Cycles containing matchings and pairwise compatible euler tours

Cycles Containing Matchings and Pairwise Compatible Euler Tours

Bi I I Jackson * DEPARTMENT OF MATHEMATKAL SClENCES

GOLDSMITHS' COL LEG€ LONDON, ENGLAND

Nicholas C. Wormald DEPARTMENT OF MATHEMATKS AND

STAT/STlCS UN/V€/ERS/TY OF AUCKLAND

NEW ZEALAND

ABSTRACT

Let M be a matching in a graph G such that d ( x ) + d ( y ) 2 IGI for all pairs of independent vertices x and y of G that are incident with M . We determine a necessary and sufficient condition for M to be contained in a cycle of G. This extends results of Haggkvist and Berman, and implies that if M is a 1-factor of G and [GI = 0 (mod 4), then M is contained in a Hamilton cycle of G. We use our results to deduce that an eulerian graph of minimum degree 2 k contains k pairwise compatible Euler tours.

1. INTRODUCTION

In the following, G will denote a finite graph without loops or multiple edges, and IG I the number of vertices of G. Under the hypothesis that d(x) + d( y) 3 lG( + 1 for all pairs of independent vertices x,y of G, Haggkvist [4] showed that, if IGl is even, then any 1-factor of G is contained in a Hamilton cycle of G . This result was generalized by Berman [2] by showing that, under the same hypothesis, any matching of G is contained in a cycle. The purpose of this paper is to prove the following generalization of Berman's theorem, which in particular gives a good characterization of the extremal graphs to Berman's theorem.

"This research was carried out while the first author was visiting the Department of Mathematics and Statistics, University of Auckland.

Journal of Graph Theory, Vol. 14, No. 1, 127-138 (1990) 0 1990 by John Wiley & Sons, Inc. CCC 0364-9024/90/010127-12$04.00

128 JOURNAL OF GRAPH THEORY

Other results on cycles containing matchings have been obtained independently by Benhocine and Wojda [ I ] and Wojda (81.

Theorem 1. Let G be a graph on n vertices and M be a matching in G such that

( I ) d ( x ) + d( y ) 2 n for all pairs of independent vertices x,y that are inci-

Then M is contained in a cycle of G unless equality occurs in ( 1 ) for at least two disjoint pairs ( x , y ) , and either

(a) for some yz E M , G - {y,z} is disconnected and has two components

(b) M is an odd cocycle of G ; or (c) M is an odd I-factor of G and KLr,Lsi2 C G - M C K,, + z2,+2. Specializing to the case when M is a I-factor and using ( l ) , we easily obtain

dent with M.

each containing edges of M ;

the extremal graphs to Haggkvist's theorem:

Corollary 1. Let G be a graph on n vertices and M be a I-factor of G such that d ( x ) + d ( y ) 2 n for all pairs of independent vertices x,y of G. Then M is contained in a Hamilton cycle of G unless at least one of the following holds:

(a) G = K , + ( K , U K z J , (b) G - M = 2 K z S t , , or

For the special case when n = 0 (mod 4), and hence M is an even 1-factor,

- (c) Kh.Zs+2 C G - M C KzS + KLr+z.

we obtain the following extension to Dirac's theorem on hamiltonian graphs:

Corollary 2. at least n / 2 . Then any I-factor of G is contained in a Hamilton cycle of G.

Let G be a graph on n = 0 (mod 4) vertices of minimum degree

Proof. The extremal graphs of parts (b) and (c) of Corollary 1 involve I-factors M of odd cardinality, whereas those in (a) cannot have minimum degree n / 2 unless s = t , and then n = 2 (mod 4).

In the final section of this paper, we shall use Corollary 1 to deduce results on the maximum number of pairwise compatible Euler tours in an eulerian graph.

