because among the four vertices of Po incident with the elements of E,,cj)U E,(,, , there are two cut vertices of H , each of which separates the remaining two vertices from each other (see the proof of Lemma VI.61, property 3), and observe that G is a simple graph and that a vertex of Po incident with some e E E, - {el, ed}, is a cut vertex of H by Lemma VI.61, property 1)). It follows from the outerplane embedding of G and the definition In addition to the of d j ) , j = 1,.. .,k, that if j < rn, then ij < .,i xi,i = 1,.. . ,r , we introduce zo and x,+~defined in G by vzo = el and ~ x , . +=~e d . Thus, B, n {xj/j = 0 , . . .,r 1) = {xi-l,xi}. Figure VI.23

+

+

VI.3. A-Trails in Plane Graphs

VI.8’7

illustrates the structure of H which follows from Lemma VI.61 and the above argument. V

V

G

H

Figure VI.23. The graphs G and H = GV,,.

The outer face F, of the block-chain H is a 1-face; and Po C bd(F,). Moreover, bd(F,) n E(Bi) # 0 for i = 1,.. . ,T 1 because E ( B i )n E(P,) # 0. This and the chosen notation allow the application of Lemma VI.60. Therefore, an A-trail of G inducing a 1-splitting in v necessarily induces a 2-splitting in each of x, ,. ..,2,. On the other hand, if we can show that Bihas an A-trail inducing a 2-splitting in and zi,i = 1,.. .,T 1, then H has an A-trail by Lemma VI.60, which is equivalent to saying that G has an A-trail inducing a 1-splitting in v. Hence, t o finish the proof of the theorem, it suffices to show that ( B i ) ~ z i - l , z i l , , has an A-trail for i = 1,.. . ,T 1. For this purpose, fix an arbitrary i E { 1,.. .,T 1) and consider B, . Depending on the position of B, in H and its structure, we can consider the following cases.

+

+

+

+

(I) v(B,)n { d J ) / j= 1,. . . ,k} = 8 . Denote for short B = Bi, x = xi-,, and y = xi.Let A,

= (z, v, t,), A, = (y, 21, t,) be two triangles with t ,, t , $ V ( G ) ,and construct a plane em-

bedding of the planar graph

13’ = B U A,

U A,

VI. Various Types of Eulerian Trails

VI.88

such that t , and t , are 2-valent in B+ and lie in the boundary of the unbounded face of B+ (see Figure VI.24). Observe that for the unbounded face F k of B it follows that

V ( b d ( F k ) )= V ( P ( x y)) , , where P ( z ,y) = b d ( F k ) n Po ; otherwise, Po U {el,ed} = bd(Fo) is a cycle in G containing a vertex of the cycle bd(Fk) in its bounded (but not in its unbounded) region, thus contradicting the fact that G is outerplane. Thus, the embedding of B+ as defined above and illustrated by Figure VI.24 is outerplane indeed. Moreover, since B is 2-connected, we can conclude by the same reasoning that P’(x,y) := b d ( F L ) - P(x,y) contains no edge other than x y (see Figure VI.24).

X

E3+

Figure VI.24. Forming the outerplane graph B+ from the outerplane graph B by adding the triangles A, and A*.

Now suppose that B has at most n - 2 vertices of valency exceeding 2; then B+ has at most n - 1 such vertices. So, applying the theorem to B+ we obtain an A-trail T+ of B+ inducing a 1-splitting in v (the 2face-coloring of B+ is induced by the 2-face-coloring of G via the induced 2-face-coloring of B). Observing that T + ,read as an edge sequence, can also be regarded as an A-trail of B:,, we can conclude that by Lemma VI.60, B has an A-trail T such that the vertex splittings defined by X,(z) and X,(y), are 2-splittings.

VI.3. A-Trails in Plane Graphs

VI.89

However, if B has precisely n- 1vertices of valency exceeding 2, induction cannot be applied to B+ as constructed above. So in this case we have to proceed differently. Consider the 2-face F2 of B (which, in fact, is also a 2-face of G) with z y E bd(F2),and form B- = B - {zy}. By Lemma VI.62 we can write the block-chain B- as the union of its blocks,

... U B , , where t =I V ( b d ( F 2 )I) -1, and B; n B3T # 8 if and only if I i - j I= 1. Furthermore, we can denote for j = 1,.. ., t - 1, B-=B,u

{W 3.} = B3

n B3~+l ;

wjE V(bd(F,)) - {z,y} by Lemma VI.62. In addition, denote w, = z and wt= y (w.1.o.g. z E B,,y E I?;). Note that since d,-

( z ) = 1( m o d 2 ) , if and only if

z E {z, y}

,

w e h a v e f o r j e J : = { l ,..., t } d,:

3

(wj-l)= d,:

3

(Wj)3

d B T ( u ) = O(mod2) if 3

1(mod 2),

u # wj-l,wj.

B being outerplane and V ( b d ( F 2 ) )= {wj/j= 0,. . . ,t } yields E(bd(F2))- {zy} = {wj-lwj/jE J}

.

Hence wj-lwjE E(B3T)for j E J, and by the above congruences, Bi := B j - {wj-lwj}is eulerian. Of course, E ( B j ) = 0 may hold true for some (but not all) j E J . Consider J, := { j E J / E ( B i )# 0) and observe that since B3T is a block of B - , it follows that BIT is 2-connected for j E J,; hence Bi is a connected eulerian outerplane graph. By the same reasoning used in proving Lemma VI.62, we can conclude that B; is a non-trivial block-chain; thus it has a cut vertex, and wj-l, wjbelong to different end-blocks of Bi and are not cut vertices of Bi. Summarizing the above applications of Lemma VI.62, we can express B as follows:

B = bd(F2)u

u B;

j€Jo

VI. Various Types of Eulerian Trails

VI.90

=x. = w 1-1

=x.=w

0

1

B = Bi

t

Figure VI.25. A block B = Bi of H , the non-eulerian blocks B; of B - {zy}, i E J, and the eulerian block-chains B; of BY - {wj-,wj},j E J,. where Bi is a non-trivial eulerian block-chain for j E J,; any two of these block-chains have at most one vertex in common, and that vertex belongs to V (bd( F,)). Figure VI.25 illustrates this structure of B. Our aim now is to find for every j E J, two sub-trails covering B; and combine them with the cycle bd(F,) so that it results in an A-trail of B as required. For this purpose let cj be an arbitrary cut vertex of B; . Just as we defined B+ from B (see Figure VI.24), we can now form for every j E J, the 2-connected outerplane eulerian graph B r by attaching

AY)

to B; two triangles A?' = (wj-l,p,tl), = (wj,p,t,) at wj-,,wj respectively, where p , t,, t , g' V ( G )and t,, t, are 2-valent in BF. We now have wj-lin place of x, w j in place of y, p in place of z1, and Bi in place of B , if we compare this construction with the one illustrated by Figure VI.24. The fact that B is 2-connected while B(i is a block-chain (and therefore, cannot contain does not matter.

"3'

lias n: < n vertices of valency exceeding 2; this derives In any case, from the fact that B- is a non-trivial block-chain. Now, by induction we can conclude that BT has an A-trail 'T inducing in c j a 1-splitting. By

VI.3. A-Trails in Plane Graphs

VI.91

Lemma VI.60, TT induces a 2-splitting in wj-17 w j and p (observe that (BT)cj,l is a block-chain with wjWl7wj,pas three of its cut vertices). ’ as an edge sequence as follows: Hence we can write T

+ =Pwj-17 T1( j ) w3-1 .

T‘ 3

7

t t 1 7

1p7pt,7t,wj,T2

(j)

7wjP

where

Ti’)and Ti’) are edge-disjoint closed trails covering Bi. T,(j)induces a 2-splitting in wj-17as does Ti’)with respect to wj. Since Ti’) and Ti’)are A-trails of the respective components of ( B i ) c,1, j it follows that two edges incident with u E V(BS)and which are not consecutive in O+(u), can never form a transition defined by any of Tij),T.j). By analogy to the proof of Lemma VI.60, we can now choose for every j E J, edges ej,f j , g j , h j with ej,fj E E w3-1 . n E ( B i ) , g j hj , E EwjnE(Bi) so that the trails 5”’’) and Ti’) (viewed as edge sequences) can be written in the form

Tij) - ej7Tl,j,fj7 and Ti’)= h j , T 2 , j , g j

.

We can look at the vertices wj,j E J U (0). If j , j + 1 J,, dB(wj)= 2; if j E J,, j 1 @ J,, TiJ) contains edges of Ewj while T,(j+’) does not

+

exist; if j

# J,,

j

+ 1 E J,,

+

Ti’’’)

contains edges of Ewj while Ti’) does

not exist; and if j , j 1 E J,, both Ti’) and T,(j+l)exist and contain edges of E w j . Moreover, if we look at O+(wj) in B , either

O+(wj) = (wj+lwj,wjwj-l)

>

VI.92

VI. Various Types of Eulerian Trails

and taking into account the above equations concerning Ti’)and O+(wj), we conclude that

is an A-trail of B inducing in wj, j E J U {0} and, therefore, especially in wo = and wt = xi a 2-splitting. In other words, (Bi)~zi-l,zsl, has an A-trail. This settles the case (I).

(11) v(B;) n { d j ) / j = 1,.. . ,k}

# 8.

Then there exists precisely one j E (1,. . . ,k} so that v(j) E V ( B i )(see the discussion preceding Figure VI.23). Again denote for short B = B i , z = zi-l,y = xi,set v = d j ) , and let P ( x , y ) = b d ( F k ) n Po be defined as in case (I). In any case, B has fewer vertices of valency exceeding 2 than G. So, if z y $! E ( B ) , we suppress TJ to obtain B* which is of the type treated in (I);and any A-trail of B* corresponds to an A-trail of B , and vice versa. Thus, for x y # E ( B ) induction applied to B’ yields an A-trail of B as required. Whence we assume zy E E(B): B contains a triangular 2-face A = ( 2 1 7 2 , Y).

If A = B , a run through A is an A-trail of B as required because of

our additional definition of a &splitting in 2-valent vertices. Assume, therefore, A # B. In this case we have precisely the situation illustrated by Figure VI.25 with t = 2, Jo = {2}, and TJ in place of w o , x in place of wl, and y having the same meaning as in Figure VI.25. Note that B - v is a 2-connected outerplane simple graph; hence B- := ( B - TJ)- {xy} = B - A is a non-trivial block-chain by Lemma VI.62. As in case (I) we find c E P ( z ,y) - {z,y} which is a cut vertex of B - , and by induction an A-trail TB of B which induces a l-splitting in c and, therefore, a 2-splitting in x and y. Thus, in all possible instances we have reduced case (11) to case (I). Theorem VI.63 now follows. It is tempting to suspect that in the hypothesis of Theorem VI.63, one can drop outerplanarity to obtain a true statement for an even larger class of eulerian graphs. Or, in more cautious terms, one could ask whether it is true that every simple 2-connected eulerian plane graph has an A-trail. In fact, when I first considered the problem of finding A-trails in plane

VI.3. A-Trails in Plane Graphs

VI.93

eulerian graphs, I thought that 2-connectedness would be a sufficient condition for a plane eulerian graph G to admit a partition {V., Vi} of V(G) - {v} for some v E V(G), such that (Gvl,,)v;,2 does not contain a subgraph homeomorphic to the graph G, of Figure VI.17. For, if one considers the problem of finding an A-trail in a connected plane eulerian graph G from an algorithmic point of view, it follows from Theorem VI.59 and the discussion of Figure VI.17 that G has an A-trail, if and only if for every v E V(G) there is a partition {V., Vi} of V(G) - {v} with the property just described. Therefore, the graph G, of Figure VI.17 characterizes the forbidden stage that one has to avoid in trying to produce an A-trail by a sequence of S-splittings, 6 E {1,2} (here, homeomorphic includes the meaning it has in the topology of the plane. In this sense, the two graphs of Figure VI.17 have to be viewed as different). Unfortunately, there exist 2-connected plane eulerian simple graphs having only 4- and 6-valent vertices and not having any A-trail. In order to see that the graph Go of Figure VI.26 is such an example, we assume first w.1.o.g. that Go is 2-face-colored with the outer face being a 1-face. is Suppose now that Go has an A-trail T. Observing that (Go)(YI,u61,2 a disconnected graph and applying Corollary VI.57, it follows from the symmetry of Go that this graph also has an A-trail inducing a l-splitting in vl. W.1.o.g. suppose T has been chosen with this property. This implies, however, with necessity that the 6-splitting induced by T in u;, 2 5 i 5 6, satisfies the congruence i G 6 (mod 2). That is, we arrive at the graph H = ((G0){UlrV3,U5},l {VZ,V4,V6}r2 as illustrated by Figure VI.27, where T , viewed as an edge sequence, is an A-trail of H as well. Now, if one applies the corresponding &splitting to every 4valent vertex of H , one obtains a graph homeomorphic to the graph G, of Figure VI.17.') No matter which &splitting is applied to every vertex of V(Go)- {v}, 6 E {1,2}, the result is either a disconnected graph or, at best, what has been termed the forbidden stage whence we can conclude that Go cannot have an A-trail.

1

Interestingly enough, a slight change in the embedding of Go yields the plane graph GT which in fact does have an A-trail T (see Figure VI.28); the checking of the validity of this assertion is left as an exercise. '1 In fact, for every splitting of the 4-valent vertices of H which does not yield a disconnected graph, the result is - homeomorphically - the same, namely G, of Figure VI.17.

VI.94

VI. Various Types of Eulerian Trails

Figure VI.26. A 2-connected plane eulerian simple graph Go with 4- and 6-valent vertices only and having no A-trail. Hence, the existence of an A-trail in a 2-connected plane eulerian graph in general depends on the actual embedding of the underlying (abstract) planar graph, a fact we have already encountered in graphs of connectivity 1 (cf. Figures VI.17, VI.26 and VI.28). So, one might ask whether for a given 2-connected planar eulerian graph G, there exists an embedding H of G on the plane such that H has an A-trail. However, S. Regner (with the help of the author) constructed 3-connected planar eulerian graphs which had no A-trails. Since for any two embeddings H,, H 2 of such a graph we have either Ozl(G) = O z 2 ( G )or Ozl(G) = OH2(G) (see Theorem 111.52), we can conclude that in the case of 3-connected

VI.3. A-Trails in Plane Graphs

VI.95

H Figure VI.27. H obtained from Go by applying a 6splitting to U 2 j - 2 + 6 , j = 1,2,3, 6 = 1,2.

planar eulerian graphs, the existence or non-existence of an A-trail in such a graph is independent of the actual embedding of that graph. On the other hand, in view of Lemma VI.60, it follows that if one has a planar 3-connected eulerian graph G with no A-trails, the planar eulerian graph H I = G' U G", where G' and G" are copies of G and G' n G" = { u } E V ( H , ) (hence K ( H , ) = l), does not admit an A-trail in any plane embedding of H , . Furthermore, if H2 = G' U G", where G', G" are as above and G' n G" = { u , zu} E V ( H 2 )(hence & ( H 2 )= 2), no embedding of H2 in the plane has an A-trail either. This conclusion follows directly from the next lemma (see also [REGN76a, Bemerkung 3.21). There, G t will denote the graph constructed from G, in the same way that B+ was

VI.96

VI. Various Types of Eulerian Trails

Figure VI.28. GF obtained from Go of Figure VI.26 by changing the embedding of the edge vlvs only: v1v6 belongs in Go to the boundary of the outer face, in GOT to the boundary of a 2-face containing v. GF has an A-trail, while Go does not.

constructed from B in the proof of Theorem VI.63 (see Figure VI.24).

Lemma VI.64. Suppose for the plane 2-connected eulerian graph G that it can be written in the form

VI.3. A-Trails in Plane Graphs

such that Gi is eulerian, K ( G ~2) 2 and G in G,+, =

G , n G j = Q if i + l < j l k ,

i = l , ..., k

VI.97 E V(G),

,

+

(putting k 1 = 1; if k = 2, then I G, n G, [=I {v,,,, v,,,} I= 2). Suppose further that G is 2-face-colored with the outer face F, being a l-face, and that E(bd(F,)) n E(Gi) # 8 for i = 1,.. . ,Ic. The following statements are equivalent,

1) G has an A-trail. 2) The notation can be chosen in such a way that

either G,, G, have A-trails inducing a l-splitting in v,,, and a 2splitting in v1,2,v k - l , krespectively, while for i = 2,. . . ,k - 1, G ihas an A-trail inducing 2-splittings in both v ~ - , , ~vi,;+,; , or Gi has an A-trail inducing 2-splittings in both ~ ~ - , , ~ , v ~for ,~+, i = 1,.. . , k - 1 (putting 2 1 ~ = , ~vk,,) and G i has an A-trail inducing a 2-splitting in each of u ~ - , , ~z ), ~ , v, ~ , where {v} = V4(Gi) - V4(G,).

Proof. We first observe that there are precisely two faces containing all vertices v ~ , ~ + i, ,= 1,.. . ,k , in their respective boundary, namely F, and another l-face, call it F,. This follows directly from the hypothesis of the lemma. Consequently, for i # j , 1 5 i, j 5 k , H = G~vi,..+l,vj,j+,},, is a disconnected graph: in order to see this, take simple open curves 1' Flu{vi,i+l, "j,j+l} and ccc F ~ u { v i , i + 1 7 "j,j+l } each joining } hence C = C, UC, is a simple and u ~ , ~ +C,, . nC, = { v , , ~ + v~ ~, , ~ + ,follows; closed curve containing open edges of G in its bounded and unbounded region. Therefore and because each Gj is eulerian, i = 1,.. ., k , it follows that H can be constructed (topologically) in such a way that C also contains open edges of H in its bounded and unbounded region and C n H = 8. We can thus conclude that H is disconnected (compare this with the proof of property 1) of Lemma VI.61). From these observations it now follows that if G has an A-trail T or if such T should be constructed from corresponding A-trails of Gi, i = 1,.. . ,k - 1, GZ respectively, T or the construction of T induces a 2-splitting in all but one vertex vi,;+,, i = 1,... ,k , at the most. Now suppose G has an A-trail T inducing a l-splitting in some v , , ~ + , . W.1.o.g. we may assume the notation chosen in such a way that i = k (if necessary, one applies a cyclic permutation to the subscripts in order to obtain i = k ) . Construct the plane graph H Z y from G by replacing

VI. Various Types of Eulerian Trails

VI.98

with two vertices x,y $! V ( G )such that E, = EUk,,n E(Gl), E, = E U k ,n ] E(G,). Hence, H,, is a non-trivial eulerian block-chain whose blocks Hiare defined by E(Gi),i = 1,.. .,k, and whose cut vertices are the vertices v++,, i = 1,.. . ,k - 1. Another way of obtaining Hxywould ,1 which are incident with edges be to identify the 2-valent vertices in GUk,] of G, (Gk), thus yielding the new vertex x (y). It follows, therefore, that ( H Z J {,], = Guk,l,l. Carrying the 2-face-coloring of G over to H,,, we can conclude from Lemma VI.60 that any A-trail TH of Hxy induces an A-trail T j in Hi= ( E ( G i ) )i, = 1,. . .,k, such that Ti induces a 2-splitting in the vertices of Hiwhich are cut vertices of Hxy.Viewing T as an edge sequence, T is also an A-trail of H,, and (Hz,)Ix,,l,l. Consequently, in this case the existence of T in G implies that G ihas an A-trail inducing 2-splittings in T J ~ - , , vj,++, ~, for i = 2,. . .,k - 1, and both G, and Gk have an A-trail inducing a l-splitting in v k , l and a 2-splitting in u , , ~ uk-1 , ,k respectively. z]k,l

Conversely, if Gi has an A-trail as described above, Hxy,as constructed above, has an A-trail TH by Lemma VI.60, and TH induces l-splittings '= ' U k , ] , l , it follows that Guk,l,, also has in x and y. Since (Hz,){z,yl,l an A-trail, i.e., G has an A-trail inducing a l-splitting in vk,,. Now suppose that G has an A-trail T inducing a 2-splitting in TI^,^+,, i = 1,.. . ,k. Similar to the notation used in the proof of Lemma VI.60, we can choose edges e i , fi,gi,hi E E(Gi),i = 1,. . .,k, such that

n E(bW,)),

e; E J%;+

gi E

Eu;,i+l

n E(b'('m))7

"

fi E Ev,-,,, W4Fl)), hi E Eu;,;+ln E("(F1));

and w.1.o.g. we may assume that the notation for defining G has been = (e:+,,g:, . . .,hi, f;+l,.. .). chosen in such a way that O+(V;,~+,)

Let us now look at the behavior of T in a fixed Gi, 1 5 i 5 k. Since T is an A-trail, the only transitions of T containing precisely one half-edge of G iare { e : ,gi-l}, { hi-l}, {gi, e:+l}, {hi,fi+l}. Therefore and since G is plane, T cannot be of the form

fl,

T = . . . ,e ; ,T,!,hi, . . . ,g i , T;",fi,. . .

(1)

where E(T;)U E(T,'I)= E(G;)- { e i , fi,gi,hi}. Consequently, T must be either of the form

T = . . . , e i , T ; , g i , . . . , hi,T;-, fi,.. .

(2)

VI.3. A-Trails in Plane Graphs

(where { g i , hi} fi! X , since T does not induce a l-splitting in of the form T = . ..,ei, T i ,fj, .. . ,hi, T:*,gj,. .. ,

VI.99

or

(3) where E ( T F ) U E(TF-) = E(T,*)U E(T,**)= E(Gi) - {e,, fi,g j , hi} (we assume w.1.o.g. that ei is the first edge of Gi in T). Thus, if T is of the form (2), Ti := ei, TF,g,, h j ,TF-, fj

is an A-trail of G j inducing a 2-splitting in each of t ~ , - ~ , ~ , v ~If ,T~ is +~. of the form (3), Ti+ := ux,e,,T:, fi7xtl,tlu,vt2,t2y,hi,T~*,gi, yv ,

(4)

where z := u , - ~ , ~y , := and v , t l , t 2 @ V(G),is an A-trail of GT inducing a 2-splitting in each of v, t ~ ~ (see , ~ Figure + ~ VI.24 with G, in place of B).

To see that T must be of the form (2) for all but one G,, and that it must be of the form (3) for the remaining G;, i E (1,. . .,k}, we observe that T cannot be of the form (2) for all i = 1,. . . ,k; otherwise, we would have

i.e., T does not cover all of E(G) since {gi.ei} E X , (note that {g:,e:+,) E X,, i = 1,...,k). We can thus conclude that T must ‘turn around’ somewhere, but because T induces a 2-splitting in every Z J ~ , + +i~ ,= 1,. . . ,k, this ‘turn’ must take place within some Gj, j E (1,. . .,k}. W.1.o.g. we can assume that the first ‘turn’ takes place for j = k if T starts with ei, i E (1,. . .,k - 1). Thus T is of the form

But then

is an A-trail of G- := G - G k . Since by definition G- is a block-chain of G if k 2 3, while by hypothesis it is 2-connected for k = 2, it follows

VI. 100

VI. Various Types of Eulerian Trails

from Lemma VI.60 for k 2 3, and from (6) for k = 2 respectively, that T induces an A-trail Tiin G, such that Tiinduces a 2-splitting in the , ~well. Moreover, Tk+, cut vertices of G- and, by (6), in vk-1,k and 2 1 ~as defined as Ti+ in (4), is an A-trail of G: as stated in the lemma.

