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The structure of transform graphs
DISCRETE MATHEMATICS ELSEVIER
Discrete Mathematics 177 (1997) 123-144
The structure of transform graphs Bill Jackson a, John Sheehan b'* aDepartment of Mathematics, Goldsmith's College, University of London, Lewisham Way, London SE14 6NW, UK bDepartment of Mathematical Sciences, University of Aberdeen, Dunbar Street, Aberdeen AB9 2TY, UK Received 27 January 1995; revised 29 May 1996 The inspiration and many of the ideas of this paper originated with George Hendry. This paper is dedicated to his memory
Abstract
The transform graph gives considerable insight into the hamiltonian path structure of a graph. In this paper the structure, and in particular the edge-density, of transform graphs is examined.
O. S u m m a r y o f contents
Let G be a finite, simple graph with vertex set V ( G ) and suppose that v is a fixed vertex o f G. Let H(v; u) be the set of hamiltonian paths h[u] with first vertex v and last vertex u. The transform graph T(G; v) o f G at v has as its vertices {h[x]: h [ x ] E H ( v ; x ) , x E V ( G ) , x # v}. Vertices h[x] and h[y] are adjacent if and only if h[x] can be 'transformed' into h[y]. This means that if
h[x]:
v'"wy..'x
then there exists a vertex w adjacent to x such that h[x] can be transformed into h[y], where h[y]:
v. . . w x . . . y
by simply reversing the last part o f the path to y. Let P be a fixed path o f G with first vertex v. Then T(G; v) can be generalized in an obvious way to T ( G ; P ) , where the hamiltonian paths all start with P. * Corresponding author. E-mail: [email protected]. 0012-365X/97/$17.00 Copyright (~) 1997 Elsevier Science B.V. All rights reserved PII S0012-365X(96)00360-3
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B. Jackson. J. Sheehan/Discrete Mathematics 177 (1997) 123-144
The transform idea has proved very useful in examining the hamiltonian path/path structure of a graph. As far as we know it really originated with P6sa [9] in his paper on hamiltonian circuits. It is used in the proof of Beck's theorem [1] on the size Ramsey number of a path and again by Thomason [2, p. 146] in his parity path theorem generalizing Smith's theorem [10]. It features prominently in the work of Hendry [6-8] which is the starting point for this paper. Let U (= k) denote the set of graphs G such that there exists some fixed vertex v with In(v;u)l = k for all uE V(G) (u ~ v). U (<~k) and U (~>k) are defined in the obvious way. Hendry proved in [6,7] that if G E U ( = 1) then T(G;v) was acyclic; furthermore in [8] it is proved that if G E U ( = 1 ) then qG = (3nG - 3)/2 where nc and qG are, respectively, the number of vertices and edges of G. In this paper we examine the structure of T(G;P). The main results are:
Theorem 3. Let G be a graph. Let T = T(G;P). Suppose that nT>>.k>~l and G E U ( <~k ). Then qT ~<(nr(4 + log k))/2 - k (logarithms are to the base 2 unless stated otherwise).
Corollary 4. Let G E U ( <<.k) (k~>l). Then, for some vE V(G), 2qG ~<(5 + logk)nG -- (logk + 7 - deg G v).
Theorem 9. Let GE U (~<2). Then either T ( G ; P ) E {P1,P2,P3, C6} or qr <~5nr - 2 (P3 is the path o f length 2 and C6 the cycle o f length 6).
Corollary 10. Let G E U ( = 2). Then, for some vE V(G), qG ~<~(7n6 + 2 deg c v - 11 ). In the final section we discuss lower bounds and some related questions.
1. Introduction All graphs in this paper are undirected, simple and finite. A graph G has vertex set V(G) and edge set E(G). We set nG = IV(G)I and q6 = IE(G)] and say that G is an (n6,qG)-graph. The suffix, referring to a particular graph G, is dropped when no ambiguity arises. Let X, Y C V ( G ) . Set E 6 ( x , Y ) = {xy E E ( G ) : x EX, y E Y} and q 6 ( X , Y ) = lEe(X, Y)[. Let v E V(G). Then the degree, deg G v, is qG({v}, V(G)). Usually, if the
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
125
set {v} is a singleton then the brackets are not included, e.g. the neighbourhood NG(v) of v in G is the set of all vertices u such that q(u, v) = 1. Now fix some vertex v E V(G). Let u E V(G). A vu-path is a path in G with first vertex v and last vertex u. Let x, yE V(G)\{v} and suppose wE V(G) is a neighbour of both x and y. Suppose that h[x] is a vx-hamiltonian path:
h[x]:
v...wy...x.
A vx-hamiltonian path is labelled throughout as h[x] (or hi[x] where i is some positive integer), i.e. the notation draws attention to the last vertex in the path. Then the vyhamiltonian path h[y] obtained from h[x] by deleting wy, inserting wx and reversing the order of the xy-sequence in h[x]: h[y]:
v. . . wx. . . y
is called a transform of h[x]; w is said to effect the transform in this case. Notice that if w does effect a transform of h[x] then h[y] is determined uniquely. Clearly, if w effects the transform from h[x] to h[y] then it also effects the transform from h[y] to h[x]. This property allows us to define the (undirected) transform 9raph T(G;v) at v. Let H(v; u) denote the set of vu-hamiltonian paths (uE V(G), u 7~ v). Then
V(T(G;v)) = {h[u]: h[ulEH(v;u), uE V(G) (u ¢ v)} and vertices h[x] and h[y] are adjacent if and only if h[y] is a transform of h[x]. These concepts may all be generalized by replacing v by a fixed path P with first vertex v. For example, let P:
VlVl
• " " /)m
(Vl = v, m f> 1). Let H(P; u) denote the set of vu-hamiltonian paths (u E V(G), u ~ v) with initial segment P. So if h[u] E H ( P ; u) we have
h[u]:
vlv2...v,,...u.
Set
H(P) = {h[u]: h[ulEH(P,u), uE V(G), u ¢ v}. Again the transform graph T(G; P) at P has
V(T(G;P)) = H(P) and vertices h[x] and h[y] are adjacent if and only if h[y] is a transform of h[x]. Let U (~l) denote the set of graphs G such that there exists some fixed vertex vE V(G) with IH(v; u)] ~
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B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
The next definitions are vital in what follows. Firstly, choose the path Q in G such that
H(P) = H(PQ) and Q is maximal with respect to this property. Here PQ denotes the concatenation of P and Q. Q is said to be the completion of P in T ( = T(G;P)) and is denoted by compr(P ). Let u* be the last vertex of PQ. We say that {H(PQul),H(PQu2) ..... H(PQut)} is the canonical decomposition of H(P) if (i) uiEN6(u*) (i = 1,2 . . . . . t), (ii) H(PQui) ¢ ~3 (i = 1,2 ..... t), (iii) H(P) = UI=I H(PQui). Observe that if [H(P)[ ~ 1 then no such ui exist, for if t = 1 then the maximality of Q is contradicted. In this case the canonical decomposition is said to be trivial. Set Vi = H(PQui); in particular, then {V1, V2. . . . . Vt} may be viewed as a partition of V(T(G;P)). Write t
Vi = U vj,
Fi =Er(Vi, V/),
ni = IV//[ and
q; =qr(Vi, Vi) ( l ~ i ~ < t ) .
j=l
j¢i
Now Lemma 1 provides key information on the structure of T.
2. Main theorem Lenuna 1. (1) Suppose that e E F / n Fj (1 <~i,j<~t, i ¢ j). Then e = h[uj]h[ui] for
some h[uj] E Vi and h[ui] E Vj. Furthermore, u* (the last vertex of PQ) effects the transform from h[ui] to h[uj]. (2) The edges of Fi (1 <~i<<.t) are independent. (3) For all uEV(G), u ~ v, suppose that IH(P;u)l<.k. Then IF~l~k (l~
(1) Suppose that eEFz. NFj. Then e joins a vertex h[x] of V/to a vertex h[y]
of Vj for some x, yE V(G)\{v}. From the definitions of V/and ~ we have h[x]:
P Q u i " " x,
h[y]:
PQuj . . . y.
So u* effects the transform and therefore x = uj, y = ui. (2) Suppose that eEF/. Then for some j (l<.j<.t, i C j), eEF/fqFj. By (1), e = h[uj]h[ui] for some h[uj] E ~ and h[ui] E Vj. Since u* effects the transform the vertex of Is/,-adjacent to h[uj] is uniquely determined. Hence any vertex of V/ is adjacent to at most one vertex of V~. The same argument proves that any vertex of Vi is adjacent to at most one vertex of Vi. (3) From (1) any vertex of ~ adjacent to a vertex of Is//belongs to H(P; ui). Since [H(P; ui)[ <<.kit follows from (2) that IF,[ ~k. []
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
127
Lemma 2. Let G be a 9raph. Let T = T(G;P) (the transform of G at P). Suppose
that nr >11. Then qT <~nr(1 + (log nr)/2).
(1)
Proof. Write n = nr and q = qr. The proof is by induction on n. Since the girth ),(T) is at least 6 (see Lemma 5(2) below) the result is easily verified for n~< 10. So now suppose that n/> 11 and consider the non-trivial canonical decomposition of H(P). Recall that ni = IV/I, q~ =qr(Vi, Vii) and, since the canonical decomposition is nontrivial, t t> 2. Suppose that (2)
1 <~nl <~n2... <~nt. Using Lemma 1(2) and (2), l
t--I
t
EI~I=2E E i=l
t-I
q(V/, Vj)~<2 ~ ni = 2(n - nt).
i=l j=i+l
(3)
i=l
From the definition of a non-trivial canonical decomposition of H(P), l
t
q = E q , + ½E IF~]• i=1
(4)
i=1
From Lemma 1(2) and Eqs. (3) and (4), t
q <<.~ qi + min{n/2, n - nt}.
(5)
i=1
Firstly, suppose that 2nt <<.n.