2. DEFINITIONS AND PRELIMINARY LEMMA

Let Q be a cycle or path in G. We shall suppose Q has been given an arbitrary orientation and use Q to denote Q with the reverse orientation. For u,v E V(Q) let v', v- denote the successor and predecessor of v on Q (v' and v- are undefined if Q is a path and v is the last, respectively first, vertex of Q). Let Q[u,v] denote the subpath of Q from u to v ( Q [ u , v ] is undefined if Q is a path and v

CYCLES CONTAINING MATCHINGS 129

precedes u). For S C V(Q) let S+ = {v+ I v E S} and S- = {v - 1 v E S}. Let M be a matching in G. An M-cycle or M-path in G is a cycle or path containing M: An M-cover in G is a pair (P,C) where P is a path, C is a cycle, V(P) n V(C) = 0 and M C E(P) U E(C).

Given an M-cover (P ,C) we shall say that a (P,C)-bypass is a segment C [ y , z ] such that y and z are adjacent to distinct end vertices of P and { yy', z-z} rl M = 0. Let V(M) be the set of vertices of G incident with M, and put Mt = V(M) fl V(M)' and M- = V ( M ) n V(M)-, taken over C U P. For v E V(C U P ) let vtM be the first vertex of M- following v in C U P and v - ~ be the last vertex of Mt preceding v in C U P (taking v + ~ = v if v E M- and v - ~ = v if v E M+).

Finally, for S C V(G) and H a subgraph of G put NH(S) = N(S) n V(H) and d H ( S ) = IMq.

Lemma 1. Let M be a matching in G and (P ,C) be an M-cover, where P = P[w,x]. Suppose G has no M-cycle. Let Q be a path of C - M having the same orientation as in C. Choose u E NJw) U NC(wtM), v E N J x ) U N J x - ~ ) , with u # v. Then

(a) {u,v} (Z W Q ) . Suppose further that {u,v} n M- = 0, p E V(C[u',utM]), q E V(C[vt,vtM]). Then

(b) pq e W); and (c) if s E N&), t E N&), and Q C C[utM,v], then either s = t o r s pre-

cedes t on Q. Proof. If any of (a), (b), or (c) fail, then we may use standard operations

on cycles to construct an M-cycle from (P,C).

3. PROOF OF THEOREM 1

Our proof is an extension if that given in [ 2 ] . Suppose M is not contained in a cycle of G. If [MI = 1 then M is a cut-edge and alternative (b) of the theorem holds. Henceforth we may assume (MI 2 2. Construct a new graph H by suc- cessively joining pairs of non-adjacent vertices of V(M) until the addition of any new such edge to H creates an M-cycle in H . Then V(M) does not induce a complete graph in H, otherwise H would contain an M-cycle. Choosing two nonadjacent vertices x,y E V(M), H + x y contains an M-cycle and thus H contains an M-path.

Proposition 1. H contains an M-cover, (P,C), with a (P,C)-bypass.

Proof. Choose an M-path Q in H . We may assume Q = Q[x,y] where x,y E V ( M ) . Let R = H - Q . Since H has no M-cycle, xy e E(H) and


NR(x) n NR( y ) = 0. Thus N,(X)- fl NQ( y ) # 0 by ( I ) . Choosing z E NQ(x)- n N Q ( y ) we have zzt E M, again since H has no M-cycle. Putting P = Q[x,z-] and C = P[z ,y ]yz gives the required M-cover with the (P,C)-bypass C [ z ' , z ] .

Choose an M-cover (P,C) with a (P,C)-bypass in H, such that IPI is minimal and subject to this condition, such that (CI is minimal. Let P = P[w,x] and let C [ y , z ] be a (P,C)-bypass with {wy,xz} C E ( H ) . Using the minimality of P it follows that

(2) NP(ytM) C M- and N p ( ~ - M ) C M+; (3) if y E N ( z - ~ ) then NP(y+'") = 0. (4) if z E N ( Y ' ~ ) then N , ( z - ~ ) = 0.

We shall say that C [ y , z ] is an irreducible (P,C)-bypass if, for T = [ y ' , z - ] ,

( 5 ) N , ( w + ~ ) C M- and N r ( Y M ) C M', (6) if y E N ( x ) then Nr(W+M) = 0, and (7) if z E N(w) then NT(x-'") = 0.

