partitioning a graph into two square-cycles

Partitioning a Graph into Two Square-Cycles

Genghua Fan

H. A. Kierstead" DEPARTMENT 0 F MATHEMATICS

ARIZONA STATE UNIVERSITY TEMPE, AZ 85287

e-mai1:fanQmath. /a.asu. edu e-mail:kierstead@math. la. asu. edu

ABSTRACT

A square-cycle is the graph obtained from a cycle by joining every pair of vertices of distance two in the cycle. The length of a square-cycle is the number of vertices in it. Let G be a graph on n vertices with minimum degree a t least $n and let c be the maximum length of a square-cycle in G. Posa and Seymour conjectured that c = n. In this paper, it is proved that either c = n or i n 5 c 5 $n. As an application of this result, it is shown that G has two vertex-disjoint square-cycles C1 and Cz such that V ( G ) = V ( c 1 ) U V(c2) . 0 1996 John Wiley & Sons, Inc.

1. INTRODUCTION

All graphs considered are simple and undirected. S(G) denotes the minimum degree of a graph G. Let P be a path (cycle). The kth power of P is the graph obtained from P by joining every pair of vertices with distance at most k in P. An edge is called a k-chord of P if it joins two vertices of distance k in P. We call the 2nd power of P a square-path (-cycle). (For technical reasons, the empty set is regarded as a square-cycle.) A hamilronian square-path (-cycle) is a square-path (-cycle) that contains all the vertices of the graph.

A classical theorem of Dirac [l] states that if G is a graph on n vertices with 6(G) 2 :, then G contains a hamiltonian cycle. In 1963 Posa (see Erdos [2]) conjectured that if S(G) 2 $n, then G contains a hamiltonian square-cycle. In 1973 Seymour [6] introduced

* Research partially supported by Office of Naval Research Grant N00014-90-J-1206.

Journal of Graph Theory Vol. 23, No. 3, 241-256 (1996) 0 1996 John Wiley & Sons, Inc. CCC 0364-9024/96/030241-16

242 JOURNAL OF GRAPH THEORY

the more general conjecture that if S(G) 2 [k / ( k+ l ) ]n , then G contains the kth power of a hamiltonian cycle. As Seymour pointed out, if his conjecture is true, the proof is likely to be very hard, because it implies the remarkably difficult theorem of Hajnal and Szemeredi [6] that any graph G with maximum degree A can be properly colored with A + 1 colors so that no two color classes differ in size by more than one. There has been progress on Posa’s conjecture. Fan and Haggkvist [3] proved that if S(G) 2 :n, then G contains a hamiltonian square-cycle. Fan and Kierstead [4] showed that for every E > 0, there exists m such that if n > m and S(G) > [(2 + ~ ) / 3 ] n , then G contains a hamiltonian square- cycle. They [5] also proved that if 6(G) 2 [(2n - 1)/3], then G contains a hamiltonian square-path.

If H is a subgraph of G , G - H denotes the subgraph obtained by deleting all the vertices of H together with all the edges with at least one end in H . For z E V ( G ) , N H (z) is the set, and d ( z , H ) the number, of vertices in H that are joined to z. We denote by zy the edge with ends z and y and define N ~ ( z y ) = N H ( z ) n N H ( ~ ) . (In the case that H = G, we simply use N(x),d(z), and N(zy).) If F is another subgraph of G, we define d ( F , H ) = C v E I / ( F ) d ( ~ , H ) . By the definition, d ( F , H ) = d ( H , F ) , and if V ( F ) n V ( H ) = 0, then d(F , H ) is simply the number of edges with one end in F and the other end in H. The complete graph on m vertices is denoted by K,,,.

The third power of a path is also called a cube-path. If P = z1z2 . . . zp is a square-path (cube-path), we say that P starts at 21x2. By definition, the existence of a square-path starting at z1z2 does not imply the existence of a square-path starting at z2z1. Here the order of the endpoints of edges is significant. This is the only case where we distinguish xy from yx for the same edge. In such a case, for clarity, we shall use the term ordered edge instead of edge. If P = zyQzw is a square-path (so Q is a square-path itself), we say that xy is connected to zw through Q, or by P.