To finish the proof, it suffices to show that if Ti is an A-trail of Gi for i = 1,. . ., k - 1, and if T t is an A-trail of G i as stated in the conclusion of the lemma, G has an A-trail. Define G- as above and apply Lemma VI.60 to G-; this results in an A-trail T - of G- which induces a 2-splitting in both vk-l,k and vk,l. Choosing the same notation for the definition of the A-trails Ti of Gi as above, we can write T - as defined in (6) and TZ as defined in (4) (with the subscript k replacing the subscript i). T as expressed by ( 5 ) can now be obtained by inserting in (6) the corresponding subtrails of (4);and T is an A-trail of G. Lemma VI.64 now follows. In fact, Lemma VI.64 can be generalized, however in one direction only. This generalization is the natural extension of Exercise IV.l .c), but again only in one direction. Lemma VI.65. (see [REGN76a, Lemma 1.41) If the connected plane eulerian graph G has an A-trail, then every block of G has an A-trail.

The proof of Lemma VI.65 and the construction of an example showing that the converse of Lemma VI.65 does not hold, is left as an exercise. Before we turn to the construction of planar %connected eulerian graphs with no A-trails, we have to study A-partitions. For this purpose, we may introduce some concepts. Definition VI.66. Let Fl, . . . ,F,, m > 1, be m distinct faces of the 2face-colored plane eulerian graph such that, for a fixed 6 E { 1,2}, Fiis a 6-face for i = 1, . . . ,m. Let us assume distinct vertices vl,,, v , , ~ ,. . . , exist with TI;,++^ E bd(F,) n bd(F;+,), i = 1,. ..,rn (with the subscripts read mod m). Then we call R = {Fl,.. . ,F,} a unicolored face-ring of G and L, := {vl,*, . . .,v,+} a complete set of links of R (if m = 2, then %,2 # 7-J2,1>* The following result relates A-partitions to face-rings and complete sets of links (see [FLEI74a, Theorem 1 and Corollary 11)Theorem VI.67. Let G be a 2-face-colored 2-connected plane eulerian graph without 2-valent vertices, and let {Vl,V,} be a partition of V ( G ) . The following statements are equivalent.

VI.3. A-Trails in Plane Graphs

(1) G has an A-trail

VI.101

T whose A-partition is { V, ,V,}.

(2) For every unicolored face-ring R of G and every complete set of links LR of R, if the elements of R are 6-faces then L , V,, 6 E (17 2).

Moreover, if we replace in (1) “A-partition” with “perfect A-partition” and in ( 2 ) “LR V,” with “V, 2 LR sf V,”, then these stronger statements are equivalent as well. Proof. (1) implies (2). Suppose for some unicolored face-ring R of G, whose elements are &faces, and for some complete set of links LR that LR V,, 6 E {1,2}. We may assume w.1.o.g. that 6 = 1. We claim that H := GLRIlis disconnected. Since R is a unicolored face-ring of G and 6 = 1, distinct 1-faces F,, . . . ,F, of G exist such that

R = {Fl, . . ,F,},

8 # bd(Fi) n bd(Fi+,) 5 V(G), i = 1,. . . ,m ,

and by definition,

Now let C, C FiU bd(Fi) be a simple open curve with Ci n bd(F’) = {u~-,,~ t ~ , ; , ~ + Then ~ } . C = ,U : C, is a simple closed curve. Precisely because Fiis a 1-face, i = 1,.. .,m, C contains in its interior (the bounded region of C), as well as in its exterior (the unbounded region of C) a positive even number of open edges of Eui,i+l, i = 1,.. .,m (otherwise, Ci and Cj+, would lie in faces of different colors, or else Ci = C,+, which implies Fi = F,+,). Moreover, C n G = L , 5 V, by assumption. W.1.o.g. suppose the notation chosen in such a way that for 0+(vi,;+,) = (ei,e;, .. . ,e & , ) , d, = d ( ~ , , ~ + ,> ) 2, precisely the open edges el,e,, . . .,en; lie in the interior of C. Then 2 5 ni = 0 (mod2) and di - ni 2 2 follows from the preceding paragraph. Then we can perform the 1-splitting in vi,++,, i = 1,. . . ,m, in such a way that ~ ( ~ .1. .,,d m i ) lie entirely in the interior of C, while dmi+’) ,.. . ,v(lCi) lie entirely in the exterior of C, where m i = $ n i and ki = i d , (see Figure VI.18). Consequently, since C n G = LR C V,, we can conclude that H is disconnected. Since H is disconnected and LR C V,, it follows that G,,, is disconnected, and further that (Gv1,1)v2,2is disconnected. On the other hand,

VI.102

VI. Various Types of Eulerian Trails

{V,, V,} being the A-partition of an A-trail T of G with v 6 containing precisely those vertices at which T induces a &splitting, 6 = 1,2, implies that (Gvl,!)v2,2 is a cycle (compare this with Corollary VI.57). This contradiction proves the implication. ( 2 ) implies ( 1 ) . Consider H = (Gvl ,,)& , which is a 2-regular graph. If

H is connected, then by Corollary VI.57, {V,,V,} is an A-partition obtained from some A-trail T inducing 6-splittings precisely in the elements of v6,6 = 1,2. In this case, the implication follows. Hence suppose H to be disconnected. Choose V, V ( G ) as small as possible such that Ho := (Gvi,l)vdf,2is disconnected, where Vd = V, n V,, V', = V, n V,. It follows that H , has a face F, such that bd(F,) is disconnected; w.1.o.g. F, is the unbounded face of H,. Furthermore, since we did not make any assumptions concerning the 2-face-coloring of G, we may now assume that F, is a 1-face. Now, if V', # 0, consider w E V:. We claim that already H , := (Gv,, ,1) Vi' - { w } ,2 has a disconnected face boundary (whence H , is disconnected): for, H , = (H1)jw1,27 and by definition of a &splitting, 5 E {1,2}, the application of the 2-splitting to a vertex of H , leaves the 1-faces of H , (homotopically) invariant (see the footnote at the end of Definition VI.54); i.e., bd(F,) is disconnected already in H,. This contradicts the choice of V,; hence V: = 0 and, therefore V, & V,. Again by the choice of V,, and since G is connected, we must have V, c VFo for VFo:= V ( ( E ( M ( F o ) ) ) Gsince ) , V, V, and H , being disconnected implies that G,,,Fo,l has a disconnected outer face. Consider C,,a component of bd(F,), define C2:= bd(F,) - C,, and let C be a simple closed curve lying in F, and such that int C 3 H,, ext C 3 H , - H I , where H , is the component of H , with H , 3 C,. The choice of V, implies that for every v E V,, some of the di) lie in int C (i.e., in H,) and some lie in ext C (i.e., in Ho - H,), i = 1,.. .,4 d c ( v ) ; otherwise, one obtains C n Gvo-{ul,l = 0 and intC n Gvo-lul,l # 0 # extC n G v o - ~ v ~ , l for some 21 E V,. W.1.o.g. we may assume that C is drawn in such a way that the transition from H, to G, viewed as a topological procedure, leaves C invariant (i-e., w.1.o.g. C is a simple closed curve in G as well compare this with Figure VI.18) and CnG = V, while (C - V,) nF(,) = 0 for every 2-face F ( 2 )of G.

C n G = V, implies that we can write V, as vO

= {'1,2

'2,3,

* * 7

vm,l)

(ill

VI.3. A-Trails in Plane Graphs

VI. 103

where vi-l,i and vi,++,divide C into two simple open curves C: and Cy such that C: n V, = {u;-,,~,v ~ , ~ + ~C:} . { v ~ - , , ~vi,++,} , lies in a uniquely determined 1-face, call it Fi, i = 1,.. . ,m. It follows from the choice of V, that

Fi#Fj

for i # j , l < i , j < r n

,

(i2

1

and that

bd(Fi) n b d ( F j ) # 0 if and only if

Ii - j

IE

(1, rn - 1)

,

(i3)

where

Otherwise, one could draw a simple closed curve C- in the plane such that C- n F(,) = 8 for every 2-face of G,C-nG= V; c V,, and (C- V,-) n Fi = 0 for at least one but not all i E (1,.. .,m}; and csing an argument similar to the proof of the first implication, these properties of C- imply that Gv- is disconnected. Consequently, the validity of (i,) 0 ’ - (i4) classifies R = {F, ,. . .,F,} as a unicolored face-ring and V, as a complete set of links of R. Since R is a set of 1-faces and V, C V,, R and V, violate the validity of (2); hence H must be connected in any case.

,

To finish the proof of the theorem suppose now that the A-partition in (1) is a perfect A-partition, and suppose in (2) that L , fulfills the stronger relation V, 2 LR V,; call the corresponding statements (1’)and (2’).

If {V,, V,} is a perfect A-partition, then, by definition, (GVl,1)V2,2is a cycle (which corresponds to T),and so is (Gvz,l)v,,2. Denote by T* the A-trail of G corresponding to the second cycle and define V;C := V,, V; :=

V,. Since (1) implies (2) a twofold application of this implication yields VgC 2 L , sf V, for every unicolored face-ring R and every complete set of links LR; i.e., LR V,, 6 = 1,2. Whence we conclude that (1’) implies (2’). Observe that {V6,V:} = {q, V,} independent of the choice of 6 E {1,2}; hence one does not have to specify the color of the elements in the unicolored face-ring R.

Suppose (2’) holds. Define Vgt as above. Since (2) implies (1) we obtain from the proof of this implication that both (Gvl,l)v2,2 and (Gv;,l)vg,2 = (Gv2,1)vl,2are cycles; i.e. {V,,V,} is a perfect A-partition. Theorem VI.67 now follows.

VI.104

VI. Various Types of Eulerian Trails

The theorem just proved indicates that if we look at G, = (V,) C G, 6 = 1,2, where (V,,V,} is an A-partition and G satisfies the hypothesis of the theorem , then G, must have a very special structure. In order t o determine this structure, let us have a look at an arbitrary cycle C of G which is not a face boundary. It follows that both int C and ext C contain (open) edges of G (where we view C as a simple closed curve in the plane). Furthermore, both int C and eztC contain 1-faces and 2-faces as well. In particular, every e E E ( C ) belongs to some bd(F,) where F, is a &face, 6 = 1,2. Hence, considering

R, = {F,/E(bd(F,)) n E(C)# 0, F, is a 6-face},

S = 1,2 ,

we conclude that R, is a unicolored face-ring having a complete set of links LR V ( C ) ,6 = 1,2. By Theorem VI.67, V ( C )If V,, and therefore, G,, 6 = 1,2. Hence,

c

if G, contains a cycle K , K is the boundary of a &face.

(4

If we suppose that G is just a 2-face-colored plane eulerian graph with K(G)= 1 and possibly 2-valent vertices, the conclusion (*) is still valid: for an A-trail of G induces an A-trail in every block of G (see Lemma VI.65), and any cycle of G, is necessarily contained in some block of G. Moreover, if a cycle C c G contains a 2-valent vertex of G, and if C is not the face boundary of a &face, R, as above can be constructed as well, with L , 2 {v E V(C)/d,(v) > 2) as a complete set of links of R,, 5 E {1,2}. In this case, L , If V,, S E {1,2}, by Theorem VI.67; hence ( L R ) C G,, and moreover, C If G,.

Summarizing these considerations we therefore arrive at the following result [REGN76a7Satz 2.11. Corollary VI.68. If T is an A-trail of the connected 2-face-colored plane eulerian graph, G, and if {V,,V,} is a corresponding A-partition of G, G, := (V,) contains a cycle C, only if C is the boundary of a 6-face, 6 E {1,2}.

We point out, however, that neither G, nor G, need to be connected. This fact is exemplified by Figure V1.29. Moreover, the converse of Corollary VI.68 does not hold even if G, is acyclic. In other words, G may have a vertex partition {V17 V,} such that G, = (V,) is acyclic, although G does not have an A-trail. This can be seen by studying the graph Go of Figure VI.26; we leave this as an exercise.

VI.3. A-Trails in Plane Graphs

VI

V4

V4

VI.105

V

1

Figure VI.29. (a) The four-sided antiprism H , (b) an A-trail T of H , and (c) the subgraphs induced by the Apartition of T.Observe that H is 4-regular and has, except for two 4-gonal faces, triangular faces only. The next result, however, shows that the converse of Corollary VI.68 is true provided G, is connected. For this result, however, we need some considerations on unicolored face-rings. Let us consider a unicolored face-ring R = {Fl,F 2 , .. .,Fm}, rn 2 2, of the 2-connected, 2-face-colored plane eulerian graph G. Suppose

6 d ( F i ) n b d ( F j ) = 0 for

li--j1>1, l s i , j s m ,

except if

{ i , j } = {l,rn}.

(*I

Since G is 2-connected bd(Fi) is a cycle for i = 1,.. .,rn. Choose the notation as in the proof of Theorem VI.67. Let LR be a complete set of links. With LR we associate a simple closed curve C = C(LR) as in the first part of the proof of Theorem VI.67, where C = U ~ l C i Ci , C Fi, and C n G = L,. Of course, C not only depends on the choice of Ci C Fi, but also to a greater degree on the choice of L , which is uniquely determined if and only if I bd(Fj)n bd(Fi+l)I= 1, i = 1,.. . ,m (e.g. if G is 3-connected). In any case, among all possible choices for L , and the associated curve C(L,), choose L(H0)and C, = C(L',o') in such a way that for every complete set of links L h

intCo nL', = 0

.

(**I

VI. 106

VI. Various Types of Eulerian Trails

Observe that if bd(Fi) n bd(Fj+l) 2 { v ~ , ~ v:,~+,}, + ~ , one obtains for every complete set of links Lh with vi,++, E Lk another complete set of links by ) {v:,~+~}. Whence we may conclude that defining Lk = ( L k - { V ~ , ~ + , } U the choice of La“’and Co satisfying (**) is possible. We note that La“’ is uniquely determined. Similarly, let La‘’ be the uniquely determined complete set of links such that C, = C(L$’) and

eztC, n Lk = 0

(* * *)

for every complete set of links Lk. By (*) and since G is 2-connected, it follows that there are cycles C,,C, c G defined by

Now we are in a position to present another result of S. Regner’s Ph.D. thesis, (see [REGN76a, Satz 2.21). Her original proof contains a minor flaw; the proof presented here differs considerably from hers insofar the details are concerned.

Corollary VI.69. Consider a vertex partition {V17 V,} for the plane, 2-connected7 2-face-colored eulerian graph G. If G, := (V,), 6 = 1,2, is connected, and if every cycle of G, is the face boundary of a &face, {Vl,V,} is an A-partition. Proof. In view of Theorem VI.67 it suffices to show that for every unicolored face-ring R and every complete set of links L,, if the elements of R are 6-faces, LR V,, 6 E {1,2}. We proceed indirectly by choosing a unicolored face-ring R with minimal I R I whose elements are &faces, and such that for some complete set of links LR,LR C V,, 6 E {1,2}; w.1.o.g. S = 1. Denote explicitly R = {F17.. . ,Fm},rn2 2, and L, = {ul,,, . . .,v,,~}. We have bd(F,) n bd(F’) = 0 for l i - j l > 1, 1 5 i , j 5 rn, unless { i , j } = {l,rn}; this follows from the choice of R. Now choose the curves Ci71 5 i 5 rn, and define the simple closed curve C with CnG = LR C V, as in the first part of the proof of Theorem VI.67. It is precisely because of C nG = LR V, and since G, is connected that either intCnV,=0

or eztCnV,=0

;

VI.3. A-Trails in Plane Graphs

VI.107

for, as an application of the Jordan Curve Theorem, every path P ( z ,y) in G joining x E int C with y E e z t C satisfies P ( z ,y) n L , # 0. W.l.o.g.,

intCnV,

=0 .

(1)

Using the considerations preceding the formulation of Corollary VI.69 and (1) we may even assume that LR and C have been chosen in such a way that L R = La"' and C = C,, thus satisfying (**). Let C, be the cycle defined in (* * * *). By (1) and the definition of C,,and because of the assumption L , C V,, we conclude C, C (Vl). Now, since every edge of C, is a boundary edge precisely of some element of R and of some 2-face7 and since R contains at least two elements, it follows that C, cannot be the face boundary of a 1-face. This contradiction to the hypothesis finishes the proof of the corollary. Consequently, as an application of Corollaries VI.68 and VI.69, a partition {V,, V,} of V ( G )where G is 2-connected7 plane and eulerian, is a perfect A-partition only if (V,) and (V,) are acyclic; and if (V,) and (V,) are trees, this partition is a perfect A-partition. We observe that the A-partition of the graph in Figure VI.29 is perfect, but the corresponding graphs (V,),(V,) are both disconnected forests, while the graph of Figure VI.26, although not having any A-trails, admits such vertex decomposition (see Exercise VI.20). Note that the graph of Figure VI.29 has two 4-gonal faces only, while its other faces are triangular. However, if G is an eulerian triangulation of the plane, a vertex partition {V,, V,} induces acyclic subgraphs (V,), (V,), if and only if these subgraphs are trees (see Proposition 111.63). So we arrive at the following result.

Theorem VI.70. A vertex partition { V,, V,} of an eulerian triangulation of the plane is a perfect A-partition if and only if (V,) and (V,) are trees. We can restate Theorem VI.70 without using the concept of an Apartition or a perfect A-partition, respectively. In order to see this we reconsider Corollaries VI.68 and VI.69 in the case of eulerian triangulations of the plane. In this case, these two corollaries are the converse of each other; for, if (VJ is disconnected, (Vj)has a cycle which is not a face boundary (see Proposition 111.63, and [FLEI74a, Corollary 2]), { i , d = {1,2). Now let D denote an eulerian triangulation of the plane which is given together with its 2-facecoloring; suppose it has an A-trail T with { V,, V,}

VI. 108

VI. Various Types of Eulerian Trails

as the vertex partition corresponding to T . Suppose (V',) contains a cycle A , S E { 1,2}; A is the (triangular) boundary of a &face by Corollary VI.68. Denote E(A) = {el, e,, e , } . By definition, T induces a &splitting in all vertices of V(A); i.e., {{ei, ei}, { e ; , e $ } , { e y , e y } } n X, = 0. In other words, any two edges of A are not consecutive in T . We say that T separates the edges of A. Similarly, we call the A-trail T nonseparating if and only if T does not separate any face boundary of D. On the other hand, if T separates the edges of some face boundary of D , the corresponding (V',) contains a cycle (which is a triangle). Hence, we can restate Theorem VI.70 as follows.

Theorem VI.70.a. For a vertex partition {Vl, V,} of the eulerian triangulation of the plane, D , the following statements are equivalent. 1)

(V,) is a tree for i = 1,2.

2) D has a non-separating A-trail.

Observe that for defining a non-separating A-trail, one need not start with a 2-face-coloring; one can also start from the vertex partition {V+, V-} related to O+(D), O - ( D ) respectively (see Lemma VI.53). Theorems VI.70 and VI.70.a are familiar. For we know that a 2-connected plane cubic graph G, has a hamiltonian cycle if and only if its dual D = D(G,) has a vertex partition {Vl,V,} such that both (V,)and (V,) are trees (Lemma 111.74). And if G, is bipartite as well, then D = D(G,) is an eulerian triangulation of the plane in which case D has multiple edges if and only if 4 G 3 ) = 2. Hence we arrive at the following equivalence concerning hamiltonian cycles and A-trails, [FLEI74a, Theorem 21. Theorem VI.71. A 2-connected, plane, cubic, bipartite graph has a hamiltonian cycle if and only if its dual has a non-separating A-trail. There is a much more direct proof of Theorem VI.71. So far unpublished to the best of my knowledge, I attribute it to W.T. Tutte.l) Because of ') I a m not sure that W.T. Tutte knows that I know that this proof is his, but his proof is the essence of a criticism of the approach to A-trails as presented in [FLEI74a]; when the original manuscript was returned t o me together with the referee's report, Tutte's name was improperly erased. Nevertheless I decided t o adhere t o the approach as presented here and, earlier, in [FLEI74a, REGN76al; for it proved fruitful to some extent, and the theory on A-trails developed so far might be of some use for finding a proof of the conjectures on A-trails stated below.

VI. 109

VI.3. A-Trails in Plane Graphs

its elegance and relative shortness we present it in detail (the essence of the proof is Tutte’s, the phrasing the author’s).

Tutte’s proof of Theorem VI.71. Let H be a hamiltonian cycle of the 2-connected, plane, cubic, bipartite graph G,. From a topological point of view, H constitutes a simple closed curve in the plane such that some of the open elements of L := E(G,) - E ( H ) lie in ezt H and some in int H . Write H as an edge sequence, H = e1,e2,. . .,e2k, denote by s(ej) a point of the open edge ej, j = 1,.. ., 2 k , and construct now a simple closed curve C as follows: for i = 1,.. .,k, let C2i-1 be a simple open curve joining s(e2i-l) and s(e2J such that

(and lying sufficiently close to the section of H containing the vertex incident with e2i-1, e2i). Similarly, for i = 1,.. . ,Ic let C2; be a simple open curve joining s(ezi) and s(e2i+l) such that c2i

- {~(ezi), s(e2;+1)} c e x t H

putting e2k+l = el. W.1.o.g

I Cjn e I<

7

1 for e E L and j = 1,.. . ,2k.

u 2k

c := cj j= 1

has the property that every e f E(G,)has precisely one point in common with C: for e E E ( H ) this follows from the definition of C, and for e E L this is a consequence of I Cj fl e I< 1 and G, being bipartite: note that e = zy divides H into two paths P1(I ,y), P2(2,y) such that Pl( 2,y) U{ e} and P 2 ( z ,y) U { e } are even cycles; and in a walk along C, if some vertex of H lies to the left of C, the next vertex in H lies to the right of C. For every j E { 1,.. .,2k}, there are faces F’ and ’F of G, such that Cj

n F’ # 0 # Cj n’F

and

Cj

nF

= 0 for every face F

# F’,

’F

and

F!3 # F!’3 if and only if C j n e # 0 for some e E L

.