Using the minimality of P, it follows that any (P,C)-bypass that is minimal with respect to inclusion is an irreducible (P,C)-bypass. Thus H has an irreducible (P,C)-bypass and we may assume that C [ y,z] is irreducible.

Let R = H - (C U P) and let S be the set of maximal paths of (C U P) - M, where each path of S keeps the same orientation as in C U P. For v E V(C) U V ( P ) let Q(v) be the path of S containing v and choose Q E S. By applying Lemma 1 to various M-covers, for example, (P.c), (F,t), ( c [ y ' , z - j , ywP[w,x]xzC[z ,y ] ) and using (2) and (5) we obtain

(8) if Q C C [ z , y ] then d , ( ~ ' ~ ) + d, (YM) I lQl and do(y'M) + d , ( ~ - ~ ) I

(9) if Q C C [ y ' , z - ] U P [ w ' , x - ] then d , ( ~ ' ~ ) + d , ( ~ - ~ ) I lQl and l Q l ; d,(-fM) + ~ Q O " ~ ) IQI;

(10) if Q = Q ( y ) then (10.1) dQ(wtM) + d , ( ~ - ~ ) I 1Ql + 1 with equality only if wtMy E E ( H ) , (10.2) d , ( ~ - ~ ) + d,(y"") 5 lQl with equality only if x-"y E E(H);

(11.1) dQ(YM) + d,(y"") I IQ1 + 1 with equality only if Y M z E E ( H ) , (1 1.2) d,(wtM) + d , ( ~ - ~ ) I lQl with equality only if wtMz E E(H);

(12.1) dQ(wtM) + d , ( ~ - ~ ) I lQl - 1, (12.2) d , ( ~ - ~ ) + d , ( ~ ' ~ ) 5 lQl + 1 with equality only if ~ y + ~ E E ( H ) ;

(11) if Q = Q(z) then

(12) if Q = Q ( w ) then

(13) if Q = Q(x) then (13.1) do(^-^) + dQ(ytM) 5 lQl - 1, (13.2) d , ( ~ + ~ ) + d , ( ~ - ~ ) I lQl + 1, with equality only if X Z - ~ E E(H);

(14) if r E R then d , ( Y M ) + d,( Y ' ~ ) I 1 and dr(w+M) + d , ( ~ - ~ ) 5 1.

(To see (14) suppose r E N ( f M ) n N( Y ' ~ ) . Replacing the path x ry edge we contradict Lemma l(a).)

-M +M by an


Summing (8)-( 14) over all paths Q E S and using the fact that simultaneous equality cannot occur in (10.2) and (12.2), or in (11.2) and (13.2) by Lemma l(a), we deduce

(15) d ( ~ ' ~ ) + d(zWM) + d ( Y M ) + d ( y t M ) 5 2n.

Since ~ ' ~ z - ~ , x - ~ y + ~ e E(H) it follows from (1) that equality must hold throughout (8), (9), (lO.l), ( l l . l ) , (12.1), (13.1), (14), and (15) and strict inequality in exactly one of (10.2) and (12.2), and one of (11.2) and (13.2). In particular, from (10.1) and (1 1.1) we deduce ~ ' ~ y , X - ~ Z E E(H) . Using minimality of P we have

(16) w = w + ~ and x = xwM.

Notation. ity ( i ) by ( i =).

Henceforth we shall denote the fact that equality holds in inequal-

Consider the following two cases.

Case I . N,-(w) n N,(x) # 0.

Proposition 2. There exists an irreducible (P,C)-bypass C [u,v] such that {u,v) n N ( W ) n ~ ( x ) z 0.

Proof. Let C ( u , v ) be a shortest (P,C)-bypass that satisfies {u , v } fl N(w) fl N ( x ) # 0. Such a bypass exists since H has a (P,C)-bypass and N,(w) n N,-(x) # 0. Without loss of generality we may assume u E N(w) fl N(x) and v E N(x) . By the minimality of T = C[u',v-] we have

(17) N,(w) U N&) C M'.