A square-cycle is optimal if (i) it is a longest square-cycle; (ii) subject to (i), it has the maximum number of 3-chords; (iii) subject to (i) and (ii), it has the maximum number of 4-chords. A square-cycle C is chord-maximum if it is optimal and, subject to this, the subgraph induced by V ( C ) has maximum number of edges among all optimal square- cycles. Let C = u1u2 . . . uCul be a square-cycle, where c = IV(C)I and u,~,+~, u,u,+2

are 1- and 2-chords of C respectively, 1 5 i 5 c. (Addition of indices is taken modulo c.) By i < 1, we mean that u,, u,+1,. . . , I L ] are around C in the order consistent with a fixed orientation of C.

Let G be a graph on n vertices with S(G) 2 i n . Section 2 is devoted to general lemmas that are needed in other sections. In Section 3, we prove that if G has no hamiltonian square-cycle, then any square-cycle in G has length at most in . In Section 4, it is showed that the longest square-cycle in G has length at least n/2. Finally, by using the two results, we establish in Section 5 that the vertices of G can be partitioned into at most two square- cycles.

2. LEMMAS

Lemma 2.1 (4, Lemma 3.21. Let C = u1u2 . . . utl,.ul be an optimal square-cycle in a graph G. Let x be a vertex in G - C.

(a) d(z , Q) 5 $ q + 1 for any segment Q = u1u2 . . . uq--luq of C, with equality only if d ( z , U I U 2 ) = d(z , uq--luq) = 2.

PARTITIONING INTO TWO SQUARE CYCLES 243

(b) d(x, C ) 5 (2c + 1)/3, with equality only if, by relabeling, d(x,uCu1u2) = 3 and Nc(x) = v(c) \ { u g k : 1 5 k 5 ( c - 1)/3}.

Remark. The second part of Lemma 2.l(a) is not contained in [4], but it follows immediately by applying the first part to the segments Q - a1a2 and Q - aq-laq.

Lemma 2.2. Let G be a graph on n vertices with 6(G) 2 i n , where n 2 7. Suppose that S is an independent set of G with ( S / 2 f and xy E E(G) with x E S. Then G has a hamiltonian square-cycle C such that xy is neither a 1-chord nor a 2-chord of C.

Since S is an independent set, we have that S(G) 5 n - IS1 and so IS1 5 n - S(G) 5 2 , which implies that (SI = n/3, S(G) = $n, and each vertex of S is joined to every vertex of G - S. Let S = {21,52,. . . , x k } , where x1 = x and Ic = IS\. Since S(G - S ) 2 S(G) - IS1 = n/3 = i ( V ( G - S)l, it follows from a classical theorem of Dirac [ l ] that G - S has a hamiltonian cycle, say a1u2 . . . uv,ul, where m = IV(G - S ) ( = 2k and a,, = y. Noting that m 2 5 (since n 2 7) and each 2, is joined to every vertex of G - S, we see that ala2x1a3a4x2a5a6x3... a23-1a2,xJa23+1a23+2... anL-1a,,xkalu2 is a

I

Lemma 2.3. Let C = u1u2 . . . ucul be a chord-maximum square-cycle in a graph G. Let W 2 V ( G - C ) with d(v, C) = (2c + 1)/3 for all v E W.

(a) If c = 4, then W is an independent set in G; (b) If c # 5 and W # 0, then there is a set S C V ( C ) with / S / = ( c - 1)/3 such that

W U S is an independent set in G and N c ( a ) = Nc(6) for every a , 6 E W U S.

Proof.

hamiltonian square-cycle in G with the required property.