This follows from the preceding paragraph. Next we choose a point vi E Cj f l F ’ and a point v[i’ E Cj fl F’, with vi = v[i’ if and only if F’ = F’.

vI.110

VI. Various Types of Eulerian Trails

Declaring the elements of the set VT := {vi,v y / j = 1,.. .,2k} as vertices we have turned C into a plane (graph theoretical) cycle T whose open edges are precisely the (topological) connected components of C - VT. By construction, every edge of T has a point in common with precisely one edge of G,, and vice versa; and this point is uniquely determined. Also by construction, for every face F of G,, there exists a topological disc I( C F such that KnC = {x E VTnF } . Hence, for every face F of G, we can contract K onto a point vF in such a way that no pair of open edges of any deformation of T (occuring during the reduction process) ever intersect. Consequently, since 2’ is a plane cycle (and thus can be viewed as an A-trail of itself), T is transformed into an A-trail of the graph D whose vertices are the above vF and whose open edges correspond bijectively to the connected components of C - VT. This A-trail of D is indeed non-separating, precisely because H is a hamiltonian cycle of G,. By construction of T and D there is not only a bijection between vF E V ( D )and the faces F of G,, but we also have for any faces F’, F” of G, that e D = V p v F t , E E ( D ) ,if and only if e E E(bd(F’))flE(bd(F‘’)) exists and the connected component of C - VT corresponding to e D has precisely one point in common with e. This, however, classifies D as the dual of G,. Conversely, suppose D = D(G,) has a non-separating A-trail To. Consider the cycle T obtained from To by appropriately splitting every v E V ( D ) into $d(v) 2-valent vertices vF belonging to the face F of G, for which v E F . By construction, every open edge of T crosses precisely one open edge of G, in precisely one point, and vice versa. Thus, for the plane simple closed curve C corresponding to T we have C n e = { s ( e ) } for every e E E(G,). C corresponds to the plane cycle C with V ( C )= { s ( e ) / e E E(G,)} such that the edges of C correspond to the connected components of C-V(C). Therefore, and because T O is nonseparating, we have for every v E V ( G ) : if E,, = {e,,,f,,,g,} C E(G,), then w.1.o.g. either s(e,)s(f,) E E ( C ) and s(e,)s(g,), s(g,)s(f,) $! E(C), or ~ ( ~ , , ~ ~ ~ s , , ~ E ,E~ ( C~) and s , )S(eu>s(f,) ~ ( f , , ) $! E(C).Now define H by E ( H ) = {e,,, f,/v E V ( G ) } . By definition, H is a 2-regular spanning subgraph of G,, and because C is a simple closed curve with C n e = { s ( e ) } for every e E E(G,), therefore H is connected. That is, H is a hamiltonian cycle of G,. This finishes the proof.

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI. 11 1

VI.3.1. The Duality between A-Trails in Plane

Eulerian Graphs and Harniltonian Cycles in Plane Cubic Graphs With Theorems VI.70, VI.70.a and VI.71 at hand it becomes clear that there is a close relationship between A-trails in connected plane eulerian graphs (respectively in eulerian triangulations of the plane, to be more precise) and hamiltonian cycles in connected planar, cubic, bipartite graphs.

Of course, not every planar, 2-connected, cubic, bipartite graph has a hamiltonian cycle.’) The graph of Figure VI.30 (a) is such an example; we leave it as an exercise to prove that this is really the smallest such example and that it is uniquely determined. If one replaces every digon of this graph with a copy of the 3-dimensional cube Q3minus one edge, one obtains a simple 2-connected plane, cubic, bipartite graph which is not hamiltonian (see Figure VI.20.(b)). It has been shown independently and almost simultaneously, [PET08la, ASAN82al that this simple graph is in fact the smallest example of this type, and that it is uniquely determined. However, the interest in hamiltonian cycles in plane cubic bipartite graphs does not originate from the equivalence with A-trails in eulerian triangulations of the plane as expressed by Theorem VI.71. Let us recall that Tait, after proving that the Four Color Conjecture is equivalent to the conjecture that every planar, 2-connected, cubic graph has a 1-factorization, conjectured that every planar, 3-connected, cubic graph is hamiltonian (see [KONISGa, p.27-281). This conjecture, stated in 1880 and aimed at an easy proof of the Four Color Conjecture, was disproved in 1946 by Tutte who conjectured subsequently that if in Tait’s conjecture one replaces ‘planar’ by ‘bipartite’, the graphs are hamiltonian, [TUTT7la]. J.D. Horton disproved Tutte’s conjecture; his first counterexample has 96 (!) vertices (later he found one with 92 vertices, [HORT82a]). More recently, M.N. Ellingham constructed a counterexample with 78 vertices, ‘1 In contrast to the case of A-trails, the existence of a hamiltonian cycle is independent of any actual embedding. On the other hand, the planar cubic bipartite graph G, has a hamiltonian cycle, if and only if for every plane embedding H3 of G,, D(H,) has a non-separating A-trail (Theorem VI.71). This is an interesting fact, for in the case of K(G,) = 2, D(H$) $ D(H{) may hold for two embeddings H i , H!/ of G,.

VI.112

VI. Various Types of Eulerian Trails

0,- ( a1.b.1 1

Figure VI.30.(a) The smallest 2-connected, plane, cubic, bipartite non-hamiltonian graph. (b) Substituting for i = 1 , 2 , 3 the digon (ui, bi) in (a) with Q3- { u i , bi}, one obtains the smallest such graph which is simple.

and in joint work Ellingham and Horton constructed a counterexample to Tutte’s conjecture on 54 vertices (for the original Horton graph, see, e.g. [BOND76a]; for the other counterexamples and more details, see [ELLI82a, ELLI83al). A short time ago, J.P. Georges and A.K. Kel’mans developed counterexamples on 50 vertices which are cyclically 4-edgeconnected, [GEOR89a, KELM88aI .I) However, the following conjecture is generally attributed to D. Barnette. According to Tutte (private communication, 1972) he has thought about this conjecture as well. Consequently, we shall call it the Barnette-Tutte Conjecture (BTC). Conjecture VI.72 (BTC). Every planar, 3-connected, cubic, bipartite graph is hamilt onian.

In view of Theorem VI.71 the BTC is equivalent to the following con’) Mr. H. Gropp drew my attention to these two articles. As he points out in [GROP89a], bipartite cubic graphs of girth 2 6 had been studied as early as 1887, but in terms of what has become known as ‘symmetric configurations’.

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.113

jecture (observe that planar, 3-connected, cubic graphs have simple triangulations of the plane as their duals, and vice versa).

Conjecture VI.73. Every simple eulerian triangulation of the plane has a non-separating A-trail. However, the next conjecture, although appearing weaker than Conjecture VI.73, is indeed equivalent to the latter, [FLEI74a].

Conjecture VI.74. Every simple eulerian triangulation of the plane has an A-trail. The equivalence between the last two conjectures (and therefore between conjecture VI.74 and the BTC) follows from the next result. There, we consider two eulerian triangulations of the plane, Do and D,, which are related to each other in the following way. For every face boundary A = (ab, bc, ca) of Do take a copy of the plane octahedron minus the outer face boundary A, = (a, b, c, ), embed it in the interior of A and identify the corresponding vertices a, b, c (see Figure VI.31).

A C

U

‘,C,’

a1

a

b

Figure VI.31. Obtaining the eulerian triangulation of the plane D,from the eulerian triangulation of the plane Do by embedding in every face of Doa copy of the graph H , (the Figure shows this operation for one face only).

Lemma VI.75. Let Do be an eulerian triangulation of the plane, and let D , be constructed from Do as described above. Then the following statements are equivalent.

VI.114

VI. Various Types of Eulerian Trails

1) Do has a non-separating A-trail. 2)

D, has an A-trail.

Proof. We first assume that Do has a non-separating A-trail and apply Theorem VI.70.a by which we can write V ( D o ) = Vf U V: such that (yo)is a tree, i = 1,2. Next consider an arbitrary face F of Do with V ( b d ( F ) )= {a,b,c}, say. Since (yo) is a tree, i = 1,2, we have V: n V ( b d ( F ) )# 8 # V: n V ( b d ( F ) ) . Hence, w.1.o.g. we can write a, b E I$ c 'E,V i where {j,k} = {1,2}. We adjoin c1 to Qo and al,bl to V j . By this extension of yo and V j , performed for every face F of Do, we obtain a vertex partition { V i , V i } of D,. It follows from the very construction of this vertex partition that ( V t )and (V:) are acyclic, hence they are trees. By Theorem VI.70.a, D, has a (non-separating) A-trail. Conversely, if D, has an A-trail, by Corollary VI.68, the corresponding A-partition {&l,V;} has the property that (5') and ( V i ) contain a cycle only if it is a face boundary of D,. Thus, we conclude for V; := V ( D o )n Vt,V: := V ( D o )n V; that ( V t ) and (V:) are acyclic. By Proposition 111.63.1) and Theorem VI.70.a, Do has a non-separating Atrail. Thus, the lemma is true. The equivalence between Conjecture VI.73 and Conjecture VI.74 now follows from Lemma VI.75 by observing that they are formulated for all simple eulerian triangulations of the plane, and that D, is simple if and only if Do is simple (note Do C 0,). The equivalence between these two conjectures is, however, matched by the corresponding equivalence in bipartite cubic graphs. For, the BTC is equivalent t o saying that every planar, 3-connected, cubic, bipartite graph has a dominating cycle. The proof of this equivalence follows along the lines of the proof of Lemma VI.75 translated into the theory of cubic graphs; it is therefore left as an exercise.

In [FLEI74a] it had been conjectured that every planar, 3-connected, eulerian graph has a n A-trail (note that in this case one need not speak of plane graphs because a change in embeddings either leaves O+(v) invariant or else replaces O+(v) with O-(v) for every vertex v). If true, this conjecture would immediately have settled Conjecture VI.74 since simple triangulations of the plane other than K3 are 3-connected. In order to construct a planar, 3-connected, eulerian graph without an

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI. 115

A-trail as S. Regner did, one proceeds in a way similar to the above construction of D, from Do. Start with a non-hamiltonian, plane, 3-connected, cubic graph G,; such graphs exist (see p. 111.64). Consider its dual D := D(G,), and denote the odd vertices of D by q ,v2, . . .,2r2k-1, 2r2k, k 2 1. Consider the set of paths P := {P1,2,. . .,P2k-l,2k}such that P2j-1,2j joins 2.’2j-1 and vzj, 1 5 j 5 k, (see Corollary V.3). For an arbitrary but fixed j , 1 5 j 5 k, denote

and for i = 1,.. . ,t , let yi denote one of the two vertices each of which defines a face boundary containing zi-,zi, and let A; be the (triangular) face boundary containing xi-, , xi,yi. Now embed a copy of the graph Hi of Figure VI.32 in the interior of A; in the same way that Ho has been embedded in the interior of A to obtain D, from Do (see Figure VI.31 above). Doing this for i = 1,.. . ,t , we obtain a plane graph D‘ having 3and 4-gond face boundaries only; and D C D’. D’ has 2k - 2 odd vertices only since

Observe that no 4-gond face boundary of D‘ contains an edge of D. Therefore, if we choose P” E P, P” # P’, we can produce the graph D” from D’ the same way we produced D’from D. Of course, a vertex corresponding to one of the above y i , may not lie in V(D);but this does not matter precisely because E(P“)n E(Q) = 0 for every 4-gond face boundary Q of D’(note: P” C 0).Now D” has 2k-4 odd vertices only. Hence, after k steps we arrive at the plane eulerian graph D ( k )having 3and 4-gonal face boundaries only; and D c D(”.Since I V(G,) I> 2, D # K3 follows; i.e. D is 3-connected, and consequently, D ( k )is 3-connected and simple (this follows from the very construction of D(’)). However, D ( k)may still contain face boundaries A which are face boundaries in D as well. For such A proceed as in the construction of D, from

VI.116

VI. Various Types of Eulerian Trails

Figure VI.32. The graph Hi.

Do by embedding H,, as illustrated in Figure VI.31. The grap.h G finally obtained has the following properties: 1) G is a plane, 3-connected, eulerian graph; 2) D 3)

c G;

No face boundary of D is a face boundary of G.

Because of these properties, if G had an A-trail, the corresponding Apartition { V,, V,} of V ( G )induces a vertex partition { V:, V,l}in D which of necessity induces acyclic graphs (V:),(V,l)(see property 3) and Corollary VI.68). Consequently, since D is a simple triangulation of the plane, they are trees. That is (see Lemma 111.74), G, with D = D(G,) has a hamiltonian cycle. This contradiction to the choice of G, shows that G has no A-trail. However, the positive outcome of this construction of G from D = D(G,) is the following general observation whose complete proof is contained in the proof of Theorem VI.91 below.

Let G, be a 3-connected, planar, cubic graph. Then a plane (even a planar 3-connected) eulerian graph G with D = D(G,) C G exists such that n o face boundary of D is a face boundary of G , and G, is (V1.A) hamiltonian i f and only i f G has a n A-trail. Moreover, we may make the following observation. In the above construction of a plane, 3-connected, eulerian graph G without A-trails having

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI. 117

properties 1)) 2), 3) it is possible to proceed in such a way that every face of D contains at the most one copy of the graph Hiof Figure VI.32. This follows from [FLEI74c, Theorem 10 and its proof] (see also the discussion on regulating sets centering around Figure VI.35 below). Another way of obtaining G with that property rests on the application of the Chinese Postman Problem. This will be done in the subsection concentrating on complexity considerations. Comparing the above construction of a plane, 3-connected, eulerian graph

G from the plane, 3-connected, cubic graph G, via D = D(G,), with Conjecture VI.74, and noting that G has 3- and 4-gonal face boundaries

only, one wonders whether it is possible to disprove this conjecture (and therefore, the BTC as well). All that one would need is a triangulation D, of the plane with all but two vertices having even valency, and the remaining two vertices lying on the same face boundary and having odd valency. W.1.o.g. one could assume that these two vertices, call them and xi,lie on the outer face boundary bd(F,). Then the outer face boundary of H;' := D, - bd(F,) would be 6-gonal as in the case of Hi in Figure VI.32, with zi--l and zibeing the only odd vertices of H;', and Hi having 3-gonal face boundaries only except for the outer 6-gon. The above construction starting from the non-hamiltonian G,, would then yield a simple eulerian triangulation of the plane G without A-trails. Unfortunately, such D, does not exist, [MOON65a, FLEI74bl. Lemma VI.76. There is no triangulation G of the plane having precisely two odd vertices z,y and such that zy E E(G).

Proof. We present two proofs; both are indirect. The first proof is due to Moon, the second proof was found by the author years ago, but has never been published before. 1) Suppose such G exists. Then H := G - {zy} is a plane eulerian graph with one 4-gonal face Q, and all other faces are 3-gonal. Consider a 2-face-coloring of H such that Q is a 1-face. Denote 3;= { F / F is an i-face}, i = 1 , 2 . Since every edge of H belongs to the boundary of precisely one 1-face and precisely one 2-face) we have

IE(H)I=

On the other hand, IE(bd(F))I= 3 for F

IE(bd(F))I= F€Fi

IE(bd(F))I

IE(bW)I= FE31

F€32

# Q yields

IE(bd(F))1-

1(rnod3) and FE32

0 (rnod3)

,

VI.118

VI. Various Types of Eulerian Trails

an obvious contradiction to the above equation. 2 ) Assuming G exists, let H = G - bd(F,) where we assume w.1.o.g that x , y E V(bd(F,)). Consider a plane embedding of the Ir‘, and denote V(I(,) = {z1,z2,y1,y2} such that {z1y1,z2y2} is a 1-factor of K4. Let Fibe a face with zjyi E E(bd(F,)),i = 1,2; Fl # F2 follows. Embed now a copy of H in each of F,, F2 respectively and identify zi with z and yi with y in the respective copy of H , i = 1,2. Identify the vertex z # x,y

of bd(F,) n H in the corresponding copy of H with the corresponding vertex zi in bd(Fj) C K4, i = 1,2 (note: z1 E {z2,y2},z2 E {zl,yl}). The new graph GI is an eulerian triangulation of the plane because for u E V(I(,) we have

x(G1) = 3 and K4 contradiction.

c

G,.

Since x(K4) = 4, we obtain the desired

Translated into the language of cubic graphs, Lemma VI.76 says that there is no plane (2-connected) cubic graph which has precisely two odd face boundaries and such that they have an edge in common. However, there is a partial solution to the BTC, [GOOD75a]. To formulate Goodey’s result, let G, be a planar, 3-connected, cubic, bipartite graph with the following properties: 1)

I E ( b d ( F ) )IE {4,6,8} for every face F of G,;

2)

IE(bd(F))I= 8 for at most one face F of G,;

3) if there exists F8 with IE(bd(F8))I= 8, then there exists an F4 with IE(bd(F4))I= 4 and bd(F8) n bd(F4) # 8.

Let B, denote the set of all G, having the above properties. It follows from Euler’s Polyhedron Formula that G, E Bo has precisely six 4-gonal face boundaries if it has no 8-gonal face boundary, and it has precisely seven 4-gond face boundaries otherwise (see Exercise VI.23). Goodey’s result can then be stated in the following way.

Theorem VI.77. Let G, E Bo be chosen arbitrarily, and let F,, be any face with I E(bd(F,,,)) I= muz such that there is a 4-gonal face F4 with bd(F,,,) n bd(F4) # 8. Then there exists a hamiltonian cycle H of G,

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.119

with E(bd(F,,,) n bd(F,)) E E ( H ) (note that for any two faces F’, F” in a 3-connected, planar, cubic graph, IE(bd(F’))n E(bd(F”)) {0,1}).

IE

Our aim is to translate Goodey’s proof of Theorem VI.77 into the theory of A-trails. For this purpose we want to explore first what it means in terms of 6-splittings of vertices in D(G,), if e E E(G,) belongs to a hamiltonian cycle H of G,. Disregard, for the moment, the fact that G, is bipartite. By Lemma 111.74, G, has a hamiltonian cycle H if and only if we can write V ( D )= V, U V, and ( y ) ,is a tree, i = 1,2, where D = D(G,). Moreover, by the same lemma, H contains precisely those edges which correspond in D to the elements of {v1v2/v1 E V,, 21, E V,}. That is, if we look at the face boundaries bd(Fj) of G, corresponding to vj E Q for fixed i E {1,2}, E ( H ) = UvjEll,E ( b d ( F j ) )- { g j , k E E(bd(Fj)r l bd(Fk))/vj,vk E q} (see the proof of Lemma 111.74). Now we use the fact that G, is bipartite, and hence D is a simple eulerian triangulation. Therefore, considering the perfect A-partition { V,,V,} of D corresponding to H c G, (see Theorems VI.70, VI.70.a and VI.71) we conclude for e* = xy, the edge corresponding to e E E ( H ) , that I {x,y } n V, ]=I (2,y} n V, I= 1. Moreover, since {V,, V,} is a perfect Aare cycles, we may assume partition, i.e., both (Dvl,1)v2,2 and (Dv2,1)yl,2 w.1.o.g. that x E V,,y E V,. This means in terms of 6-splittings that if D is 2-face-colored, D has a non-separating A-trail inducing a l-splitting in x and a 2-splitting in y. Now let I,denote the set of all 2-face-colored simple eulerian triangulations D of the plane (with the outer face being a l-face), having the following properties:

1’) d,(v) E {4,6,8} for every 2’)

21

E V(D);

d,(v) = 8 for at most one v E V ( D ) ;

3’) if there exists vs E V ( D )with d,(v,) 4-valent vertex in D.

= 8, then v8 is adjacent to a

As above, it follows from Euler’s Polyhedron Formula that D E 7, has precisely six 4-valent vertices, if it has no 8-valent vertex; otherwise it has precisely seven 4-valent vertices (see Exercise VI.23). We note that both B, and 7, are infinite sets; this we conclude from [GRUN67a, p.254, Theorem 21 or Lemma 111.58 and the fact that the elements of B, and 7, are in l-l-correspondence by dualization. Hence, we arrive at the next result on A-trails.

VI. 120

VI. Various Types of Eulerian Trails

Figure VI.33. The eulerian triangulation of the plane, D, , obtained from the eulerian triangulation of the plane, Do, by a W,-extension (possibly a = c # d or b = d # c ) . T h e o r e m VI.77.a. Let D E 7, be chosen arbitrarily, let vmuz E V ( D ) be of maximal valency and chosen in such a way that it is adjacent to a 4-valent vertex w. Then D has a non-separating A-trail inducing a l-splitting in umaz and a 2-splitting in w. Before we prove Theorem VI.77.a we must establish some other results. For this purpose, we define a special class of eulerian triangulations of the plane. Let D be a triangulation of the plane, and let A = (a, b, c ) be a face boundary in D. Embed the graph H , as depicted in Figure VI.31 in the corresponding face of D (i.e, A is replaced with an octahedron 0,);call the resulting graph D,.We say D, results f r o m D b y a n octahedral extension (shortly 0,-extension) at A. To define a second operation assume the existence of a 4-valent vertex z, E V ( D ) and consider the 4-wheel W, defined by N*(z,); denote N(z,) = { a , b, c, d } following O+(z,) = (az,, bz,, czo,dt,) (W, can be degenerate, i.e., a = c or b = d ) . Define the triangulation of the plane D, by

We say D, results f r o m D by a W,-extension with the 4-valent vertex t o as its center (see Figure VI.33).

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.121

Now let n be a positive integer and define Do = D , D, = D’, and D i results from Di-l by an 0 6 - or W4-extension, i = 1,.. . ,n. Then we say D’ results from D by a sequence of 0 6 - and/or W4extensions. The validity of the following statements is a consequence of the definition of the above extensions: 1) D’ is eulerian if and only if D is eulerian; 2 ) If I V ( D )I> 4,then D is simple if and only if D’ is simple.

However, the condition I V ( D )I> 4 cannot be omitted; the octahedron can be viewed as obtained from the eulerian triangulation of the plane consisting of two 4- and two 2-valent vertices, by a W4-extension. If IV(D)I > 4 and if a = c or b = d , D has two pairs of multiple edges: two edges ba and, correspondingly, two edges zoa or two edges zob. However, only one of these two pairs of multiple edges can disappear by a FIT4extension. This becomes clear from the following observations. C Firstly, if Di results from D,-lby an 06-extension, E(Di_,) E(D,), i E (1,. . . ,n}. Therefore, if I V(D,-l) I> 3, then independent of the valencies of a , b, c in Di-l (where A = ( a , b, c); see Figure VI.31), D j - { a , b, c} is disconnected for every j with i 5 j 5 n (if Di-l has a 2-valent x E { a , b, c } , then already D j - ( { a , b, c } - {x}) is disconnected, and if none of a , b, c is 2-valent in Di-l, then d D , ( y ) 2 6 for y E { a , b, c} which implies that none of a , b, c can be the center of a W4-extension in D,, i 5 k 5 n).

Secondly, if D, results from D j - l by a W,-extension and if d DI.- 1 ( a ) > 4,d D‘-1 . ( c ) > 4, then Q = ( a , b , c , d ) is a separating 4-gon in every Dj, i - 1 5 j 5 n. The valency condition concerning a and c implies the existence of a vertex x 6 { a , b , c , d , z o } and, moreover, E ( Q ) c E ( D j ) ,i - 1 5 j 5 n. These considerations lead us to the following structural result. In formulating it, we follow the notation of Figures VI.31 and VI.33.

Lemma VI.78. Let D be a simple eulerian triangulation of the plane resulting from the octahedron 0 6 by a (non-empty) sequence of 0,- and/or W4-extensions. Then at least one of the following statements is true. 1) D contains a separating triangle A = ( u , b , c ) and O6 3 A as an induced subgraph such that DL := D - H , can be obtained from 0 6 by a (possibly empty) sequence of 0,- and/or W,-extensions ( H , := 0, - A);

VI. 122

VI. Various Types of Eulerian Trails

2 ) D contains (at least) two 4-valent vertices zb, tg such that the corre-

sponding 4-wheels W’ ,W4/‘have the following properties: d g ( t i ) = d D ( t i ) = d D ( t y ) = dD(z!J = 4

;

E(W;)n E ( w ~ = )0 ; each of

DI, := ( D - {ti,t k } )

U { a l z ~c’th> , and

0 6 (defined analogously)

can be obtained from 0, by a (possibly empty) sequence of W,-extensions.