If v E N(w) n N(x) then Nr(w) U Nr(x) C M-, hence Nr(w) U N T ( X ) = 0 and C[u,v] is irreducible. Thus we may assume that v t! N(w). Suppose C[u,v] is not irreducible. Then by (17) there exists r E NT(w) f l M'. Choosing t to be the last such vertex on C [ u , v ] , we deduce that C[t ,v] is an irreducible (P,C)- bypass and, by minimality of C[u ,v ] , f t! N(x) . Thus strict inequality holds in (10.2) for C[ t , v ] = C [ y , z ] , and hence we must have (12.2=). It follows that ~ t + ~ E E(H) , contradicting the fact that v e N(w) if t'M = v, and contradicting (17) otherwise.

By Proposition 2 we may suppose that the irreducible (P,C)-bypass, C [ y , z ] , satisfies y E N(w) f l N(x) and z E N ( x ) . Using Lemma l(a) it follows that NQ(y,(w) = { y } and thus by (10.1 =),

(18) N Q ( z - ~ ) = V(Q) for Q = Q( y ) .

Choose Q E S. Using (6) and (3) we have

(19) if Q C C [ Y ' ~ , Z - ~ ] U P[w',x-] then dQ(w) + d , ( ~ ' ~ ) 5 lQl.


Suppose Q C C [ z , y ] , s E NQ(w) , t E NQ(ytM). Using Lemma l(b) we deduce either s = t or precedes s on Q . However, if the latter occurs then P[w,x]xzC[~ , t ] t? , '~C[ y + M , ~ - M ] ~ - M y c [ y,s]sw is an M-cycle in H . Thus s = t and

(20) if Q C C [ z , y ] then dQ(w) + ~ Q ( Y ' ~ ) 5 lQl.

Furthermore, using (6), (3), and Lemma 1 we have

(21) if Q = Q ( y ) then dQ(w) = I and dp(W) + ~ Q ( , v ' ~ ) 5 l Q l ; (22) if Q = Q ( z ) then dQ(w) 5 1 and d e ( w ) + d e ( J ' M ) 5 lQl + 1 with

(23) if Q = Q ( w ) = w then d,(w) + dQ( J + ~ ) = 0; (24) if Q = Q ( x ) = x then d w + de(y'') 5 1; (25) if v E R then d,(w) + d,(y

equality only if wz, E Q H ) ;

!M ) 5 1.

Summing (19)-(25) we obtain

(26) d ( w ) + d( y 'M) 5 n

It follows from ( 1 ) that equality must hold throughout (19)-(26). In particular, by (22),

(27) wz,ytMz E E ( H ) .

Thus, by replacing P by k or C by C ' = C [ z , y ] y ~ - ~ ~ [ z ~ ~ , y ' ~ ] y + ~ z we may reverse the roles of w and x, or of J + ~ and fM. Furthermore, it follows from the minimality of (C I that IC 'I = lCi.and thus

(28) y+M = y+, f M = ;-

Using (6) and (9=) we deduce

(29) NQ(z- ) = V ( Q ) for all Q E S, Q C C [ y ' , z - ] .

Since, by Lemma l(b), z - t @ E ( H ) for all t E v ( C [ ~ - ~ , y ] ) - {y}, it follows from (18) that

(30) y-M = J .

Reversing the roles of y' and z- in (18). (291, and (30) gives

(31) N p ( y + ) = V ( Q ) for Q = Q ( z ) , (32) N Q ( y + ) = V ( Q ) for all Q E S. Q C C [ v ' , z - l , (33) Z t M = Z .

Proof. Suppose N J x ) @ N J w ) and choose u E N,(x) - N J w ) . By (7), u E V ( A ) . Let Q = Q(u) . Using Lemma l(a) and (8=) we deduce that Ne(x) =


V(Q) , N,(w) = 0. By Lemma l(b), NQ(y') C {u-"}. Thus d,(w) + dQ(y') 5 1. This contradicts (20=) and hence N J x ) C N J w ) . Reversing the roles of x and w, we deduce (a).