Proof. (a) If the statement is not true, let 2, y E W with xy E E(G). Since d(v , C) = 3 for all v E W, we have that 2 5 INc(xy)( 5 3. If i N ~ ( x y ) ( = 3, we have a K5; if INc(xy)l = 2, we have a square-cycle C' with V(C') = V ( C ) U {x, y}. In either case, we have a contradiction to the maximality of c. This proves (a).

(b) Since W # 0, c = 1 (mod 3), and since c 2 5, we have that c 2 7. Let x E W. By Lemma 2.l(b), with a proper labeling we have that Nc(x) = V ( C ) \ S, where S = {u3k : 1 5 k 5 ( c - 1)/3}. Let C, be the square-cycle obtained from C by replacing u3, with x, that is, C, = u1u2 ..'u3t-2u3t-1xu3t+iu3t+2 ...u,.ul, where 1 5 t 5 ( c - l ) /3 . First, we have that U3?& $! E(G) since, otherwise, &u1xu2u4u3u5ug . . . u, would be a square-cycle of length c + 1. And analogously ~ , - 1 ~ , - 4 $! E(G). Now, since u,x is a 3-chord in C1 and by the optimality of C, we have that

ucu3 E E(G) , and analogously, u,-1u2 E E(G). (2.1)

Next, we have that uc-lu3 q! E(G) since, otherwise, uc-1ucu3ulu22u4u5.~ . u, -~ would be a square-cycle of length c + I. Moreover, since xu7 is a 4-chord in C1, it follows from the optimality of C that ~ 3 1 1 7 E E(G). Then u 3 t ~ 3 t + 4 E E(G) follows from the optimality of C and from the fact that xugt--q and Z U ~ , + ~ are 4-chords in C,, 2 5 t 5 ( c - 4)/3. We have seen that u3U6 q! E(G). If ~ 3 ~ ~ 3 ~ ~ ~ E E(G) for some t , 2 5 t 5 ( c - 4)/3, then u,2~~2u~(u~ugug)(~~u~u~)~~~ (u3t+lu3tu3t+2)uL3t+3U3tf4 . . . U , - ~ U , is a square-cycle of length c + 1. Therefore we have that

c - 4 for all t , 1 5 t 5 -.

3 u3tu3t+3 $! E(G)


We note that C,, 1 5 t 5 (c - 1)/3, has the same length as C, and by (2.2), has as many 3-chords as C does. Since C is chord-maximum, d(u3,, C) 2 d(z, C), and since Ct is also optimal, we have that d ( u S t , C) = d ( z , C) = (2c + 1)/3. We claim that Nc(wgt,) = Nc.(z). If this is not true, then u g t is joined to some u 3 k E S. We may suppose that k > t and choose k - t to be as small as possible. By (2.2), k 2 t + 2, and by the minimality of k - t lu3tusk-:3 $! E(G) . Since d(u3t,Ct) = (2c + 1 ) / 3 , N ~ , ( u ~ ~ ) has the property described in Lemma 2.l(b). Noting that u3tu3/c E E(G) and uS tu3ku3 @ E(G), we see that d(uS t , u 3 k - - 2 u ~ k - - 1 ~ 3 k ) = 3. It follows from (2.1) (with Ct , uSt in the position of C, z) that u:jk--3u:$k- E E(G) , contradicting (2.2). Therefore, NC(uSt) = Nc(x) for all ulgt E S, which implies that S U {z} is an independent set.

