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Graph whose edges are in small cycles
Discrete Mathematics 94 (1991) 1l-22 North-Holland
11
raph whose edges are in small cycles Hong-Jim Lai* West Virginia University, Morgantown, WV 26506, USA Received 15 June 1987 Revised 24 January 1990 Abstract Lai, H.-J., Graph whose edges are in small cycles, Discrete Mathematics 94 (1991) 11-22. Paulraja (1987) conjectured the following: (i) If every edge of a 2-connected graph G lies in a cycle of length at most 4 in G, then G has a dominating closed trail. (ii) If, in addition, S(G) 2 3, then G has a closed spanning trail. Collapsible graphs are defined and studied by Catlin (1988). Catlin showed that if H is a collapsible subgraph of G, then G has a spanning closed trail if and only if G/H, the graph obtained from G by contracting H, has a spanning closed trail. Catlin (1987) conjectured that a graph satisfying the hypothesis of (ii) is collapsible. In this paper, all three conjectures are proved.
Introduction We shall use the notation of Bondy and Murty [l] except for contractions. A graph may have multiple edges but not loops. A spanning closed trail of G is 4!ed a spanning eulerian subgraph (SES) of G. The collection of graphs that have an SES is denoted by 9.9. Note that K, E 95’. If a closed trail C of G satisfies E(G - V(C)) = #, then C is called a dominating eulerian subgraph (DES) of G. Let G be a graph. An block of G that has exactly one cut-vertex of G is called an end block of G. A block B is acyclic if B = K2. For a subset X c E(G), the contraction G/X is the graph obtained from G by identifying the ends of each edge in X and then deleting the resulting loops. If H is a subgraph of G, then we write G/H for G/E(H). Let W be a subgraph of G. If for some t 2 2, \V = K2,,, then W is called a W,,,-subgraph Of G. Let c = ~1~22)3u4~5~(,~1 be a 6-cycle. Define @ t0 bc the graph with V(0) = V(C) and E(O) = E(C) U (~2~5). * This is part of the author’s Ph.D. Dissertation, done in Wayne State University under Dr. Paul A. Catlin. 0012-365X/91/$03.50 @ 1991 .-- Elsevier Scicncc Publishers B.V. All rights rcscrvcd
12
H.-J. Lai
Let H be a subgraph of G. Define H’= G[E(G) - E(H)] and A,(H) = V(H) n V(W). The vertices in A,(H) are called the vertices of attachment of H in G. If K(H) 2 2 and IA,(H)1 = 2, then H is called a 2-b/0& of G. A graph G is an edge-disjoint union of K+ if there is a partition El, &, . . . , Ek of E(G) such that G[E,] z KZ,ni’~(ni a 2 and 1 s i s k). Note that when i #j, we may have ni # nj. We consider the following conditions. Any edge of G is an m-cycle of G, m s 4;
(I)
G is an edge-disjoint union of K$s.
(2)
Let %= {G: G satisfies (1) with K(G) 2 2) and %, = {G E 5%S(G) 2 3). In [6-81, Paulraja raised the following two conjectures. Conjecture 1. If G E F&, then G E 9Y. Conjecture 2. If G E 59, then G has a DES. A graph G is collapsible if for every even subset R c V(G), G has a subgraph CR(called the R-subgraph of G) such that G - E(r,) is connected and R is the set of odd-degree vertices of rR. The collection of all collapsible graphs is denoted by VZ’. Let HI, Hz, . . . , HC be all the maximal collapsible subgraphs of G. Denote by (G), the graph of order c obtained from G by contracting HI, . . . ,HC to c distinct vertices. We call (G), the reduction of G and HI, Hz, . . . ,HC the preimages of the vertices of (G), . A graph G is reduced if G = (G), . In [2], Catlin showed that (G), is well defined and unique. He also proved the following theorem. Theorem A (Catlin [2]). Let G be a graph. (i) y G E %JZ?,then G E 9.2. (ii) G E 9’9 if and only if ( G)l E %!?. (iii) If H E %S? is a subgraph of G, then G E %9 if and only if
Conjecture 3. If G E %,, then G E KY, By (i) of Theorem A, we have the following proposition.
Graph whose edges are in small cycles
13
Proposition 1. Conjecture 3 implies Conjecture 1. For each i = 1,2, . . . , define Di(G) =
{v E V(G):
deg&)
= i},
and D;(G) = (v E V(G): de&v)
a i}.
roposition 2. Conjecture 1 implies Conjecture 2. Proof. Assume that Conjecture 1 holds. Let G E 9. Since an SES is always a DES, we may assume that G $ XZ and G E %- 9&. Thus 5(G) = 2. Let D,(G) = {xi’ xz’ . . . ,x,). Since K(G) 2 2, no vertex in D,(G) is the common end of multiple edges;
(3)
and by (1) and K(G) 3 2, no paths in G contain three consecutive vertices in D,(G).
(4)
Choose yl,y2,...,y, such that ei = Xiy’ E E(G),
(1 s i s m), and such that if for some i, j with XiXjE E(G) then yi = Xj and xi = Yj. Note that the latter can happen only when Xi and Xj are the internal vertices of a path of G of length 3. By (3) and (4), the multiplicity of each ei is one in G. Let el be an edge with the same ends as ei but el$ E(G), (1 s i s m). Obtain a new graph G’ by adding 1ei, e;, . . . ,el,} to G. It follows that G’ E % and so G’ E sP.9, by Conjecture 1. Let C’ be an SES of G’. Since G $99, we may assume that for some k s m, { ei,
e;, . . . ,el) U {elf e2, . . . ,ek} G W’);
(5)
and that ei $ E(C’), (j > k). Choose C’ so that (5) is satisfied and that k is as small as possible. Let C = C’ - {e,, . . . ,ek, e;, . . . ,e;}. Then every vertex of C has even degree in C. If ei = X~X’E E(C), for some 1 s s, t G m, then since x, and x’ are in D,(G), x, and x’ are in D2(C’) also. It follows that e, $ E(C’), contrary to the choice of C’. So if X~X,E E(C’), then e, E E(C’) and ei $ (C’). This implies that C is connected, and that x1, x2, . . . ,xk form an independent set in G and are the only vertices not in V(C). Thus C is a DES of G. 0
Reductions
Notation P (see Fig. 1). Let W be a K 2,,-subgraph of G and iet H be a subgraph of G containing W. Denote LJ2(W)
=
{Yb Y29
l
l
l
,YJ*
H.-J. Lai
G Fig. 1. Notation 1 with V, = {y, , y2}.
G/x
Let V1 be a subset of D,(W) such that IVJ = 2, let V2= V(W) - V1 and let it = ( VI, V2) denote this partition of V(W). Denote by H/JC the graph obtained from H by identifying all vertices of V1 to form a single vertex Q, by identifying all vertices of V2 to form a single vertex u2, and by joining vl, v2 with a new edge % = 2f,v2, so that E(H) - E(W) = E(H/z)
- {e=}.