Choose Q E S , Q C A. Using Lemma l(a), Proposition 3(a), and (8=) it follows that lQl = 2, and that N,(w) = NQ(x) = {v}, say. Using Lemma l(b), the fact that the roles of y' and z- may be reversed, and (20=) we deduce Np(y ' ) = N,(w) = N,(z-). Since this holds for all Q E S with Q C A, (b) and (c) follow.

Proposition 4. Let e be an edge of M contained in C[z ,y] = A. Then d,(w) + d,(x) E {0,41.

Proof. Suppose the Proposition is false. Using Proposition 3(a) we deduce that there exists e E M n E(A) such that N,(w) = N,(x) = {v}, say. Without loss of generality assume that v E M-. Let u = v'. Then u # y and hence Q(u) C A. Using (8=) for Q = Q(u) , it follows from Proposition 3(a), 3(b) that u ' E N(w) rl N ( x ) . We shall show

(34) N(u) c N A ( W ) .

First note that N p ( u ) = 0 by minimality of P. Furthermore, by (14=), R C N(w) U N(z- ) and thus NR(u) = 0 by minimality of P and Lemma l(b). Suppose t E NT(u) for T = C[y ' , z - ] . Reversing the segment C [ y',z-] of C if necessary, we may assume t @ M'. Thus by (33), t e V(Q(z) ) , and hence, by (18) or (29), z-t- E E ( H ) . The edges ur and z-t- contradict Lemma l(c), for Q = Q ( t ) , and thus N A u ) = 0. Finally, suppose s E NA(u) - N(w). If s E M', we may use ( 8 = ) for Q = Q ( s ) and Proposition 3(a),(b) to deduce st E N(w) n N ( x ) , and now the edge us contradicts Lemma l(b). Hence s E M-, and, as above, we deduce that s- E N(w) fl N ( x ) . Thus C[s- ,v] is a (P,C)-bypass. Since it must contain an irreducible (P,C)-bypass we may assume, relabeling if necessary, that C [ y , z ] C C [ s - , v ] . By Proposition 3, u' € N ( z - ) , since u' E N(w). Now the edges su, z-u' contradict Lemma l(a) (with Q = Q(u) and P = C[s,z-1). Thus (34) holds.

Using (34) and Proposition 3(c) we deduce that d(u) < min{d(w),d(y')}. It now follows from (26) that either d(u) + d(w) or d(u) + d( y') contradicts (1).

We are now ready to complete the discussion of Case 1. We first suppose that C[z , y ] has length one. If some vertex s of C[y ' , z - ] is joined to a vertex t of P by a path B [s , t ] , all of whose internal vertices belong to R, then, reversing either P or the segment C [ y ' , z - ] of C if necessary, we may assume { s , t } fl M - = 0. Thus by (28), s z V ( Q ( y ) ) , and the M-cover ( P [ t ' , x ] , C[s' ,y]ywP [w,t]B [t,s]k[s,y']y's') contradicts the minimality of P (note that y's' E E ( H ) by (32)). Hence no such B[s, t ] exists and thus P and C[y ' , z - ] are each contained in a component of H - { y,z}. It follows that G - { y,z} has two components, each containing an edge of M and thus alternative (a) of the Theorem holds for G.


Finally we suppose that C[z.y] has length at least 2 . For i E {0,4} let E, = {e E M f l E ( C ) 1 d,(w) + d,(x) = i}. Using Proposition 4 and Lemma l(a) it follows that the edges of E, and E, alternate along C[z,y], and hence by Propo- sition 3(b), H has an irreducible bypass of length three. Modifying our choice of C [ y , z ] if necessary such that it has length three, it follows that lQl = 2 for all Q E S with Q C C , and that the edges of E4 and E, alternate around C . Fur- thermore, by minimality of P, I f I = 2. Thus [MI = 21E,I + l and IMI is odd. For i E (0.4) put V, = V ( E , ) . Applying Proposition 3 with C[y',z-] equal to each edge of E, in turn, we deduce that the only edges joining vertices of V, U {w,x} are edges of M and each vertex of V, U {w,x} is adjacent to every vertex of V,. Thus K2s,,+Z C ( H - R ) - M C K , + K 2 + > where lE41 = s.