(6)
From (1) (using induction), (5) and (6), t
q <~~ ni(1 + (log ni)/2) + n/2 i=1
= ( 3 n ) / 2 + 1 (i=~lnilogni)
~<(3n)/2 + l(n log(n/2)) = n(1 + l(logn)).
(7)
Finally, suppose that 2nt > n.
(8)
From (1) (using induction) and (5) t
q <~~ ni(1 q- ½(log ni)) + n - nt. i=1
(9)
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128
From (9) and the note at the end of this proof
g<~2n + ~1 ( n l o g n - n ) = 2n + ½ ( n l o g n - 2n) = n(1 + ½(logn)). The induction is now complete.
[]
Note • Let n >t 11. Let p = (n l, n2 . . . . . n t ) (t >~2) be a partition o f n with 1 <~n l <~n2 <~ "'" <~nt-1 ~<(n/2) < nt. Then l
ni logni -- 2nt <~n logn -- 2n.
(10)
i=l
Proof. Set l
g(p ) = ~
ni
log n i --
2nt.
i=l
Suppose that for some i, 1 <<.i<~t- 1, n i ~ 2 . Let p+ = (nl,n2 . . . . . n i - 1,ni+l . . . . . nt-1, n, + 1 ). Write x = ~ I - ~ ni and f ( x ) = g(p+) - 9(P) = (x - 1 ) log(x - 1 ) - x logx + ( n - x + 1 ) l o g ( n - x + 1 ) - (n - x ) l o g ( n - x ) 2. Now consider f to be a real-valued function over the interval 2<~x<~(n/2). Then f is strictly decreasing. Therefore as x increases 9 strictly decreases, until f changes, if it does change, from positive to negative; in which case 9 strictly increases. Furthermore, for all partitions p such that x = 1 or x = ~n/2J, (10) is satisfied. The proof is complete. [] T h e o r e m 3 . Let G be a 9raph. G E U ( <~k). Then
Set T = T ( G ; P ) .
qr ~<(nr(4 + log k))/2 - k.
Suppose
that nr>>.k>>.l and
(11)
Proof. Very slight modifications are needed to the notation used in Lemma 2. Again we set n = nr and q = qr and proceed by induction on n. As in Lemma 2 we may assume that n I> 11. When n = k, (11 ) follows immediately from Lemma 2. So now assume that n > k and consider, as in Lemma 2, the non-trivial canonical decomposition o f H ( P ) . Suppose that (see (2) o f Lemma 2)
1 ~ n l <~n2~ . " <~ns<~k < ns+l <~ns+2~ "'" <~nt
(12)
(in the extreme cases: (i) ni ~ k, 1 <~i<<.t, we set s = 0 and delete the relevant inequalities).
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
129
From the definition of a non-trivial canonical decomposition of H(P): t
l
|
q = ~ q i 4- ~ ~ [fi]. i=1
(13)
i=1
Set l
o= I E IFil i=1
Since G E U (<~k) from Lemma 1(2), (3) and Eq. (12)
0<. l
ni
+½((t-s)k).
(14)
From Lemma 2, (l 1) (using induction) and (13) l
q <. ~ ni(1 + (logni)/2) + ~ i= 1
(nj(2 + (logk)/2) - k) + 0
(15)
j=s +1
= (n(4 + log k)/2) - k - e, where s
= (t - s - 1)k - 0 + l ~ ni(2 + logk - log ni).
(16)
/=1
So from (15) the result follows if e~>O. From (14) and (16),
g>~ ½(t - s-- 2)k + il k n i ( 1 + l o g k
logni).
(17)
i=I
Therefore, using (12) and (17), e>~0 when t - s>~2. Now suppose that t - s -- 1. Then, using exactly the same inequality as in (3) of Lemma 2, 0 4 ~ ni.
(18)
i=1
From (12), (16) and (18), e>~0. Finally, we may assume that
t-s=O.
(19)
Again using the same arguments as in (3) and (5) of Lemma 2 (or directly using Lemma 1(2))
0 <~min{n/2,n - ns}.
(20)
Firstly, suppose that
n>~2k.
(21)
130
B.
Jackson, £ Sheehan / Discrete Mathematics 177 (1997) 123-144
From (12), (16), (19)-(21) $
e >i 1 ~ n i ( 1 + l o g k - l o g n i ) - k i=1
>/ ½((1 + log k)n - (logk)n) - k />0. So now suppose that n < 2k.
(22)
Firstly, assume that n ns > -. (23) 2 From (12), (16), (19), (20), (23) and the note immediately following Lemma 2,
> / - k + ½(logk)n- ½ En, logni-2n, i=1
~> - k + ½(logk)n - ½(n logn - 2n) />0. This final inequality follows, using elementary calculus, from (22) and since k<.Nn. Finally, we may assume that
ns
n
(24)
From (12), (16), (19), (20), (22) and (24), s
/> - k + ½n(1 + logk) - 1 ~ ni log ni i=1
1
/> - k + ½n(1 + l o g k ) - inlog
n
~>0. This completes the proof. Example.
qr~
[]
Ks+ 1 E U (-~- k ) where k = (s - 1)!(s~>2). Using Stirling's formula
1(
(log nT)nT \ log log(nr) J
whereas the upper bound of Theorem 3 is asymptotically (log nr)nr/2 log2. Here natural logarithms are used rather than our convention of logarithms to the base 2. Theorem 3 does of course provide information on the density of edges in the graph G itself; if h[x]EH(P;x) and T---- T(G;P) then degr(h[x])= qc(x, V ( G ) \ V ( P ) ) - 1, (x E V(G), x # v). For example:
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
131
Corollary 4. Let G be a graph such that G E U ( = k)(k >>.1). Then, for some v E
V(G), qc~< ½((5 + logk)nG - (logk + 7 - degGv)). Proof. Let vE V(G) be such that
IH(v;u)l
= k for all uE V(G)(u ~ v). Then setting
r = T(G; v), 2qr=
~
degr(h(u))=k
h(u)EH(v,u)
u~v
~
(degcu--1)
uE V(G)
u~v
From (25) and Theorem 3, 1
qG = ~ q r + ½(no + degav - 1) <~ ½((5 + l o g k ) n a - (logk + 7 - degav)).
(26) []
In particular, when k --- 2 we have
qG<~3nG -- ½(8 -- degGv ). We show how this bound may be improved in the next section (see Corollary 10).
3. The case k = 2
Let G be a graph and v, x E V(G). Let P be a fixed path in G with first vertex v. Recall that elements of H ( P ; x ) are labelled h[x] (or hi[x] where i is some positive integer), i.e. the notation draws attention to the last vertex x in the path. L e m m a 5. Let G be a graph. Set T = T(G;P).
(1) Let hl[x],h2[x] be distinct elements of H(P;x). dr(hl[x],h2[x]), between hi[x] and h2[x] in T is at least 3. (2) The girth, y(T), of T is at least 6.
Then
the distance,
Proof. (1) Suppose that h[x] EH(P;x). By definition of a transform, if w effects the transform of h[x] into h[y] for some y E V(G) then y ~ x. So dr(hl[x],h2[x])>~2. Suppose that dr(hl[x],h2[x]) = 2. Then there exists y E V(G) such that dr(h[y], hi[x]) = 1 (i = 1,2). Suppose now that wi effects the transform from h[y] to hi[x]. Then, by the definition of a transform, wi immediately precedes x in h[y], i --- 1,2. This means that Wl = w2 and hence hi[x] = h2[x] which is not the case. The proof is complete.
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
132
(2) Let Cm be a cycle of length m in T. Set Cm = (hl[xl],h2[x2] . . . . . hm[xm]) where hi[xi] EH(P;xi), xi E V(G), i = 1,2,...,m. Choose P* so that PP* is the maximal path common to all hi[xi], i = 1,2 ..... m. Let u* be the last vertex of PP*. We use this choice of P* and u* again in Lemma 6 and Result A.1. By the choice of P*, u* effects a transform in the cycle at least once; without loss of generality, assume u* effects the transform from hm[xm] to hi[x1]. Because Cm is a cycle h][xl] is also transformed by a sequence of m - 1 transforms into hm[xm]. So u* must again effect one of these transforms since hl[xl] begins PP*xm and hm[xm] begins PP*xl. Choose the smallest i (1 ~
:
PP*xm " " x l
hi[xi] : hi+l[xi+l] :
P P * x m " " xi PP*xi "'" Xm
hm[xm]
PP*x] "'" Xm
:
Notice here that since u* effects the transform from hi[xi] to hi+l[xi+l], Xi+l = Xm. Since i ~ {1,m - 1}, from Lemma 5(1), dr(hi+l[xi+l],hm[xm])>~3 and the proof is complete. [] Comment. A referee points out that, using the techniques of Result A.1 below, the assumption G E U ( ~<2) in Lemma 5 is unnecessary.
Lemma 6. Let G be a graph and suppose that G E U (~<2). Set T = T( G; P). Suppose
that T ~ C6. Then there exists X = {xl,x2,x3} c_ V(G) such that G has the spanning subgraph indicated in Fig. 1. Moreover, (i) the 6-cycle C6 in T is C6 = (h 1[Xl ], h2 Ix2], h3 Ix3 ], h4[xl ], h5 Ix2], h6 [x3 ])
for some hi[xi],hi+3[xi]En(e;xi), i= 1,2,3, (ii) q(X,(V(P*)\{u*}) U {w}) = 0, where P * = compv(P ), and w and u* are the last vertices of P and PP*, respectively.
V
X1 LI~X~
W P
)
4" N,
p,
Fig. 1. Spanning subgraph of G (Lemma 6).
)
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
1! "
n tx2) . L/
:I
!
V I
'~ " --,.,.~j V3
V:
Fig. 2. T ~
C6
!
:1
,3"/7 h (x3)
133
(Lemma 6).
Proof. Suppose that T ~ C6 = (h I [x1],h2[x2] . . . . . h6[x6]) for some hi[xi] EH(P;xi), i = 1,2 . . . . . 6. Set P* = c o m P v ( P ) and let u* be the last vertex of PP* (P* plays exactly the same role as P* in Lemma 5(2)). Suppose that the canonical decomposition of H(P) is
{H(PP*xl),H(PP*x2) ..... H(PP*xt)}
(t >~2).