06-

and/or

Proof. By hypothesis, there exists a sequence of eulerian triangulations of the plane, Do,D,, . . .,D,, such that Do = 0 6 , D, = D , and D iresults from D,-,by an 0 6 - or W4-extension, i = 1,.. . ,n. Suppose n = 1. If D results from 0 6 by an O,-extension, both sides of the (uniquely determined) separating triangle A = ( a , b, c) are graphs isomorphic to Ho := O6 - A. Hence statement 1) of the lemma prevails. If, however, D results from 0 6 by a w4-extension, D is isomorphic to the 6-sided double pyramid. Hence we can partition V,(D) (which defines the 6-gonal base of this double pyramid), {Vl,V t } ,such that PI := (Vi) and P” := ( V t ) are paths of length two. Now, W‘ := (Vi U V 6 (D)) and W4/‘ := ( V t U V , ( D ) ) are 4-wheels with the properties expressed in statement 2 ) of the lemma ( t b , 2: are the 2-valent vertices of PI, PI‘ respectively, and ti,z4 t;) are the end vertices of P‘ (P”)).Now we deduce the validity of the lemma if n = 1.

(tr,

Hence consider the case n > 1, and suppose first that Diresults from D,-, by an 0,-extension for some i E (1,.. .,n}. W.1.o.g. suppose for the triangle A = (a, b, c) c Di-l where this 06-extension is performed, that A # bd(F,) where F, is the outer face of Di-,. Consequently, the graph Ho used to perform the 0,-extension at A, lies in the bounded region of A. Moreover, A separates Diand d D i ( z )2 6 for any z E { a , b, c } . Thus, every 0 6 - or W,-extension performed after the 06-extension at A, takes place entirely either ‘outside’ or ‘inside’ A. Now, if not all of these latter extensions take place on the same side of A, then consider the last one which takes place outside and the last one which takes place inside of A. Call the correspondingly resulting graphs D j and D,, i < j , k 5 n.

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.123

If each of D j and D , results from W,-extensions of Dj-,, Dk-l respectively, statement 2) of the lemma holds for the corresponding 4-wheels Wii) and W i k )with Wi := Wi’), W t := W i k ) :this follows from the choice of j and k and the discussion preceding the formulation of Lemma VI.78, and by defining the sequences D,*,. . .,D:-l for DA, respectively D,**,. . ., for 0 6 by

D:=D,,

r = O ,...,j - 1 ,

and D,**=D,,

s=O

,..., k - 1 ,

DT = (D,+, - { z y ) , z p ) } ) u { u ( j ) z p ) , c ( j ) z p ) }r, = j , . . . ,n - 1, D,**= (D,+, - { z ,

9

22(k)})

U

{dk))zi”,c(’)zik)}, s = k,. . .,n - 1,

where zf),zi’), I!), d‘), c(‘) correspond to z,, zl, z2, u, c in Figure VI.33 for I = j,k. Observe that d D , ( z p ) ) = dDr(zp)) = d D.(’1 ( I c ) dDs()z$”) = 4 for every r E { j , . . .,n } , respectively for every s E {k,. . . ,n}. Hence suppose w.1.o.g. that D j results from Dj-, by an 0,-extension. Denote by H f ) N H, the graph used for this 0,-extension performed at the triangle A ( j ) . Since the octahedron 0;’ = A ( j )U Hi’) belongs to every D, because of the choice of j, r = j,. . .,n, we conclude that DA := D - ~ i j results ) from 0, by a sequence of 0,- and/or ~,-extensions, namely, Do,*..>Dj-1, Dj+l - H o ( j l , . . . , D n - &) Since we can use the same argument if all extensions following the 0,extension at A take place outside A (with A and Ho in place of A ( j )and Hi’)), and because of the cases solved already, we have to assume that all these later extensions take place inside A.

In particular, we may assume that D, does not result from D,-, by an 0,-extension (otherwise, for D := D,, DA := D, - H i n ) , Hi”) 11 H,, statement 1) of the lemma holds). Denote by W2 the 4-wheel associated with the W4-extension yielding D, from DnVl. Now take the minimal i E { 1,.. .,n } such that Di results from Di-l by an 0,-extension. If i = 1, H,* := D , - (H, U A ) N H , and so H,* C D,,t = 2 , ..., n; and since H , U A I I0,, we can define

D , * : = H , U A and Df-,:=D,-H,*,

t = 2 ,...,n

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VI. Various Types of Eulerian Trails

to obtain D;-, = D - H,* =: DL resulting from 0 6 by a sequence of 0,and/or W4-extensions.

If i = 2, D, is isomorphic to the 6-sided double pyramid. Denote V4(D,) = {u1,v2 ,...,21,) according to O+(w) = ( w z ~ ~ , w v. .,w'u6) ~,. where V ( ( D 1)= {w, z}. W.1.o.g. this notation has been chosen in such a way that A = (215,216,w)(= ( u , b , c ) ) . But then Wi = ( { ~ ~ , 2 ) ~ , 2 1 ~ , w , 2 } ) is a 4-wheel lying entirely outside A, and Wi c D j , j = 1,.. . ,n. Hence the W4-extension yielding D , from D,-, (see above) implies that PV: lies entirely inside A; E(W i )n E(W l )= 0 follows. With 2r2, zll ,'u3, v6,u4 assuming the role played in Figure VI.33 by z o , zl, z 2 ,a, c respectively, we define

This yields DO+ 2! 0 6 , D:-, = D; (as defined in statement 2 ) of the lemma ), and Dfresults from DJ-, by the same extension yielding Dj+l from D j , j = 1,.. .,n - 1.

To finish the proof of the lemma we have to consider the case where either i > 2, or D results from 0 6 by a sequence of W,-extensions. Assume first that D j is a (4 2j)-sided double pyramid, where j = i - 1 if i > 2 exists, and j = n if D := D, results from 0, by n W4-extensions. We either find Wi,W[ similar to the case i = 2, or else we can take any two edge-disjoint 4-wheels Wi,W t of the (4 2n)-sided double pyramid. In both cases, Wi and W: satisfy statement 2) of the lemma.

+

+

Thus we finally consider the case when there is some k < i, k 5 n respectively, such that D j is a (4 2j)-sided double pyramid for 1 5 j 5 k - 1, while D, is not a double pyramid. Note that D, is a 6sided double pyramid in any case. It follows that D, contains a 4-gon Q = ( w ,ur, ~, v r + 2 ) , where {w, z} = V(Dk-1) - v4(D,-1)7 ~ D I ; - (21,) ~ = d,,(v,) - 2 = 4 for s E { T , T 2}, and where we assume the notation chosen in such a way that vtvt+, E E(D,-,) for t = 1,. . . ,2k+2 (putting v2,+3= 21,). W.1.o.g. T = 1. Thus Q = (w, x,2ry, v3) is a separating 4-gon containing no 4-valent vertex in D j , j 2 k 2 2. Thus, every 0,- or W4extension yielding D j + , from D j , j = k, .. .,n - 1, takes place entirely either inside or outside Q. Arguing along the same lines as before (with Q in place of A) we may conclude that all these extensions take place on the same side of Q. On the other hand, v2 and v5 are in D, the centers of 4, = ( N * ( v 5 )respectively, ) lying on different wheels W,"' = ( N * ( v 2 ) )WA5)

+

+

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.125

sides of Q. Consequently we have Wi c D j and 1 V4(Wi)nV4(Dj) I= 3 for

some Wi E {W,‘”’,W i 5 ) }and j = k,. . .,n. Denote by W[ the 4-wheel corresponding to the W4-extension yielding D, from D,-,.D:, defined correspondingly, results from 0 6 by the sequence, Do,. . .,Dn-,= 0 6 . As for D;, we denote Wi = ( { z ; , zi, z;, b’, d’}) and a’, c‘ in analogy to the notation of Figure VI.33, where {b’d’} = {v,, v3} and {u’, c’} = {w,x} if Wi = W i 2 ) ,while {b‘,d‘} = { w , ~and } {a’,c’} = {v3,v7} if Wi = W J 5 ) (putting v7 = q if k = 2). Now define D;:= Dj for O < j < k - l - 6 , for k - 6 6 I < n - 1, Or:= ( I l l + , - { z ~ , t ~U} {a’z~,c’z~} ) where S = 0 if Wi = W i 2 )and 5 = 1if Wi = W l 5 ) .We obtain D; = D i - , resulting from 0 6 by a sequence of 0 6 - and/or W4-extensions since D; results from D;-, by the same extension yielding Dj+l from D j , j = k - 5,... ,n - 1. Note that for k = 2 and 6 = 1 0; results from D,* by a W4-extension, and so does D, with respect to Do but the centers of these W4-extensions are different. Lemma VI.78 now follows. We now use this lemma to prove a result on A-trails which generalizes a special case of Goodey’s result.

Theorem VI.79. Let D be a 2-face-colored eulerian triangulation of the plane, D being either the octahedron itself or arising from the octahedron by a sequence of 0,- and/or W,-extensions. Let A = (x,y, z ) be an arbitrarily chosen face boundary with x,y, z E V ( D ) .D has a non-separating A-trail whose (perfect) A-partition {V,, V2}satisfies x E V,,y, .z E V,. Proof. W.1.o.g. A is the boundary of the outer face F, of D , and F, is a 1-face (note that perfect A-partitions are independent of the 2-facecolorings). If D is the octahedron, then it has a non-separating A-trail 2’; e.g., taking the graph of Figure VI.20.b) and defining V, = V’,V, = V”, we have I V,n V ( A )I= 1, I V, n V ( A )I= 2. Because of the symmetries of the octahedron we may therefore assume w.1.o.g x E V,, y, z E V,. Thus the theorem holds in the case of the octahedron. Now let D satisfy the hypothesis of the theorem and suppose D is not the octahedron. By the very construction of D and because of Lemma VI.78 we distinguish between the following two main cases. (1) D satisfies statement 1) of L e m m a VI.78. Retaining the notation of Figure VI.31 with D in place of D, we define

D;:= D - Ho

VI.126

VI. Various Types of Eulerian Trails

in which A, := (a, b, c) is a face boundary. Because of the symmetry of the possible situation of A in A, U H,, we assume w.1.o.g. that if A C A, U H,, then either r = a l , y = b,, z = cl, orr=al,y=a,~=c,,orx=c,y=a,z=cl orr=bl,y=b,z=c. Now, applying induction to Dh we obtain w.1.o.g. (by symmetry if A $ AouH,, or because of the freedom to choose if A c A, UH,) a E vjo and b, c E V l , { j , k} = {1,2}, for the perfect A-partition { V f ,V:} of V(Dh) corresponding to some non-separating A-trail of DI,. Define

v , = ~ ~ U { b , , c , } , V , = V ~ U ( a 1 } ,{ j , k } = { l , 2 }

.

Consequently, { vj, V,} is a perfect A-partition as required if A $ AoUHo, or if j = 2 , k = 1 and {x,y,z} # { b , , b , c } . On the other hand, if A c A, U H , and j = 1,k = 2, {V,,V,} is a perfect A-partition as required provided {x,y, z } = {bl, b, c } ; in the other cases we use the fact that ( V f )and (Va),6 = 1,2, are trees in D; therefore, for V; := V, and V; := V,, {V;, V;} is a perfect A-partition as required. Finally, in the case j = 2, k = 1 and {r,y, z } = { b , , b, c}, we define V;, V; as before to obtain a desired perfect A-partition {V;, V;}. This finishes case (1). (2) D satisfies statement 2) of Lemma VI.78. Then D contains a 4wheel W, as depicted in Figure VI.33 (ii) with A W,. By retaining the notation of this figure with D in place of D, and DI, in place of Do, we define the reduction of D to 0; and apply induction to DI, to obtain {qo, V:} as above. Depending on the distribution of a, to,c in Vf and V: we consider two cases.

If for { j , k} = {1,2}, either { a , zo, c} C 50,or { a , z,} C Yo and c E V: (the case {c, z o } C and a E ‘V needs no extra consideration because of symmetry), then { b , d } c V’ follows of necessity. In both cases we define vj := vjo U {zl, z,}, Vk := V i , { j ,k} = {1,2}

5’

{%, V,} = {V,, V,} of D as required. Hence we have to consider the case { a , c} C vj”, z, E Vf . Then { b , d } p

to obtain a perfect A-partition

vjo because {vjo,Vf}is an A-partition. By symmetry we may assume

w.1.o.g. that b E %O, d E V’. To obtain a perfect A-partition (6,V,} = {V,, V,} of D as required by the theorem, we define in this case

3 := vj” u { z o } , v, := ( V i - (2,))

u {zl,z,}, { j ,k} = {1,2}

.

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.127

Observing that these considerations cover d cases regardless of A n W4 = the application of Lemma VI.78, A p W4 in any case), we now conclude the validity of the theorem.

8 or A n W, # 0 (because of

Translated into the theory of cubic graphs, Theorem VI.79 is - by dualization - equivalent indeed to the following result on hamiltonian cycles (see also the considerations following the statement of Theorem V1.77).

Corollary VI.80. Let G, be a (3-connected) planar, cubic, bipartite graph which can be derived from the cube by a sequence of one or both of the following operations: 1) Replace a vertex with three mutually adjacent 4-gons; 2) Replace a 4-gon Q with three 4-gons two of which are not adja-

cent.

Then, given any e E E(G,),there exists in G, a hamiltonian cycle not containing e . The proof of the next corollary follows immediately from parts of the proof of Theorem VI.79; therefore we leave it as an exercise to the reader.

Corollary VI.81. Let Do and D, be eulerian triangulations of the plane such that D, results from Do by a sequence of 0,- and/or W4extensions. Then Dohas a non-separating A-trail if and only if D, has a non-separating A-trail. We omit the translation of Corollary VI.81 into the theory of cubic graphs; it can be done the same way by which Corollary VI.80 has been obtained from Theorem VI.79. There is a procedure similar to the W,-extension to create a new eulerian triangulation of the plane D, from a given eulerian triangulation of the be given as in Figure plane Do. Namely: let the vertices a, b, c, d E V(Do) VI.33, and suppose ac E E(Do). Subdivide ac with two new vertices, z1 and z2 say, and join both t l and z2 by an edge to each of b and d. The resulting graph D, is an eulerian triangulation of the plane indeed. We call this construction of D,from Do the weak W,-extension (see Figure VI.34 below). Translated into the theory of plane cubic graphs, the weak W4-extension corresponds to the replacement of an edge e E E(G,)with Q2in such a way that the parity of I E(bd(F))I two adjacent 4-gons Ql, remains unchanged for every face F of G,. That is, if G, is bipartite, then so is the new graph.

VI. Various Types of Eulerian Trails

VI. 128

However, if we rephrase Lemma VI.78 in order to cover the graphs obtained from 0, by a sequence of 0,- and/or weak W,-extensions, the resulting statement is false. That is, starting from 0, one can produce an eulerian triangulation of the plane D by a sequence of weak W4extensions such that N ( v ) n V4(D) = 0 for every v E V,(D) except for one pair of adjacent elements of V,(D) (We leave it as an exercise to construct such an example). This fact is no surprise, however, if one observes that in Corollary VI.81, the relation between A-trails in Do and A-trails in D, is bijective (see also the proof of Theorem VI.79). This observation does not hold any longer if one replaces in Corollary VI.81 the term ‘W4-extension’ with ‘weak W,-extension’. For, while it is true that a non-separating A-trail of Do can be extended to a non-separating A-trail of D , (obtained from Do by a sequence of 0,- and/or weak W4extensions), the converse may not be true for some non-separating A-trail of D,. Figure VI.34 illustrates this consideration.

DO

Figure VI.34. D, obtained from Do by a weak W4extension. If D, has a perfect A-partition {V,,V,} as indicated by the vertex labels 1 and 2, then for the corresponding partition (V’,V,} of V(D,),the graph (Vf)has a cycle containing ac, while (V:) is disconnected.

w e note that, starting from 0 6 , one reaches only certain eulerian triangulations of the plane by 06-,W4- and weak W,-extensions. For, in

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.129

general, an eulerian triangulation of the plane need not contain adjacent 4-valent vertices. In fact, if G, is a plane, 3-connected, cubic graph of girth 5, one can find an independent set of edges, E,, such that the cubic graph G$resulting from G, by replacing every e E E, with a 4-gon Q,, is plane and, in addition, bipartite (see Figure VI.35). Such E, is called a regulating set (of edges), and every l-factor L of G, is a regulating set; for it yields in Gg a 2-factor Q* consisting of even boundary cycles, namely Q* = { Q J e E L C E(G,)} (by Theorems 111.66 and 111.67 a plane cubic graph is bipartite if and only if it has a 2-factor consisting of even boundary cycles). On the other hand, a regulating set of G, need not be a subset of a l-factor L c E(G,).’) In any case, translated into the language of eulerian graphs, the transition from G, to G$ (with the help of a regulating set E,) means transforming Do := D(G,) into an eulerian triangulation of the plane. This transformation is achieved by subdividing every e* = ac E E(D,) which corresponds to e E E, c E(G,), with a vertex zo and joining zo to both b and d where a , c, b and a , c, d are the vertices of the boundary triangles containing e* (see Figure VI.34). Moreover, since E, is an independent edge set, no pair of edges e * , f* € E ( D o ) corresponding to e, f E E,, belongs to a face boundary in Do. That is, the above transformation of Do can be performed simultaneously for all corresponding e*. This is also why in the above construction of a plane 3-connected eulerian graph without A-trails, one can proceed in such a way that every face of D contains at most one copy of the graph Hiof Figure VI.32. We point out that the concept of a regulating set has been introduced and studied in [FLEI74c]. We now present the translation of Goodey’s proof of Theorem VI.77 into the theory of A-trails. We note however, that our proof generalizes two special cases of Goodey’s proof which have been treated in [GOOD75a] in an “as can easily be seen” manner. This generalization has been expressed in Theorem VI.79. Proof of Theorem VI.77.a. We first observe a general structural property of D which follows directly from the fact that D is a simple triangulation of the plane.

a) If P4 is a path with I(P4)= 3 and V(P4)C V4(D), P4 is an induced path, and V(P4)C N ( v 8 )where d(v,) = 8, or else D is the 6-sided double ‘1 We note that Gg may be 3-connected although K(G,)= 2. This is the case, however, if and only if all cut sets Es of size 2 in G, belong to some 2 implies E, f l Es 1 in any regulating set Eo of G, (note that Es case; one way of proving this fact is, e.g., by dualization of Lemma VI.76).

I

I=

I

I#

VI. Various Types of Eulerian Trails

VI. 130

G;

63

Figure VI.35. Transforming the plane cubic graph G, into

a plane cubic bipartite graph G: by replacing the edges e of a certain independent edge set E, with 4-gons Q,.

pyramid, D

II

0, respectively.

Noting that the theorem is true for 0, E 7, by Theorem VI.79 we proceed by induction and distinguish between several cases depending on the distribution of the 4-valent vertices of the chosen D E I,. Case 1. D results from an O,-extension of the eulerian triangulation of the plane Do. Then Do contains a boundary triangle A = (a, b, c) at which this 0,-extension has been performed. Suppose first that A ( D ) < 8. Then d,(a) = d,(b) = d,(c) = 6 and therefore, in Do = D - H , (for H,, see Figure VI.31) we have d D o ( u )= d D o ( b ) = dDo(c) = 4. That is, Do results from an 0,-extension of some eulerian triangulation D' of the plane; a.s.0. Therefore, D results in this case from 0, by a sequence of 0,-extensions. We conclude that in this case Theorem VI.79 implies the validity of Theorem VI.77.a. On the other hand, if A ( D ) = 8, the esistence of a non-separating A-trail in D is secured by induction and Corollary VI.81; and the proof of Theorem VI.79 shows how a perfect A-partition {V', V:} of V(D,) can be extended to such a partition {V,,V,} of V ( D ) . In particular, if ug with d D ( u 8 ) = 8 lies in A, w.1.o.g. ug = u,,, = a and w = al (see the statement of Thcorem VI.77.a); whence we conclude by induction, and since the relations d D o ( u )= 6 = A ( D , ) , d D o ( b )= 4,a , !$ V(D,) hold, that a E V', b E V:.

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.131

Therefore, we may define V,, V, in such a way that a, E V,, b, E V,, and c1 E V, if and only if c E where { j , k} = { 1,2}. Consequently, { V,,V,} is as required in any case.

6

Case 2. D results from a W4-extension of the eulerian triangulation of the plane Do. By induction and Corollary VI.81, D has a non-separating

A-trail since Do has one; and we know from the proof of Theorem VI.79 how to extend a perfect A-partition {V:, V . } of V ( D o )to such a partition {Vl, V,} of V ( D ) if TI,,,^ E E(Do - { z o } ) . If, however, {v,,,, w} n V(W4) # 8, we proceed as follows. If A ( D ) < 8, we conclude in a way similar to Case 1that D results from 0, by a sequence of W,-extensions, and therefore, Theorem VI.77.a follows from Theorem VI.79 in this case as well. If however, A ( D ) = 8, we perform different reductions depending on the position of v,, = and w in W, U { a , c} (we use the notation of Figure VI.33). Since d(v8) = 8, it follows that v8 # z i , i = 0,1,2; hence, by symmetry, we assume w.1.o.g. that TI^ E { a , d } since we necessarily have us E { a , b,c,d}. Because of property a), if min{d(a),d(c)} = 4 it follows that v8 = d since D 3 P4 3 { z o , I,, 2,). Whence we also may assume w.1.o.g. that d(a) = 4 if min{d(a),d(c)}= 4.

(i) d(a) = 4. Then TI,,,= us = d, and we necessarily have the situation as described in Figure VI.36, where

(note that d(b) = 6 of necessity), and therefore

E E ( D ) , N ( u - ) n N(c)= { d , b, b-}, N ( d ) n N ( a - ) = { a , C, dl}, ~ ( dn)N(c)= {z,, a-, d 2 } . CU-

Noting that we necessarily have d(a-) = d(c) = 6, we conclude that either D is the graph depicted in Figure VI.36, or else A = (d, ,d,, b-) is a separating triangle of D. Also, for the chosen 4-valent vertex w adjacent to TI,,,we have w E ( a , zo, z,, z 2 } . In both cases we reduce the side of A containing zo to the octahedron, thus obtaining a smaller graph D, with A ( D , ) = 6 (see Figure VI.36). Consequently, either D, N 0,, or else it results from 0, by a sequence of 0,-extensions (see Case 1). In any case, we may assume by Theorem VI.79 that b- E Vt,d,, d, E V; holds for a corresponding perfect A-partition {V:, of V ( D , - V ( i n t A ) ) .

sl}

VI. Various Types of Eulerian Trails

VI.132

D

D,

Figure VI.36. 4, 218 = d.

Reducing D to D , in the case d ( a ) =

Whence we may extend {V:, V;} to a perfect A-partition V ( D )as required by the theorem by defining

{V,, V,}

of

Note that d = vma, E V, while w E V,. This solves case (i).