And finalIy let H’ = (H/n) - e,. Theorem Gjn
B (Catlin [3]). Let W, Ed and G/z then G E %Z.
be defined in Notation 1. If
E %.Z,
Lemma 1. If G’ = (G/a) - e, E %Z, then G E %.Z’. Proof. It follows from Theorem B and (iii) of Theorem A.
Cl
It is easy to see the following. Lemma 2 ([Sj). If G E %, then (G): has girtJi 4 and (G), satisfies (1). Lemma 3 ([5]). Let B be a block of(G),. (i) If GE%& then BE%$; (ii) rf G E 59, then B E 5% Lemma 4. If G is a counterexample to Conjecture 3 with 1V( G)I minimized then G is reduced. _ oof. Let G satisfy the hypothesis of Lemma 4. If G is not reduced, then (G),, the reduction of G, has smaller order than G. By Lemmas 2 and 3, each block of (G), is in %I and so by the minimality of G, each block of (G), is in CZ’. Thus by (v) and (iii) of Theorem A, G E %Z, a contradiction. 0
Graph whose edges are in small cycles
15
Notation 2. Let C = x1x2x3&$x1be a 4-cycle of G. Define: G* = [G -
{&X2,
X3X4)1/{&X4,
-x2x3},
and denote by q and ~2 the vertices of G* to which Xix4 and respectively. For convenience, we regard V(G*) = [V(G) - V(C)] u {u,, Lemma 5 [4-S].
~2x3
are contracted,
E(G*) U E(C) = E(G).
~21,
If G” E %JZ, then G E %.Z.
Lemma 6. Let G be a counterexample to Conjecture 3 with 1V(G)1 minimized. Then G does not have a subgraph isomorphic to 0. Proof. By contradiction, suppose that G has 0 as a subgraph. Let C = ~1~2~3~&&j~] denote the 6-cycle Of @ and let C, = ~~~-$~&j~~ and C2 = x2x3x4x5x2 be the two 4-cycles contained in 0. Define ~(l)=({%%),
b2d6))
and
@)=(b2,x4),
{x39x5))*
Define G’ = G/n(l) and G2 = G/z(2). Let vi and vi denote the vertices of G’ to which {x1; x5} and {x2, x6) are mapped, respectively; and let v:, v$ be the vertices of G2 to which {x3, x5} and {x2, x4} are mapped, respectively. Let el = vi in G2 and C2 = x3v~v~x4x3 in v: and e2 = v: vs. We shall regard C1 =x&_&q G’ throughout the proof of this lemma. Since G E gI, we have 6(G’) 23 and S(G2) 2 3, and both G’ and G2 satisfy (1) . Claim 1. K(G~) = K(G~) = 1, and all cut vertices of G’ are in {v:, vi} and all cut-vertices of G2 are in (v:, v$>. Proof. Clearly G’ is connected. If K(G’) 3 2, then G1 E %LY?follows by the minimality of G. Hence G E 9Z.Zby Theorem B, a contradiction. Thus K(G ‘) = 1. Since K(G) = 2, the cut-vertices of G’ must be in {v :, v:}. The proof for G2 is similar. Cl Claim 2 (see Fig. 2). For i E { 1,2}, if vi is a cut-vertex of G’ and if L’ is an end-block of G’ containing vi but not vi, then L’ E %Y. A similar result holds when we replace v’, by vi.
Fig. 2. Claim 2 of Lemma 6.
H.-J. Lai
16
proof. Without ioss of generality, we assume i = 1 and V: is cut-vertex of G’ with L1 an end-block of G’ containing vt but not vi. Let H = G[E(L’)]. Then by K(G) 2 2, we have A,(H) = {x1, xs}, Since G satisfies (l), either H satisfies (1) or there is a vertex y E V(H) with yx,, yx, E E(H). If H satisfies (1), then L’ E +?Iand so L’ E %Z, by the minimality of G. Hence we assume that there is a vertex y E V(H) such that yx,, yxs E E(H). By K(G) 3 2, y is not a cut-vertex of G. Hence either x1 or x5 is adjacent to some vertex of V(H) - {y}. kt follows that 6(L1) 3 3. Since (1) holds for G, (1) holds for L’ also. Thus L’ E %.%’by the minimality of G. 0 For i E { 1, 2}, let B’ be the block of G’ containing both vi and vi. Claim 3. Either vi or vi has degree less than 3 in B’. Proof. By Claim 2, all blocks of G’ other than B’ are in Cee4p. Clearly B’ satisfies (1) and K(B~) 2 2. If 6(B’) 3 3, then B’ E %Z’ follows from the minimality of G and so by (v) of Theorem A and by Theorem B, G E %.Z, a contradiction. Hence 6(B’) < 3. Since S(G) 3 3, every vertex in V(B’) - {v:, v’,} has degree at least 3 in B’. Hence either V: or v’, has degree less than 3 in B’. Cl Claim 4. For i, j E { 1, 2)) if vi is not a cut-vertex of G’, then vi has degree at least 3 in B’.
Proof. Without loss of generality, we assume that i = j = 1. Since the degree of xi is at least 3 in G, xl is incident with an edge e’ E E(G) - E(O). Since vi is not a cut-vertex of G’, all the edges incident with vi in G’ are in B’. Since we map xl and xs to vt, and since e’ E E(G) - E(O), vi has degree at least 3 in B’. Cl The proof of Lemma 6 will now be divided into the following cases. Case 1: For some i E { 1,2}, v:, vi are cut vertices of G’. Without loss of generality, we assume i = 1 and that there are end blocks L] and Li in G’, where B’ $ (Et, I,:}, with A&L,‘) = {vi) (1 c j s 2). Let Hj = G[E(L,‘)], (1 f j d 2). Then HI and Hz are connected, and A,(H,) = {x1, x5} and A&Hz) = {x2, x,}. Note that HI and Hz also induce subgraphs in G2 (which we also call H, and H2), with A&H,) = {x,, v:} and A&H,) = {x,, us}. Therefore, the block B2 of G2 containing v: and vif also contains C, , Hl and H2, and so v: and vz have degree at least 3 in B2, contrary to Claim 3. Case 2: The only cut-vertex of G’ is vi, (1 d i 6 2). Let L’ be an end block of G’ that does not contain vi and has v’, as its only vertex of attachment in G’, and let H’ = G[E(L’)]. Then H’ is connected with A,(H’) = {xl, x,}. Thus xs is incident with an edge errE E(H’). Since H’ can be regarded as a subgraph of B2, e” is an edge incident with v: in B2. Hence v: is incident with e” and two edges in C., and so VThas degree at least 3 in B2. Since
Graph whose edges are in small cycles
17
Fig. 3. Case 3 of Lemma 6.