As usual set V~= H(PP*xi) (i = 1,2 . . . . . t). Since T ~ C6 the degree of each vertex of T is 2 and since, by Lemma 1(2), the edges of F~ = Er(V/, Vi) are independent it follows that IV,l>~2 ( i = 1,2 . . . . . t). Now suppose that t = 2 . Since T-~C6 and G E U (~<2), by Lemma 1(3) IF~I = IF21--2. Hence by Lemma 1(1) there exist exactly two elements from H(P;xi) in V,- ( i = 1,2). So by Lemma 5(1), IV,I =3 (i= 1,2) and qr(V~, VD~< 1. Hence, qr~<4 which is not the case. It follows that t - - 3 and [V/] = 2 ( i = 1,2,3). Again using Lemmas 1 and 5 in exactly the same way we find (see Fig. 2) that C6 = (h I [Xl ], h 2 Ix2], h 3Ix3], h 4 Ix4], h 5 Ix5], h 6 Ix6] ), where hi[xi] e H(P;xi) and V1 = {h5[x2],h6[x31}, h2[x2]}. We have hl[xl]:
PP*x3""Xl,
h2 [x2]:
PP*x3...x2
V2 = {h4[xll, h3[x3]},
[:3 = { h l [ X l ] ,
and so on. Let ~,fl be ordered subpaths of hl[xl] such that h l[xl]:
PP*x3~x2flxl.
Then, following through the sequence of transforms in the cycle h 1[Xl]:
PP*x3~x2flXl,
h2[x2]:
PP*x3~xl fix2,
C6, w e
have
B. Jackson, J. Sheehan ~Discrete Mathematics 177 (1997) 123-144
134
x! v n
u r ,'~ u. e"
PQui
'
:>
<~
' Pi
x3
Fig. 3. Lemma 8 (Proof).
h3 [x3]:
PP*x2~xl~x3,
h4[xl]:
PP*x2~x3~xl,
hS[x2]:
PP*Xl~X3~X2,
h6[x3]: ee*xl~x2~x3,
hl[xl]:
PP*xa~x2~xl.
It follows that ~ and/~ are both trivial and G contains the spanning subgraph of Fig. 1. Furthermore, if q(X, (V(P*)\{u*})U { w } ) ~ 0 then using such an edge would yield another element of H(P) different from the vertices of the cycle, which is impossible. Lenuna 7 (Hendry [6]). Let G be a graph and P a path in G with first vertex v such that [H(P; u)[ ~< 1 for all u E V(G) (u ~ v). Then T(G; P) is acyclic. Proof. The proof is contained in [6]. It is implicitly proved in Lemma 5(2).
[]
Notation (Lemmas 8 and Theorem 9). We return to the notation of the introduction. Set T = T ( G ; P ) and Q=compr(P ). Let
(H(PQul),H(PQu2) ..... H(PQur))
(t~>2)
be the (non-trivial) canonical decomposition of H(P) where u is the last vertex of PQ. Now set T~.=T(G, PQui) and Pi*=compr(PQui) (see Fig. 3). Let u* be the last vertex of PQuiPz.*. Lemma 8. Let G be a graph and suppose that G E U (~<2). Suppose that for some i (1 <~i<~t) in the (non-trivial) canonical decomposition ofH(P), Ti ~- C6. Then T is monocyclic. Proof. Set X = {Xl,X2,X3} C V(G) and U=(V(Pi*)\{u*})U {ui}. Then, by Lemma 6 (see Figs. 1 and 3), G has a spanning subgraph as shown in Fig. 3, where
q(X, U) : 0.
(27)
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
135
Now assume that q ( u , X ) = 0, u being the last vertex of PQ. Then, from (27), u* is a cut-vertex of G\(V(PQ)\{u}) which implies that T --- T~ which is not the case since t>~2. Hence, q(u,X)>~l. If q(u,X)>~2 then IH(P;ui)l>~4. Hence,
q(u,X)=l.
(28)
Without loss of generality, we may suppose that UXl E E(G) (see Fig. 3) and that
ux2, ux3 ([ E( G). We now show that t = 2 and the canonical decomposition of H(P) is
(H(PQui),H(PQxl ) ).
(29)
Assume that H ( P Q y ) ¢ O for some y ENG(u)\{ui,xl}. Then, from (28), y E V(P/*) (y~:ui). Choose any h[z] E H(PQy) (z E V(G)). Then since G E U (~<2) and [H(PQui; x j ) l = 2 (l~
(H(PQxlx2 ), H(PQxlX3 ) ).
(30)
Again from (27)
H ( PQxl x2 ) = H (PQx lX2X3U*),
(31)
H(PQxlx3 ) = H(PQxlX3X2U* ).
(32)
By Lemma 1(1) any edge between H(PQxlx2) and H(PQxlx3) would be of the form h[x3]h[x2], where h[xj] E H(P;xj) ( j = 2 , 3 ) . But IH(PQui;xj)l = 2 and hence, since G E U (~<2), from (29), IH(PQxl;xj)I = 0 , j = 2 , 3 . This contradiction proves that no such edges exist. Suppose that x E V(G) and h[x] E H(PQxlx2;x). Then, from (31), h[x] is of the form h[x]:
PQxlx2x3u* . . . x
and so each such h[x] is associated with an element h I [x] of the form h 1[x]:
PQxlX3X2U*... x
which belongs to H(PQxlx3;x) and conversely. Since G E U (~<2) this implies IH(PQxlXfix)l ~< 1 for all xE V(G) ( j = 2 , 3 ) . Therefore, by Lemma 7,
T(G;PQxlXj) is acyclic.
( j = 2,3)
(33)
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
136
T(G; PQxj x3) S/'~
I
/~N
\/
',, %
"',
\
/
/
T(G; PQx1x2)
T(G; PQui)
", \,2,
...3 ,:
--..
-Tx;),:
/
\
1
~h(
1 I I
Fig. 4. T(G;P) (Lemma 8 (Proof)).
H
Fig. 5.
Now notice (see Fig. 3) that there exist h 1[ui] E H(PQxlx2; Ui) and h 2[ui] E H(PQxj x3; ui) given by hl[ui]:
PQxlx2x3 Pi* ui,
(34)
h2[ui]:
PQxlx3x2 Pi* ui,
(35)
where in hl[ui], u (the last vertex of PQ) effects a transform into hi[x1] and in h2[ui], u effects a transform into h2[xl], for some hi[x]] E H(PQui) ( i = 1,2). These edges h 1[ui]h i Ix] ] and h2[ui]h 2Ix1] between H(PQxl ) and H(PQ ui) are, since G c U ( ~<2) and using Lemma 1(3), the only such edges; one edge exactly joining an element of H(PQx]x2) to an element of H(PQui) and one edge exactly joining an element of H(PQx]x3) to an element of H(PQui) and these edges are independent. The situation is pictured in Fig. 4. Since T(G;PQui) is monocyclic and T(G;PQx]xj) ( j = 1,2) is acyclic these final remarks show that T is monocyclic. [] Theorem 9. Let G be a graph and P some fixed path in G. Suppose that G E U (~<2). l f T(G;P) is non-empty then either T(G;P)E{P],P2,P3, C6} or
qr ~<45-nr- 2.
(36)
Proof. In the appendix (Result A.1) it is proved that T ~ C7 and T T H (see Fig. 5). So, using Lemma 5(2), the theorem is true for nr<~7.
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
137
Now assume that nr>~8 and proceed by induction on nr. Firstly, set n = n r and q = qr. In the (non-trivial) canonical decomposition of H ( P ) set V/= H(PQui), ni = I vii and qi = q(V/, V/), Ti = T(G;PQui), i = 1,2 . . . . . t. If T/~ C6 for some i (1 <~i<<,t) then, by Lemma 8, T is monocyclic and, since n>~8, (36) is satisfied. Therefore, suppose that Ti ~ C6, i = 1,2 . . . . . t. For j (1 ~
q <~ ~ q i + ½ ( 2 t - a l ) i=1
5n ) ~< ~ - - - 2 t
=
(54 ) ---2
3
1
1
1
+ ~a| + ~a2+~a3W~(2t-al) -(t-2)+~al+~a2+~a3
~< (~4-~ - 2 ) - (t - 4)/2.
(37)
(38)
From (38) the induction is complete if t~>4. Therefore we may assume that 2~
(39)
Then, from (37), the induction is complete if ~>0. Suppose that t = 3. Then al + a 2 + a 3 43.
(40)
If a2~~0. If a 2 ~ 3 then n = 6 which is false. If a 2 = 2 and al + a3 = 1 then n ~<7 which is false. From (40) the remaining possibility is a2 = 2 and al +a3 = 0 in which case, from (39), e~>0. Now suppose that t = 2. Then al+a2+a3~<2.
(41)
Equality cannot hold in (41) otherwise n~<6. From (39) we may assume that al+a2+a3=l
(42)
otherwise most certainly e/> 0. Therefore from (42) and since n >~8 we may assume, without loss of generality, that nl ~<3 and n2>~5.
138
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
Hence, since G E U (~<2), from Lemmas 1, 5(1), (36), (42) and induction
) )
q <~ ql + q2 + l
<~
- 4
=
- 2
3
+ 1 + -~al + ~a2 + -~a3 , l - 1 + ~al + ~a2 + -~a3
5n <~-~- - 2 .
The induction is now complete.
[]
Comment. The wheel W4 with four spokes (and the hub as distinguished vertex) is in U ( = 2 ) and satisfies n r = 8 , q r = 8 . In general, ~ E U ( = 2 ) and n r = q r = 2 n so n = 4 is the only instance for the wheel where the bound of Theorem 9 is exact. Corollary 10. Let G E U (=2), G ~ K 4 . Then, f o r some vE V(G),
qG ~<¼(7no + 2 deg 6 v - 11 ). Proof. We have nr = 2(nG - 1). Now use Eq. (26) and Theorem 9. Hendry [6,8] shows that if GE U (= 1) then q~ = 3(n~ - 1). The situation is much more complicated for U (~>k) (k > 1). We discuss the general case in the final part of this paper.