Suppose R # 0 and choose v E R . By (141, v E N ( w ) U N(z-). Replacing C by C[z,y]ywxz if necessary, we may assume v E N ( z - ) - N(w). Reversing the roles of w and x we have v qi N ( x ) and thus, by (14=), v E N(yt) . Since C[y,z] is an arbitrary irreducible (P,C)-bypass, we deduce that V, C N(v). This contradicts Lemma I(b) and thus R = 0 and K,,Zr+2 C& - M C K , + z2st2. Using ( I ) , it follows that K,,Zst2 C G - M C K , + Kz,+> and hence alternative (c) of the Theorem holds.

-

Case 2 . N,-(w) fl Nc(x) = 0 (for every M-cover (P,C) that has a bypass, and for which (PI is minimal and, subject to this condition, (CI is minimal).

Proposition 5.

(a) For each Q E S with Q C C we have V ( Q ) C N(w) or V ( Q ) C N ( x ) . (b) 1PI = 2 .

Proof. Since N,(w) fl N,(x) = 0 we have strict inequality in (10.2) and (11.2). Hence (12.2=) and (13.2=) hold and { W ~ ' ~ , X Z - ~ ) C E ( H ) . Using NJw) fl N,(x) = 0, (8=) , and Lemma l(a) we deduce that Proposition 5(a) holds for all Q C C[z,y]. Similarly using Lemma l(b), ( lO.l=), and ( l l . l = ) we deduce Proposition 5(a) holds for Q E { Q ( y ) , Q ( z ) } . Thus C [ Z - ~ , ~ ' ~ ] contains an irreducible (P,C)-bypass of length three. Modifying our choice of C [ y,z] such that it has length three, it follows that Proposition 5(a) holds for all Q E S. Furthermore, by the minimality of P we have I f I = 2.

For v E {w,x} let S,. = { Q E S( V ( Q ) C V ( C ) fl N(v)}

Proposition 6. The paths of S,. and S, alternate around C

Proof. Suppose the proposition is false. Then, without loss of generality, we may choose a sequence of consecutive paths Qo, Q1, . . . , Q R t I such that Qo,Qg+l E S,, Q , E S, for 1 I i I g, and g 2 2. Let Q, = C[u,,v,l, 0 I i I g + 1. By Lemma l(b), t e N(u , , , ) for all t E V(QI) - {v,}. Con- sidering the M-cover (vgugt I ,C [v,+ Iug]uKwxvR+ I ) and applying Proposition 5(a) we deduce V ( Q , ) C N(v,).


We shall show d(x) + d(u,) contradicts (1). Choose Q E S.

(35) If Q = QR then dQ(x) = 0 and dQ(x) + do(ua) 5 l Q l - 1. (36) If Q E S,. - {Q,} then dQ(x) = 0 and d&) + ~ Q ( u , ) 5 lei. (37) I f Q E S , , Q = C [ u , v ] then Np(u,) C {v} by Lemma l (b ) and

v @ NQ(ug) by Lemma l(c) (using the edges vuR,v,va). Thus dQ(ug) = 0 and d&) + dQ(un) 5 I Q I .

( 3 8 ) If Q = Q(x) = x then d&) + d,(u,) = 0. (39) If Q = Q(w) = w then dQ(x) + dQ(u,) = 2. (40) If r E R then, by Lemma l(b), d,(x) + d,(u,) 5 1.

Summing (35)-(40) gives d(x) + d(u,) < n. This contradicts (1) and com- pletes the proof of Proposition 6.