V$ is not a cut-vertex of G2, it follows by Claim 4 that r.~$has degree at least 3 in B2, contrary to Claim 3. Case 3 (see Fig. 3): The only cut-vertex of G’ is v: and the only cut-vertex of G2 is v;. Let L’ be an end block of G’ that contains vi but not Ug_i and let H’ = G[E(L’)], (16 i < 2). Since L’ is an end block of G’, (1 s i s 2), and since K(G)> 2, &AH’) = {xi, xs)
and
&(H2)
=
{xzt
x,>.
Now we show that G - E(O) has a component H (say) with A,(H) = (~3, x6). Since 6(G) 2 3, xj is incident with an edge that is not in E(O). By K(G) 2 2, there is an (x1, xi)-path P in G - E(@) for some j E (1, 2,4, $6). If j = I or j=5, then in G’, v : has degree at least 3 in B ’ . By Claim 4, vi has degree at least 3 in B’, contrary to Claim 3. Ifj=2 or-j = 4, then in G2, us has degree 3 in B2, and a similar contradiction arises. Hence j = 6 and so A,(H) = {x1, x6}. Claim 5. There is no y’ E V(G) - {x2, xs} such that either y ‘x6, y ‘x, E E(G) or y’x,, y’xj E E(G). Proof. If y’x4, y’x, E E(G) and y’ #xs, then since G has a connected subgraph H with A,(H) = {x1, x,}, in G’, both vt and vi have degree at 23 in B’, violating Claim 3. The proof when y’x,, y’x, E E(G) is similar. 0 Subcase 3.1: There is a vertex y E V(H’) with yxl, yxs E E(G). Then xl yxsx2x1 is a 4-cycle in G. Let n(3) = ({y, x2}, {x,, x,}) and let G” = G/X(~) Denote by vi and vz the vertices of G’ to which {x,, x5} and (y, x2) are mapped, respectively. Since x&v~xqxJ is a 4-cycle in G”, and since O1 = G[(x,, x2, x3, x4, xs, y}] = 0, we can apply the previous proofs to G” to make the following.
H.-J. Lai
18
Claim 3’. Let B3 be the block in G3 that contains v: and vz. Then either VTor vz has degree less than 3 in B3.
Proof. Note that the path P’ = x6v~v~.x3 and the (x3, x&path P in G - E(O) 3 The block B3 in G3 containing v: and vz form a cycle in G3 containing VTand v2. contains P, P’, H2 and the edge v&. Hence v: and us have degree 23 in B3, contrary to Claim 3’. Cl Subcase 3.2: There is no y E V(H’)
with yx,, yxs E E(H’).
x1x2, x5xg})/{x1x6, x2x5}, and let VT, vz denote the vertices of LetG*=(G-{ G* to which x1x6 and x2x5 are contracted, respectively. The component H of G - E(O) containing x6 does not contain x 2. Hence, by the assumption of this subcase and by Claim 5, no 4-cycle of G contains exactly 2 edges of E(C,). Since G is reduced (and hence is simple), no 4-cycle of G has exactly 3 edges in E(C,). By Claim 3, no 4-cycle of G contains x5x6 and no other edges in E(CI); and no 4-cycle of G contains x1x2 and no other edges of E(C,). Hence G* satisfies (1). Since K(G) 3 2, every cut-vertex of G* other than v: and vz must separate VT and vz. Since H’ is connected, there is a (x1, x5)-path Q in H’ that is disjoint from the (x3, x6)-path P. These paths Q and P, together with $x3, form a cycle in G * containing VT and v z. Hence only VT or vg can be a cut-vertex of G*. If VT and vz are not cut-vertex of G *, then K(G *) 3 2. By the assumption of this subcase and by A&H’) = { xl, x5), (1) holds for H’. Hence xl has degree 82 in H’ and so VT has degree 23 in G*. Similarly, x5 has degree 22 in H’ and so vz has degree 23 in G*. It follows by the minimality of G that G* E W.9 and so by Lemma 5, G E W’, a contradiction. If VT is a cut-vertex of G”, then G * has an end block BT containing v T but not vz. Let B1 = G[E(BT)]. Since K(G) 3 2, A&B,) = {x,, x6}. Hence x1 is incident with an edge el E E(B,) and x6 is incident with an edge e6 E E(B,). Since el, e6 $ E(O), so in G’, the vertices vi, vi have degree 2 3 in B’, contrary to Claim 3. If vz is a cut-vertex of G*, then G* has an end block Bq that contains vz but not VT. Let B2 = G[E(B,“)]. Since K(G) 22, A,(B,) = {x2, x,} and so in G’, vt and vi have degree 23 in B’ contrary to Claim 3. This proves Lemma 6. Cl
Main result Theorem 1, [f G E +!I*,then G E KY?. By contradiction, let G be a counterexample to Theorem 1 with IV(G)1 minimized. Immediately from Lemmas 4 and 6 and from definitions, we have the following observations.
Graph whose edges are in small cycles
19
Lemma 7 ([S]). Each of the following holds. (i) G i.~reduced and satisfies (2). (iij k” -Wrk u”,maxknal K,,,-subgraph of G, (t 2 2), then &(W)
c D,*(G).
(6)
(iii) If L and J are two subgraphs of G with IV(J) n V(L)] s 1 and J U L = G, then both J and L satisfy (2). (iv) If W is a maximal I&-subgraph of G and if L is a connected subgraph of G - E(W) with A,(L) c V(W), then (2) hold for L. (v) If C = ~1~2~3~4x1 is a maximal K2,2-subgraph of G and if G - E(C) has an (xl, x&path PI and an (x2, x3)-path P2, then V(P,) n V(P,) # 8.
Lemma 8. G cannot have a maximal K2,2-subgraph. Proof. By contradiction, let C =x1x2x3x4x1 be a maximal K2,2-subgraph of G. Define G*, vl, v2 as in Notation 2. Then G* satisfies (2) and by (6), S(G*) 2 3. By Lemma 5 and by the minimality of G, K(G*) < 1. By (v) of Lemma 7, G* is connected. Hence K(G*) = 1. Case 1: There is a cut-vertex z $ {vl, v2) in G*. Then by K(G) 3 2, z must separate v1 and v2 in G*. By (v) of LeriLr;la 7, we may assume that z is in every (x1, x&path and in every (x2, x3)-path of G - E(C). Hence there are connected subgraphs HI, H2, H3, H4 of G - E(C) such that A,(H,) = {z, xi} and G-E(C)=H,UH,UH,UH,.
Let b,* be the graph obtained from HI U H4 by identifying xl and x4, J* be the graph obtained from H2 U H3 by identifying x2 and x3. Note that G* = J* U L*. By (iv) of Lemma 7, G - E(C) satisfying (2), and so by (iii) of Lemma 7, both HI U H4 and H2 U H3 satisfy (2). Thus L* and J* satisfy (2). By (2) and by the fact that z is a cut-vertex of H, U H4, the degree of z in L* is at least 4 and so by (6), S(L*) B 3. Similarly, 6(J*) 3 3. It follows that J*, L* E %, and so by the minimality of G, J*, L* E %S. By (v) of Theorem A, G* E %6p and so by Lemma 5, G E %.Z, a contradiction. Case 2: There is no cut-vertex in G * not in { vl, v2}.