4. Lower bounds and a construction
The graphs U (= 1) have been studied extensively in [5-8]. An example of a (infinite) family of such graphs, each with distinguished vertex v is shown in Fig. 6. a e b a~ e b
a e b
V V
a e b
V
V
.......
V
Fig. 6.
V
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
a
b
a
139
b
K
K* Fig. 7.
In this family of graphs e = ab is a 'forced edge'; every hamiltonian path in H(v) must include the edge e. Generally, elements of U (= 1) always contain forced edges. Now consider the graph K E U ( = 2 ) shown in Fig. 7 with distinguished vertex v*. By deleting the edge e of, say, the second graph in Fig. 6, deleting v* of K and identifying a and b we obtain the graph K* in Fig. 7. It is easy to check that K* E U (= 2). In this construction any element K E U (= 2) with distinguished vertex v* may be used provided degg(v*)= 2. In the same way any element of U (= k), using the forced edge construction, is associated with an element of U (= 2k). Let s be a non-negative integer. Then, [5,7], it is easy to construct arbitrarily large (in terms of vertices) elements of U (= 1) with at least s forced edges; now perform the construction simultaneously at s of these edges to obtain an element of U (= 2s). Hence there exist elements G of U (= 2s) which have asymptotically 3nG/2 edges. The best known lower bound is: Result 11 (Chartrand and qc >~¼(5no - 1)(no >/4).
Nordhaus
[3]).
Let
GEU(>>.k)(k>~I).
Then
Proof. Let G E U (~>k) (k>~l). Then no vertex, other than v, of G has more than one neighbour of degree exactly two. [] Comment 1. The family of graphs G constructed in [4] belong to U (>/1) and have [(5nG - 1)/41 edges (including a 'large' number of forced edges). Using the forced edge construction shows that the bound of Result 11 is asymptotically sharp. Conunent 2 (A generalization of U ( = k ) ) . The set U* ( = k ) (k>~l) is defined as follows. G E U* ( = k) if and only if for all u, v E V(G) (u ~ v) there exist exactly k uv-hamiltonian paths. Clearly, G E U* ( = k ) if and only if G E U ( = k ) and every vertex of G is distinguished.
140
13. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
Fig. 8.
Result 12. Let G E U* ( = k ) (k~>l). Then (1) G contains exactly lk(deg v) hamiltonian circuits for each v E V(G), (2) G is regular of degree 2s/k, where s is the number of hamiltonian circuits of G, (3) each ed#e of G belongs to exactly k hamiltonian circuits, (4) n6 is odd. Proof. (1)-(3) follow immediately from the definition of U* (--k). Part (4) is a consequence of Eq. (25). [] Example (i) Suppose that G E U* (=1). Then G E U ( = 1 ) and from [6], qa = 3(nG - 1). Hence, G has a vertex of degree 2 and, since G is regular, G TM/£3. (ii) Suppose that G E U* (= 2) and G is regular of degree 3. Then G must have property P(2) where: P(2):
G contains exactly 3 hamiltonian circuits; each edge belongs to exactly two of these circuits.
Notice however that the 3-regular graphs in Fig. 8 all belong to U (= 2) and have property P(2) but only the first two belong to U* (=2). Incidentally, the second condition of P(2) is not guaranteed by Smith's theorem [10] which allows the possibility that an edge belongs to no hamiltonian circuit. (iii) We conjecture that U* (= 3) = ~b. Suppose that G E U* (= 3). Then, from Result 12, G is regular of even degree so the smallest (in terms of degree) such graph G is regular of degree 4. Let P(3) be the property P(3):
G has exactly 6 hamiltonian circuits; each edge belongs to exactly three of these circuits; G has an odd number of vertices.
From Result 12 if G E U* (=3) and G is regular of degree 4 then G has P(3). Do there exist any regular graphs of degree 4 which have property P(3)?
141
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
Final comment. We end with conjectures (which, more modestly, should perhaps be only questions). Conjecture (Strong). Suppose that k ( > 1) is an odd integer. Then U ( = k ) = 0. Conjecture (Weak). Suppose that p is an odd prime. Then U ( = p ) = 0.
Comment. Many obvious conjectures/questions arise concerning the possible structures of transform graphs. As an indication of such we wildly conjecture: Conjecture. T(G;P) ~ C2k+l (k~> 1).
Acknowledgements We would like to thank Michael Sheard for many helpful comments.
Appendix A Result A.1. Let G be a graph and P a fixed path in G (with first vertex v). Then T(G;P) ~ C7 and T(G;P) ~ H (see Fig. 5).
Proof. Set P* = c o m p r ( P ) and W = PP*. Suppose that u is the last vertex of W and that the canonical decomposition of H(P) is (H(Wul),H(Wu2) ..... H(Wut))
(t ~>2).
Set T = T(G;P), Vi = H(Wui), qi = q(Vi, Vi) and ni = I/I/I (i = 1,2 . . . . . t) with
nl <~n2 <~ .'. <.nt.
[]
Case 1 (C7): Suppose that T --- C7. Then from L e m m a 1(2), nl >~2. If t = 2 then 2~~2; from Lemmas 1(1) and 5(1) it follows that if nl = 2 then ql = 0 and if nl = 3 then ql ~< 1 and in either case using L e m m a 1(2), T ~ C7. It follows that t = 3 and nl = n2 = 2, n3 = 3. Using Lemmas 1 and 5(1) in the same way the structure of T must be, up to a relabelling, as shown in Fig. 9 where h 1[ul] E H(P; ul )M V3, h3[u2] E H(P; u2)M V3, ha[u3] E H(P; U3)('] V2, hS[ul] E H ( P ; u l ) N V2, h6[u2] E H(P;u2)fq V1, hT[u3] E H(P;u3)N V1, h2[x] E H(P;x) A V3 for some x E V(G) and the cycle is C7 = ( h 1 [u I ], h 2 Ix], h 3 [u 2 ], h 4 [/d3 ], h 5 [u 1], h 6 [u2 ], h 7 [u3 ] ). Now for some (ordered) subpaths if, fi and 7 we have either h 1[ul]:
Wu3~u2~xyul
(A.1)
142
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
g 6
%
5
t h (u2)/~'-'--'---"'~ I
,
T':
'
I I ~, h 7(1"13)
'
I ~
tl
." h3(u2)
t
,"T '
t
h (uO ~
I e"' I / ~ " ~
t ~
't~ h4(t13) I ! k
Vl
V2
/ V3
Fig. 9. T(G;P) ~- C7 (Result A.I). or hl[ul]:
Wu3 ~c flu2 ~u l .
(A.2)
In the first case, following through the sequence of transforms in the cycle C7, yields
h 1 [ul] :
Wu3~u2~x7ul,
h 2 Ix]
:
Wu3~u2flul~x,
h3[u2] :
Wu3~ ful flu2,
h4[u3].•
rVu~~u~~x~u3,
hS[ul]:
Wu2flu3Y~ul ,
h6[u2] :
WU 1 ])xO~u3 flu2,
hT[u3] :
Wu~~x'du~~u~,
h I [Ul] :
Wu3 flU2~Of~Ul.
So &"and/~ are both trivial. Therefore, u 2 is adjacent to x in G and using x to effect a transform of h3[u2] yields an element of H ( P ) which is not a vertex of the cycle. This is impossible. In the second case, following through the sequence of transforms in C7, yields
hi[u1]:
rVu3~u2~ul,
h:[x]:
Wu3~u~~u2flx,
h3[u2] :
Wu3~ul 7'-xflu2,
ha[u3] :
Wu2 flX~Ul ~u3,
hS[ul]:
Wu2flxTu3~ul,
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
h2(u2)
f\
I i
, h6(u3)
143
!I I
v
t ~,
h3(u3) -
~,
/ ;
hl(ul)
I
h(x)
I I I ! |
l
I
!
,s k
\
t d
V1
V2
V3 Fig. 10. T(G;P)~-H (Result A.1).
h6[u2] :
WUl ~u3~x~u2,
h7[u3] :
WUl ~u2flx~u3,
h l[ul]:
Wu3yx~u2~ul.
So 8 and ~ are both trival. Therefore, ul is adjacent to u2 in G and using u2 to effect a transform of hS[ul] yields an element of H(P) which is not a vertex of the cycle. This is impossible. Therefore, T ~ C7. Case 2 (T -~ H): Now suppose that T = H. Using Lemmas 1 and 5(1) in exactly the same way the structure of T must be, up to a relabelling, as shown in Fig. 10 where V1 = {h5[u2],h6[u3]}, V 2 = {h3[u3],h4[ul]} and //3 = {hl[ul],h2[uz],h[x]} where h[x] E H(P;x) for some x E V(G) and h[x]hl[Ul] E E(T). The 6-cycle in H is (hl[ul],h2[u2],h3[u3],h4[ul],hS[uz],h6[u3]) where for some (ordered) subpaths and /3, h I [Ul]:
Wu3~u2flUl.
Following through the sequence of transforms in the cycle yields h 1 [Ul]:
Wu3~u2~ul,
h2[u2] :
Wu3~ulflU2,
h3[u3] :
Wu2~ul~u3,
h4[Ul]"
Wu2flu3~ul,
hS[u2]:
Wul ~du3 3u2,
h6[u3]"
WUl ~u2/~u3,
hl[ul] :
Wu3~u2~u I .
B. Jackson, J SheehanIDiscrete
144
Mathematics 177 (1997)
123-144
So Z and fi are both trivial. Since h[x] is adjacent to h’[ul] in T there exists a transform of h’[ul]
to h[x]. However,
the only transforms
and h[x] is equal to neither. This contradiction
of h’[ul]
in T are h~[u~] and he[u3]
shows that T 9 H.