(ii) d ( a ) > 4. We reduce D to Do, the graph from which D results by a

W,-extension. We first consider the cases where u, = u8 = a (whence w = zl),or u, = 218 = d and w = zl, and choose in Do in both cases zo = w. While a = v,, necessarily holds in the first case (here we have d D o ( a )= S), we have the freedom to choose in the second case (here we have A(Do) = 6); whence let d = vmar in the second case. By induction, V ( D o )admits a perfect A-partition {Vp,V:} with u, Vp, and zo E V;. In both of the above cases, we define

vl:=vp,

v2:=v;u{zl,z2}

if b , d E

Vl:= Vp u { z o } ,

V2:=(V; - { z 0 } ) u {z1,z2}

if b E V:, d E V:

and if b,d E V; (thus v,,

E

VP,

= a ) , we define

v,:= v .u { z o ,z , } , v, := {V,” - { z o } }u (2,)

.

,

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.133

The case v,, = a and b E &O, d E V: is symmetric to the case b E V;, d E yoand thus needs no extra consideration. Hence, to finish (ii) we are left with the consideration of the case v,,, = 218 = d and w = zo. Let b- be the uniquely defined vertex satisfying b- E N ( b ) - { a , c, zo, zl,z 2 } . We have d D o ( b )= 4 and d D o ( b - ) E {4,6}. Observing that A ( D o )= 6 we obtain a perfect A-partition {Vf, V;} with b- E Vp, b E V: either by induction or by Theorem VI.79 (depending on whether dDo(b-) = 6 or d D o ( b - ) = 4). Note that since the four vertex sets { a , b, c, d } , { a , b-, c, d } , { a , b, c, z o } and { a , b-, c, z o } define a 4-gon each, we cannot have { a , c , d } c Ko or { a , z o , c } c yo for any i E {1,2}. Consequently, if { a , c } C yo, { z O , d } C yowhere { i , j } = {1,2}. On the other hand, if a E yo,c E vj” for i # j, we may assume by symmetry that a E V., c E V;. This implies zo E V f , d E V;; in this case, we know from the proof of Theorem VI.79 how to extend {Vf,V;} to a perfect A-partition {q,q} of V ( D ) , and by defining V, := V . , V, := V; we obtain an appropriate A-partition {V,, V,} of V ( D ) .So we are left with the case { a , c } C W.1.o.g. i = 2 (otherwise we permute Vp and V:; thus b E V: may hold). Hence d , z o E Vf. If b E V., then V, := ( V . - { z o } ) U {zl}, V, := V: U { z o , z 2 } yields an A-partition as required by the theorem. For the case b E V; we define V, := (Vp - { z o } ) U { z1,z 2 } , V, := V i U { z o } t o obtain the desired result. This finishes the considerations of (ii) and thus settles Case 2.l)

KO.

Case 3, I N ( w )nv@) I= 1. Hence D results from a weak W,-extension of the eulerian triangulation of the plane Do. Using the notation of Figure VI.34 and because of Cases 1, 2 handled already, we conclude d ( z ) 2 6 for z E ( a , b, c, d } ; by symmetry we assume w.1.o.g. that if d ( z ) = 8 , then z E {a,b}. Hence v,, E { a , b } , and by symmetry we assume w = z1 in any case.

‘1 This last case v, = v8 = d , w = zo (translated into its dual form) requires in Goodey’s proof a different type of reduction because he does not have the dual equivalent of Theorem VI.79 at hand. Note that in this case we may have dDo(b) = dDo(b-) = 4 which does not permit to conclude inductively b- E V., b E V:. On the other hand, one has to avoid the situation a , c , d E &’, zo,b E V:; for this would necessarily yield d,zo E V,, z,, z2 E V,. Of course, one could also apply Theorem VI.79 in this case to obtain the desired result; but since dDo(d) = such that d E V:, a, zo E d D o ( u )= 6 one cannot invoke induction concerning Theorem VI.77.a to obtain d E Vp, a E V;.

v;

VI. 134

VI. Various Types of Eulerian Trails

We define the eulerian triangulation of the plane D, by identifying b and d in D - {z,, z,, ab, bc}, calling this new vertex z . Thus dDl ( c ) = 4 in any case, while d,,(z) E {6,8}, d D l ( a ) E {4,6} depending on the existence and position of an 8-valent vertex in D. Consequently, in D, we denote u,,, = z (either because of necessity or because we have the freedom to choose), and choose w = c. Assuming first that D, is simple we invoke induction to obtain a perfect A-partition {V., V - }of D, with z E V . , c E V-. This induces V; := (V: - { z } ) U { b , d } , V', := V i - a partition of V ( D )- {z1,z2} such that (V,) is a forest with at most two components, while (V,l)is a tree. Now we define

V,:= Vi U {z,},

v,:=v;,

V,:= V', U {z,},

V2:=V,l U {z,} V2:=V', u {z,,

4:" Vi U (2,)

2,)

if b = v,, if a E V:, if a = u,,,

and a E V,l, and a E V,l.

In all cases, {V,, V,} is an A-partition of V ( D )as required by the theorem. To finish Case 3 it remains to be shown that D,must be a simple graph. This is tantamount to showing that bd 6E ( D ) and that N ( b ) n N ( d ) = { a >71'

'27

'}'

Suppose first that bd E E(D). Then each of the sides of the triangle A* = (b, z,, d ) defines a triangulation of the plane D*for which A* is a face boundary. However, dD. (I,) E 1 (mod 2), and since all vertices of D*- V(A*) are even vertices of D*, precisely one of b and d is an odd vertex in D*. Thus D* is a triangulation of the plane with precisely two odd vertices; and they are adjacent. This contradiction to Lemma VI.76 implies bd 6 E ( D ) . Suppose there exists z E N ( b )n N ( d ) - { a , z,, z 2 ,c}. Since d is 6-valent either A = ( c , d , z ) or A = ( a , d , z ) is a face-boundary of D. Consider A, = ( t , b , z ) with E ( A ) n E(A,) = { t z } where t E { a , c } depending on which case prevails, and denote by D* the triangulation of the plane for which A, is a boundary triangle and which does not contain z,, z 2 .D* $ K3 i because of min{dD(a),d,(c)} > 4. Since we have in any case d,, ( t ) 0 (mod 2), we conclude d,, ( b ) d,, (z) E 0 (mod 2) by Lemma VI.76. Denoting by y the vertex satisfying { t , y } = { a , c } we conclude from d, y, z,, z2 6 V ( D * )and the above congruences that {d,.(b),d,.(2),d,,(t)} E {2,4}. Since d,,(z) = 2 implies that b and t separate D* and thus also D, we must have d,.(z) = 4;

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.135

and since d f!, V ( D * ) we conclude d , ( x ) = 6. By the same to= d,.(t) = 4 follows. Moreover, since umaz E { a , b } ken, d,,(b) and y,z1,z2 $! V ( D * )we conclude u, = b and d,(b) = 8. Since d D ( a ) > 4, d,(a) = 6 follows. Hence suppose w.1.o.g. that t = c. Define {b,} = N ( b ) - (('7 ' 1 7 '2) zl, t2, c7 2)ND* ( b ) ) , {d,} = N ( d )Because of the preceding considerations on the degrees of b and z and since d is 6-valent, these equations are well-defined; moreover, zd, ,xb, E E ( D ) - E ( D * )and zd E E ( D ) imply d, = b,. That is, A, = (a,b,d,) is a triangle; and it must be a face boundary because N(b) - { a , d,} lies on the same side of A,. Consequently, d,(a) = 4; otherwise, a and d, separate D. However, d,(a) > 4. This contradiction finishes Case 3.l)

N ( w ) n V,(D) = 0. Using the notation of Figure VI.33(i) = with D in place of Do, we have w = zo. Assume w.1.o.g. that v, a; hence d D ( a ) E {6,8} and d,(x) = 6 for x E { b , c , d } . Define D, by identifying b and d in D - {zo,ab,bc}, and as in Case 3, call the new vertex z . Again, D, is an eulerian triangulation of the plane with = 47 dDl (.) = 8. d,l ( a ) E j47 6), dDl Assuming first that D , is simple we apply induction to D, to obtain an of V ( D , ) with z E V:, c E V;. Similar appropriate partition {Vi,V2].} to Case 3, this yields a partition {V:, V,l}of V ( D )- { z o } with b, d E V:, c E V,l. Consequently, if we define Case 4.

then {V,, V,} is a desired A-partition of V ( D ) . Thus, to finish Case 4 and hence the proof of Theorem VI.77.a it remains to be shown that D, is simple. We proceed similar to the corresponding considerations in Case 3 by assuming that there exists a vertex x E '1 One could have proceeded differently by considering D** := (D-D*)U A, and D* separately, applying induction t o D** and Theorem VI.79 to D* (observe that A ( D * )5 6 and that d,.(s) = 4 for s E V(A,)). However, in

this instance I preferred the approach presented because it did not rely on Theorem VI.79. Unfortunately, Goodey omits the corresponding part in his proof (which is admittedly rather sketchy) and restricts connectivity considerations t o the dual of Case 4 below where he uses the dual form of Lemma VI.76. Moreover, even his restricted considerations are incomplete and contain a flaw (see the footnote at the end of the proof).

VI.136

VJ. Various Types of Eulerian Trails

N ( b )n N ( d ) - ( a , zo, c } (one shows as before that bd in place of I,).

# E ( D ) by using zo

We first show that xt $Z E ( D ) for at least one t E {a,c}. Suppose to the contrary that xa,xc E E ( D ) . Then at least one of the triangles A, = (x,d,a),and A, = (x,b,a) is not a face boundary of D , and the same is true with respect to the triangles A3 = ( x , d , c ) and A4 = (2, b, c); otherwise, d,(a) = 4, d,(c) = 4 respectively, must hold. Since b,d E & ( D ) we assume w.1.o.g that A3 is not a face boundary of D. But then we find the triangulation of the plane D* $ K3 as in the connectivity consideration in Case 3 with A3 being a face boundary of D*; and zo # V ( D * ) . By Lemma VI.76, by K ( D ) 2 3, and since b,CE V6(D),d,(Z) E {4,6},weconcluded,.(d) = d,,(C) = d,,(z) = 4; thus d D ( x )= 6. Consequently, A, is a face boundary of D and, therefore, A, separates D. Especially, the side of A, not containing t o ,contains two or four elements of E,, two elements of Ed, but no element of E, since we have d,(x) = 6 = d,. (x)+ I {xb,xa} I. That is, a and d separate D , a contradiction. Suppose now xu,xc # E ( D ) . Then the side of the 4-gon Q1:= (a, b, x,d ) not containing z o , contains precisely one element eb E Eb and precisely one element ed E Ed, and the same is true with respect to the 4-gon Q2 := (c,b, x,d). This follows from b, d E V,(D). Denote eb = bub, ed = dud. It follows from the distribution of the elements of Eb and Ed with respect to the sides of Q,, that xub,xud E E ( D ) . To see that ub = u d , we consider the triangulation D- of the 4-gon Q- := (b, x,d , zo) defined by the side of Q- not containing a. By the above considerations we have d D - ( b ) = d,-(d) = 4, d D - ( z o ) = 3. Hence d,-(z) f 1(rnod2). Let vb,v d be defined the same way we defined u b , ud above, but with respect to Q,. Observe that

Since and therefore, (b,c,vb) and (d,c,vd) are face boundaries. 2 3, we conclude v b # ud, hence o + ( c) = d,(c) = 6 and K ( D ) ( ~ v ~ , ~ d , ~ ~ ~ , ~for b ,some c v ~s, E c sv(D-). ) This in turn implies to6 and d,-(x) = 1(mod2) that d D - ( x ) = 5. Since gether with d,(x) ub and u d on the one hand and vb,vd on the other hand lie on different sides of Q-,U b = ud follows of necessity. But then we necessarily have

<

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI. 137

contradicting dD(U) E {6,8} (observe that (a, d , u d )and (a, b, ub)are face boundaries of 0).Thus

xu E E ( D ) if and only if x c @ E ( D ).

(*I

Now we conclude from Lemma VI.76 and d,(c) = 6 that if xc E E ( D ) , then precisely one of A3 and A4 is not a face boundary and the side of Q1 = ( a , b , x , d )not containing zo has nothing in common with either Eb or Ed (note b , d E V,(D)). Thus, xu E E ( D ) , a contradiction to (*). Whence xu E E ( D ) , xc $ Ei ( D ) follows of necessity. If d D ( a ) = 6, then we reach the same contradiction by the symmetrical argument; i.e., d,(a) = 8. In any case, d,(a) E {6,8} and xu E E ( D ) imply that at least one of A, = ( x , d , a ) and A , = (x,b , a ) is not a face boundary; in fact, d,(x) 5 6 and Lemma VI.76 imply that A, is a face boundary if and only if A, is not a face boundary. Now we argue symmetrically to the above case x c E E ( D ) to obtain the contradiction x a , x c E E ( D ) . Thus we reach the conclusion that x does not exist; i.e. ~ ( 0 2 3. ~ Theorem ) VI.77.a now follows.')

As a matter of self-criticism it should be pointed out, however, that Theorem VI.77.a (and thus Goodey's result) has been stated incorrectly in [FLEI83b, Theorem 4.41; for, that part of the hypothesis where the 8valent vertex is required to be adjacent to a 4-valent vertex, has been '1 In his connectivity considerations, Goodey makes the following mistake: translated into our language, he confuses the vertex pairs { a , c } and { b , d } claiming that x' E N ( a ) fl N ( c ) - {b, d o , d } # 8 is impossible. His proof is incomplete, unnecessarily complicated and makes use of the fact that a graph cannot have just one odd vertex (in its dual form). To see that X' can exist indeed consider 0, with its 4-gond base denoted by a, b, C , d in some cyclic order, and denote the 'top' ('bottom') vertex of this double pyramid by z0 (2'). Apply an O,-extension to the face boundary (a,d , x') and a weak r*v,extension t o the 4-gon (a,x', C , b) having bx' as diagonal. The graph D' thus obtained satisfies d D , ( a ) = 8, d,c,x' E v6(0'). But d,,(b) = 4. A W4extension of D' applied t o the 4-wheel defined by a , b, c and the two 4-valent vertices created by the above weak W4-extension, establishes a counterexample t o Goodey's connectivity considerations. Moreover, Goodey fails t o consider the case x E N ( b ) n N ( d ) - {a,zo, c}. Goodey's flaw rests solely on the fact that he only considers the case dD(U) = 6; the formal argument is correct. But he wrongly concludes that X' does not exist; for it can exist if d D ( a ) = 8, as our example shows. We note, however, that Goodey's result has been proved anew in [PET077a].

VI. 138

VI. Various Types of Eulerian Trails

omitted; and the theorem's conclusion was stated in a weaker form; namely, that D just has an A-trail. Unfortunately, I was not able to generalize Theorem VI.77.a to achieve the same conclusion as Theorem VI.79. On the other hand, this generalization would follow if one assumes the validity of any of the Conjectures VI.72 - VI.74. To see this we formulate another conjecture on A-trails (respectively, A-partitions) and then show that it is equivalent to Conjecture VI.73 (and therefore, also to Conjecture VI.74). Conjecture VI.82. Let D be a simple, 2-face-colored7eulerian triangulation of the plane, and let A = (x,y, z ) be a face boundary of D. Then D has a perfect A-partition {V,, V,} with x E 5 ,y, z E V,. Proposition VI.83. (see also [REGN76a7Satz 3.2 and Satz 3.31). Conjecture VI.82 and Conjecture VI.73 are equivalent. Proof. Since non-separating A-trails and perfect A-partitions are equivalent concepts in the case of plane eulerian triangulations, Conjecture VI.82 implies Conjecture VI.73. Hence it sufEices to show the converse.

So, let D and x,y, z be chosen as stated in Conjecture VI.82, and let A, = (2, y, w), w # z , be the face boundary of D satisfying E ( A ) n E ( A , ) = {xy}. Let D' and D" be two copies of D with x', y', z', (x",y", z"), denoting the vertices of D' (D") corresponding to z,y, z , and define H' := D' - A', H" := D" - A", where A' and A" are the face boundaries corresponding to A. We now embed H' in int A and H" in int A, in such a way that after identifying y' with z , z' and y" with x, x' and 2' with y, and x" with w, the resulting graph D,is an eulerian triangulation of the plane. Consequently, D' N H'U A and D" N H" UA,. Suppose D,has a perfect A-partition {V:, V;}. Denote by {Vf, V:}, {V', V i } , {V", V:} respectively, the vertex partitions of D, D', D" respectively, induced by

P l l 7 v;1.

'1 Non-separating A-trails were defined for triangulations of the plane only, but as such it is no problem to extend this definition to arbitrary plane eulerian graphs. The equivalence between these two concepts in the general case is not given. For, while a perfect A-partition defines two non-separating A-trails, it is not true in general, even in the 4-regular case, that the A-partition {V,, V,} corresponding to a non-separating A-trail is a perfect A-partition. However, if (V,) is connected for S = 1,2, this {V,, V,} is, in fact, a perfect A-partition (see Corollary VI.69).

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.139

5'

x,y E implies z E V i where {j,k} = { 1 , 2 } . But then x" E V , , y",z" E which is tantamount to saying that D has a perfect A-partition {V;,V;} with x E V,, y,z E vj*, {j,k} = {172}. Hence if k = 1, then V. := V:, s = 1 , 2 , and if k = 2 , then V' := V; and V, := V;, yield a perfect A-partition {V,, V2}in D with 3: E V,, y, z f V,. Chsidering {x,y} 6 ql, j = 1 , 2 , we assume w.1.o.g. x E V,: y E V:. So, if z E V;, then for V, := V f , V, := V . , {&, V,} is a perfect Apartition as required. If, on the other hand, z E V' holds, then we conclude x' E Vi, y', 2' E V,l. This is tantamount to saying that D has a perfect A-partition {V;, V;} with y, z E V;, x E V;; hence {Vl,V2} defined by V, := V;, V, := V; is as required. The proposition now follows. We leave it as an exercise to translate Conjecture VI.82 into a stronger (but equivalent) version of the BTC (Conjecture VI.72).

So, on the one hand, we have obtained (basically minor) partial solutions of Conjecture VI.73 (or, equivalently, the BTC,') Conjecture VI.74, respectively), and on the other hand, we have been able to deduce a stronger equivalent version of that conjecture, namely Conjecture VI.82. The equivalence between A-trails in eulerian triangulations of the plane and hamiltonian cycles in plane bipartite, cubic graphs has been the basis for the considerations of this subsection (and led to the even more general equivalence expressed in (V1.A); see the discussion preceding Lemma VI.76); and this equivalence can be viewed as sufficient motivation to study A-trails in general plane eulerian graphs. But historically, the study of A-trails has emanated from another hamiltonian problem in plane, bipartite, cubic graphs. Kotzig (see, e.g. [KOTZ62a764al) has studied and constructively determined those cubic graphs G, which admit a l-factorization {&, L,, L,} such that L, U L j is a hamiltonian cycle of G, for i # j,1 5 i , j _< 3; he also proved that apart from the multigraph on two vertices, there is no plane bipartite, cubic graph which admits a l-factorization of the above type [KOTZ62a7Theorem31. This led me to the following question: let G, be a 2-connected plane, bipartite, cubic graph, and let L = {L,, L,, L 3 } be

'1 It has been shown in (HOLTSSa] that the BTC holds for graphs with fewer than 66 vertices.

VI.140

VI. Various Types of Eulerian Trails

its natural 1-factoTization (see Theorem 111.67). When does G, contain a hamiltonian cycle H with L C E ( H ) for some L E L ? (F) To answer this question, suppose w.1.o.g. L = L,, and let Q = L, U L,. By definition of C, Q is the 2-factor of G, whose elements are precisely the boundaries of the 1-faces of G, in the 3-face-coloring underlying the definition of L. Form G := G3/Q in the natural topological sense by contracting every bd(F) € Q onto a vertex of F without allowing intersections of open edges in the course of the contraction procedure. Consequently, for every v F E V ( G )its cyclic order O+(v,) is determined by the cyclic order in which the half-edges of L incident to vertices in bd(F) occur in bd(F), and where vF is the vertex of G corresponding to 6d(F) C G,. By construction and since G, is bipartite, G is a connected plane, eulerian graph. The above construction of G from G , leads to the following simple result whose proof follows from the above considerations and is, therefore, left as an exercise. Lemma VI.84. Let G , be a 2-connected plane, bipartite, cubic graph, and let L be its natural 1-factorization. Let L E L be chosen, and let Q := E(G,) - L. Furthermore, let G = G3/Q be constructed as above. Then the following statements are equivalent. 1) G, has a hamiltonian cycle containing L.

2) G has an A-trail. Thus, the original question (F) is equivalent to asking which connected plane, eulerian graphs G admit an A-trail. As we have seen before, there are even such G with K ( G )= 3 which do not admit A-trails. The example G constructed had only 3- and 4-gonal face boundaries, but for E V ( G )- V,(G), d(v) was fairly large. In contrast, the graph Go of Figure VI.26 has 4- and 6-valent vertices only, K(G,) = 2, but it even has 8gonal faces. One can lower the face sizes by replacing with 2-gons those 2-connected subgraphs of Go which axe defined by the components of Go-V6(G,) and the edges joining them to the 6 - d e n t vertices (see Figure VI.37). The eulerian multigraph G' obtained is 6-regular and has 2-, 3-, and 4-gonal face boundaries only. But, it has no A-trail for the same reason why Go has no A-trail. However, on the grounds of Lemma VI.84 we can re-interpret G' as G' = G$/Q6 where QS is a set of hexagonal face boundaries of Gi;and the other face boundaries of Gi are 4-,6-, and 8-gonal. Noting that G$ has precisely two 8-gonal face boundaries we observe that D = D(G$) is a simple eulerian triangulation with 4-,

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI. 141

6-, and precisely two 8-valent vertices. That is, D ‘almost’ belongs to or, in other words, Gi ‘almost’ belongs to B,,(see the definitions of I,and B, preceding the statements of Theorem VI.77.a, Theorem VI.77 each of whose elements respectively). Yet, although G$ has a 2-factor Q6, is a hexagon, Gk does not have a hamiltonian cycle containing the 1-factor E(G$)- Qs. However, we pose the following question: f o r G, E a, does G, have a hamiltonian cycle H with L c H where L is a 1-factor in the natural 1-factorization of G, ? One might ask the same question concerning the cubic graphs whose duals have been treated in Theorem V1.79. In terms of A-trails in eulerian graphs, we are led to the following question: i f G is a 2-connected, plane, eulerian graph with V,(G) = 0 for n > 6, and i f I E ( b d ( F ) )I< 5 for every face F and I E(bd(F’)) I= 4 f o r at most one face F’, does then G have a n A-trail ? The graph G’ of Figure VI.37.a) shows that one cannot allow for more than one such face F‘. Also, this G’ shows that specifications concerning the 2-factor Q are not promising in general (for there, Q = Q6 is a set of hexagons). However, it is true that if a (not necessarily planar) connected cubic graph G, has a 2-factor Q4each of whose cycles is a 4-gon, then G, has a hamiltonian cycle H 2 E(G,) - Q4 [KOTZ68c, Theorem 61. This follows as an application of Corollary VI.6. It is therefore left to the reader to prove this result. The above question (F), respectively the equivalence expressed in Lemma VI.84 can just as well be treated in terms of A-trails in eulerian triangulations of the plane.