Without loss of generality, let v1 be a cut-vertex of G*. Then G * has a nontrivial connected subgraph B * with A&B*) = {v,} and v2 $ V(B*). Let B = G[E(B*)]. Then A,(B) = ( x I, x,}. By K(G) 2 2, B is connected. By (v) of Lemma 7, every (x2, x3)-path in G - E(C) uses x1 or x4.
(7)
Since G* has no cut-vertex not in {v, , v2}, G * - E( B *) has a cycle containing ZJ: and v2. Let H* be the block of G* that contains both v1 and v2 and let H = G[E(H*)]. Then A,(H) 5 V(C) and so by (iv) of Lemma 7, H satisfies (2). Without loss of generality we assume that x2 E V(H).
H.-J. Lai
20 Claim.
X,EV(H).
Suppose not, we assume that x3 $ V(H). By 5(G) 2 3, x3 is incident with an edge in G - E(C). Note that this edge is an edge on G*. Let K* be a block of G* containing v2 but not ul, and let K = G[E(K*)]. By K(G) 2 2, x2, x3 E V(K). Hence there is an (x2, r3)-path in K, contrary to (7). This proves the Claim. Cl proof.
Since H* contains ul, either x1 or x4 is in V(H). If x4 $ V(H), then by (7), x1 is a cut-vertex of H and so x1 E D;(H); if x4 E V(H), then V(C) E D,*(H). Hence in either case, u1 and u2 have degree 33 in H* and so by the minimality of G, H* E %.Z. Note that (G[E(H) b?(C)])* =H*. By Lemma 5, G is not reduced, contrary to Lemma 4.0 Lemma 9 ([5]). D3(W
If
G
has a K2,t-subgraph W with t 2 3, then t = 3 and
SD,(G)-
Proof. The proof uses essentially the same technique as in the proof of Lemma 8 and is routine. Cl
Notation 3 (see Fig. 4). Define D to be the union of two copies of K2.3’~ WI and W2such that VW,) n VW,) =
o,(K)
n D2W.2)
=
{Y 1 c_D,(G).
G/41)
G” + y Fig. 4. Notation 3.
Graph whose edges are in small cycles
21
Denote 4(W)
= {YIP ~2, Y)
and
&(w,)
= (~3, ys, y}.
Suppose that H is a subgraph of G containing D. Let v,* = {Yl, Y2L
vL2 =
wm
-
Kl
and
Jd(l)
=
wll,
v12)*
For convenience, we let ul, y denote the vertices of G/X( 1) to which VII and VI2 are mapped, respectively. Denote ezo) = uly and H’ = (H/n(l)) - e,o). We regard W2as a subgraph of H’. Let h={Y,,Yd9
v22=
WC)
-
v21
and
42)
=
051,
v,,)*
denote the vertices of H’/n(2) to which V21 and V22 are mapped, respectively and let eXt2)= u2y. Define H” to be the nontrivial component of H’/42) - ezc2,. Note that
Let
u29 y
G’= By
(iii)
(Gh(l))l~@)
of Theorem
- {y).
A, by Theorem
B, and by K(G) 22,
we have the
following. Lemma 10([S]). Let D, G’, ul , u2 be defined in Notation 3. Then: (i) If G’k %5?, then G E %.9. (ii) Neither ul nor cc2is a cut-vertex of G”. Since G E 9&, G’ satisfies (1). By (6), S(G”) 2 3. Since G is a minimum counterexample to Theorem 1, by (i) of Lemma 11, K(G’) 6 1. This, together with K(G) > 2, implies K@")=
1.
(8)
Lemma 11 ([S]). Zf H is a subgraph of G with lAG( Ii contains a 2-block of G.
= 2, then either H = K2 or
Proof. This follows from the fact that G E 5!& Cl Lemma I2 ([S]). Zf H is a minimal 2-block of G, then: (i) H cannot have D as a subgraph. (ii) H cannot have a &-subgraph W with (9
D2(W) c D:(G). roof. The proof is routine minimality of H. El G has a 2-block.
with the help of Lemmas
10 and 11, and the
22
H.-J. Lai
proof.By (i)of Lemma 7, and by Lemmas 8 and 9, G is an edge-disjoint union . . of K&s such that rf W 1s a K 2,3-subgraph of G, then Q(W) E q,(G). Hence G either has D as a subgraph or has a subgraph W z K2,3 satisfying (9). Case I: G has D as a subgraph. We shall use the notations in Notation 3 and let G”, uI , u2 be defined as in Notation 3. By (S), by (ii) of Lemma 10, and by K(G) 2 2, G” has a cut-vertex z $ (u,, u2} separating u1 and u 2. Let B” be an end block of G” that contains u, but not u2, and let B = G[E(B”)]. Then A&B U W,) = {z, y} and so by Lemma 11 withH=BUW,, Ghasa2-block. 0 Case 2: G has a subgraph W 5 KZS3satisfy@ (9). Define G’, vl, u2 as in Notation 1 with t = 3. By (9), 6(G’) 2 3. It is easy to see that rc(G’) = 1 and G’ has a cut-vertex c $ {ul, v2} that separates ul and u2 in G’. Let B’ be an end block of (3’ that contains u1 but not u2 and let Cl B = G[E(B’)]. Then A,(B) = {z, y3) an d so by Lemma 11, G has a 2-block. Proof of Theorem 1. By Lemma 13, G has a minimal 2-block H. By Lemma 7, 8 and 9, H must be an edge-disjoint union of K2,3’~.Since H is a 2-block, and since 6(G) 2 3, H$K2,3. Hence either H has D as a subgraph or H has a subgraph W = K2,_tthat satisfies (9), contrary to Lemma 12. 0
References [1] J.A. Bondy and U.S.R. Murty, Graph Theory with Applications (Elsevier, Amsterdam, 1976). [2] P.A. Catlin, A reduction method to find spanning eulerian subgraphs, J. Graph Theory 12 (1988) 29-44. [3] P.A. Catlin, Supereulerian graph, Collapsible graphs and 4-cycles, Congr. Numer. 56 (1987) 223-246. [4] H.-J. Lai, Reduction towards collapsibility, submitted. [S] H.-J. Lai, Contractions and eulerian subgraphs, Ph.D. Thesis, Wayne State University, 1988. [6] P. Paulraja, On graphs admitting spanning eulerian subgraphs. Ars Combin. 24 (1987) 57-65. [7] P. Paulraja, Research Problem. 85, Discrete Math. 64 (1987) 109. [8] P. Paulraja, Research Problem. 86, Discrete Math. 64 (1987) 317.