Next we show that Nc(y) = N c ( z ) for all y E W \ {z}. By Lemma 2.l(b), d ( y , ~ , u , + ~ , u , + ~ ) = 3 for some u., E V ( C ) and N(y) = V ( C ) \ S’, where S’ = {u,+:$~ : i 2 l} and /S‘I = (c - 1)/3. If j = c, then Nc(y) = N c ( z ) . Suppose that this is not the case. For 1 5 t 5 (c - 4)/3, if j = 3t or 3t + 1, then, as seen in (2.1), uju3+3 E E(G) , contradicting (2.2). Thus, j E {c - 2, c - 1,1,2} u {3t + 2 : 1 5 t 5 ( c - 4)/3}. It is easily verified that for any such value of j, S’ n S = 0. So S U {x} and S’ u {y} are two independent sets, and moreover, S C Nc(y) and S’ C Nc(x). Let D be the graph induced by V(C)U{z,y) , plustheedgexyif i t isnot inE(G). Sod(w,D) = (2c+1)/3+1 = $lV(D)l for all v E SuS’u{z ,y} . Note that (SU{Z}/ = IS’u{y}/ = (c+2)/3 = i lV (D) l . We see that d ( v , D ) 2 (SU{z}(+(S’U{y}( = $lV(D)I for all v E V(D)\ (SUS’u{z ,y}) . There- fore, 6 ( D ) > { l V ( D ) ( . Note that IV(D)l 2 c + 2 > 9. It follows from Lemma 2.2 that D has a hamiltonian square-cycle that does not contain the edge zy, and so it is a hamiltonian square-cycle of length c + 2 in G. This contradiction shows that Nc(y) = N<:(x) for all

Finally we need to show that W is an independent set. If this is not true, let x , y E W and zy E E(G). Since N c ( z ) = Nc(y) and c 2 7, we may replace a vertex of S by the edge zy and obtain a square-cycle of length c + 1. Thus W is an independent set and the

Y E W \ { x I .

proof is completed. I

Lemma 2.4. Let G be a graph on n vertices with S(G) 2 $n and let C = u1u2 ”.u,:u1 be an optimal square-cycle in G. Then either c = n or c 2 5.

For n 5 9, the result can be easily verified. We assume that n 2 10. If c = 3, then G contains no K4. This means that N ( u v ) is an independent set for any edge uv E E(G). Since S(G) 2 $n, we have that (N(uv) l 2 n/3, and the by Lemma 2.2, c = n. If c = 4, then d(v, C) 6 3 for all v E V(G - C), and by Lemma 2.3(a), S = {z E V ( G - C ) : d(x, C) = 3) is an independent set. Moreover,

Prooj

8 -n 5 d ( C , G) 6 2(n - c) + IS1 + 12 = 2n + 4 + IS/, 3

which implies, since n 2 10, that /St > and d(z ,C) 5 c - 1, we see that d(x) 5 n - 1 - IS/, and thus IS1 5 n - 1 - d(x) 1.

- 1. Let z E S. Since S is an independent set - 1, I a contradiction. This proves the lemma.

Lemma 2.5. Let C = u1u2 . . . u,ul be a chord-maximum square-cycle in a graph G. Let xy be an edge in G - C.


(b) If d(xy,C) = i c and c # 3, then d ( ~ y , u ~ u , + ~ ) I 3 for all consecutive U ~ , U , + ~ E V ( C ) and there are s , t such that x E N ( U , ~ U , ~ + ~ ) , ~ E N ( U ~ U ~ + ~ ) , and s + 2 5 t 5 s + 4.

Proof. (a) If c = 1 (mod 3), by Lemma 2.l(b) and Lemma 2.3, d(zy, C) 5 (4c - 1)/3; if c + 1 (mod 31, by Lemma 2.l(b), d(zy, C) 5 4c/3.

(b) By the given conditions, c = 0 (mod 3) and c 2 6, and it follows from Lemma 2.l(b) that d ( x , C ) = d(y,C) = $c. Let Q be any segment of C with IV(Q)I = 3k + 2. Then, for any a E (5, y}, by Lemma 2.l(a) d(a , C - Q) 5 $(c - IV(Q)\) + 1. Noting that c - IV(Q)I = 1 (mod 3), we have d(a ,C - Q) 5 $ ( c - IV(Q)I) + $, and therefore