We are now ready to complete the proof of Case 2. It follows from Proposi- tion 6 that IMl = 21S,I + 1 and hence [MI is odd. For u E { W J } let V,, be the union of V(Q) over all Q in S,. Suppose H contains a path B [ s , t ] where s E V, U { w } , r E V, U {x}, B [ s + , t - ] C R , and sr @ M. (We allow the possibility B [ s , t ] = s t . ) By Lemma l(a), s E V, and r E V,. Without loss of generality we may suppose s @ M - . Using Lemma l(b), it follows that s E M + and t E M - . Let v = s-. By Proposition 6, {v ,v - } C N ( x ) . Considering the M-cover (vs,C [s' ,v-]v-xwst) and applying Proposition 5(a) it follows that Q ( r ) C N(s) or Q(t) C N(v) . The former alternative contradicts Lemma l(b), while the latter alternative contradicts Lemma l(a) (applied to the aforementioned M-cover with B [ s , t ] contracted to an edge).

Hence we may suppose no such path B [s , t ] exists. Choose uv E M. Suppose H contains a path D [u,v] with 0 # D[u',v-] C R . Replacing C by C[z,y]ywxz if necessary, we may assume uv E E(C) . By Proposition 6, C[u-,v'] is an irreducible (P,C)-bypass. Choosing r a vertex of D[u' ,v -1 , it follows from Lemma l(a) that r e N(w) U N ( x ) . Since equality holds in (14), we have r E N ( y f ) f l N(z- ) for each irreducible (P,C)-bypass C [ y , z ] . This contradicts Lemma l(b) and hence no such path D [ a , v ] exists.

It follows that the only paths from V, U {w} to V, U {x} in H are the edges of M. Thus H - M is disconnected and hence also G - M is disconnected. Using (1) and the fact that IM I is odd, we now easily deduce that M is an odd cocycle of G and thus alternative (b) of the theorem holds.

Finally, to deduce the bound on the number of disjoint pairs satisfying equality in (1) we note that the number of such pairs in (a), (b), and (c) is at least 2, IMI, and 1/2(1M/ + I), respectively.

4. 1 -FACTORIZATIONS AND COMPATIBLE EULER TOURS

The following conjecture appeared in [3]:

Conjecture 1. Let F , , F,, . . . , F,-2 be 1-factors of K2,. Then there exists a 1-factorization L , , L,, . . . ,L,- , of K2, such that L, U F, is a Hamilton cycle of K , for 1 5 i 5 2m - 2 .


Conjecture 1 was verified in [3] for the special case when F,, F,, . . . , F,-, are identical. We next use Theorem 1 to give further evidence in support of this conjecture.

Theorem 2. Let F,, F,, . . . , F, be I-factors of K,, k 5 m . Then there exist edge-disjoint 1-factors L I , L z , . . . , L k of K,, such that F, U L, is a Hamilton cycle of K, for 1 5 i 5 k .

Proof. LfF, = F, = - 1 - = F,, then the theorem follows from [3, Theorem 31. Hence suppose F, # Fk and choose e E Fk - F,. Choose a 1-factor L , of K , such that e E L , and F, U L , is a Hamilton cycle of Kz,. Suppose we have constructed edge-disjoint 1-factors L , , L, , . . . , L, such that L, U F, is a Harnil- ton cycle of K , for 1 5 i 5 t < k . Let

We need to find a 1-factor L,, , of H such that F,, , U L,, , is a Hamilton cycle of H. Since each vertex of H has degree 2m - 1 - t or 2m - t , the existence of L,, , will follow immediately from Corollary 1 unless m = k = t + 1. Thus IH 1 = 2k and each vertex of H has degree k or k + 1. Furthermore, H has at least two vertices of degree k + 1 since e E L , n Fk. It can now be easily checked that H cannot satisfy altemati; (a), (b), or (c) of Corollary 1, and thus the required I-factor L, of H exists.