11
raph whose edges are in small cycles Hong-Jim Lai* West Virginia University, Morgantown, WV 26506, USA Received 15 June 1987 Revised 24 January 1990 Abstract Lai, H.-J., Graph whose edges are in small cycles, Discrete Mathematics 94 (1991) 11-22. Paulraja (1987) conjectured the following: (i) If every edge of a 2-connected graph G lies in a cycle of length at most 4 in G, then G has a dominating closed trail. (ii) If, in addition, S(G) 2 3, then G has a closed spanning trail. Collapsible graphs are defined and studied by Catlin (1988). Catlin showed that if H is a collapsible subgraph of G, then G has a spanning closed trail if and only if G/H, the graph obtained from G by contracting H, has a spanning closed trail. Catlin (1987) conjectured that a graph satisfying the hypothesis of (ii) is collapsible. In this paper, all three conjectures are proved.
Introduction We shall use the notation of Bondy and Murty [l] except for contractions. A graph may have multiple edges but not loops. A spanning closed trail of G is 4!ed a spanning eulerian subgraph (SES) of G. The collection of graphs that have an SES is denoted by 9.9. Note that K, E 95’. If a closed trail C of G satisfies E(G - V(C)) = #, then C is called a dominating eulerian subgraph (DES) of G. Let G be a graph. An block of G that has exactly one cut-vertex of G is called an end block of G. A block B is acyclic if B = K2. For a subset X c E(G), the contraction G/X is the graph obtained from G by identifying the ends of each edge in X and then deleting the resulting loops. If H is a subgraph of G, then we write G/H for G/E(H). Let W be a subgraph of G. If for some t 2 2, \V = K2,,, then W is called a W,,,-subgraph Of G. Let c = ~1~22)3u4~5~(,~1 be a 6-cycle. Define @ t0 bc the graph with V(0) = V(C) and E(O) = E(C) U (~2~5). * This is part of the author’s Ph.D. Dissertation, done in Wayne State University under Dr. Paul A. Catlin. 0012-365X/91/$03.50 @ 1991 .-- Elsevier Scicncc Publishers B.V. All rights rcscrvcd
12
H.-J. Lai
Let H be a subgraph of G. Define H’= G[E(G) - E(H)] and A,(H) = V(H) n V(W). The vertices in A,(H) are called the vertices of attachment of H in G. If K(H) 2 2 and IA,(H)1 = 2, then H is called a 2-b/0& of G. A graph G is an edge-disjoint union of K+ if there is a partition El, &, . . . , Ek of E(G) such that G[E,] z KZ,ni’~(ni a 2 and 1 s i s k). Note that when i #j, we may have ni # nj. We consider the following conditions. Any edge of G is an m-cycle of G, m s 4;
(I)
G is an edge-disjoint union of K$s.
(2)
Let %= {G: G satisfies (1) with K(G) 2 2) and %, = {G E 5%S(G) 2 3). In [6-81, Paulraja raised the following two conjectures. Conjecture 1. If G E F&, then G E 9Y. Conjecture 2. If G E 59, then G has a DES. A graph G is collapsible if for every even subset R c V(G), G has a subgraph CR(called the R-subgraph of G) such that G - E(r,) is connected and R is the set of odd-degree vertices of rR. The collection of all collapsible graphs is denoted by VZ’. Let HI, Hz, . . . , HC be all the maximal collapsible subgraphs of G. Denote by (G), the graph of order c obtained from G by contracting HI, . . . ,HC to c distinct vertices. We call (G), the reduction of G and HI, Hz, . . . ,HC the preimages of the vertices of (G), . A graph G is reduced if G = (G), . In [2], Catlin showed that (G), is well defined and unique. He also proved the following theorem. Theorem A (Catlin [2]). Let G be a graph. (i) y G E %JZ?,then G E 9.2. (ii) G E 9’9 if and only if ( G)l E %!?. (iii) If H E %S? is a subgraph of G, then G E %9 if and only if
Conjecture 3. If G E %,, then G E KY, By (i) of Theorem A, we have the following proposition.
Graph whose edges are in small cycles
13
Proposition 1. Conjecture 3 implies Conjecture 1. For each i = 1,2, . . . , define Di(G) =
{v E V(G):
deg&)
= i},
and D;(G) = (v E V(G): de&v)
a i}.
roposition 2. Conjecture 1 implies Conjecture 2. Proof. Assume that Conjecture 1 holds. Let G E 9. Since an SES is always a DES, we may assume that G $ XZ and G E %- 9&. Thus 5(G) = 2. Let D,(G) = {xi’ xz’ . . . ,x,). Since K(G) 2 2, no vertex in D,(G) is the common end of multiple edges;
(3)
and by (1) and K(G) 3 2, no paths in G contain three consecutive vertices in D,(G).
(4)
Choose yl,y2,...,y, such that ei = Xiy’ E E(G),
(1 s i s m), and such that if for some i, j with XiXjE E(G) then yi = Xj and xi = Yj. Note that the latter can happen only when Xi and Xj are the internal vertices of a path of G of length 3. By (3) and (4), the multiplicity of each ei is one in G. Let el be an edge with the same ends as ei but el$ E(G), (1 s i s m). Obtain a new graph G’ by adding 1ei, e;, . . . ,el,} to G. It follows that G’ E % and so G’ E sP.9, by Conjecture 1. Let C’ be an SES of G’. Since G $99, we may assume that for some k s m, { ei,
e;, . . . ,el) U {elf e2, . . . ,ek} G W’);
(5)
and that ei $ E(C’), (j > k). Choose C’ so that (5) is satisfied and that k is as small as possible. Let C = C’ - {e,, . . . ,ek, e;, . . . ,e;}. Then every vertex of C has even degree in C. If ei = X~X’E E(C), for some 1 s s, t G m, then since x, and x’ are in D,(G), x, and x’ are in D2(C’) also. It follows that e, $ E(C’), contrary to the choice of C’. So if X~X,E E(C’), then e, E E(C’) and ei $ (C’). This implies that C is connected, and that x1, x2, . . . ,xk form an independent set in G and are the only vertices not in V(C). Thus C is a DES of G. 0
Reductions
Notation P (see Fig. 1). Let W be a K 2,,-subgraph of G and iet H be a subgraph of G containing W. Denote LJ2(W)
=
{Yb Y29
l
l
l
,YJ*
H.-J. Lai
G Fig. 1. Notation 1 with V, = {y, , y2}.
G/x
Let V1 be a subset of D,(W) such that IVJ = 2, let V2= V(W) - V1 and let it = ( VI, V2) denote this partition of V(W). Denote by H/JC the graph obtained from H by identifying all vertices of V1 to form a single vertex Q, by identifying all vertices of V2 to form a single vertex u2, and by joining vl, v2 with a new edge % = 2f,v2, so that E(H) - E(W) = E(H/z)
- {e=}.