0
References [I] J. Beck, On size Ramsey numbers of paths, trees and cycles, I, J. Graph Theory 7 (1983) 115-129. [2] B. Bollobas, External Graph Theory (Academic Press, New York, 1978). [3] G. Chartrand and E. Nordhaus, Graphs hamiltonian-connected from a vertex, in: G. Chartrand et al., eds, The Theory and Applications of Graphs (Wiley, New York, 1981) 189-201. [4] C. Knickerbocker, P. Lock and M. Sheard, The minimum size of graphs hamiltonian-connected from a vertex, Discrete Math. 76 (1989) 277-278. [S] C. Knickerbocker, P. Lock and M. Sheard, On the structure of graphs uniquely hamiltonian-connected from a vertex, Discrete Math. 88 (1991) 33-48. [6] G. Hendry, Graphs uniquely hamiltonian-connected from a vertex, Discrete Math. 49 (1984) 61-74. [7] G. Hendry, Ph.D. Thesis, Aberdeen University, 1985. [8] G. Hendry, The size of graphs uniquely hamiltonian-connected from a vertex, Discrete Math. 61 (1986) 57-60. [9] L. P&a, Hamiltonian circuits in random graphs, Discrete Math. 14 (1976) 359-364. [IO] W. Tutte, On hamiltonian circuits, J. London Math. Sot. 21 (1946) 98-101.
Discrete Mathematics 177 (1997) 123-144
The structure of transform graphs Bill Jackson a, John Sheehan b'* aDepartment of Mathematics, Goldsmith's College, University of London, Lewisham Way, London SE14 6NW, UK bDepartment of Mathematical Sciences, University of Aberdeen, Dunbar Street, Aberdeen AB9 2TY, UK Received 27 January 1995; revised 29 May 1996 The inspiration and many of the ideas of this paper originated with George Hendry. This paper is dedicated to his memory
Abstract
The transform graph gives considerable insight into the hamiltonian path structure of a graph. In this paper the structure, and in particular the edge-density, of transform graphs is examined.
O. S u m m a r y o f contents
Let G be a finite, simple graph with vertex set V ( G ) and suppose that v is a fixed vertex o f G. Let H(v; u) be the set of hamiltonian paths h[u] with first vertex v and last vertex u. The transform graph T(G; v) o f G at v has as its vertices {h[x]: h [ x ] E H ( v ; x ) , x E V ( G ) , x # v}. Vertices h[x] and h[y] are adjacent if and only if h[x] can be 'transformed' into h[y]. This means that if
h[x]:
v'"wy..'x
then there exists a vertex w adjacent to x such that h[x] can be transformed into h[y], where h[y]:
v. . . w x . . . y
by simply reversing the last part o f the path to y. Let P be a fixed path o f G with first vertex v. Then T(G; v) can be generalized in an obvious way to T ( G ; P ) , where the hamiltonian paths all start with P. * Corresponding author. E-mail: [email protected]. 0012-365X/97/$17.00 Copyright (~) 1997 Elsevier Science B.V. All rights reserved PII S0012-365X(96)00360-3
124
B. Jackson. J. Sheehan/Discrete Mathematics 177 (1997) 123-144
The transform idea has proved very useful in examining the hamiltonian path/path structure of a graph. As far as we know it really originated with P6sa [9] in his paper on hamiltonian circuits. It is used in the proof of Beck's theorem [1] on the size Ramsey number of a path and again by Thomason [2, p. 146] in his parity path theorem generalizing Smith's theorem [10]. It features prominently in the work of Hendry [6-8] which is the starting point for this paper. Let U (= k) denote the set of graphs G such that there exists some fixed vertex v with In(v;u)l = k for all uE V(G) (u ~ v). U (<~k) and U (~>k) are defined in the obvious way. Hendry proved in [6,7] that if G E U ( = 1) then T(G;v) was acyclic; furthermore in [8] it is proved that if G E U ( = 1 ) then qG = (3nG - 3)/2 where nc and qG are, respectively, the number of vertices and edges of G. In this paper we examine the structure of T(G;P). The main results are:
Theorem 3. Let G be a graph. Let T = T(G;P). Suppose that nT>>.k>~l and G E U ( <~k ). Then qT ~<(nr(4 + log k))/2 - k (logarithms are to the base 2 unless stated otherwise).
Corollary 4. Let G E U ( <<.k) (k~>l). Then, for some vE V(G), 2qG ~<(5 + logk)nG -- (logk + 7 - deg G v).
Theorem 9. Let GE U (~<2). Then either T ( G ; P ) E {P1,P2,P3, C6} or qr <~5nr - 2 (P3 is the path o f length 2 and C6 the cycle o f length 6).
Corollary 10. Let G E U ( = 2). Then, for some vE V(G), qG ~<~(7n6 + 2 deg c v - 11 ). In the final section we discuss lower bounds and some related questions.
1. Introduction All graphs in this paper are undirected, simple and finite. A graph G has vertex set V(G) and edge set E(G). We set nG = IV(G)I and q6 = IE(G)] and say that G is an (n6,qG)-graph. The suffix, referring to a particular graph G, is dropped when no ambiguity arises. Let X, Y C V ( G ) . Set E 6 ( x , Y ) = {xy E E ( G ) : x EX, y E Y} and q 6 ( X , Y ) = lEe(X, Y)[. Let v E V(G). Then the degree, deg G v, is qG({v}, V(G)). Usually, if the
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
125
set {v} is a singleton then the brackets are not included, e.g. the neighbourhood NG(v) of v in G is the set of all vertices u such that q(u, v) = 1. Now fix some vertex v E V(G). Let u E V(G). A vu-path is a path in G with first vertex v and last vertex u. Let x, yE V(G)\{v} and suppose wE V(G) is a neighbour of both x and y. Suppose that h[x] is a vx-hamiltonian path:
h[x]:
v...wy...x.
A vx-hamiltonian path is labelled throughout as h[x] (or hi[x] where i is some positive integer), i.e. the notation draws attention to the last vertex in the path. Then the vyhamiltonian path h[y] obtained from h[x] by deleting wy, inserting wx and reversing the order of the xy-sequence in h[x]: h[y]:
v. . . wx. . . y
is called a transform of h[x]; w is said to effect the transform in this case. Notice that if w does effect a transform of h[x] then h[y] is determined uniquely. Clearly, if w effects the transform from h[x] to h[y] then it also effects the transform from h[y] to h[x]. This property allows us to define the (undirected) transform 9raph T(G;v) at v. Let H(v; u) denote the set of vu-hamiltonian paths (uE V(G), u 7~ v). Then
V(T(G;v)) = {h[u]: h[ulEH(v;u), uE V(G) (u ¢ v)} and vertices h[x] and h[y] are adjacent if and only if h[y] is a transform of h[x]. These concepts may all be generalized by replacing v by a fixed path P with first vertex v. For example, let P:
VlVl
• " " /)m
(Vl = v, m f> 1). Let H(P; u) denote the set of vu-hamiltonian paths (u E V(G), u ~ v) with initial segment P. So if h[u] E H ( P ; u) we have
h[u]:
vlv2...v,,...u.
Set
H(P) = {h[u]: h[ulEH(P,u), uE V(G), u ¢ v}. Again the transform graph T(G; P) at P has
V(T(G;P)) = H(P) and vertices h[x] and h[y] are adjacent if and only if h[y] is a transform of h[x]. Let U (~
126
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
The next definitions are vital in what follows. Firstly, choose the path Q in G such that
H(P) = H(PQ) and Q is maximal with respect to this property. Here PQ denotes the concatenation of P and Q. Q is said to be the completion of P in T ( = T(G;P)) and is denoted by compr(P ). Let u* be the last vertex of PQ. We say that {H(PQul),H(PQu2) ..... H(PQut)} is the canonical decomposition of H(P) if (i) uiEN6(u*) (i = 1,2 . . . . . t), (ii) H(PQui) ¢ ~3 (i = 1,2 ..... t), (iii) H(P) = UI=I H(PQui). Observe that if [H(P)[ ~ 1 then no such ui exist, for if t = 1 then the maximality of Q is contradicted. In this case the canonical decomposition is said to be trivial. Set Vi = H(PQui); in particular, then {V1, V2. . . . . Vt} may be viewed as a partition of V(T(G;P)). Write t
Vi = U vj,
Fi =Er(Vi, V/),
ni = IV//[ and
q; =qr(Vi, Vi) ( l ~ i ~ < t ) .
j=l
j¢i
Now Lemma 1 provides key information on the structure of T.
2. Main theorem Lenuna 1. (1) Suppose that e E F / n Fj (1 <~i,j<~t, i ¢ j). Then e = h[uj]h[ui] for
some h[uj] E Vi and h[ui] E Vj. Furthermore, u* (the last vertex of PQ) effects the transform from h[ui] to h[uj]. (2) The edges of Fi (1 <~i<<.t) are independent. (3) For all uEV(G), u ~ v, suppose that IH(P;u)l<.k. Then IF~l~k (l~
(1) Suppose that eEFz. NFj. Then e joins a vertex h[x] of V/to a vertex h[y]
of Vj for some x, yE V(G)\{v}. From the definitions of V/and ~ we have h[x]:
P Q u i " " x,
h[y]:
PQuj . . . y.
So u* effects the transform and therefore x = uj, y = ui. (2) Suppose that eEF/. Then for some j (l<.j<.t, i C j), eEF/fqFj. By (1), e = h[uj]h[ui] for some h[uj] E ~ and h[ui] E Vj. Since u* effects the transform the vertex of Is/,-adjacent to h[uj] is uniquely determined. Hence any vertex of V/ is adjacent to at most one vertex of V~. The same argument proves that any vertex of Vi is adjacent to at most one vertex of Vi. (3) From (1) any vertex of ~ adjacent to a vertex of Is//belongs to H(P; ui). Since [H(P; ui)[ <<.kit follows from (2) that IF,[ ~k. []
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
127
Lemma 2. Let G be a 9raph. Let T = T(G;P) (the transform of G at P). Suppose
that nr >11. Then qT <~nr(1 + (log nr)/2).