(2.3) 2 2 1

d(a, Q) = -C - d(a, C - Q ) 2 -IV(Q)I - - = 2k + 1. 3 3 3

Suppose, to the contrary, that d(xy, U , U , + ~ ) = 4 for some i. By relabeling, we may assume that d(xy, u1u2) = 4. By the maximality of c, either d(u,, xy) = 0 or d(u3 , xy) = 0. Without loss of generality, we assume that d(u3,zy) = 0. By (2.3) with Q = u3u4,d(u4,zy) = 2. Then d ( u 5 , x y ) = 0 for otherwise a longer square-cycle results by replacing by xy. Again by (2.3) with Q = u5u6, d(u6, xy) = 2. We claim that c 2 9. If this is not true, then c = 6 and {a1, u2, u4, u6, x, y} induces either K6 or K6 minus the edge ~ 4 ~ 1 , which gives a square-cycle with at most one 3-chord missing. Thus, since it is optimal, C contains all the 3-chords, except for at most one. Then the graph induced by V ( C ) U {z,y} has a square-cycle of length at least 7, a contradiction. Therefore c 2 9, as claimed. If d(u,,xy) = 0, by (2.3) with k = 2 and Q = u,-2uc-1u[~u1u2u3u4u5, we obtain that d(u,-2u,-l, xy) = 4 and ~,-2u,-~zyu~u2 . . . tLc-3uc-2 is a square-cycle of length c + 1. Thus we assume that d(u,, zy) 2 1. By symmetry, we may assume that ucx E E(G). Note that d(u3, zy) = d(u5, zy) = 0. By (2.3) with Q = u3u4u5u6u7, we have that d(u6u7, zy) = 4. Thus C' = u,ulu2zu4yu62L7"'uc-1u[: is a square-cycle of length c. We note that zu,.,xu6,yu2 are 3-chords in C'. By the choice of C, there must be at least three 3- chords incident with u3 and u5. So either u,u3 or u5u8 is a 3-chord in C. In the former case, u,u1u3th2u4xy'zL6u7.. . u,-~u,, and in the latter case ulu22yuqu6ugu7ug.. . u,ul, is a square-cycle of length c + 1. This contradiction shows that d(zy, U , U ~ + ~ ) 5 3 for all consecutive ui, u ,+~ E V(G) .

Since d(z ,C) = i c , we may assume that 17: E N(u1u2). By the above result, y q! N(uluZ) . Without loss of generality, we suppose that yu2 # E(G). By (2.3) with Q = u2u3u4ugu6, we have that y E N ( ~ ~ u j + ~ ) , 3 5 j 5 5. This completes the proof of Lemma 2.5. I

Lemma 2.6. Let C = u1u~ . . . u,ul be a chord-maximum square-cycle in a graph G. Let T be a triangle in G - C. Then d ( T , C) 5 2c - 1.

Pruuj If c = 3, then d(v, C) I 2 for all 'u E V ( T ) , which gives that d ( T , C) I 6. But, if d ( T , C) = 6, then, since G has no K4, the subgraph induced by V ( T ) u V ( C ) is a square-cycle of length 6, contradicting the maximality of c . Thus d(T, C) 5 5 = 2c - 1. We assume therefore that c 2 4. Let x, y, z be the three vertices of T. If the assertion is not true, then by Lemma 2.5,

(2.4) 2 3

d(x, C ) = d(y , C ) = d ( z , C ) = -c.


and d(T, uiu,+l) 5 4 for all consecutive u i , ui+l E V ( C ) . By (2.41, d(T, C) = 2c, and it follows that d(T, U , U , + ~ ) = 4 for all consecutive ui, u ,+~ E V ( C ) . We note that c E 0 (mod 3 ) by (2.4), and hence c 2 6.