Theorem VI.85. Let G, be a 2-connected, plane, bipartite, cubic graph with its natural 1-factorization L,and let D = D(G,) be given together with the 3-vertex-coloring {Cl,C,,C,} corresponding to the (natural) 3-face-coloring of G, (see Theorem 111.67). The following statements are equivalent.

1) G, has a hamiltonian cycle H containing L , E C.

2) V(D)has a perfect A-partition {V,, V,} satisfying the equations v, = c;UC,, v, = cl’uc,, ucl’=

c;

c,.

Proof. We consider H as a simple closed curve in the euclidean plane. For every e E L,, since e E E(H)nE(bd(F,))nE(bd(F,)) for some 2-face F2 and 3-face F3 of the natural 3-face-coloring of G,, we conclude that F, E int H if and only if F3 E e s t H .

VI. Various Types of Eulerian Trails

VI.142

a)

G'

b)

G;

Figure VI.37. a) A 6-regular multigraph G' without Atrail having no n-gonal face boundaries for n > 4, and having precisely two 4-gond face boundaries. b) The cubic bipartite graph G$ satisfying G' = G $ / Q , (&, is defined by the boundaries of the faces marked Q,). Gi is hamiltonian; but G$ has no hamiltonian cycle H with E(GL) - Qs c E(H). Suppose for el, fl E L , that the corresponding i-faces F', F" with el E E(bd(F')),fi E E(bd(F")), lie on different sides of H,iE (2,3}. By choosing el and fl as lying on H as close to each other as possible, we may assume that el and fi are joined by an edge g in H . W.1.o.g. g E L, E L. Denote by e 3 , f 3 E L, the edges adjacent to g and el , f l respectively. The open edges ei,fi cannot lie on the same side of H ; otherwise, E(bd(F'))nE(bd(F"))# 0, contradicting the fact that F' and F" are both i-faces. On the other hand, e:, fi lying on different sides of H implies that {e,, g, fi} c E(bd(F*))and {el,9,f,} c E(bd(F**))for

VI.3.1. Duality between A-Trails and Hamiltonian Cycles

VI.143

some faces F* and F**. However, since in the natural 3-edge-coloring C every face boundary contains edges of only two classes of C, while each of bd(F*) and bd(F**)contains edges of all three classes of C,we obtain a contradiction which implies that either a l l 2-faces lie in ezt H , or they all lie in int H . Suppose w.1.o.g that the 2-faces lie in int H . It follows from the argument at the beginning of the proof that the 3-faces lie in ext H . Denoting by Ci (Cy)the necessarily non-empty set of 1-faces lying in int H ( e z t H ) , and by further denoting V, := V(T,),V, := V ( T 2 )where T’,and T, are the trees in D(G,) defined by the faces of G, lying in i n t ( H ) , e z t ( H ) respectively, we obtain V, = Ci U C,,V, = Cr U C, where Ci U C; = C,,Ci n Cy = 0 by definition. Moreover, since G, is bipartite, D := D(G,) is an eulerian triangulation of the plane and so {V,, V,} is a perfect A-partition of V ( D )by Theorem VI.70. Conversely, suppose the perfect A-partition {V,, V,} of V ( D ) is given as described in Z), and let Ti = ( y ) i, = 1,2, be the corresponding trees. For each i E { 1,2}, the edges of G, belonging to exactly one face boundary among the faces represented in y, define a hamiltonian cycle H in G,, with H resulting from V, as well as from V,. Since

V, = Ci u C,,i.e., C, n V, = 0 ,

and since

L, = { e E E(bd(F,)) n E(bd(F,))/Fj is a j-face, j = 2,s) we conclude that every edge of L , belongs to precisely one boundary of a face represented in V,. That is, L, c E ( H ) . This finishes the proof of the theorem. Consequently, questions on the existence of A-trails in plane eulerian graphs c m , in principle, be treated as questions on the existence of special types of A-trails (perfect A-partitions, respectively) in eulerian triangulations of the plane, or by the same token, can be treated as questions on the existence of special types of Hamiltonian cycles in plane, bipartite, cubic graphs. This leads us to the question why one should deal with A-trails in plane eulerian graphs instead of considering right away hamiltonian problems in plane, bipartite, cubic graphs (particularly in view of the equivalence between the BTC and Conjecture VI.73, as well as between Goodey’s result (Theorem VI.77) and Theorem VI.77.a). History s e e m to justify the ‘hamiltonian approach’, and in section VI.2 we proved a

VI. 144

VI. Various Types of Eulerian Trails

result on hamiltonian decompositions to solve an eulerian problem. However, as we shall see in the context of compatible cycle decompositions, dealing with certain problems in eulerian graphs and translating corresponding results into the theory of cubic graphs has proved to be a very fruitful approach indeed. In the particular case of A-trails it is my opinion that the ‘eulerian approach’ may eventually be just as fruitful. This opinion results not only from many years of experience in dealing with both eulerian and hamiltonian problems, but also from a methodological difference concerning A-trails and hamiltonian cycles in the respective classes of graphs. Namely: finding eulerian trai2s of whatever kind always involves aZZ edges of the graph, while finding a hamiltonian cycle requires the proper arrangement of certain edges and rejecting, so to say, the remaining edges. The latter problem becomes, therefore, more complicated if the graph under consideration has only vertices of low degree (in particular if it is 3-regular). As a support for this point of view, compare Ore’s Theorem (Theorem 111.75) with the BTC (Conjecture VI.72)). We finish this subsection by exhibiting a relation between arbitrary plane 2-connected graphs and certain eulerian triangulations of the plane having a non-separating A-trail ([REGN76a, Satz 5.1]), respectively plane bipartite cubic graphs having a hamiltonian cycle. Namely: let G be plane and 2-connected, and consider GA := S(G) U R ( S ( G ) )

.

It follows from this definition that GA is an eulerian triangulation of the plane (note that S(G) is bipartite); and it is a simple graph precisely because of K ( G )2 2. Again by definition, the 3-vertex-coloring of G, (which is uniquely determined up to permutation of the color classes) is given by the partition {V,, V,, V,} of V(G,) where V1 := V ( G ) ,

V, := V ( S ( G ) )- V(G),

V3 := V ( R ( S ( G ) )) V(S(G))

(note that V, is an independent set already in S(G), V, is an independent set by definition of S(G), and V, is an independent set because every face of S(G) contains precisely one vertex of R(S(G))). Observing that V, is a set of 4-valent vertices in GA, we may conclude from Theorem VI.67 that H := ((GA)v,,1)v,,2 is connected and has only 2- and 4-valent vertices; hence it has an A-trad TH which can be interpreted as an A-trail T of GA. It follows from this definition of T that T is non-separating. Hence G, := D(GA) is a hamiltonian, plane, bipartite, cubic graph with 4 G 3 ) = 3.

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI.145

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs Returning to the discussion preceding Theorem VI.85 we are faced with the question into which direction one should go to determine classes of plane eulerian graphs admitting A-trails. As we have seen above, and posed equivalently, specifications as to the degrees of the vertices of a color class in the 3-vertes-coloring of an eulerian triangulation of the plane do not seem to be promising, unless every vertex of the color class is 4-valent. But there is a parameter which plays an essential role in many graph theoretical problems, and especially in the theory of planar graphs connectivity. As we have seen before, 3-connected, planar, eulerian graphs exist which do not admit A-trails. Although, in our construction, the triangulation of the plane we started from can be 5-connected (as there are non-hamiltonian, planar, cyclically 5-edge-connected, cubic graphs), the outcome is always a 3- but not 4-connected, planar, eulerian graph. I have tried to modify the construction in order to obtain a 4-connected, planar, eulerian graph without A-trails which satisfies the equivalence in statement (V1.A) (see the discussion following Lemma VI.75)) but without success. Hence I pose the following conjecture. Conjecture VI.86. Every planar, 4-connected, eulerian graph has an A-trail. The above lack of success is not the only reason for posing this conjecture. For it follows from Tutte’s ‘Bridge Theorem’ (Theorem 111.70) that every planar 4-connected graph is hamiltonian (Corollary 111.71). And this is a very strong property indeed. Moreover, it has been shown that the problem of finding a hamiltonian cycle in a 4-connected planar graph, is polynomial. More precisely, an algorithm based on the proof of Tutte’s ‘Bridge Theorem’ (Theorem 111.70) has been developed whose “time and space-complexity . . . is overvalued in O ( n3) ”,where n is the number of edges, [GOUY82a, Theorem 21. This is in sharp contrast to the fact that deciding the existence of hamiltonian cycles in planar, 3-connected, cubic graphs is an NP-complete problem, [GARE76a]; and this fact will play a certain role in our considerations on the complexity of determining whether a planar, eulerian graph has an A-trail. Moreover, finding hamiltonian cycles in bipartite cubic planar graphs is an NP-complete problem as well, [PLES83a]; however, this result does include all such 2-connected

VI.146

VI. Various Types of Eulerian Trails

graphs (which are not so interesting in view of the BTC and, via dualization, of little significance to the consideration of A-trails). On the other hand, it was shown in [NISH83a] that the problem of determining a hamiltonian walk (shortest closed walk covering all vertices) in triangulations of the plane, can be solved in O ( p 2 )time, and that the length of a hamiltonian walk is at most ( p - 3) provided p 2 9. Thus, in the case of 4-connected triangulations of the plane, one can find a hamiltonian cycle even in O ( p 2 )time. More recently, a linear algorithm for finding a hamiltonian cycle in such graphs has been developed in [ASANS4a], and in the same year N. Chiba and T. Nishizeki published such algorithm for arbitrary 4-connected planar graphs (see, e.g., [NISHSSa, p.182- 1841. The existence of hamiltonian cycles in 4-connected triangulations of the plane follows from Tutte’s result, but was proved already by H. Whitney in the thirties (see Corollary 111.72). Note that a triangulation of the plane need not be Hamiltonian - just apply an 0,-extension for every face of such a triangulation). These considerations can be viewed as positively supporting Conjecture VI. 86.‘1

3

A hamiltonian cycle H in a simple plane graph G defines two simple 2connected outerplane graphs G , ,G, satisfying G,UG, = G and G , nG, = H . 2 ) For, viewing H as a simple closed curve in the plane and a plane graph as a certain point set, one can define G , and G, by

V ( G , ) = V(G,) = V ( G ) E(G,)= { e E E(G)/e n int H = 0 ) , E(G,)= {fE E ( G ) / f n ext H = 0) . Now, if both G, and G, are eulerian, they have by Theorem VI.63, Atrails T, and T,, respectively. Hence one is inclined t o believe that this suffices to construct an A-trail T of G starting from T, and T,. A. Rosa once told me during a stroll that in his opinion (and I hope I quote him correctly), there are two categories of conjectures: firstly those where “everyone knows that they must be correct, but one cannot yet prove them”, and secondly those “which create surprise if they are correct”. I believe that this is a very good description of how (how many ?) mathematicians relate to conjectures (in my own research I was faced with both situations). As far as Conjecture VI.86 is concerned, I would place it in-between these two categories. 2, Of course, for precisely one of G , and G,, the ‘outer face’ is the bounded iE region defined by H ; but it is no problem t o re-embed the corresponding Gi, { 1,2}, without altering O+(Gi)- simply start with an embedding of G on the sphere.

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI.147

Unfortunately, for a 4-connected, planar, eulerian graph G to have a hamiltonian cycle H does not guarantee an easy way of finding an A-trail in G . This general statement is illustrated by the following discussion in which we say that the hamiltonian cycle H of G has the A-property if any pair of edges { e , f } adjacent in H belongs to a face boundary of G (or, equivalently, e’ and f’ are neighbours in O+(v) where e‘,f’ E Ec; cf. [REGN76a, Definition 4.2.1.1). As we shall see below if G admits such an H having the A-property, it is possible indeed to combine Atrails T, and T, in G , and G,, respectively, to obtain an A-trail in G. In fact, in this case one can start from arbitrary A-trails. However, we first study A-partitions in outerplane eulerian graphs. The following result, [REGN76a, Satz 4.1.2.1, gives a simple and precise characterization of such A-partitions (this result will be more relevant, however, in establishing a formula for the number of A-trails in outerplane graphs. - See Volume 2, Chapter IX).

Theorem VI.87. Let G be a simple 2-face-colored, 2-connected, outerplane, eulerian graph whose outer face F, is a 1-face. Then the following statements are equivalent. 1) G has an A-trail.

2) V ( G )admits a partition {V,, V,} such that I V(bd(F,))nV1 I= 1 for every 1-face F, # F,. Proof. If G is a cycle, then the equivalence between 1) and 2) is vacuously true; for in this case, F, # F, does not exist. Hence we assume in the following discussion that A ( G ) > 2 holds. That is, at least one 1-face Fl # F, exists. a) Let T be an A-trail of G , and let F, # F, be arbitrarily chosen. Moreover, let { V,, V,} denote any A-partition defined by 2’. Since Fl is a 1-face we must have I V(bd(F,))n V, I> 0; this follows exclusively from the fact A ( G ) > 2; for in this case, V(bd(F,))n V, = 8 implies that G,,, is disconnected with (E(bd(F,)))as a component. Suppose now V(bd(F,))nVlZI {v,w}. Then R = {F,, F,} is a unicolored face-ring with L , = {v,w} as a complete set of links. The elements of R are 1-faces, and L R C Vl follows from the assumption. By Theorem VI.67, however, this contradicts the fact that {Vl, V,} is an A-partition. b) Suppose the partition {Vl, V,} of V ( G )satisfies I V(bd(F,))n V, I= 1 for every 1-face F, # F,. Consider an arbitrary unicolored face-ring R

VI.148

VI. Various Types of Eulerian Trails

whose elements are &faces for fixed 6 E { 1,2}, and let LR be a complete set of links of R. By Theorem VI.67, it suffices to show LR v6.Suppose to the contrary that for some such R and LR we have LR C v6. Suppose first 6 = 1. Since I R I> 1 it follows that Fl E R for some Fl # F,, and by definition I V(bd(F,)) n LR I= 2. Consequently, I V ( b d ( F l ) )n V, I> 1 which violates 2). Now suppose 6 = 2. Among the possible choices for R and LR consider one with minimal I LR 1; i.e., L C LR implies that there is no unicolored face-ring R' having L as a complete set of links. We want to show that LR = V ( b d ( F * ) )for some 1-face F* # F,.

IE

It follows from the minimality of LR that I bd(F') n bd(F") n LR {0,1} for different elements Ft7F" E R. Hence we can espress R and LR in the following form:

bd(Fi)n bd(Fj)= 0 for j - i > 1, 1 5 i

< j 5 m, { i , j } # (1, m } .

As in the first part of the proof of Theorem VI.67, we construct a simple closed curve C with m

GnC=L,

.

and C - L R C U F , i= 1

The unbounded region of C, ext C, contains F,, and since G is outerplane joining it follows that int C n V ( G ) = 0. Consequently, the path ? J ~ - ~ and , ~ ?J;,;+~ in bd(Fi) whose open edges lie in int C must satisfy pi,i+l

and therefore,

int

{'i-I,i?

'i,i+I}

7

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI.149

Hence,

(LR) C int C U LR and ( L R )is a cycle.

This, A(e) = l ( e E E ( G ) ) ,and the minimality of LR imply that

(intC - E( ( L R ) ) )n G = 0

.

That is, adjacent edges in ( L R ) are consecutive in O+(v), where v is their common end-vertex. Thus ( L R )is a face boundary bd(F*). Since E ( ( L R )n ) E(bd(Fj))# 8 for i = 1,.. . ,m, it follows of necessity that F* is a l-face; and F* # F, because ( L R )n ezt C = 0. Thus, we have found a l-face F* with

V(bd(F*))= LR

V,,

i.e.,

IV(bd(F*))n V, I= 0 .

This contradiction to statement 2) finishes the case 6 = 2; whence we conclude that LR V,, 6 E { 1,2}, for every unicolored face-ring R whose elements are 6-faces, with LR being a complete set of links of R. By Theorem VI.67, G has an A-trail whose A-partition is {Vl,V,}. Now we prove the result indicated at the end of the discussion preceding Theorem VI.87.

Theorem VI.88. ([REGN76a, Satz 4.2.1.1). Let G be a simple plane, eulerian graph with a hamiltonian cycle H having the A-property. Then G has an A-trail. Proof. We can write G as the union of two 2-connected, outerplane, eulerian graphs G , and G, with G,nG2 = H (see the discussion preceding Theorem VI.87). Let T, and T2 be A-trails of G,,G, respectively; by Theorem VI.63, T, and T2 exist (we start with a 2-face-coloring of G and apply Theorem VI.63 to Gi whose 2-face-coloring is induced by that of G, i = 1,2. Thus, the face of G j whose boundary contains all vertices of G, has either color 1 or 2 depending on the value of i E {1,2}). Because of the A-property of H we have for arbitrary v E V ( G )

Consider now {V:, V i } ,an A-partition corresponding to Ti in Gi, i = 1,2, and define

Wj’ := T/j; n ( V ( G i )- V . ( G J ) for arbitrary i , j E {1,2}

.

VI. Various Types of Eulerian Trails

VI. 150

This definition together with V j n Vj = 8 and (*) yield

~ j n’W: = 8 (observe that Wj

( j , i ) # (k,z), { i , j , k, Z} C_ {I,2)

for

2 V,(G,)

(**)

for { i , k } = {1,2}).

Consequently, {PVj/i,j = 1,2} is a partition of V ( G )- V,(G); therefore, W j := Wj’ U Wj”, j = 1,2, defines a partition of V ( G )- V,(G). Now we observe that a 2-valent vertex 21, cannot belong to any complete set of links LR of any unicolored face-ring R; for 21, belongs to precisely one &face, 6 = 1,2. Hence, if we can show for such R and LR that LR g W,, provided the elements of R are &faces, 6 E {1,2}, then for any V: V,(G) an A-partition {V,, V,} of V ( G )is obtained by defining V, := W , U V:, V, := PV, U (V,(G) - V:) (see Theorem VI.67).

s

So, let R and LR be as above. If all elements of R are &faces of Gi, i E {1,2}, then the existence of Tiimplies LR g Wi by Theorem VI.67, 6 E {1,2}. Since W i 2 V,(Gj) for { i , j } = {1,2}, LR g W, follows. Hence we have to assume that some elements of R are faces of G, and some are faces of G,; and all of them are &faces, 6 E { 1,2}. We further specify the situation by assuming w.1.o.g. that the “outer” face F 1 ( F 2 ) of G, (G,) has color 6 (6 1) where we put 6 + 1 = 1 if 6 = 2 (see footnote 2, in the discussion preceding Theorem VI.87). Moreover, denote R = {F,,. . . ,F,; m > 1) and LR = ( V ~ , ~ , V .~ ., ~ , (see the proof of Theorem VI.87, part b)). By relabeling the elements of R and LR if necessary we may assume w.1.o.g. that bd(F,) 2 G,, bd(F,) 2 G,. Define

+

i, = min{i/l < i 5 m A bd(F,)

s G,}

and

i, = i, - 1 .

Then we obtain in G, a unicolored face-ring R‘ whose elements are 6faces, and a corresponding complete set of links L‘ = LR, by defining

R’ := {F,,. . .,Fil ,F1}, L‘ = {2r1,,,

. . ., v ~ , , ~ ~ , v .~ , ~ }

We have L‘ E LR and L’ V(G,). If we had L , C W,, this and W i C V,(G,) would imply L’ E W i C V J . But Theorem VI.67, applied to G,, says that the existence of TIand its A-partition {V:, V i } implies L‘ V i . Thus, we must have L , g W,. Again by Theorem VI.67 and the above definition of V, and V,, this means that {V,, V,} is an A-partition of G; i.e., G has an A-trail. Theorem VI.88 now follows.

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI. 1.51

However, not every A-trail T of G is obtainable from A-trails T, and T, of G , and G, respectively, as described in the proof of Theorem VI.8g That is, some A-partition {V,, V,} of V ( G )may not be an ,4-partition of V ( G , ) = V ( G )for i = 1,2. This can be seen already in the case of 0, which we can view as obtained from the cycle c 6 = (u,, . . . , 216) by adding A, = (u,,u3, v5) and A, = ( u 2 ,u4,216). Embed A, in the bounded region of c6; thus A, lies in the unbounded region of c6. We assume the outer face of the 2-face-colored 0 6 to be a 1-face. Defining H := c6, G , := H U A,, G, := H U A2, and V, := {u,, v3,2 ] 6 } , V, := {u,,u4, us} we conclude that although {V,, V,} is a (perfect) A-partition of V ( G ) ,{V,, V,} is not an A-partition of V ( G , ) = V ( G ) ,nor of V(G,) = V ( G )(note that H has is disconnected, and so is ( G 2 ) { v z , V 6 } , l . the A-ProPertY). For, (G1){v12vs},2 However, if an A-partition {V,, V,} of V ( G )is an A-partition of V ( G i )as well, i E {1,2}, we may say that the latter is an induced A-partition of the former. Correspondingly, we can say that the A-trail Tiof Gi is induced by the A-trail T of G. The graph of Figure VI.38 shows an eulerian triangulation D of the plane together with a hamiltonian cycle H such that the corresponding graphs G,, G, satisfying G , U G, = D and G , n G, = H (see the discussion preceding Theorem VI.87), are eulerian. We discuss this graph D: H does not have the A-property since ( V ( G , ) - V,(G,)) n (V(G,) - V9(G3)) - - = { a , b , c , d , e , f } . We know from Theorem VI.63 that Gi has an A-trail Ti, i = 1,2. We want to show that no choice of T, and T2 permits extension of these A-trails to an A-trail T of G in a way similar to the one in the proof of Theorem VI.88. For this purpose, it suffices to show that T, does not induce the same type of splitting in V, := { a , b, c, d, e , f } as T,. To be more precise, let G , and G, have their 2-face-colorings carried over from a 2-face-coloring of D. Among G , and G, let G , be defined as the graph whose outer face F, is the unbounded region of H (where H is viewed as a simple, closed, plane curve), and suppose w.1.o.g. the 2-face-coloring of D is chosen in such a way that F, in G , is a 1-face. Whence we may conclude that the “outer” face int H of G, is a 2-face. As in the proof of Theorem VI.88, let {Vl,V j } be the A-partition of the arbitrarily chosen A-trail Tiof G;, i = 1,2. We want to show that the equations V, n V: = V, n V:, V, n V; = V, n Vz cannot hold. Suppose a E V t . Since d, e , f are boundary vertices of 1-faces of G , each of which contains a as boundary vertex as well, { d , e, f} C V$ follows by Theorem VI.87. By the same theorem however, { e , f} C V$ is impossible since {e,f} c bd(F,) for some 2-face F2 of G, and because the “outer”

VI. 152

VI. Various Types of Euierian Trails

Figure VI.38. An eulerian triangulation of the plane D with hamiltonian cycle H defining two 2-connected7 outerplane, eulerian graphs G,,G2 with D = G , U G,, H = G , n G,. D has no A-partition which induces A-partitions in both G , and G, (note that (V(G,)- V,(G,)) n (V(G,) -

V,(G,)) # 0).

face of G, is a 2-face as well. Therefore, the above equations cannot hold if a E V:. Now suppose a E V l nV;. Since b and c are boundary vertices of 2-faces of G, each of which also has a as a boundary vertex, we have { b , c} c Vf by Theorem VI.87. On the other hand, { b , c} E V(bd(F,))for some 1-face of G,; thus, by Theorem VI.87 we must have { b , c } V:. Consequently, also in the case a E V: we may conclude that the above equations cannot hold. However, we leave it as an exercise to check that D has an A-trail indeed (we note, though, that n(D) = 3).