(1)
Proof. Write n = nr and q = qr. The proof is by induction on n. Since the girth ),(T) is at least 6 (see Lemma 5(2) below) the result is easily verified for n~< 10. So now suppose that n/> 11 and consider the non-trivial canonical decomposition of H(P). Recall that ni = IV/I, q~ =qr(Vi, Vii) and, since the canonical decomposition is nontrivial, t t> 2. Suppose that (2)
1 <~nl <~n2... <~nt. Using Lemma 1(2) and (2), l
t--I
t
EI~I=2E E i=l
t-I
q(V/, Vj)~<2 ~ ni = 2(n - nt).
i=l j=i+l
(3)
i=l
From the definition of a non-trivial canonical decomposition of H(P), l
t
q = E q , + ½E IF~]• i=1
(4)
i=1
From Lemma 1(2) and Eqs. (3) and (4), t
q <<.~ qi + min{n/2, n - nt}.
(5)
i=1
Firstly, suppose that 2nt <<.n.
(6)
From (1) (using induction), (5) and (6), t
q <~~ ni(1 + (log ni)/2) + n/2 i=1
= ( 3 n ) / 2 + 1 (i=~lnilogni)
~<(3n)/2 + l(n log(n/2)) = n(1 + l(logn)).
(7)
Finally, suppose that 2nt > n.
(8)
From (1) (using induction) and (5) t
q <~~ ni(1 q- ½(log ni)) + n - nt. i=1
(9)
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
128
From (9) and the note at the end of this proof
g<~2n + ~1 ( n l o g n - n ) = 2n + ½ ( n l o g n - 2n) = n(1 + ½(logn)). The induction is now complete.
[]
Note • Let n >t 11. Let p = (n l, n2 . . . . . n t ) (t >~2) be a partition o f n with 1 <~n l <~n2 <~ "'" <~nt-1 ~<(n/2) < nt. Then l
ni logni -- 2nt <~n logn -- 2n.
(10)
i=l
Proof. Set l
g(p ) = ~
ni
log n i --
2nt.
i=l
Suppose that for some i, 1 <<.i<~t- 1, n i ~ 2 . Let p+ = (nl,n2 . . . . . n i - 1,ni+l . . . . . nt-1, n, + 1 ). Write x = ~ I - ~ ni and f ( x ) = g(p+) - 9(P) = (x - 1 ) log(x - 1 ) - x logx + ( n - x + 1 ) l o g ( n - x + 1 ) - (n - x ) l o g ( n - x ) 2. Now consider f to be a real-valued function over the interval 2<~x<~(n/2). Then f is strictly decreasing. Therefore as x increases 9 strictly decreases, until f changes, if it does change, from positive to negative; in which case 9 strictly increases. Furthermore, for all partitions p such that x = 1 or x = ~n/2J, (10) is satisfied. The proof is complete. [] T h e o r e m 3 . Let G be a 9raph. G E U ( <~k). Then
Set T = T ( G ; P ) .
qr ~<(nr(4 + log k))/2 - k.
Suppose
that nr>>.k>>.l and
(11)
Proof. Very slight modifications are needed to the notation used in Lemma 2. Again we set n = nr and q = qr and proceed by induction on n. As in Lemma 2 we may assume that n I> 11. When n = k, (11 ) follows immediately from Lemma 2. So now assume that n > k and consider, as in Lemma 2, the non-trivial canonical decomposition o f H ( P ) . Suppose that (see (2) o f Lemma 2)
1 ~ n l <~n2~ . " <~ns<~k < ns+l <~ns+2~ "'" <~nt
(12)
(in the extreme cases: (i) ni ~
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
129
From the definition of a non-trivial canonical decomposition of H(P): t
l
|
q = ~ q i 4- ~ ~ [fi]. i=1
(13)
i=1
Set l
o= I E IFil i=1
Since G E U (<~k) from Lemma 1(2), (3) and Eq. (12)
0<. l
ni
+½((t-s)k).
(14)
From Lemma 2, (l 1) (using induction) and (13) l
q <. ~ ni(1 + (logni)/2) + ~ i= 1
(nj(2 + (logk)/2) - k) + 0
(15)
j=s +1
= (n(4 + log k)/2) - k - e, where s
= (t - s - 1)k - 0 + l ~ ni(2 + logk - log ni).
(16)
/=1
So from (15) the result follows if e~>O. From (14) and (16),
g>~ ½(t - s-- 2)k + il k n i ( 1 + l o g k
logni).
(17)
i=I
Therefore, using (12) and (17), e>~0 when t - s>~2. Now suppose that t - s -- 1. Then, using exactly the same inequality as in (3) of Lemma 2, 0 4 ~ ni.
(18)
i=1
From (12), (16) and (18), e>~0. Finally, we may assume that
t-s=O.
(19)
Again using the same arguments as in (3) and (5) of Lemma 2 (or directly using Lemma 1(2))
0 <~min{n/2,n - ns}.
(20)
Firstly, suppose that
n>~2k.
(21)
130
B.
Jackson, £ Sheehan / Discrete Mathematics 177 (1997) 123-144
From (12), (16), (19)-(21) $
e >i 1 ~ n i ( 1 + l o g k - l o g n i ) - k i=1
>/ ½((1 + log k)n - (logk)n) - k />0. So now suppose that n < 2k.
(22)
Firstly, assume that n ns > -. (23) 2 From (12), (16), (19), (20), (23) and the note immediately following Lemma 2,
> / - k + ½(logk)n- ½ En, logni-2n, i=1
~> - k + ½(logk)n - ½(n logn - 2n) />0. This final inequality follows, using elementary calculus, from (22) and since k<.Nn. Finally, we may assume that
ns
n
(24)
From (12), (16), (19), (20), (22) and (24), s
/> - k + ½n(1 + logk) - 1 ~ ni log ni i=1
1
/> - k + ½n(1 + l o g k ) - inlog
n
~>0. This completes the proof. Example.
qr~
[]
Ks+ 1 E U (-~- k ) where k = (s - 1)!(s~>2). Using Stirling's formula
1(
(log nT)nT \ log log(nr) J
whereas the upper bound of Theorem 3 is asymptotically (log nr)nr/2 log2. Here natural logarithms are used rather than our convention of logarithms to the base 2. Theorem 3 does of course provide information on the density of edges in the graph G itself; if h[x]EH(P;x) and T---- T(G;P) then degr(h[x])= qc(x, V ( G ) \ V ( P ) ) - 1, (x E V(G), x # v). For example:
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
131
Corollary 4. Let G be a graph such that G E U ( = k)(k >>.1). Then, for some v E
V(G), qc~< ½((5 + logk)nG - (logk + 7 - degGv)). Proof. Let vE V(G) be such that
IH(v;u)l
= k for all uE V(G)(u ~ v). Then setting
r = T(G; v), 2qr=
~
degr(h(u))=k
h(u)EH(v,u)
u~v
~
(degcu--1)
uE V(G)
u~v
From (25) and Theorem 3, 1
qG = ~ q r + ½(no + degav - 1) <~ ½((5 + l o g k ) n a - (logk + 7 - degav)).
(26) []
In particular, when k --- 2 we have
qG<~3nG -- ½(8 -- degGv ). We show how this bound may be improved in the next section (see Corollary 10).
3. The case k = 2
Let G be a graph and v, x E V(G). Let P be a fixed path in G with first vertex v. Recall that elements of H ( P ; x ) are labelled h[x] (or hi[x] where i is some positive integer), i.e. the notation draws attention to the last vertex x in the path. L e m m a 5. Let G be a graph. Set T = T(G;P).
(1) Let hl[x],h2[x] be distinct elements of H(P;x). dr(hl[x],h2[x]), between hi[x] and h2[x] in T is at least 3. (2) The girth, y(T), of T is at least 6.
Then
the distance,
Proof. (1) Suppose that h[x] EH(P;x). By definition of a transform, if w effects the transform of h[x] into h[y] for some y E V(G) then y ~ x. So dr(hl[x],h2[x])>~2. Suppose that dr(hl[x],h2[x]) = 2. Then there exists y E V(G) such that dr(h[y], hi[x]) = 1 (i = 1,2). Suppose now that wi effects the transform from h[y] to hi[x]. Then, by the definition of a transform, wi immediately precedes x in h[y], i --- 1,2. This means that Wl = w2 and hence hi[x] = h2[x] which is not the case. The proof is complete.
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
132
(2) Let Cm be a cycle of length m in T. Set Cm = (hl[xl],h2[x2] . . . . . hm[xm]) where hi[xi] EH(P;xi), xi E V(G), i = 1,2,...,m. Choose P* so that PP* is the maximal path common to all hi[xi], i = 1,2 ..... m. Let u* be the last vertex of PP*. We use this choice of P* and u* again in Lemma 6 and Result A.1. By the choice of P*, u* effects a transform in the cycle at least once; without loss of generality, assume u* effects the transform from hm[xm] to hi[x1]. Because Cm is a cycle h][xl] is also transformed by a sequence of m - 1 transforms into hm[xm]. So u* must again effect one of these transforms since hl[xl] begins PP*xm and hm[xm] begins PP*xl. Choose the smallest i (1 ~
:
PP*xm " " x l
hi[xi] : hi+l[xi+l] :
P P * x m " " xi PP*xi "'" Xm
hm[xm]
PP*x] "'" Xm
:
Notice here that since u* effects the transform from hi[xi] to hi+l[xi+l], Xi+l = Xm. Since i ~ {1,m - 1}, from Lemma 5(1), dr(hi+l[xi+l],hm[xm])>~3 and the proof is complete. [] Comment. A referee points out that, using the techniques of Result A.1 below, the assumption G E U ( ~<2) in Lemma 5 is unnecessary.