And finalIy let H’ = (H/n) - e,. Theorem Gjn
B (Catlin [3]). Let W, Ed and G/z then G E %Z.
be defined in Notation 1. If
E %.Z,
Lemma 1. If G’ = (G/a) - e, E %Z, then G E %.Z’. Proof. It follows from Theorem B and (iii) of Theorem A.
Cl
It is easy to see the following. Lemma 2 ([Sj). If G E %, then (G): has girtJi 4 and (G), satisfies (1). Lemma 3 ([5]). Let B be a block of(G),. (i) If GE%& then BE%$; (ii) rf G E 59, then B E 5% Lemma 4. If G is a counterexample to Conjecture 3 with 1V( G)I minimized then G is reduced. _ oof. Let G satisfy the hypothesis of Lemma 4. If G is not reduced, then (G),, the reduction of G, has smaller order than G. By Lemmas 2 and 3, each block of (G), is in %I and so by the minimality of G, each block of (G), is in CZ’. Thus by (v) and (iii) of Theorem A, G E %Z, a contradiction. 0
Graph whose edges are in small cycles
15
Notation 2. Let C = x1x2x3&$x1be a 4-cycle of G. Define: G* = [G -
{&X2,
X3X4)1/{&X4,
-x2x3},
and denote by q and ~2 the vertices of G* to which Xix4 and respectively. For convenience, we regard V(G*) = [V(G) - V(C)] u {u,, Lemma 5 [4-S].
~2x3
are contracted,
E(G*) U E(C) = E(G).
~21,
If G” E %JZ, then G E %.Z.
Lemma 6. Let G be a counterexample to Conjecture 3 with 1V(G)1 minimized. Then G does not have a subgraph isomorphic to 0. Proof. By contradiction, suppose that G has 0 as a subgraph. Let C = ~1~2~3~&&j~] denote the 6-cycle Of @ and let C, = ~~~-$~&j~~ and C2 = x2x3x4x5x2 be the two 4-cycles contained in 0. Define ~(l)=({%%),
b2d6))
and
@)=(b2,x4),
{x39x5))*
Define G’ = G/n(l) and G2 = G/z(2). Let vi and vi denote the vertices of G’ to which {x1; x5} and {x2, x6) are mapped, respectively; and let v:, v$ be the vertices of G2 to which {x3, x5} and {x2, x4} are mapped, respectively. Let el = vi in G2 and C2 = x3v~v~x4x3 in v: and e2 = v: vs. We shall regard C1 =x&_&q G’ throughout the proof of this lemma. Since G E gI, we have 6(G’) 23 and S(G2) 2 3, and both G’ and G2 satisfy (1) . Claim 1. K(G~) = K(G~) = 1, and all cut vertices of G’ are in {v:, vi} and all cut-vertices of G2 are in (v:, v$>. Proof. Clearly G’ is connected. If K(G’) 3 2, then G1 E %LY?follows by the minimality of G. Hence G E 9Z.Zby Theorem B, a contradiction. Thus K(G ‘) = 1. Since K(G) = 2, the cut-vertices of G’ must be in {v :, v:}. The proof for G2 is similar. Cl Claim 2 (see Fig. 2). For i E { 1,2}, if vi is a cut-vertex of G’ and if L’ is an end-block of G’ containing vi but not vi, then L’ E %Y. A similar result holds when we replace v’, by vi.
Fig. 2. Claim 2 of Lemma 6.
H.-J. Lai
16
proof. Without ioss of generality, we assume i = 1 and V: is cut-vertex of G’ with L1 an end-block of G’ containing vt but not vi. Let H = G[E(L’)]. Then by K(G) 2 2, we have A,(H) = {x1, xs}, Since G satisfies (l), either H satisfies (1) or there is a vertex y E V(H) with yx,, yx, E E(H). If H satisfies (1), then L’ E +?Iand so L’ E %Z, by the minimality of G. Hence we assume that there is a vertex y E V(H) such that yx,, yxs E E(H). By K(G) 3 2, y is not a cut-vertex of G. Hence either x1 or x5 is adjacent to some vertex of V(H) - {y}. kt follows that 6(L1) 3 3. Since (1) holds for G, (1) holds for L’ also. Thus L’ E %.%’by the minimality of G. 0 For i E { 1, 2}, let B’ be the block of G’ containing both vi and vi. Claim 3. Either vi or vi has degree less than 3 in B’. Proof. By Claim 2, all blocks of G’ other than B’ are in Cee4p. Clearly B’ satisfies (1) and K(B~) 2 2. If 6(B’) 3 3, then B’ E %Z’ follows from the minimality of G and so by (v) of Theorem A and by Theorem B, G E %.Z, a contradiction. Hence 6(B’) < 3. Since S(G) 3 3, every vertex in V(B’) - {v:, v’,} has degree at least 3 in B’. Hence either V: or v’, has degree less than 3 in B’. Cl Claim 4. For i, j E { 1, 2)) if vi is not a cut-vertex of G’, then vi has degree at least 3 in B’.
Proof. Without loss of generality, we assume that i = j = 1. Since the degree of xi is at least 3 in G, xl is incident with an edge e’ E E(G) - E(O). Since vi is not a cut-vertex of G’, all the edges incident with vi in G’ are in B’. Since we map xl and xs to vt, and since e’ E E(G) - E(O), vi has degree at least 3 in B’. Cl The proof of Lemma 6 will now be divided into the following cases. Case 1: For some i E { 1,2}, v:, vi are cut vertices of G’. Without loss of generality, we assume i = 1 and that there are end blocks L] and Li in G’, where B’ $ (Et, I,:}, with A&L,‘) = {vi) (1 c j s 2). Let Hj = G[E(L,‘)], (1 f j d 2). Then HI and Hz are connected, and A,(H,) = {x1, x5} and A&Hz) = {x2, x,}. Note that HI and Hz also induce subgraphs in G2 (which we also call H, and H2), with A&H,) = {x,, v:} and A&H,) = {x,, us}. Therefore, the block B2 of G2 containing v: and vif also contains C, , Hl and H2, and so v: and vz have degree at least 3 in B2, contrary to Claim 3. Case 2: The only cut-vertex of G’ is vi, (1 d i 6 2). Let L’ be an end block of G’ that does not contain vi and has v’, as its only vertex of attachment in G’, and let H’ = G[E(L’)]. Then H’ is connected with A,(H’) = {xl, x,}. Thus xs is incident with an edge errE E(H’). Since H’ can be regarded as a subgraph of B2, e” is an edge incident with v: in B2. Hence v: is incident with e” and two edges in C., and so VThas degree at least 3 in B2. Since
Graph whose edges are in small cycles
17
Fig. 3. Case 3 of Lemma 6.