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI.153

So, while in general it is not possible to deduce from some A-trail of G A-trails Ti in Gi, i = 1 , 2 (where Gi is defined as above with respect to a hamiltonian cycle H ) , even if H has the A-property, we do achieve the following interesting result, [REGN76a, Satz 4.2.21. We present a proof differing entirely from Regner's. Theorem VI.89. Suppose for the 2-face-colored, plane, eulerian graph G that it has a hamiltonian cycle H defining two outerplane eulerian graphs G,, G, satisfying G, U G, = G , G, n G, = H (see above). If G has an A-trail T inducing an A-trail Tiof Gi, T induces an A-trail T j in G, as well, { i , j } = {1,2}. Proof. We assume w.1.o.g. that F,, the unbounded face of G, (with bd(F,) = H ) , is a l-face in the 2-face-coloring of G, induced by the 2-face-coloring of G, and that G has an A-trail T inducing an A-trail T9 in G,. Let {V,, V,} be the A-partition of V ( G )= V(G,) corresponding to T ,T2respectively. We have to show that {V,, V,} is an A-partition of V(G,) as well. For this purpose we first observe that G1 := GLT1,, is a connected outerplane graph whose outer face F&, is a l-face with E(bd(F,)) E(bd(Fk))and such that E(bd(F&,))n E ( B ) # 8 for every cycle B of G1 (Lemma VI.5S, Theorem VI.59). If we can show that every non-trivial block of G1 - E(G,) is an end-block of GI, then G1 - (G, E ( H ) )has precisely one non-trivial component, namely (Gl)vl ,,, which satisfies the hypothesis of Lemma VI.58. Consider the graph

Go is well defined because the "outer" face F" of G, is a 2-face in the induced 2-face-coloring of G, (since F, is a l-face), and therefore E ( H ) = E(bd(F")); so, the construction of Go from (G2)vl,1amounts to adding diagonals of H lying in G,, i.e., adding the elements of E(G,) E ( H ) . Also, since T, appears as a run through the boundary of the uniquely determined l-face of (G2)v1,,, it follows by definition that Go is connected. Moreover, E(bd(F")) C E((G,)"] ,,) implies that 1) (G2)V1,1contains (E(bd(F,))) as a block;

therefore

2) Go has a block B" 5) (E(bd(F"))) with B"

N

G,;

VI. 154

VI. Various Types of Eulerian Trails

3) every block C of Go - B" is a cycle with E ( C ) E E(G,) - E ( H ) and therefore, C is an end-block of Go (see 2)). We relabel the vertices of B" according to their original labels in G; then we obtain B" = G, and the isomorphism

For every block C of Go other than B" (see 3)) we have C f l B" C V,. Hence, the unique vc E CnB" is not affected by the transition from Go to Since every z E V(C - vc) is 2-valent in Go and therefore also in (GO),, ,,, it follows that every block C of Go - B" is an end-block as unlabeled graphs, of Finally, if we view G, and then the isomorphism (*) becomes an identity (in (*), the isomorphism can be chosen so as to act as the identity on E(G1) = E((Go)v,,l). Consequently, every non-trivial block of G1 - E(G,) is an end-block of G1 whose edge set belongs to G, - H . So,

which satisfies the hypothesis of Lemma VI.58 because G1 also does (the left side of the last equation only expresses the deletion of certain endblocks of G' which renders a connected graph and leaves unchanged the embedding properties espressed in that lemma). It follows that (Gl)vl has an A-trail. Consequently G, has an A-trail. The theorem now follows.

,,

Maybe, the above considerations on A-trails in plane, hamiltonian, eulerian graphs will be of help in eventually proving or disproving Conjecture VI.86. I do not know, however, which of the plane, 4-connected, eulerian graphs admit a hamiltonian cycle H having the A-property or at least an H such that the graphs G,, G, defined by int H U H and ext H U H , are eulerian.

So, let us consider Conjecture VI.86 for the case of eulerian triangulations D.The constructions performed before (see also the considerations preceding Conjecture VI.86) do not permit one to draw the conclusion that A-trails and non-separating A-trails are equivalent concepts in 4connected, eulerian triangulations of the plane (compare this with Conjectures VI.73, VI.74 and the discussion leading up to Lemma VI.75). Modify Conjecture VI.86 for triangulations of the plane to the extent

VI.3.2. A-Trails and Hamiltonian Cycles in Eulerian Graphs VI.155

that one replaces ‘A-trail’ with ‘non-separating A-trail’. It follows from Theorem VI.71 and the relation &(D(G,))= XJG,) for connected plane cubic G, # K4, that this modified conjecture is equivalent to the BTC (Conjecture VI.72) for the case of cyclically 4-edge-connected, planar, bipartite, cubic graphs. But what does the existence of a (not necessarily non-separating) A-trail in D = D(G,) imply for G, ? The answer to this question is given in the next theorem. Recall that a dominating cycle in a graph G (or Tutte cycle if G is cubic) is a cycle C such that E(G - V ( C ) )= 0.

Theorem VI.90. Let G3 be a connected, plane, bipartite, cubic graph with bipartition {V.,V,}, and let D(G,) denote its dual. The following statements are equivalent. 1) D(G,) has an A-trail. 2) G, has a Tutte cycle C such that V’ := G,

- V ( C )satisfies

a) V, n V‘ C int C if and only if V, n V’ c ezt C ;

b) V l n V ’ n e z t C = O o r V l n V ’ n i n t C = 8 . The proof of Theorem VI.90 can be obtained by modifying Tutte’s proof of Theorem VI.71; it is therefore left as an exercise. Note that in Theorem VI.90, the equivalence between Tutte cycles of a special type in G, and A-trails in D(G,) does not require X,(G,) = 4. On the other hand, Tutte’s ‘Bridge Theorem’ guarantees the existence of a dominating cycle in planar, cyclically 4-edge-connected, cubic graphs (Corollary 111.73). So, in the case of X,(G,) = 4, the equivalence expressed in Theorem VI.90 boils down to the question whether in the non-empty set of Tutte cycles of G, one can find an element satisfying a) and b). We also note that regarding non-separating A-trails, it suffices to prove Conjecture VI.73 for 4-connected triangulations; this follows from Proposition VI.83, and it is tantamount to saying that it suffices to prove the BTC for the corresponding cyclically 4-edge-connected graphs (see Exercises VI.25 and VI.30).

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VI. Various Types of Eulerian Trails

VI.3.3. How to Find A-Trails: Some Complexity Considerations and Proposals for Some Algorithms Complexity considerations, although a very important part of today’s development of graph theory (in particular with regard to applications to various problems in operations research, for example), do not constitute an essential part of this book. Nevertheless, here and in other chapters such considerations will be performed depending on the importance I attribute to a (solved or unsolved) problem; however, I have restricted my considerations on this topic to such questions as whether a given problem is (at least) NP-complete or whether it can be solved/decided in polynomial time.l) In what follows, we basically restrict ourselves to 3-connected, planar, eulerian graphs. For, as we have seen before, a 2-connected, planar, eulerian graph G may have plane embeddings G , and G, such that G, has no A-trail while G, does (see Figures VI.26 - VI.28). Thus, we would have t o consider embedding problems as well, respectively we would have to check all plane embeddings concerning the existence or non-existence of A-trails. On the other hand, 3-connected, planar graphs G are uniquely embeddable on the plane (see Theorem 111,52); hence (see the discussion preceding Lemma VI.64), the existence of an A-trail in such a G which is also eulerian, is independent of any concrete embedding of G on the plane (or equivalently, 3-dimensional sphere). Nevertheless, for practical purposes we shall always consider some embedding of G. On the grounds of Regner’s construction of a planar, 3-connected, eulerian graph without A-trails (respectively statement (V1.A) preceding Lemma VI.76), and because of [GARE76a], we can formulate and prove the following result.

Theorem VI.91. The problem of determining whether a given connected, plane, eulerian graph has an A-trail, is an NP-complete problem. ’) I know that this is a very subjective point of view, and I foresee the disappointment and criticism of some of my colleagues. However, given the material I had to choose from, a thorough treatment of complexity and/or algorithms would have occupied a large part of the book. Moreover, I frankly admit that I am not a specialist on these topics. Hence I prefer to leave them to more competent researchers.

VI.3.3. A-Trails: Complexity Considerations, Algorithms

VI. 13'7

This is true even if the problem is restricted to 3-connected graphs.l)

Proof. By statement (VI.A), if G, is a planar, 3-connected, cubic graph, a planar, 3-connected, eulerian graph G with D(G,) c G exists such that no face boundary of D(G,) is a face boundary of G , and G has an A-trail if and only if G, has a hamiltonian cycle. Since the hamiltonian problem for G, is NP-complete, [GARE76a], it is (more than) sufficient to show that the A-trail problem for G 3 D(G,), is polynomially equivalent to the hamiltonian problem for G,, and that G can be constructed from G, in polynomial time. Since it follows from Theorems 111.89 and 111.90 that deciding whether a given graph is planar, and if so, finding an actual plane embedding, can be done in polynomial time, we may assume that G,, G respectively, are actually plane graphs. The problem of finding a 2-face-coloring of G can also be solved in polynomial time (Theorem 111.91). Hence, we may assume that the 2-face-coloring of G is given as well. Assume an A-trail T given in G 3 D(G,). A single run through T tells us at every vertex w which is reached (be it that 21 is reached in this run for the first time or not), whether 21 E V, or 21 E V, where {Vl,U,} is the A-partition of T. For, together with the embedding of G we can determine the face boundaries of D(G,) in polynomial time (Theorem 111.90~)~ so we know whether edges of E, consecutive in T belong to the boundary of a 1-face or 2-face of G. Thus, {V,,V,} is obtained in polynomial time from T . Having stored V ( D ( G , ) ) we obtain 5':= K n V ( D ( G , ) ) , such that (y')is a tree, i = 1,2, in polynomial time P 2 ( P G ) . Knowing the bijection wF E V ( D ( G , ) ) t+ bd(F) C G , for every face F of G, (which requires polynomial space and time only), we form Go = UVFE"{ bd(F) c G,. This and deleting those edges of Go which belong to bd(F') and bd(F") for F' # F", vF,uF,, E E((V:)), can be done in polynomial time P3(pG3), and results in H , a hamiltonian cycle of G,. Since, p G L C0nSt.pG3 (see below) P ( p G 3 ) := CpG) P 2 ( P G ) P 3 ( p G 3 ) is the polynomial in pG3 =I V(G,) I representing the time required to obtain the hamiltonian cycle H of G, from the A-trail T of G.

+

+

The first part of this theorem has been proved in [BENT87a], where the authors speak of 'non-intersecting eulerian trails' what we call A-trails. Also, their constructions involve multiple edges and cut vertices in most instances. It is also worth noting that the motivation for this paper is derived from a problem in flame cutting.

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VI.158

Conversely, suppose a hamiltonian cycle H C G , be given. Then it takes to reach the vertex partition {V;,Vi} of D(G,) polynomial time Pi(pG3) such that (V;’) is a tree, i = 1 , 2 (V,l corresponds to the set of faces F of G, with F C int H ) . In order to obtain an A-partition {V,,V,} of V ( G ) such that V;‘ = V; n V ( D ( G , ) ) ,i = 1,2, we choose a particular way of constructing a plane, 3-connected7eulerian graph G with D(G,) C G and such that no face boundary of D(G,) is a face boundary of G. It has been noted in the discussion preceding Lemma VI.76 that the construction of G is not unique, and that it can be done in such a way that every face of D(G,) contains precisely one copy of either Hi of Figure VI.32 or of H,, of Figure VI.31. This construction will be performed as the last step of the proof; hence assume first that G with this property is given. Then - fG3 +5pG 3 -< 6pG, , where the second inequality follows from Euler’s PG < polyhedron formula, and the first inequality is due to the fact that a face of D(G,) (which corresponds to a vertex of G,, and vice versa) contains either 3 or 5 vertices of V ( G )- V(D(G,)). Having stored D(G,) and {V,l,V,l}, we can extend this partition to { v,,V,} as follows. a) The face F of D(G,) contains precisely three vertices; then it contains Ho. Following the notation of Figure VI.31, we may by symmetry assume w.1.o.g. that a , b E V,l,c E V,l. In this case we define { a l , b , } c V,, c1 E V,. Note that there are altogether only 6 choices for the distribution of a, b, c in V,l and V,l.

b) The face F of D(G,) contains precisely five vertices, so it contains Hi. We follow the notation of Figure VI.32, and having again six choices for the distribution of xi-l, xi,yi in V; and V,l,we assume by symmetry w.1.o.g. that either { Z ~ - ~ , Z ~c } V,l and yi E V,l, or {xjal,gi} c V[ and xi E Vi. In any case, let the vertex of G, uniquely determined by N ( z i - , ) n N(y,) n F , belong to V,; let the vertex ti of G , uniquely determined by N ( x i - , ) n N ( x i ) n F , belong to V, unless A = (xi-l, ti,xi) is the boundary of a 2-face and yi E V,l in which case we let ti belong to V,; and let the corresponding vertex of N ( z i ) n N(yi) n F belong t o V,. As for the remaining two vertices ui-, E N ( z i - , ) and ui E N ( z i )of V ( H J n F , define ui-,,ui E V,. Since

(V;’)

is a tree, i = 1 , 2 , and because of the way we extended { V:, V,l} to { V,,V,}, it follows from these considerations that { V,,V,} is an A-partition which can be obtained from {V;,V,l} in polynomial

VI.3.3. A-Trails: Complexity Considerations, Algorithms

VI.159

time Pi(pG3). So, together with the 2-face-coloring of G we can perform the transition from {V,, V2}to a cycle T which stands for an Atrail of G, by a step by step procedure. Namely: define GO := G, and Gi := (Gi-l)jviI,6, i = 1,...,p G , where V ( G ) = {zl, ,...)upG}, and the value of 6 E {1,2} is determined by zli E V,. Since this transition from Gi-l to Gi can be done in polynomial time, we reach T = GPG from {V,, V2}in polynomial time Pi(pG). Again, since pG = const.pG3, we reach the A-trail T of G from the given hamiltonian cycle H of G, in polynomial time P ' ( p G 3 ) = P i ( p G 3 ) Pi(pG3) Pi(pG). Thus, the problem of finding an A-trail in G 3 D(G,) is polynomially equivalent to the problem of finding a hamiltonian cycle in G,.

+

+

Finally, let us show that the construction of G with the property that every face of D(G,) contains precisely one copy of either Hi or H,,can be done in polynomial time. First we observe that the transition from G, to D(G,) can be done in polynomial time Ql(pG3). We anticipate now some of the theory on the Chinese Postman Problem (CPP) which we shall develop in Volume 2, Chapter VIII. Namely: given any connected graph H , there exists an eulerian supergraph H I with the following properties: 1) V ( H , ) = V ( H ) , E( H , )

2 E(H);

2) if x and y are not adjacent in H, they will not be adjacent in H,;

3) for every cycle C, of H , we have

Expressed more descriptively, 1)- 3) says that H I is obtained from H by replacing certain edges of H with an edge of multiplicity 2 in such a way that H , is eulerian and such that for every cycle C of H , the number of edges of C which are replaced in H , with a multiple edge is at most +Z(C). As we shall see in the discussion on the CPP, the construction of H , from H can be done in polynomial time. Note that in this construction,

=

d H I - - E ( H ) ( u ) dH(2r)(mod2) for every

zl

E V ( H )= V ( H , ) .

Applied to the construction of G from H = D(G,) the above amounts to saying that we can label the edges of D(G,) in polynomial time

VI. Various Types of Eulerian Trails

VI. 160

with 0 and 1 in such a way that in D(G,) for E(D(G,))/X(e)= 1) (with X denoting the label function)

Q2(I)GJ)

I E, n El 1-

d D ( G 3 ) ((mod2) ~) for every

and such that for every cycle

El :=

E V(D(G,))

{e E

(1)

C of D(G,)

In particular, (2) holds for the triangular face boundaries A of D(G,). That is, for every such A, I E ( A )nEl {0,1}. Following the notation of Figures VI.31 and VI.32 we write V ( A ):= { a , b, c } if E ( A )n El = 8, and V(A) := { Z ~ - ~ , X ~if ,E~( A ~ )} n El # 0 with the notation chosen in such a way that w.1.o.g. { Z ~ - ~ Z=~E} ( A )n El. Among the two face boundaries containing Z ~ - ~ Z ; ,choose one and fix it; denote it as above with A. For each of the A chosen with E ( A ) n E, # 8 embed the graph Hiof Figure VI.32 in the corresponding face. For every other face boundary A = ( a ,b, c ) embed H, of Figure VI.31 in the corresponding face. Denote by G the graph resulting from D(G,) this way. This requires polynomial time Q , ( p G 3 )only, because for each of the pG3 face boundaries of D(G,) the embedding of H,, Hi respectively, can be done in constant time. Thus, the entire transition from G, to G can be done in polynomial time Q(PG,) = Qi( P G ~ ) Q,(PG~ 9 3 6 3 ).~ ~ Because of (1) and the very construction of G from D(G,) it follows that G is plane and eulerian; and it is 3-connected because G, is 3-connected. This finishes the proof of the theorem.

+

+

We note that in the second step of the proof of Theorem VI.91 it is not necessary to extend explicitly {V;,V,l)to {Vl, V2}. It suffices to construct G = (Gv:,1)v;,2 which is connected and 4-regular, and then find algorithmically an A-trail of G in polynomial time by successively applying the Splitting Lemma, say (see Volume 2, Chapter X). As for the complexity of the A-trail problem in planar, 4-connected graphs, it can very well be that it parallels the phenomenon concerning the complexity of the hamiltonian problem. For as we have noted before, in 4-connected planar graphs, a hamiltonian cycle can be found in polynomial time. This implies that a Tutte cycle in a planar cubic

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VI. 161

graph G, with X,(G,) 2 4 can be found in polynomial time; in this case, L(G,) is a planar 4-connected graph (see Corollary 111.73). Of course, this does not guarantee that a Tutte cycle of the type described in Theorem VI.90, can be found in polynomial time.,) In fact, there is some evidence of the likelihood of the above parallelism. In addition to the facts quoted in the discussion following Conjecture VI.86 we may make the following observations (we restrict ourselves to outlines of proofs, the details are left as exercises). 1) An A-trail of a 2-connected outerplane simple graph can be found in polynomial time.

We use Theorem VI.87 to develop a procedure for constructing an A-trail in a 2-connected outerplane simple graph G. We assume G to be 2-facecolored with the outer face F, of G being a l-face. The procedure (an algorithm, so to say) amounts to successively marking vertices of valency exceeding 2 in such a way that ultimately every l-face F, # F, contains exactly one marked vertex. Denoting by V, the set of marked vertices and V, := V ( G )- V,, the partition {V,, V,} will then satisfy Theorem VI.87, and therefore, (Gvl,l)v,,2 will be a cycle representing an A-trail of ‘1 On the other hand, it has been claimed that the hamiltonian cycle problem is solvable in polynomial time for arbitrary edge graphs (see [GARE79a, p.199, [GT37]], where the authors refer to [LIUC68a]); hence the Tutte cycle problem would lie in P (see Lemma 111.69). This claim must be erroneous. In order t o see this, we first consider L(G,) for a 2-connected cubic graph G,. Next we introduce a new vertex vP+ifor every vertex vi of G, and add the new edge v ~ v ~ + i ~=, 1,.. . , p . Denote by H the graph thus obtained. Finally we consider L ( N ) and S(G,)UL(G,). The latter graph can be viewed as obtained from S(G,) by letting V(L(G,)) = V,(S(G,)) and then adding E(L(G3)). If we proceed thus L ( H ) and S(G3)UL(G3) are isomorphic graphs. With this visualization of L ( H ) ,it now follows by a straightforward argument that there is a l-l-correspondence between hamiltonian cycles of G, and such cycles in L ( H ) . Moreover, the above construction, the transformation of a hamiltonian cycle of G, into the corresponding hamiltonian cycle of L ( H ) and the inverse transformation can be performed in polynomial time. Whence we may conclude that the hamiltonian cycle problem for L ( H ) is NP-complete even if G, is planar and 3-connected (cf. [GARE76a]). We note that these arguments are a slight modification of Bertossi’s considerations in [BERTSla] who dealt with the analogous error concerning the hamiltonian path problem [GARE79a, p.199, [GT 3911. - Incidentally, I could not find in Liu’s book the claim quoted in [GARE79a].

VI. 162

VI. Various Types of Eulerian Trails

G. The starting point, however, is the construction of the weak dual of G

D"(G) := D(G) - f, where f, is the vertex of D ( G ) ,the dual of G, corresponding to F,. It has been shown, [FLEI74d, Theorem 1 and Corollary], that D"(G) is a tree for 2-connected7outerplane graphs. Hence, in our case T" := D"(G) is a tree whose vertex bipartition {Vlw,VT}following the 2-face-coloring of G is such that the end-vertices of T" belong to V2". Choose vo E l.;U and orient the edges of "'2 in such a way that T" is transformed into an out-tree T," with V(T,") = V(T") and root vo. For every v E Vlw let F, # F, denote the l-face of G corresponding to v, and for any l-face F # F, of G let v F denote the corresponding vertex of Vlw.We observe that the following initial steps can be done altogether in polynomial time Po(n)where n =I V ( G )I: a) Establishing the face boundaries of G and classifying them according to whether they belong to a l-face or to a 2-face.

b) Establishing for every v E V ( G ) the set L, of l-faces Fl with v E V(bd(F1)). c ) Constructing the plane out-tree T," with root vo and vertex bipartition { Vlw,V?}. We first mark the root vo E V(T,"). Then we mark an arbitrary vertes v1 E bd(FVo),and let v1 belong to V, (note that Fvo is a l-face by the choice of vo). Consider next L,,,and mark v F E Vy for every F E Lvl - { F v 0 } . At this stage as well as at any later stage of this marking process, the following property is fulfilled precisely because G is a 2connected outerplane graph:

If P(vo,w ) is a path in T O W joining the root vo with w E V?, and if w has been marked, all elements of V ( P ( v o,w))nVlwhave been marked. Next we choose an unmarked vertex

(*> E Vlw subject to the condition that

all elements of V(P(vo,v) - v) n Vlwhave been marked already. (**) We mark this chosen v. Precisely because G is outerplane and simple it follows that there exists v2 E V(bd(F,)) such that

VI.3.3. A-Trails: Complexity Considerations, Algorithms

VI. 163

d(v2)2 4 and for every F E Lv2 - { F u } , vF E V(T,") has n o t been marked yet. (* * *I We mark v2. Then we mark in T," every vertex vF # v for F E Lv2-{ F,}. We observe that because of (*) the elements of Vlwalready marked together with the elements of V2wadjacent from these marked vertices, define a subtree T' of T," with root v,,. Therefore, if TI # T,",we choose z E V;U subject to condition (**) (with z in place of v) in order to mark x and to find v3 E V(bd(F,))satisfying (* * *) (with v3 in place of v2), and mark us. Then we mark in T," the vertex v F # z for every F E Lvj - {F,}. We observe that the subtree T" of T," defined after this new step of the marking procedure in the same way that TI was defined above, satisfies TI c TI' T,"

c

Therefore, choosing repeatedly a vertex of Vlwsubject to (**) and a vertex of G satisfying (* * *) and marking the latter vertex, eventually leads to an out-tree T ( i )satisfying

T'

c TI' c . . .