Lemma 6. Let G be a graph and suppose that G E U (~<2). Set T = T( G; P). Suppose
that T ~ C6. Then there exists X = {xl,x2,x3} c_ V(G) such that G has the spanning subgraph indicated in Fig. 1. Moreover, (i) the 6-cycle C6 in T is C6 = (h 1[Xl ], h2 Ix2], h3 Ix3 ], h4[xl ], h5 Ix2], h6 [x3 ])
for some hi[xi],hi+3[xi]En(e;xi), i= 1,2,3, (ii) q(X,(V(P*)\{u*}) U {w}) = 0, where P * = compv(P ), and w and u* are the last vertices of P and PP*, respectively.
V
X1 LI~X~
W P
)
4" N,
p,
Fig. 1. Spanning subgraph of G (Lemma 6).
)
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
1! "
n tx2) . L/
:I
!
V I
'~ " --,.,.~j V3
V:
Fig. 2. T ~
C6
!
:1
,3"/7 h (x3)
133
(Lemma 6).
Proof. Suppose that T ~ C6 = (h I [x1],h2[x2] . . . . . h6[x6]) for some hi[xi] EH(P;xi), i = 1,2 . . . . . 6. Set P* = c o m P v ( P ) and let u* be the last vertex of PP* (P* plays exactly the same role as P* in Lemma 5(2)). Suppose that the canonical decomposition of H(P) is
{H(PP*xl),H(PP*x2) ..... H(PP*xt)}
(t >~2).
As usual set V~= H(PP*xi) (i = 1,2 . . . . . t). Since T ~ C6 the degree of each vertex of T is 2 and since, by Lemma 1(2), the edges of F~ = Er(V/, Vi) are independent it follows that IV,l>~2 ( i = 1,2 . . . . . t). Now suppose that t = 2 . Since T-~C6 and G E U (~<2), by Lemma 1(3) IF~I = IF21--2. Hence by Lemma 1(1) there exist exactly two elements from H(P;xi) in V,- ( i = 1,2). So by Lemma 5(1), IV,I =3 (i= 1,2) and qr(V~, VD~< 1. Hence, qr~<4 which is not the case. It follows that t - - 3 and [V/] = 2 ( i = 1,2,3). Again using Lemmas 1 and 5 in exactly the same way we find (see Fig. 2) that C6 = (h I [Xl ], h 2 Ix2], h 3Ix3], h 4 Ix4], h 5 Ix5], h 6 Ix6] ), where hi[xi] e H(P;xi) and V1 = {h5[x2],h6[x31}, h2[x2]}. We have hl[xl]:
PP*x3""Xl,
h2 [x2]:
PP*x3...x2
V2 = {h4[xll, h3[x3]},
[:3 = { h l [ X l ] ,
and so on. Let ~,fl be ordered subpaths of hl[xl] such that h l[xl]:
PP*x3~x2flxl.
Then, following through the sequence of transforms in the cycle h 1[Xl]:
PP*x3~x2flXl,
h2[x2]:
PP*x3~xl fix2,
C6, w e
have
B. Jackson, J. Sheehan ~Discrete Mathematics 177 (1997) 123-144
134
x! v n
u r ,'~ u. e"
PQui
'
:>
<~
' Pi
x3
Fig. 3. Lemma 8 (Proof).
h3 [x3]:
PP*x2~xl~x3,
h4[xl]:
PP*x2~x3~xl,
hS[x2]:
PP*Xl~X3~X2,
h6[x3]: ee*xl~x2~x3,
hl[xl]:
PP*xa~x2~xl.
It follows that ~ and/~ are both trivial and G contains the spanning subgraph of Fig. 1. Furthermore, if q(X, (V(P*)\{u*})U { w } ) ~ 0 then using such an edge would yield another element of H(P) different from the vertices of the cycle, which is impossible. Lenuna 7 (Hendry [6]). Let G be a graph and P a path in G with first vertex v such that [H(P; u)[ ~< 1 for all u E V(G) (u ~ v). Then T(G; P) is acyclic. Proof. The proof is contained in [6]. It is implicitly proved in Lemma 5(2).
[]
Notation (Lemmas 8 and Theorem 9). We return to the notation of the introduction. Set T = T ( G ; P ) and Q=compr(P ). Let
(H(PQul),H(PQu2) ..... H(PQur))
(t~>2)
be the (non-trivial) canonical decomposition of H(P) where u is the last vertex of PQ. Now set T~.=T(G, PQui) and Pi*=compr(PQui) (see Fig. 3). Let u* be the last vertex of PQuiPz.*. Lemma 8. Let G be a graph and suppose that G E U (~<2). Suppose that for some i (1 <~i<~t) in the (non-trivial) canonical decomposition ofH(P), Ti ~- C6. Then T is monocyclic. Proof. Set X = {Xl,X2,X3} C V(G) and U=(V(Pi*)\{u*})U {ui}. Then, by Lemma 6 (see Figs. 1 and 3), G has a spanning subgraph as shown in Fig. 3, where
q(X, U) : 0.
(27)
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
135
Now assume that q ( u , X ) = 0, u being the last vertex of PQ. Then, from (27), u* is a cut-vertex of G\(V(PQ)\{u}) which implies that T --- T~ which is not the case since t>~2. Hence, q(u,X)>~l. If q(u,X)>~2 then IH(P;ui)l>~4. Hence,
q(u,X)=l.
(28)
Without loss of generality, we may suppose that UXl E E(G) (see Fig. 3) and that
ux2, ux3 ([ E( G). We now show that t = 2 and the canonical decomposition of H(P) is
(H(PQui),H(PQxl ) ).
(29)
Assume that H ( P Q y ) ¢ O for some y ENG(u)\{ui,xl}. Then, from (28), y E V(P/*) (y~:ui). Choose any h[z] E H(PQy) (z E V(G)). Then since G E U (~<2) and [H(PQui; x j ) l = 2 (l~
(H(PQxlx2 ), H(PQxlX3 ) ).
(30)
Again from (27)
H ( PQxl x2 ) = H (PQx lX2X3U*),
(31)
H(PQxlx3 ) = H(PQxlX3X2U* ).
(32)
By Lemma 1(1) any edge between H(PQxlx2) and H(PQxlx3) would be of the form h[x3]h[x2], where h[xj] E H(P;xj) ( j = 2 , 3 ) . But IH(PQui;xj)l = 2 and hence, since G E U (~<2), from (29), IH(PQxl;xj)I = 0 , j = 2 , 3 . This contradiction proves that no such edges exist. Suppose that x E V(G) and h[x] E H(PQxlx2;x). Then, from (31), h[x] is of the form h[x]:
PQxlx2x3u* . . . x
and so each such h[x] is associated with an element h I [x] of the form h 1[x]:
PQxlX3X2U*... x
which belongs to H(PQxlx3;x) and conversely. Since G E U (~<2) this implies IH(PQxlXfix)l ~< 1 for all xE V(G) ( j = 2 , 3 ) . Therefore, by Lemma 7,
T(G;PQxlXj) is acyclic.
( j = 2,3)
(33)
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
136
T(G; PQxj x3) S/'~
I
/~N
\/
',, %
"',
\
/
/
T(G; PQx1x2)
T(G; PQui)
", \,2,
...3 ,:
--..
-Tx;),:
/
\
1
~h(
1 I I
Fig. 4. T(G;P) (Lemma 8 (Proof)).
H
Fig. 5.
Now notice (see Fig. 3) that there exist h 1[ui] E H(PQxlx2; Ui) and h 2[ui] E H(PQxj x3; ui) given by hl[ui]:
PQxlx2x3 Pi* ui,
(34)
h2[ui]:
PQxlx3x2 Pi* ui,
(35)
where in hl[ui], u (the last vertex of PQ) effects a transform into hi[x1] and in h2[ui], u effects a transform into h2[xl], for some hi[x]] E H(PQui) ( i = 1,2). These edges h 1[ui]h i Ix] ] and h2[ui]h 2Ix1] between H(PQxl ) and H(PQ ui) are, since G c U ( ~<2) and using Lemma 1(3), the only such edges; one edge exactly joining an element of H(PQx]x2) to an element of H(PQui) and one edge exactly joining an element of H(PQx]x3) to an element of H(PQui) and these edges are independent. The situation is pictured in Fig. 4. Since T(G;PQui) is monocyclic and T(G;PQx]xj) ( j = 1,2) is acyclic these final remarks show that T is monocyclic. [] Theorem 9. Let G be a graph and P some fixed path in G. Suppose that G E U (~<2). l f T(G;P) is non-empty then either T(G;P)E{P],P2,P3, C6} or
qr ~<45-nr- 2.
(36)
Proof. In the appendix (Result A.1) it is proved that T ~ C7 and T T H (see Fig. 5). So, using Lemma 5(2), the theorem is true for nr<~7.
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
137
Now assume that nr>~8 and proceed by induction on nr. Firstly, set n = n r and q = qr. In the (non-trivial) canonical decomposition of H ( P ) set V/= H(PQui), ni = I vii and qi = q(V/, V/), Ti = T(G;PQui), i = 1,2 . . . . . t. If T/~ C6 for some i (1 <~i<<,t) then, by Lemma 8, T is monocyclic and, since n>~8, (36) is satisfied. Therefore, suppose that Ti ~ C6, i = 1,2 . . . . . t. For j (1 ~
q <~ ~ q i + ½ ( 2 t - a l ) i=1
5n ) ~< ~ - - - 2 t
=
(54 ) ---2
3
1
1
1
+ ~a| + ~a2+~a3W~(2t-al) -(t-2)+~al+~a2+~a3
~< (~4-~ - 2 ) - (t - 4)/2.
(37)
(38)
From (38) the induction is complete if t~>4. Therefore we may assume that 2~
(39)
Then, from (37), the induction is complete if ~>0. Suppose that t = 3. Then al + a 2 + a 3 43.
(40)
If a2~
(41)
Equality cannot hold in (41) otherwise n~<6. From (39) we may assume that al+a2+a3=l
(42)
otherwise most certainly e/> 0. Therefore from (42) and since n >~8 we may assume, without loss of generality, that nl ~<3 and n2>~5.