V$ is not a cut-vertex of G2, it follows by Claim 4 that r.~$has degree at least 3 in B2, contrary to Claim 3. Case 3 (see Fig. 3): The only cut-vertex of G’ is v: and the only cut-vertex of G2 is v;. Let L’ be an end block of G’ that contains vi but not Ug_i and let H’ = G[E(L’)], (16 i < 2). Since L’ is an end block of G’, (1 s i s 2), and since K(G)> 2, &AH’) = {xi, xs)
and
&(H2)
=
{xzt
x,>.
Now we show that G - E(O) has a component H (say) with A,(H) = (~3, x6). Since 6(G) 2 3, xj is incident with an edge that is not in E(O). By K(G) 2 2, there is an (x1, xi)-path P in G - E(@) for some j E (1, 2,4, $6). If j = I or j=5, then in G’, v : has degree at least 3 in B ’ . By Claim 4, vi has degree at least 3 in B’, contrary to Claim 3. Ifj=2 or-j = 4, then in G2, us has degree 3 in B2, and a similar contradiction arises. Hence j = 6 and so A,(H) = {x1, x6}. Claim 5. There is no y’ E V(G) - {x2, xs} such that either y ‘x6, y ‘x, E E(G) or y’x,, y’xj E E(G). Proof. If y’x4, y’x, E E(G) and y’ #xs, then since G has a connected subgraph H with A,(H) = {x1, x,}, in G’, both vt and vi have degree at 23 in B’, violating Claim 3. The proof when y’x,, y’x, E E(G) is similar. 0 Subcase 3.1: There is a vertex y E V(H’) with yxl, yxs E E(G). Then xl yxsx2x1 is a 4-cycle in G. Let n(3) = ({y, x2}, {x,, x,}) and let G” = G/X(~) Denote by vi and vz the vertices of G’ to which {x,, x5} and (y, x2) are mapped, respectively. Since x&v~xqxJ is a 4-cycle in G”, and since O1 = G[(x,, x2, x3, x4, xs, y}] = 0, we can apply the previous proofs to G” to make the following.
H.-J. Lai
18
Claim 3’. Let B3 be the block in G3 that contains v: and vz. Then either VTor vz has degree less than 3 in B3.
Proof. Note that the path P’ = x6v~v~.x3 and the (x3, x&path P in G - E(O) 3 The block B3 in G3 containing v: and vz form a cycle in G3 containing VTand v2. contains P, P’, H2 and the edge v&. Hence v: and us have degree 23 in B3, contrary to Claim 3’. Cl Subcase 3.2: There is no y E V(H’)
with yx,, yxs E E(H’).
x1x2, x5xg})/{x1x6, x2x5}, and let VT, vz denote the vertices of LetG*=(G-{ G* to which x1x6 and x2x5 are contracted, respectively. The component H of G - E(O) containing x6 does not contain x 2. Hence, by the assumption of this subcase and by Claim 5, no 4-cycle of G contains exactly 2 edges of E(C,). Since G is reduced (and hence is simple), no 4-cycle of G has exactly 3 edges in E(C,). By Claim 3, no 4-cycle of G contains x5x6 and no other edges in E(CI); and no 4-cycle of G contains x1x2 and no other edges of E(C,). Hence G* satisfies (1). Since K(G) 3 2, every cut-vertex of G* other than v: and vz must separate VT and vz. Since H’ is connected, there is a (x1, x5)-path Q in H’ that is disjoint from the (x3, x6)-path P. These paths Q and P, together with $x3, form a cycle in G * containing VT and v z. Hence only VT or vg can be a cut-vertex of G*. If VT and vz are not cut-vertex of G *, then K(G *) 3 2. By the assumption of this subcase and by A&H’) = { xl, x5), (1) holds for H’. Hence xl has degree 82 in H’ and so VT has degree 23 in G*. Similarly, x5 has degree 22 in H’ and so vz has degree 23 in G*. It follows by the minimality of G that G* E W.9 and so by Lemma 5, G E W’, a contradiction. If VT is a cut-vertex of G”, then G * has an end block BT containing v T but not vz. Let B1 = G[E(BT)]. Since K(G) 3 2, A&B,) = {x,, x6}. Hence x1 is incident with an edge el E E(B,) and x6 is incident with an edge e6 E E(B,). Since el, e6 $ E(O), so in G’, the vertices vi, vi have degree 2 3 in B’, contrary to Claim 3. If vz is a cut-vertex of G*, then G* has an end block Bq that contains vz but not VT. Let B2 = G[E(B,“)]. Since K(G) 22, A,(B,) = {x2, x,} and so in G’, vt and vi have degree 23 in B’ contrary to Claim 3. This proves Lemma 6. Cl
Main result Theorem 1, [f G E +!I*,then G E KY?. By contradiction, let G be a counterexample to Theorem 1 with IV(G)1 minimized. Immediately from Lemmas 4 and 6 and from definitions, we have the following observations.
Graph whose edges are in small cycles
19
Lemma 7 ([S]). Each of the following holds. (i) G i.~reduced and satisfies (2). (iij k” -Wrk u”,maxknal K,,,-subgraph of G, (t 2 2), then &(W)
c D,*(G).
(6)
(iii) If L and J are two subgraphs of G with IV(J) n V(L)] s 1 and J U L = G, then both J and L satisfy (2). (iv) If W is a maximal I&-subgraph of G and if L is a connected subgraph of G - E(W) with A,(L) c V(W), then (2) hold for L. (v) If C = ~1~2~3~4x1 is a maximal K2,2-subgraph of G and if G - E(C) has an (xl, x&path PI and an (x2, x3)-path P2, then V(P,) n V(P,) # 8.
Lemma 8. G cannot have a maximal K2,2-subgraph. Proof. By contradiction, let C =x1x2x3x4x1 be a maximal K2,2-subgraph of G. Define G*, vl, v2 as in Notation 2. Then G* satisfies (2) and by (6), S(G*) 2 3. By Lemma 5 and by the minimality of G, K(G*) < 1. By (v) of Lemma 7, G* is connected. Hence K(G*) = 1. Case 1: There is a cut-vertex z $ {vl, v2) in G*. Then by K(G) 3 2, z must separate v1 and v2 in G*. By (v) of LeriLr;la 7, we may assume that z is in every (x1, x&path and in every (x2, x3)-path of G - E(C). Hence there are connected subgraphs HI, H2, H3, H4 of G - E(C) such that A,(H,) = {z, xi} and G-E(C)=H,UH,UH,UH,.
Let b,* be the graph obtained from HI U H4 by identifying xl and x4, J* be the graph obtained from H2 U H3 by identifying x2 and x3. Note that G* = J* U L*. By (iv) of Lemma 7, G - E(C) satisfying (2), and so by (iii) of Lemma 7, both HI U H4 and H2 U H3 satisfy (2). Thus L* and J* satisfy (2). By (2) and by the fact that z is a cut-vertex of H, U H4, the degree of z in L* is at least 4 and so by (6), S(L*) B 3. Similarly, 6(J*) 3 3. It follows that J*, L* E %, and so by the minimality of G, J*, L* E %S. By (v) of Theorem A, G* E %6p and so by Lemma 5, G E %.Z, a contradiction. Case 2: There is no cut-vertex in G * not in { vl, v2}.