After this i-th step, the set V, C V ( G )consisting precisely of the marked vertices of V ( G ) ,and V, := V ( G )- V, defines a vertex partition. We note that it is precisely because of (* * *) that statement 2) of Theorem VI.87 is fulfilled. Hence {V,, V,} is an A-partition of V(G). We note in passing that the preceding algorithmic considerations together with Theorem VI.87 amount to another proof of Theorem VI.63 independent from the proof established earlier. As for the complexity of the above marking procedure we observe that after the j-th step, 1 5 j < i, the choice for v E VlWsubject to (**) is arbitrary in the set of sources of T," - V ( T ( j ) ) .So, constructing T ( j ) and choosing and marking such v among the sources of T," - V ( T ( j ) ) can be done in polynomial time Pl(n). Since the marking procedure in Tow can be combined with marking the corresponding elements in the sets Lu, the discovery of a vertex vj+2 E V ( G )satisfying (* * *) in the ( j l)-th step can be done in polynomial time P2(n).Hence, the time required to construct the A-partition {V,,V,} is at most P-,(n)+n(P,(n)+ P2(n)).Since the transition from {V,,V,} to (GVl,1)V2,2 can be done in polynomial time as well, it follows that finding an A-trail in G can be done in polynomial time.

+

VI. 164

VI. Various Types of Eulerian Trails

On the grounds of 1) we arrive at the following statement.

2) If G i s a simple, plane, eulerian graph with a given hamiltonian cycle H having the A-property, a n A-trail of G can be found in polynomial time. By using the notation of the proof of Theorem VI.88 we may conclude from 1) that A-trails T, and T2 of GI and G,, respectively, can be found in polynomial time (where G, U G, = G, G, n G, = H ) . Hence the corresponding A-partitions {Vj,V i } of Ti,i = 1 , 2 , can be found in polynomial time. Since the four vertex sets

W; := vj‘ n ( V ( G i )- V2(Gi)), i , j E {1,2} can be found in polynomial time, we therefore arrive at Wj := ryj’ U Wj”, j = 1,2, in polynomial time. Since (Gw,,1)w2,2can then be obtained in polynomial time, and since this graph is a cycle (see the proof of Theorem VI.88), it corresponds to an A-trail of G obtained in polynomial time. We observe, however, that the above considerations are based on a given hamiltonian cycle H of G. Hence the question arises: how complex is the problem of deciding whether a given 4-connected, planar, eulerian graph has a hamiltonian cycle with the A-property ? Of course, in 2) we did not need any assumptions on K ( G )(however, K ( G )2 2 since G is hamiltonian), but in view of Conjecture VI.86 and Tutte’s theorem on hamiltonian cycles (Corollary 111.71) we restrict ourselves in the above question to 4-connected graphs. It should be noted that if G is any eulerian triangulation of the plane (4-connected or not, simple or not), this problem belongs to P ; for, in such G, if e, f E E ( G )are incident with v E V(G)and O+(v) = (. . . ,e(v), f(v), . . .), e and f either determine a unique hamiltonian cycle H having the A-property with e, f E E ( H ) , or else H does not exist at all (Exercise VI.32). Thus, it can very well be that the above question has a ‘polynomial’ answer as well. Note that for a triangulation D of the plane, a hamiltonian cycle having the A-property is a special type of a left-right path as defined in [SHAN75a]. We leave it to the reader to check the complexity of finding an A-trail in a graph covered by either of the Theorems VI.77.a and VI.79 (Exercise VI.33), and now consider the general case.

An A-Trail Algorithm For Arbitrary Plane Eulerian Graphs

VI.165

An A-Trail Algorithm for Arbitrary Plane Eulerian Graphs Our starting point is a 2-connected, plane, eulerian graph Go other than a cycle and, indirectly, Theorems VI.59 and VI.67. However, instead of considering unicolored face-rings we combine a vertex-marking procedure with a sequence of 1-splittings. '1 Step 0. Consider Go together with a 2-face-coloring such that the outer face Fomof Go is a 1-face. Denote V< = V(bd(F,")) - V,(Go), and let x1 E V< be arbitrarily chosen. Set i = 1. Step 1. Form G i := ( G j - l ) { z i l , l . Denote by Fi" the outer face of Gi, and let the 2-face-coloring of Gi be induced by the 2-face-coloring of

Gi-1Step 2. Mark those so far unmarked elements of V(bd(Fim))which are cut vertices of Gi,and let denote the set of unmarked vertices in V(bd(Fim))- V,(Gi).

v

Step 3. If Step 1.

If

V;.. # 8, then choose any xi+l

E

v. Set i = i + 1 and go to

v = 0, then continue.

Step 4. Set V,.* = {z/x E ( E ( G i )- E(bd(F;"))) U ( V ( G i )- V2(Gi)); x is unmarked if it is a vertex }. If V,.* = 8, then go to Step 7.

If

v*# 8, then continue.

Step 5. Search for the largest integer j

< i such that v;" - { z ~ + #~ }8.

If such j does not exist, then go to Step 8. If j exists, continue. Step 6. Mark xj+l and set

3.

= v;" - {x~+~}. Set i = j and go to Step

Step 7. Go has an A-trail. Step 8. Go has no A-trail. '1 For the case of eulerian triangulations of the plane, this algorithm has been outlined already in [FLEI88a].

VI. Various Types of Eulerian Trails

VI. 166

In fact, this algorithm does more than decide whether Go has an Atrail: for, if = 0 = V,.* for some i > 0 (which is the necessary and sufficient condition for reaching Step 7), then Gi satisfies the hypothesis of Lemma VI.58. Hence a run through bd(F;”) describes an A-trail To of Go. Moreover, an A-partition {Vf, V;} in Go corresponding to To is, in terms of the algorithm’s notation, defined by

v:

= {xl, ...,Xi} , V : = {x E V(G)/z has been marked or x E V,(Go)}

.

We note that the construction of G iand the determination of V,*,Ty* respectively, can be done very fast. This is also true for the determination of the integer j < i in Step 5. However, the algorithm on the whole is very slow. This becomes clear by studying the Steps 4,5 and 6: V;. = 0 and # 0 means that, for the time being, one cannot continue the ‘greedy approach’ expressed in Steps 1, 2 and 3; i.e., one has reached a deadlock. At this point, each block B of G isatisfying BnV,**# 0 defines a unicolored face-ring R (whose elements are 2-faces) with a complete set of links L, := V(bd(F,”)) - V2(B), where F r is the unbounded face of B (whose embedding is induced by that of G i ) and b d ( F p ) c bd(F;”). For, = 0 means that all elements of V(bd(F,“)) - V2(Gi)have been already marked; since this and B nV;,.* # 0 imply int bd(F,”) n V,.* # 0, we conclude that there are (at least two) 2-faces F2 with E(bd(F2))n E(bd(F,”)) # 0, whence R is defined by the set of these 2-faces. We note, in addition, that a marked vertex of G, need not be a cut vertex of Gi (see Step 6). Steps 5 and 6 are means of getting out of this deadlock. That is, one has to go back to some G j , j < i (for practical reasons one chooses the largest j ) and choose some vertex E - {zj+l} to which one applies a 1-splitting to obtain G[i+l. At this point has become a marked vertex, and all the algorithmic work in establishing Gj+l,. . .,G,is to no avail. That is, the marking and splitting procedures in these graphs are undone. But having shifted xj+l from to the set of marked vertices of G, does not say anything about the ‘final fate’ of xj+l in an A-partition {Vf,V;} of Go that might exist. For, at a later stage one might reach j‘ < j ; and continuing with Step 1 for i = j‘ 1 might result in the formation of a Vc, k > j’ containing the above again. We note however, that xj+l remains a marked vertex as long as the above Vj* has not been eshausted by Step 6.

v*

+

The working of the above algorithm (and its slowness) becomes even clearer if Go = D is an eulerian triangulation of the plane. For, in this

An A-Trail Algorithm For Arbitrary Plane Eulerian Graphs

VI.167

case the algorithm amounts to constructing connected graphs D, (induced by the vertices to which one applies a 1-splitting) and D, (induced by the marked vertices) such that

c

c

and for 6 = 1,2, D6 has a cycle only if = bd(I"6) where F6 is a &face (see Corollaries VI.68 and VI.69 and the subsequent discussion). And at every stage of the algorithm, the graph induced by the vertices to which one applies a 1-splitting, has this property. We leave it as an exercise to translate this algorithm into an algorithm for finding hamiltonian cycles in plane cubic bipartite graphs (it's the obvious algorithm !). However, in view of the equivalence between the Conjectures VI.73 and VI.74 as expressed by Lemma VI.75 one can modify the above algorithm in order for it to decide whether a given plane eulerian graph Go has a non-separating A-trail (Exercise VI.34.b)). If Go has such an A-trail, the outcome of this modified algorithm is a connected outerplane eulerian graph G i(for some i > 0) whose blocks are the boundaries of the 2faces F2 of Go and such that bd(F2) has more edges than it contains cut vertices of Gi.Restricting ourselves to eulerian triangulations Do of the plane this means that each block of Dicontains at most two cut vertices o f D i (*). Statement (*) is trivially true for i = 0, and for i = 1 as well; hence one is tempted to suspect that (*) can be preserved in all the graphs Do,D,, . .. ,Diof the modified A-trail algorithm. This "suspicion" leads us to the following more general conjecture where we consider simple eulerian triangulations of an n-gon in the plane. In this 0 (mod 3) must hold; and for every such n, triangulations of case, TI this type do exist (see e.g. [FLEI74b] and note that Lemma VI.76 is a consequence of this fact).

Conjecture VI.92. Let a positive integer n G O(mod3) be chosen, and let Do be a simple eulerian triangulation of an n-gon in the plane together with a 2-face-coloring of Do such that the outer face Fom of Do is a 1-face. Then (distinct) vertices xk E V(D,) - V2(D0)exist such that D, := (Dk-l)~zkl,l, k = 1,.. .,i, has the following properties: a)

xk

E V(bd(F,"_,))- V2(D,-1) (F,"_, is the outer face of D,-l);

b) every block of D, contains at most two cut vertices of D,; c) if k = i, then every block of D, is a (triangular) boundary of a 2-face.

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VI. Various Types of Eulerian Trails

Thus, because of property b) Conjecture VI.92 is a generalization of Conjecture VI. 73.

VI.3.4. Final Remarks on Non-Intersecting Eulerian Trails and A-Trails, and another Problem As we have noticed at the beginning of this section) non-intersecting eulerian trails and A-trails are in principle different concepts; they coincide however, for G with A(G) 5 4. This basic difference between these two types of eulerian trails can also be demonstrated topologically. Recall that an eulerian trail T of the connected eulerian graph can be determined by a sequence of splitting operations on vertices of valeiicy exceeding 2 such that the graph obtained after the last splitting operation in that sequence, is a cycle (see also Corollary V.13). On the other hand, a connected eulerian graph G on q edges can be viewed as obtained from a q-gon Cq by a homomorphism which acts bijectively on the edge sets of G and Cq (see Remark V.12). We apply these considerations to an arbitrary connected) plane, eulerian graph G and a non-intersecting eulerian trail T of G. Denote by C, the plane cycle corresponding to T. Moreover, similar to the concept of a &splitting, 5 = 1,2, we consider, in the construction of C, from G, the k-fold application of the splitting procedure to obtain k 2-valent vertices vl, . . . z'k from the 2k-valent vertex v, k > 1, as a simultaneous procedure to replace v with these k 2-valent vertices. Note that this replacement procedure can be done in such a way that vl,. . . vlclie in a (geometrical) circle K ( v ,E ) with center v and radius E , and x ( v , E ) contains no other vertices of G, C, respectively) and E ( v , E ) n ( E ( G )- E,) = 0. Note that C, is a simple closed curve since T is non-intersecting. Thus G can be viewed as obtained from a regular q-gon in the plane Cq( q =I E(G)I) as follows: 1L cq CT 4 G )

)

where and

cp is a homotopic deformation

$ is a continuous mapping of the plane

&2

onto itself

VI.3.4. Final Remarks

VI. 169

such that

$(x) = v for x E r ( v , E ) and v E V ( G )- V2(G), $ : €2 -

u

K(v, E)

--+

€2

- V ( G )is bijective.

E V(G )-Vz ( G )

Since an A-trail is a special type of non-intersecting eulerian trail, the above is true, of course, if 5" is an A-trail. However, if T is an A-trail, we can assume CT as having the additional property that for every v E V ( G )- V2(G)

there exists 6 < E such that

r ( w , 6)n CT = {q,. . .,vk}.

(A41

Using planarity we can replace (A) with

precisely one of K ( v , E ) n int CT and K ( v ,E ) n e x t C, is a region. (A')

If we consider a 2-face-coloring of G (with F" being a l-face as usual) and the induced coloring of CT,K ( v ,E ) n int CT is a region if and only

if T induces a 2-splitting in v.

In the case of a non-intersecting eulerian trail which is not an A-trail, then for at least one v E V ( G )- V2(G),neither of these two sets defined in (A') is connected.

So we have a way of topologically constructing all connected, plane, eulerian graphs (compare this with [ABRH79a, Theorem 31) as well as those graphs among them which admit an A-trail. The A-trail problem can thus be re-interpreted as the problem of describing various classes of these graphs in terms of graph theoretical parameters. We observe that the above considerations can, in principle, be extended to arbitrary graphs embedded in some (orientable or non-orientable) surface F ,except where 2-face-colorings are involved. In particular, property (A) remains valid as the topological means for distinguishing A-trails from the other non-intersecting eulerian trails. Note that the concept of an A-trail was originally introduced via O+(v), O-(v) respectively, and did not require the concept of 6-splittings. However, for different nonintersecting eulerian trails TIand T2 of the same G , the cycles C, and CTz may not be homotopic.

As for A-trails in connected, plane, eulerian digraphs D and/or mixed graphs M it makes sense to consider this problem only if D is canonically oriented, and/or if M can be canonically oriented (it can easily be

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VI. Various Types of Eulerian Trails

decided, whether M admits such orientation; for, a canonical orientation is uniquely determined by just one given arc). To consider the A-trail problem in a canonically oriented digraph D is, however, tantamount to considering the problem in the underlying graph G. For, if G has an Atrail T , either T or T-l corresponds to an A-trail in D; and an A-trail of D corresponds to an A-trail of G. This follows directly from the relation between an A-trail and 6-splittings, 6 = 1,2. However, it might be of interest to study (in the contest of Theorem VI.33) those in-trees of D which correspond to A-trails of D. Another problem (which has not been treated explicitly in this chapter) considers arbitrary connected eulerian graphs G drawn in the plane in such a way that open edges do not contain points corresponding to vertices, two open edges intersect in at most one point at which they must cross each other, and no three open edges intersect in the same point. Such a drawing of G defines a certain O+(G) from which one deduces a special system of transitions X , by defining t E X,(v) f-t t = {ei, e i + i } , 1 5 i 5 k = k, = $d(v), and v E V(G) is arbitrarily chosen. However, X , = X , for some unique trail decomposition S of G. Whence the following questions arise: 1) For a given graph G, does there exist a drawing of G in the plane such that X , = X , for some eulerian trail T of G ? 2 ) How many drawings of G exist such that some eulerian trail T of G satisfies X , = X , (where X , relates to a particular drawing) ? 3) Which trail decompositions S of G, satisfying X , = X , can be obtained this way ? These questions have been raised in [HARB89a] where question 1) is answered in the affirmative for G = Ii2n+l, n EN.

VI.4. Exercises Exercise VI.1. Prove Corollary VI.2, part 2), by applying the Splitting Lemma (i.e., without using Theorem VI.l). Exercise VI.2. Construct a graph G and define a partition system P(G) satisfying Corollary VI.4.2) such that G has no P(G)-compatible path/cycle decomposition S in which the number of path elements of S equals half the number of odd vertices of G. Exercise VI.3. Prove Corollaries VI.5 and VI.6 by using the Splitting Lemma only.

VI.4. Exercises

VI.171

Exercise VI.4. Prove Lemma VI.7 and Lemma VI.8 by using the Splitting Lemma. Exercise VI.5. Prove Corollaries VI.9 and VI.10. Exercise VI.6. Show that the digraph of Figure VI.2 has no A*compatible eulerian trail. Exercise VI.7. Prove Corollary VI.13 (hint: for v with d,(v) > 6, shorn first that if I X * ( v )I= i d D ( v ) + l , then there exists CiE X * ( v ) ,i = 1,2,3, with I CiI= 2 and (At)+ 3 Ci (A:)-. Then apply the Splitting Lemma in a manner similar to the proof of Theorem VI. 11). Exercise VIA. Show that D, of Figure VI.4. has an X(D,)-compatible eulerian trail if and only if D has an X(D)-compatible eulerian trail, where X ( D ) is given and X ( D , ) = X ( D ) U X ( s , ) U X ( s z ) . Exercise VI.9. a) Construct connected eulerian digraphs D with the property that V ( D ) consists of 4-valent vertices and an odd number of 6-valent vertices. b) Taking two graphs of the type described in a), construct from it an eulerian digraph with an even number of arcs and a 4-valent cut vertex. Exercise VI.10. Prove Corollary VI.19. Exercise V I . l l . Prove Lemmas VI.27 and VI.28. Determine those graphs G whose valencies are multiples of four such that these two lemmas are also true for such G. Exercise VI.12. Show that the subdigraphs D, and Di = D* - DA of the digraph D* of Figure VI.7 satisfy statement 4)a) of Theorem VI.35, while statement 4)b) holds for Di only. Exercise VI.13. Prove Corollary VI.43. Exercise VI.14. Construct a 4-regular simple digraph which has two but not three pairwise compatible eulerian trails (hint: proceed similar to the discussion centering around Figure VI.14). Exercise VI.15. Prove Lemma VI.58. Exercise VI.16. Prove: if G satisfies the hypothesis of Lemma VI.58, then there exists a connected, outerplane, simple graph H having a subgraph H‘ homeomorphic to G, such that H has no A-trail. If one drops the property “simple”, then H can be constructed in such a way that H’ spans H .

VI.172

VI. Various Types of Eulerian Trails

Exercise VI.17. Construct a 2-connected, outerplane, eulerian graph having no A-trail. Exercise VI.18. a) Construct an A-trail in the graph GT of Figure VI.28. b) Show that GT has no A-trail inducing a 1-splitting in v1 if the outer face of GT is a 1-face in the 2-face-coloring of GT. c ) Find other plane embeddings of the abstract graph underlying Go and GT, which admit A-trails. Exercise VI.19. Prove Lemma VI.65 and show by example that the converse of this Lemma is not true. Exercise VI.20. Show that the plane graph Go of Figure VI.26 (which has no A-trail) does have a vertex partition {Vl, V,} such that (V,) is acyclic, S = 1,2. Exercise VI.21. Prove that the graph of Figure VI.30 (a) is the smallest planar, 2-connected, cubic, bipartite non-hamiltonian graph, and that it is uniquely determined. Exercise VI.22. Prove: the following statements are equivalent. 1) Every plane, 3-connected, cubic, bipartite graph has a hamiltonian cycle (BTC). 2) Every plane, 3-connected, cubic, bipartite graph has a dominating cycle. Exercise VI.23. a) Prove: if G, (D) is a plane, bipartite, 3-connected, cubic graph (simple eulerian triangulation of the plane) whose face boundaries are 4-, 6-, or 8-gonal (whose vertices are 4-, 6-, or 8-valent), then G,(D) has precisely six 4-gonal face boundaries (Cvalent vertices) if it has no 8-gonal face boundary (8-valent vertex); it has precisely seven 4-gonal face boundaries (4-valent vertices) if it has precisely one 8-gonal face boundary (8-valent vertex). b) Construct an eulerian triangulation of the plane with 4-, 6-, and 8valent vertices only such that it has precisely one 8-valent vertex v; and v is not adjacent to any 4-valent vertex. Exercise VI.24. a) Prove Corollary VI.81. b) Starting from the octahedron, construct an eulerian triangulation of the plane D by a sequence of weak W4-extensions such that D has precisely one pair of adjacent 4-valent vertices. Exercise VI.25. Translate Conjecture VI.82 into the theory of cubic graphs to obtain a stronger (but equivalent) version of Conjecture VI.72.

VI.4. Exercises

VI.173

Exercise VI.26. Prove Lemma VI.84. Exercise V1.27. Prove: if G, is a connected 3-regular graph having a 2-factor Q4 consisting of 4-gons only, then G, has a hamiltonian cycle H 3 E(G,)- Q4 (hint: use Corollary VI.6 and suppress those 4-gons in

Q4 which either have diagonals or which contain an edge of multiplicity 2).

Exercise VI.28. Show that the graph D of Figure VI.3S has an A-trail. Exercise VI.29. Prove Theorem VI.90. Exercise VI.30. Show that it suffices to prove Conjecture VI.73 for the corresponding 4-connected graphs (hint: apply Proposition VI.83). Exercise VI.31. a) Concerning finding an A-trail in a 2-connected

outerplane simple graph, work out the details of the proposed algorithm (in particular, show that properties (*) and (* * *) are valid). b) Check in detail the complexity considerations on this algorithm (how large is the exponent of Pi(,) for i = 0,1,2 ?).

Exercise VI.32. Show: a) If G is a(n) (eulerian) triangulation of the plane and e,f E E ( G ) are adjacent in a face boundary of G, then there is at most one hamiltonian cycle H of G having the A-property with e, f E E ( H ) ;and determining whether this H exists or not, can be done in linear time (note that G is already embedded in the plane). b) Using a) show that the problem of deciding the existence of a hamiltonian cycle with the A-property in a(.) (eulerian) triangulation of the plane, lies in P. Exercise VI.33. Study the complexity of finding an A-trail in D , where D is an eulerian triangulation of the plane as described in Theorems VI.77.a and VI.79. Exercise VI.34. a) Translate the A-trail algorithm for arbitrary plane

eulerian graphs into an algorithm for finding a hamiltonian cycle, a Tutte cycle respectively, with the properties expressed in Theorem VI.90, in a plane cubic bipartite graph (hint: use Lemma VI.75, Theorem VI.90 and their respective proofs). b) Modify this A-trail algorithm in such a way that it decides whether a given connected plane eulerian graph has a non-separating A-trail.