# A cycle structure theorem for hamiltonian graphs

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A cycle structure theorem for hamiltonian graphs
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JOURNAL OF COMBINATORIAL THEORY, Series B 45, 99-107 (1988) A Cycle Structure Theorem for Hamiltonian Graphs* E. F. SCHMEICHEL Department of Mathematics and Computer Science, San Jose State University, San Jose, California 95192 AND S. L. HAKIMI Department of Electrical Engineering and Computer Science, University of Calijornia, Davis, California 95615 Communicated by the Managing Editors Received October 10, 1986 An n-vertex graph is called pancyclic if it contains a cycle of length 1 for every 1 such that 3 < I< n. We establish the following result: Let C be a hamiltonian cycle in an n-vertex graph G. Suppose C contains consecutive vertices X, y such that d x)+d y)>n. Then G is either (i) pancyclic, (ii) bipartite, or (iii) missing only an (n - I)-cycle. Moreover, (iii) can occur only if G has a very explicit structure “near” x and y. This result can be used to show that three well-known hamiltonian degree con- ditions (due to Chvatal, Fan, and Bondy) actually imply that a graph is essentially pancyclic. 0 1988 Academic Press, Inc. We consider only finite undirected graphs without loops or multiple edges. Our terminology and notation will be standard except as indicated. A good reference for undefined terms is . Various sufficient conditions for a graph to be hamiltonian have been given in terms of the vertex degrees of the graph. Three such conditions (due to Chvatal, Fan, and Bondy) are given below. THEOREM A . Let G be a graph on n 3 3 vertices with vertex degree sequence d, <d,< ... <d,,. If d,<k<n/2 implies d,-,>n-k, then G is hamiltonian. * Supported in part by the National Science Foundation under Grants IRS-8121741 and ECS 85-l 1211. 99 0095-8956/88 \$3.00 582b/45/1-7 Copyright 0 1988 by Academic Press, Inc. All rights of reproduciion m any form reserved.  100 SCHMEICHEL AND HAKIMI THEOREM B . Let G be a 2-connected graph on n vertices. Iffor all vertices x and y, distance (x, y) = 2 implies max(d(x), d(y)) > n/2, then G is hamiltonian. THEOREM C [S, 81. Let G be a 2-connected graph on n vertices. If for all vertices x, y and z, we have x, y, z independent implies d(x) + d(y) + d(z) > sn - 1, then G is hamiltonian. In fact, each of these conditions implies substantially more about the cycle structure of G than the mere fact that G is hamiltonian. Let us call an n-vertex graph pancyclic if it contains an I-cycle for every 1 such that 3 < 1 d n. We then have the following rest&s. THEOREM A’. Let G be a graph on n > 3 vertices with vertex degree sequence d,<dd2< ... <d,,. Ifdk<kknn/2 implies dndkan-kk, then G is paucyclic or bipartite. THEOREM B’. Let G be a 2-connected graph on n vertices. ff for ali ver- tices x and y, distance (x, y) = 2 implies max(d(x), d(y)) 3 n/2, then G is either panwk K,,2.nlZ, K,p n,2 e, or the graph shown in Fig. 1 below. THEOREM C'. Let G be a 2-connected graph on n vertices. If for all ver- tices x, y, and z, we have x, y, z independent implies d(x) + d(y) + d(z) 2 n - 1, then G is either pancyclic, K,,,2,n,2, Kn,2,n,2 e, or C,. Although proofs of Theorems A’ and B’ have appeared previously (in [9, 21, respectvely), the arguments used were difficult and somewhat ad hoc. The purpose of this paper is to establish a basic cycle structure theorem for hamiltonian graphs (Theorem 1) which can then be used to give straightforward proofs of Theorems A’, B’, and C’. Details of these proofs which use Theorem 1 will appear in [l] to avoid making the present paper unduly long. A brief word about our notation. If the vertices of G are x1, x2, . x,, we will write (i, j) E G rather than the more customary (vi, vi) E E(G). ~112 vertices 1;:-“- FIGURE 1  CYCLESTRUCTURETHEOREM 101 Similarly, we will denote the degree of xi by simply d(i), and the cycle (xi, xi, ...2 xk, xi) as simply (i, j, . k, i). Indices throughout are to be taken as modulo n. STATEMENT AND PROOF OF THE MAIN RESULT We will establish the following cycle structure theorem for hamiltonian graphs. THEOREM 1. Let G be a graph on the n 3 3 vertices x,, x2, .l,, x, with hamiltonian cycle C= (1, 2, . n, 1). Suppose that d(1) + d(n) 3 n, with say d( 1) 6 d(n). Then G is either (i ) pancyclic, (ii) bipartite, or (iii) missing only an (n - 1)-cycle. Moreover, if (iii) holds, then d(n - 2), d(n - l), d(2), d(3) <n/2, and G has one of two possible adjacency structures near x, and x,. In the irst structure, the vertices x,-?, x,- ,, x,, x,, x2, x3 are independent except for edges of C, and (n, n - 3), (n, n -4), (1, 4) (1, 5) E G (see Fig. 2). The second structure (which can occur only if d(1) < d(n)) is identical to the first except that (n, 3)gG and (1, 5)\$G. The actual proof will be preceded by three lemmas. Let I be any integer such that 3 < I < n. If 2 <k < n - 1 and G contains both of the edges (n, k) and (1, f,(k)), where f,(k) = n-(l-k)+ 1, if 2<k61-2 k-11-3, if I-l<k<n-1, then G contains an I-cycle (see Fig. 3). Note that ft is one-to-one. LEMMA 1. Zf d( 1) + d(n ) > n, then G is pancyclic. FIGURE 2  102 SCHMEICHEL AND HAKIMI n-Cl-k) +l @ Q--J+3 n 1 n 1 FIGURE 3 Proof: If d( 1) + d(n) > n, then for any f, 3 < 16 n, we must have (n, k), (l,f,(k))~ G for some k, and G would contain an I-cycle. 