Without loss of generality, let v1 be a cut-vertex of G*. Then G * has a nontrivial connected subgraph B * with A&B*) = {v,} and v2 $ V(B*). Let B = G[E(B*)]. Then A,(B) = ( x I, x,}. By K(G) 2 2, B is connected. By (v) of Lemma 7, every (x2, x3)-path in G - E(C) uses x1 or x4.
(7)
Since G* has no cut-vertex not in {v, , v2}, G * - E( B *) has a cycle containing ZJ: and v2. Let H* be the block of G* that contains both v1 and v2 and let H = G[E(H*)]. Then A,(H) 5 V(C) and so by (iv) of Lemma 7, H satisfies (2). Without loss of generality we assume that x2 E V(H).
H.-J. Lai
20 Claim.
X,EV(H).
Suppose not, we assume that x3 $ V(H). By 5(G) 2 3, x3 is incident with an edge in G - E(C). Note that this edge is an edge on G*. Let K* be a block of G* containing v2 but not ul, and let K = G[E(K*)]. By K(G) 2 2, x2, x3 E V(K). Hence there is an (x2, r3)-path in K, contrary to (7). This proves the Claim. Cl proof.
Since H* contains ul, either x1 or x4 is in V(H). If x4 $ V(H), then by (7), x1 is a cut-vertex of H and so x1 E D;(H); if x4 E V(H), then V(C) E D,*(H). Hence in either case, u1 and u2 have degree 33 in H* and so by the minimality of G, H* E %.Z. Note that (G[E(H) b?(C)])* =H*. By Lemma 5, G is not reduced, contrary to Lemma 4.0 Lemma 9 ([5]). D3(W
If
G
has a K2,t-subgraph W with t 2 3, then t = 3 and
SD,(G)-
Proof. The proof uses essentially the same technique as in the proof of Lemma 8 and is routine. Cl
Notation 3 (see Fig. 4). Define D to be the union of two copies of K2.3’~ WI and W2such that VW,) n VW,) =
o,(K)
n D2W.2)
=
{Y 1 c_D,(G).
G/41)
G” + y Fig. 4. Notation 3.
Graph whose edges are in small cycles
21
Denote 4(W)
= {YIP ~2, Y)
and
&(w,)
= (~3, ys, y}.
Suppose that H is a subgraph of G containing D. Let v,* = {Yl, Y2L
vL2 =
wm
-
Kl
and
Jd(l)
=
wll,
v12)*
For convenience, we let ul, y denote the vertices of G/X( 1) to which VII and VI2 are mapped, respectively. Denote ezo) = uly and H’ = (H/n(l)) - e,o). We regard W2as a subgraph of H’. Let h={Y,,Yd9
v22=
WC)
-
v21
and
42)
=
051,
v,,)*
denote the vertices of H’/n(2) to which V21 and V22 are mapped, respectively and let eXt2)= u2y. Define H” to be the nontrivial component of H’/42) - ezc2,. Note that
Let
u29 y
G’= By
(iii)
(Gh(l))l~@)
of Theorem
- {y).
A, by Theorem
B, and by K(G) 22,
we have the
following. Lemma 10([S]). Let D, G’, ul , u2 be defined in Notation 3. Then: (i) If G’k %5?, then G E %.9. (ii) Neither ul nor cc2is a cut-vertex of G”. Since G E 9&, G’ satisfies (1). By (6), S(G”) 2 3. Since G is a minimum counterexample to Theorem 1, by (i) of Lemma 11, K(G’) 6 1. This, together with K(G) > 2, implies K@")=
1.
(8)
Lemma 11 ([S]). Zf H is a subgraph of G with lAG( Ii contains a 2-block of G.
= 2, then either H = K2 or
Proof. This follows from the fact that G E 5!& Cl Lemma I2 ([S]). Zf H is a minimal 2-block of G, then: (i) H cannot have D as a subgraph. (ii) H cannot have a &-subgraph W with (9
D2(W) c D:(G). roof. The proof is routine minimality of H. El G has a 2-block.
with the help of Lemmas
10 and 11, and the
22
H.-J. Lai
proof.By (i)of Lemma 7, and by Lemmas 8 and 9, G is an edge-disjoint union . . of K&s such that rf W 1s a K 2,3-subgraph of G, then Q(W) E q,(G). Hence G either has D as a subgraph or has a subgraph W z K2,3 satisfying (9). Case I: G has D as a subgraph. We shall use the notations in Notation 3 and let G”, uI , u2 be defined as in Notation 3. By (S), by (ii) of Lemma 10, and by K(G) 2 2, G” has a cut-vertex z $ (u,, u2} separating u1 and u 2. Let B” be an end block of G” that contains u, but not u2, and let B = G[E(B”)]. Then A&B U W,) = {z, y} and so by Lemma 11 withH=BUW,, Ghasa2-block. 0 Case 2: G has a subgraph W 5 KZS3satisfy@ (9). Define G’, vl, u2 as in Notation 1 with t = 3. By (9), 6(G’) 2 3. It is easy to see that rc(G’) = 1 and G’ has a cut-vertex c $ {ul, v2} that separates ul and u2 in G’. Let B’ be an end block of (3’ that contains u1 but not u2 and let Cl B = G[E(B’)]. Then A,(B) = {z, y3) an d so by Lemma 11, G has a 2-block. Proof of Theorem 1. By Lemma 13, G has a minimal 2-block H. By Lemma 7, 8 and 9, H must be an edge-disjoint union of K2,3’~.Since H is a 2-block, and since 6(G) 2 3, H$K2,3. Hence either H has D as a subgraph or H has a subgraph W = K2,_tthat satisfies (9), contrary to Lemma 12. 0
References [1] J.A. Bondy and U.S.R. Murty, Graph Theory with Applications (Elsevier, Amsterdam, 1976). [2] P.A. Catlin, A reduction method to find spanning eulerian subgraphs, J. Graph Theory 12 (1988) 29-44. [3] P.A. Catlin, Supereulerian graph, Collapsible graphs and 4-cycles, Congr. Numer. 56 (1987) 223-246. [4] H.-J. Lai, Reduction towards collapsibility, submitted. [S] H.-J. Lai, Contractions and eulerian subgraphs, Ph.D. Thesis, Wayne State University, 1988. [6] P. Paulraja, On graphs admitting spanning eulerian subgraphs. Ars Combin. 24 (1987) 57-65. [7] P. Paulraja, Research Problem. 85, Discrete Math. 64 (1987) 109. [8] P. Paulraja, Research Problem. 86, Discrete Math. 64 (1987) 317.