Case 1. There is some s such that d(u, , T ) = 1, say d(ul, T ) = 1. Then d ( u 2 , T ) = 3, which implies that d(u3, T ) = 1, and then d(u4, T ) = 3 , . . . , d ( u ~ k + l , C ) = 1, d ( ~ ~ ~ , G) = 3 , . . . . We may suppose that u1x E E(G). If c = 6, then u6U1tL2xyzu4u6 is a square-cycle of length 7, a contradiction, thus c 2 9. If u1 and us are joined to distinct vertices of T , then replacing u3 by T gives a square-cycle of length c + 2 . Thus ul and u5 are joined to the same vertex x. By the same reason, u3 and u7 are joined to the same vertex in T. Since x cannot be joined to the four consecutive vertices, the vertex in T joined to u7 must be distinct from x, say it is y. Then u1U.m4zyu6u7 . . . u,.u1 is a square-cycle of length c + 1, thus case 1 leads to a contradiction.

d(ui,T) = 2 for all u, E V ( C ) . By (2.4) and Lemma 2.5(b), we may assume that d ( u l , zy) = 2 and d ( u 2 , yz) = 2 .

u,y E E(G) . Noting that y, by the maximality of c, cannot be joined to four consecutive vertices, we have that u 3 y @ E ( G ) and so d(ug , xz) = 2 . If u,x E E(G) , then uu,:u1xyzu2u~~~"uc is a square-cycle of length c + 3. Thus we assume that ucz E E(G). If ulz E E(G) , then u,u1yu2zu3u4. "u, is a square-cycle of length c + 2; otherwise d(u~q, xy) = 2 , then ulu2yzZu3u4 . . . ucul is a square-cycle of length c + 3.

u,y @ E(G) and so d(u, ,xz) = 2 . Clearly, u3z @ E(G) for otherwise, as before, u,ulzyzuzu:3 . . .uc is a square-cycle of length c + 3. So d ( u g , x y ) = 2 . Since y cannot be joined to four consecutive vertices, u4y @ E(G) and SO d(u4 ,xz ) = 2 . If U , . - ~ X E E(G), then ~ ~ - ~ u , x z y u ~ u ~ u 4 ~ ~ . u,.-1 is a square-cycle of length c + 2; otherwise d(u,-l,yz) = 2 , then u,-1u,zxyu1u2. . . u , - ~ is a square-cycle of length c + 3. In either

I

Lemma 2.7. Let G be a graph on n vertices and let C be a chord-maximum square-cycle of length c in G. Set H = G - C. If H f 0 and d(v,G) 2 $n for all w E V ( H ) , then E ( H ) f 0 and any edge of H is contained in a K4 in H.

Let h = / V ( H ) / . By Lemma 2.l(b), S ( H ) 2 (2h - 1)/3 > 0 and so E ( H ) # 8. For any e E E ( H ) , by Lemma 2.5(a), d ( e , H ) 2 t h , which implies that e is contained in a triangle, say T, in H . By Lemma 2.6, d ( T , H ) 2 2h + 1 and so T is contained in a K4 in H . I

Lemma 2.8. Let G be a graph on n vertices with b(G) 2 $71 and let C be a chord- maximum square-cycle of length c in G, where c < n. Set H = G - C and h = /V (H) I . Let Q be a chord-maximum square-cycle of length q in H . If q < h, then q I h - 4.

E V ( H - Q ) , then by Lemma 2.7, H - Q contains a K4, and we are done. Assume that there is y E V ( H - Q) such that d ( y , H ) < gh, which together with Lemma 2.l(b) implies that d(y, C) = (2c + 1)/3 (so c = 1 (mod 3)) . Let H' = H - y. For any w E V ( H ' ) , if d(w, C) = (2c + 1)/3, then by Lemma 2.3, vy @ E(G). This means that b ( H ' ) > i ( h - 1). If q < ( h - l) , then the result follows by the arguments we just used. Otherwise, q = h - 1 and d ( y , Q) = (2q + 1)/3 (so q = 1 (mod 3)). If q 2 5, by applying Lemma 2.3(b) to y and Q, we have an independent set S1 C V ( Q ) suchthat ISl] = (q-1)/3,S1u{y}isanindependentset,andNQ(a) = N Q ( ~ ) , implying d(a , C) = (2c + 1)/3, for all a E S1. If q = 4, let S1 = { z } , where z is the only vertex in Q not joined to y. (Note that d(z, C) 2 $n - 3 = (2c + 1)/3 and hence d(z , C) = ( 2 c + 1)/3 by Lemma 2.1(b).) In either case, S1 U {y} is an independent set with

Case 2.