We next consider an application to the theory of compatible Euler tours. We shall use the terminology of [6]. It was shown in [6] that if G is an eulerian graph such that each vertex of V4(G) has degree at least 2 k , and X , , X , , . . . ,X,-, are transition systems for G, then G has an Euler tour that is compatible to each X,, 1 5 i 5 k - 2 . As an immediate corollary it was deduced that G contains k - 1 pairwise compatible Euler tours, and conjectured that G contains 2k - 2 pairwise compatible Euler tours. This conjecture has been verified for a certain family of eulerian graphs in [3], where it was also pointed out that the general conjecture would follow from Conjecture 1.

We shall use Theorem 2 to improve the bound of k - 1 pairwise compatible Euler tours from [6].

Theorem 3. degree at least 2k. Then G has k pairwise compatible Euler tours.

Let G be an eulerian graph such that each vertex of V4(G) has

Proof. We shall use a simplified version of the proof technique in [3]. We proceed by induction on IV4(G)l. If VJG) = 0 then the Theorem is trivially true, hence suppose VJG) # 0 and choose v E V4(G). Let d(v) = 2rn and choose a splitting of G at v, G*, such that G* is connected and hence eulerian (such a splitting can be obtained by splitting along the transitions at v induced by an arbitrary Euler tour of G). By induction, G* contains pairwise compatible Euler tours, TY, T: , . . . , T f . Let T, be the Euler tour of G induced by TY. Then


TI, T,, . . . , Tk are pairwise compatible at every vertex of V(G) - {v}. We shall “redirect” each T, at v to obtain compatibility at v.

Let G, be the graph obtained by splitting G at each vertex of V - {v} along the transition system corresponding to T,, and let F, be the transition system at v in G, containing the transitions at v in G, that are cocycles in G, for 1 5 i 5 k. Consider each F, as a 1-factor of KZm, where the vertices of K2,,, are labelled with the edges incident with v in G. If we can find edge-disjoint 1-factors L , , L,, . . . .Lk of Kh such that F, U L, is a Hamilton cycle of Kh for each i, 1 I i 5 k, then redirecting each T, along the transitions at v corresponding to L, gives the required set of k pairwise compatible Euler tours of G. Thus Theo- rem 3 follows from Theorem 2.

Finally, we use Corollary 2 of this paper to strengthen Theorem 3 when k is even.

Theorem 4. Let G be an eulerian graph such that each vertex of VJG) has degree at least 2k, for some even integer k. Let X , , X , , . . . ,Xk-, be transition systems for G. Then G has an Euler tour that is compatible to X, for all i, l s i s k - 1 .

Proof. As pointed out in [6, Remark 31, this follows from Corollary 2, using the proof technique of [6].

5. PROBLEMS

5.1. We feel the degree condition of Corollary 1 can be weakened for the special case of regular graphs:

Conjecture 2. Let G be a 3-connected, k-regular graph on at most 3k - 2 vertices and M be a 1-factor of G. Then M is contained in a Hamilton cycle of G unless M contains an odd cocycle of G.

5.2. Haggkvist and Thomassen [ 5 ] have shown that any matching of (k - 1) edges in a k-connected graph is contained in a cycle. There remains the following conjecture of Lovasz [7]:

Conjecture 3. Then M is contained in a cycle of G unless M is an odd cocycle of G.

Let M be a matching of k edges in a k-connected graph G.

References

[I] A. Benhocine and A. P. Wojda, Graphs containing every matching in a cycle. Discrete Math., to appear.


[2] K. A. Berman, Proof of a conjecture of Haggkvist on cycles and indepen-

[3] H. Fleischner, A. J . W. Hilton, and B. Jackson, On the maximum number

[4] R. Haggkvist, On F-Hamiltonian graphs. Graph Theory and Related Topics.

[ S ] R . Haggkvist and C. Thomassen, Circuits through specified edges. Dis-

(61 B. Jackson, Compatible Euler tours for transition systems in Eulerian graphs.

171 L. Lovasz, Problem 5. Period Math. Hungar. 4 (1974) 82. [8] A. P. Wojda, Hamiltonian cycles through matchings. Demonstratio Math.

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Cycles containing matchings and pairwise compatible euler tours