138
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
Hence, since G E U (~<2), from Lemmas 1, 5(1), (36), (42) and induction
) )
q <~ ql + q2 + l
<~
- 4
=
- 2
3
+ 1 + -~al + ~a2 + -~a3 , l - 1 + ~al + ~a2 + -~a3
5n <~-~- - 2 .
The induction is now complete.
[]
Comment. The wheel W4 with four spokes (and the hub as distinguished vertex) is in U ( = 2 ) and satisfies n r = 8 , q r = 8 . In general, ~ E U ( = 2 ) and n r = q r = 2 n so n = 4 is the only instance for the wheel where the bound of Theorem 9 is exact. Corollary 10. Let G E U (=2), G ~ K 4 . Then, f o r some vE V(G),
qG ~<¼(7no + 2 deg 6 v - 11 ). Proof. We have nr = 2(nG - 1). Now use Eq. (26) and Theorem 9. Hendry [6,8] shows that if GE U (= 1) then q~ = 3(n~ - 1). The situation is much more complicated for U (~>k) (k > 1). We discuss the general case in the final part of this paper.
4. Lower bounds and a construction
The graphs U (= 1) have been studied extensively in [5-8]. An example of a (infinite) family of such graphs, each with distinguished vertex v is shown in Fig. 6. a e b a~ e b
a e b
V V
a e b
V
V
.......
V
Fig. 6.
V
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
a
b
a
139
b
K
K* Fig. 7.
In this family of graphs e = ab is a 'forced edge'; every hamiltonian path in H(v) must include the edge e. Generally, elements of U (= 1) always contain forced edges. Now consider the graph K E U ( = 2 ) shown in Fig. 7 with distinguished vertex v*. By deleting the edge e of, say, the second graph in Fig. 6, deleting v* of K and identifying a and b we obtain the graph K* in Fig. 7. It is easy to check that K* E U (= 2). In this construction any element K E U (= 2) with distinguished vertex v* may be used provided degg(v*)= 2. In the same way any element of U (= k), using the forced edge construction, is associated with an element of U (= 2k). Let s be a non-negative integer. Then, [5,7], it is easy to construct arbitrarily large (in terms of vertices) elements of U (= 1) with at least s forced edges; now perform the construction simultaneously at s of these edges to obtain an element of U (= 2s). Hence there exist elements G of U (= 2s) which have asymptotically 3nG/2 edges. The best known lower bound is: Result 11 (Chartrand and qc >~¼(5no - 1)(no >/4).
Nordhaus
[3]).
Let
GEU(>>.k)(k>~I).
Then
Proof. Let G E U (~>k) (k>~l). Then no vertex, other than v, of G has more than one neighbour of degree exactly two. [] Comment 1. The family of graphs G constructed in [4] belong to U (>/1) and have [(5nG - 1)/41 edges (including a 'large' number of forced edges). Using the forced edge construction shows that the bound of Result 11 is asymptotically sharp. Conunent 2 (A generalization of U ( = k ) ) . The set U* ( = k ) (k>~l) is defined as follows. G E U* ( = k) if and only if for all u, v E V(G) (u ~ v) there exist exactly k uv-hamiltonian paths. Clearly, G E U* ( = k ) if and only if G E U ( = k ) and every vertex of G is distinguished.
140
13. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
Fig. 8.
Result 12. Let G E U* ( = k ) (k~>l). Then (1) G contains exactly lk(deg v) hamiltonian circuits for each v E V(G), (2) G is regular of degree 2s/k, where s is the number of hamiltonian circuits of G, (3) each ed#e of G belongs to exactly k hamiltonian circuits, (4) n6 is odd. Proof. (1)-(3) follow immediately from the definition of U* (--k). Part (4) is a consequence of Eq. (25). [] Example (i) Suppose that G E U* (=1). Then G E U ( = 1 ) and from [6], qa = 3(nG - 1). Hence, G has a vertex of degree 2 and, since G is regular, G TM/£3. (ii) Suppose that G E U* (= 2) and G is regular of degree 3. Then G must have property P(2) where: P(2):
G contains exactly 3 hamiltonian circuits; each edge belongs to exactly two of these circuits.
Notice however that the 3-regular graphs in Fig. 8 all belong to U (= 2) and have property P(2) but only the first two belong to U* (=2). Incidentally, the second condition of P(2) is not guaranteed by Smith's theorem [10] which allows the possibility that an edge belongs to no hamiltonian circuit. (iii) We conjecture that U* (= 3) = ~b. Suppose that G E U* (= 3). Then, from Result 12, G is regular of even degree so the smallest (in terms of degree) such graph G is regular of degree 4. Let P(3) be the property P(3):
G has exactly 6 hamiltonian circuits; each edge belongs to exactly three of these circuits; G has an odd number of vertices.
From Result 12 if G E U* (=3) and G is regular of degree 4 then G has P(3). Do there exist any regular graphs of degree 4 which have property P(3)?
141
B. Jackson, J. SheehanlDiscrete Mathematics 177 (1997) 123-144
Final comment. We end with conjectures (which, more modestly, should perhaps be only questions). Conjecture (Strong). Suppose that k ( > 1) is an odd integer. Then U ( = k ) = 0. Conjecture (Weak). Suppose that p is an odd prime. Then U ( = p ) = 0.
Comment. Many obvious conjectures/questions arise concerning the possible structures of transform graphs. As an indication of such we wildly conjecture: Conjecture. T(G;P) ~ C2k+l (k~> 1).
Acknowledgements We would like to thank Michael Sheard for many helpful comments.
Appendix A Result A.1. Let G be a graph and P a fixed path in G (with first vertex v). Then T(G;P) ~ C7 and T(G;P) ~ H (see Fig. 5).
Proof. Set P* = c o m p r ( P ) and W = PP*. Suppose that u is the last vertex of W and that the canonical decomposition of H(P) is (H(Wul),H(Wu2) ..... H(Wut))
(t ~>2).
Set T = T(G;P), Vi = H(Wui), qi = q(Vi, Vi) and ni = I/I/I (i = 1,2 . . . . . t) with
nl <~n2 <~ .'. <.nt.
[]
Case 1 (C7): Suppose that T --- C7. Then from L e m m a 1(2), nl >~2. If t = 2 then 2~
Wu3~u2~xyul
(A.1)
142
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
g 6
%
5
t h (u2)/~'-'--'---"'~ I
,
T':
'
I I ~, h 7(1"13)
'
I ~
tl
." h3(u2)
t
,"T '
t
h (uO ~
I e"' I / ~ " ~
t ~
't~ h4(t13) I ! k
Vl
V2
/ V3
Fig. 9. T(G;P) ~- C7 (Result A.I). or hl[ul]:
Wu3 ~c flu2 ~u l .
(A.2)
In the first case, following through the sequence of transforms in the cycle C7, yields
h 1 [ul] :
Wu3~u2~x7ul,
h 2 Ix]
:
Wu3~u2flul~x,
h3[u2] :
Wu3~ ful flu2,
h4[u3].•
rVu~~u~~x~u3,
hS[ul]:
Wu2flu3Y~ul ,
h6[u2] :
WU 1 ])xO~u3 flu2,
hT[u3] :
Wu~~x'du~~u~,
h I [Ul] :
Wu3 flU2~Of~Ul.
So &"and/~ are both trivial. Therefore, u 2 is adjacent to x in G and using x to effect a transform of h3[u2] yields an element of H ( P ) which is not a vertex of the cycle. This is impossible. In the second case, following through the sequence of transforms in C7, yields
hi[u1]:
rVu3~u2~ul,
h:[x]:
Wu3~u~~u2flx,
h3[u2] :
Wu3~ul 7'-xflu2,
ha[u3] :
Wu2 flX~Ul ~u3,
hS[ul]:
Wu2flxTu3~ul,
B. Jackson, J. Sheehan/Discrete Mathematics 177 (1997) 123-144
h2(u2)
f\
I i
, h6(u3)
143
!I I
v
t ~,
h3(u3) -
~,
/ ;
hl(ul)
I
h(x)
I I I ! |
l
I
!
,s k
\
t d
V1
V2
V3 Fig. 10. T(G;P)~-H (Result A.1).
h6[u2] :
WUl ~u3~x~u2,
h7[u3] :
WUl ~u2flx~u3,
h l[ul]:
Wu3yx~u2~ul.
So 8 and ~ are both trival. Therefore, ul is adjacent to u2 in G and using u2 to effect a transform of hS[ul] yields an element of H(P) which is not a vertex of the cycle. This is impossible. Therefore, T ~ C7. Case 2 (T -~ H): Now suppose that T = H. Using Lemmas 1 and 5(1) in exactly the same way the structure of T must be, up to a relabelling, as shown in Fig. 10 where V1 = {h5[u2],h6[u3]}, V 2 = {h3[u3],h4[ul]} and //3 = {hl[ul],h2[uz],h[x]} where h[x] E H(P;x) for some x E V(G) and h[x]hl[Ul] E E(T). The 6-cycle in H is (hl[ul],h2[u2],h3[u3],h4[ul],hS[uz],h6[u3]) where for some (ordered) subpaths and /3, h I [Ul]:
Wu3~u2flUl.
Following through the sequence of transforms in the cycle yields h 1 [Ul]:
Wu3~u2~ul,
h2[u2] :
Wu3~ulflU2,
h3[u3] :
Wu2~ul~u3,
h4[Ul]"
Wu2flu3~ul,
hS[u2]:
Wul ~du3 3u2,
h6[u3]"
WUl ~u2/~u3,
hl[ul] :
Wu3~u2~u I .
B. Jackson, J SheehanIDiscrete
144
Mathematics 177 (1997)
123-144
So Z and fi are both trivial. Since h[x] is adjacent to h’[ul] in T there exists a transform of h’[ul]
to h[x]. However,
the only transforms
and h[x] is equal to neither. This contradiction
of h’[ul]
in T are h~[u~] and he[u3]
shows that T 9 H.
0
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