Case 2a.

Case 2b.

case we have a contradiction. This completes the proof of Lemma 2.6.

Proof.

Proof. By Lemma 2.4, c 2 5. If d(v, H ) 2 $h for all


IS11 = (q- l ) /3andd(w,C)= ( 2 c + 1 ) / 3 f o r a l l v ~ S1u{yj . ItfollowsfromLemma2.3 that there is an independent set S2 C V ( C ) such that IS2 J = ( c - 1)/3 and S2 U S1 U { y j is an independent set, which has cardinality n/3, and by Lemma 2.2, c = n, a contradiction. This completes the proof. I

Lemma 2.9. Let G be a graph on n vertices and let P = ~ 1 ~ 2 ~ 3 . . . uP-1uP be a square- path starting at u1u2 and ending at uP-luP. Set H = G - P and h = IV(H)I. Let A = {u E V ( H ) : d(v, G) 2 (5n + 1)/6}. If p > 2h - 5, then G has a square-path P’ starting at u1u2 and ending at uP-luP with V(P’) = V ( P ) u A.

We use induction on /A / . If A = 0, there is nothing to prove, and thus we assume that A # 0 and let a E A . We note that d(a ,P) 2 (5n + 1)/6 - (h - 1) = z ( p + 1) + ( p + 5 - 2h)/12 > i ( p + 1), which implies that there are four consecutive vertices u2,uz+l,u,+2,ut+3 on P that are all joined to a, so that Q = u1u2 . . . u,uz+lau~,+Zu2+3~,-I~~ is a square-path. Let H’ = H - Q and A’ = A \ { u } .

I

Proof:

3

Obviously IV(Q)I > 21V(H’)J - 5, and the result follows by induction.

Lemma 2.10. Let G be a graph on n vertices with a square-path P = u1u2 . . . ~ , - 3 ~ , - 2

such that G - P is a single edge, denoted by xy. Suppose that u1 $ N(zy). If d(u1, G) + d(y,G) 2 $n - 1 and d(u1u2,G) + d(xy,G) 2 (10n - 8)/3, then G has a hamiltonian square-path starting at sy and ending at ~ , - 3 ~ , , - 2 .

Let B = {u,u,+~ : u, E N(zy) and u,+1 E N ( u I z L ~ ) , 1 5 i I n - 3). For each U,U,+~ E B and i + n - 3 (note that u1 @ N(zy)), if both u,-1 E N(y) and u,+2 E N(u l ) , then zyu,u,-1 . . . u2u1u,+1u2+2 ‘ . . ~ , - 3 ~ , - 2 is a square-path with the required property. Thus we assume that this is not the case. Then d(ul ,P) + d(y, P ) I ((V(P)I - 1) + (V(P) / - (If?/ - 1) = 2n - 4 - IB(. Since ul $! N ( x y ) , we have d(u1,xy) I 1, and obviously d(y , x) = 1. It follows that

Proof:

5 -n - 1 I d ( u l , G) + d(y, G) I d(u1, P ) + d(y , P ) + 2 I 2n - 2 - JBJ, 3

which gives that ( B ( 5 n/3 - 1. We shall prove that this is impossible. Let D = {u2 : u,+~ E N(ulu2), 1 5 i 5 n - 3). Clearly, ID1 = JNp(u1u2)J = IN(uIu2) \ {z,yjI . Since u1 @ N(xy), we have I{x,y}nN(u1u2)1 5 1. Noting that IN(u1u2)I 2 d(u1u2,G) -n , we obtain that (Dl 2 d(ulu2, G) - n - 1. By definition, B = {ZL,U,+~ : u, E D n N(zy)} and hence

which gives that JBJ 2 (472 - 11)/3 - ID U N(zy)J. Since u1 4 D U N(zy), ID U N(xy)I 5 IV(P)I - 1 = n - 3, and it follows that IBI 2 (n - 2)/3. This contradiction completes the proof of Lemma 2.10. I