1 Thus we assume in the remainder of the paper that d( 1) + d(n) = ~1. Since fi is one-to-one, it follows that if G does not contain an I-cycle for some 1 such that 3 < I< n, then exactly One of the pairs (n, k), (1, f,(k)) will be an edge in G, for each k such that 2 < k < n - 1. We will refer to this fact as the l-correspondence principle in the remainder of this paper. LEMMA 2. Suppose G is neither pancyclic nor bipartite. Then G contains a 3-cycle (n, i, i + 1, n), for some . Proof: If d(n) > n/2, this is immediate. Otherwise, since d(1) + d(n) = n and d( 1) d d(n), we must have n even and d( 1) = d(n) = n/2. The only way the lemma could fail is if the edges incident to u, are all edges of the form (n, odd). Thus G contains a cycle of any even length. We now consider two cases. Case 1. G contains a 3-cycle (1, j, j+ 1, 1) for some j. If j= 2 or n - 1, it is easy to see G would be pancyclic, a contradiction. Thus we suppose 3 <j< n - 2. We can then obtain a cycle of any odd length 1~ 5 as follows: Choose an odd integer s such that s < j < j+ 1 < s + I - 3 d n - 1, and consider the (I - 1 -cycle (n, s, s + 1, . s + I - 3, n). Replace the edge (j, j + 1) in this cycle by the path j, 1, j+ 1 to obtain an I-cycle. Thus G would be pancyclic, a contradiction.’ Case 2. G does not contain a 3-cycle (1, j, j+ 1, 1) for some j. Since d( 1) = n/2, the edges incident to vi must be exactly those of the form (1, even). Since G is not bipartite, G must contain an (odd, odd) edge or an (even, even) edge (otherwise the even and odd vertices would form a bipartition). But this edge, together with the edges of C and the edges incident to v, and u,,, allows us to get a cycle of any length in G. Again we have the contradiction that G is pancyclic. 1 For future reference, we record the following easy consequence of the above argument. COROLLARY. rf G is neither pancyclic nor bipartite, and d( 1) = d(n) =  CYCLE STRUCTURE THEOREM 103 n/2, then G contains 3-cycles (n, i, i + 1, n) and (1, j, j+ 1, l), for some i and j. Lemma 3. If G is neither pancyclic nor bipartite, then G contains a 3-cycle (1, k, n, 1 ), for some k. ProoJ By Lemma 2, G contains a 3-cycle (n, i, i+ 1, n) for some i. If G does not contain a 3-cycle (1, k, n, 1) for some k, then since d( 1) + d(n) = n it follows that every vertex in G is adjacent to exactly one of the vertices v1 or u,,. We now use this fact to show that G contains an I-cycle for every I such that 3 <I < n. (Hence G would be pancyclic, a contradiction.) We have two cases. Case 1. I-l<i. Then 2,<i- (I-3),<i- 1, since I> 3. We know x~~(~-~) is adjacent to either x1 or x,. In the former instance, (n, i, i- 1, . i- (I- 3), 1, n) is an l-cycle in G. In the latter instance, (n, i + 1, i, . i- (I- 3), n) is an I-cycle in G. Case 2. I-l>i+l. We know xl-, is adjacent to either x1 or x,. In the former instance, (1, 2, . i, n, i+ 1, i+2, . I- 1, 1) is an I-cycle in G. In the latter instance, (1, 2, ..., I - 1, n, 1) is an I-cycle in G. 1 LEMMA 4. If G contains either of the edges (1, n - 1) or (n, 2), then G is pancyclic. Proof: Suppose first that (1, n - 1) E G and that G does not contain an I-cycle for some I such that 3 < I < n - 1. (Clearly G contains a 3-cycle and an (12 - l)-cycle.) Note that (n, n -2) E G, since otherwise (1, n-I+ II) E G by the I-correspondence principle, and (1, n - I + 1, n - I + 2, . n - 1, 1) would be an I-cycle in G. On the other hand, (n, l-2)\$ G since otherwise (1, 2, . . . I- 2, n, n - 1, 1) would be an Z-cycle in G, and (n, I - 1) G since otherwise (1, 2, . I- 1, n, 1) would be an l-cycle in G. If (n,i)EG and (n,i+l)\$G for some i, Z<idn-2, then by the I-correspondence principle we have ( 1, i - 1 - 4) E G, and thus ( 1, i - I+ 4, i- I+ 5, . i, n, n - 1, 1) would be an I-cycle in G. It follows’ that for some s, l<s<n -2, all the pairs (n, n- l), (n, n -2), . (n, s) are edges in G, while all the pairs (n, s - 1 ), (n, s - 2), . . (n, I - 2) are missing from G. If (n, i) E G and (n, i- 1) \$ G for some id I- 3, then by the Z-correspon- dence principle (1, n - I + i) E G, where n - I+ i < n - 3, and so (1, 2, . i, IZ, n - 2, n - 3, . n - I + i, 1) would be an l-cycle in G. It follows from this and the above paragraph that x, is adjacent to precisely x1, x2, . x, and -‘c,, x s+ r, ,.., x, _, for some r and s, where s - r = n - d(n). (See Fig. 4.)

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