3. AN UPPER BOUND FOR LONGEST SQUARE-CYCLES

Theorem 3.1. Let C = u1u2 - - .ucu l be a chord-maximum square-cycle of a graph G with n vertices. Suppose that G - C f 0 and d(v,G) 2 $n for all ‘u E V ( G - C) . Then c 5 $n, with equality only if there is a triangle T in G - C such that d ( T , G) = 2n.


We first prove two additional lemmas and then go to the proof of Theorem 3.1. Recall that a cube-path contains all the 2- and 3-chords. Define an almost-cube-path to be a square-path that has at most one 3-chord absent.

Lemma 3.2. Let P = x1x2 ~ ~ ~ x p - - 2 x p - l x p be a cube-path, where p 2 5. Then x1z2 is connected to each of the ordered edges {xp--lxp, xpzp-l, xp-2xp, xpxp--2} by a square- path P’ with V(P’) = V(P) , and to each of the ordered edges {xp--2xp--1,xp-1xp-~} by a square-path P” with V(P”) C V ( P ) and IV(P”)I = p - 1. Moreover, all of these square-paths are almost-cube-paths, possibly except for the one connecting x1 x2 to X ~ X ~ - ~ .

Trivially P itself is an almost-cube-path connecting x122 to xp-lxp. Since P is a cube-path, each of the following is an almost-cube-path: 21x2 . . . Xp-3zp-22pXp-l,21~2 . . . xp~3xp-1xp-2xpr x1x2 . . . xp-3xP--2xp-~, and x152 .. ~xp-3xp-lxp-2. Clearly, X ~ Z Z

I

Lemma 3.3. Let C = u1u2. . . u,ul be an optimal square-cycle of G and let xy and zw be two independent edges in G - C. Suppose that d ( ~ y , u ~ u ~ ) = 4 and xy is connected to Z7U and wz by square-paths of at least p vertices in G - C, at least one of which is an almost-cube-path.

Proof.

. . . ~ ~ - 3 % ~ - 1zpxp-2 is a square-path. This proves the lemma.

(a) If c > 2p + 3, then ~ ( z w , C) 5 $c - $ p + !. (b) If c 5 2p+ 3, then d ( z w , C ) 5 c + 2.

Proof. Let Q = u3u4 . . . up and Q’ = ucu,_ . . . u9t be two segments of C starting at u3u4 and U , U , _ ~ , respectively, such that (i) V(Q) n V(Q’) = 0 and both IV(Q)l 5 p + 1 and IV(Q’)I 5 p + 1; (ii) d(u,, zw) 5 1 and d(u,t,zw) 5 1; (iii) subject to (i) and (ii), IV(Q)I + IV(Q‘)I is as large as possible.

Letu, ~ V ( Q ) , 3 < i < q - l . S i n c e ( V ( Q ) ( ~ p + l , i f d ( u , , z w ) =2, thend(u,+l ,zw) = 0, for otherwise zw or wz is connected to ~ , u , + ~ by a square-path and we may replace the segment u3u4 . . . u,- by a square-path in G - C to obtain a longer square-cycle. Taking into account that d ( u g , zw) 5 1, we have that

d(zw,Q) F IV(Q)I1 (3.1)

with equality only if d ( ~ , - ~ , zw) = 2 whenever d(u,, zw) = 0 with u, E V(Q). Similarly,

d(z.u~, Q’) F IV(Q’)I, (3.2)

with equality only if d ( ~ , + ~ , zw) = 2 whenever d(u,, zw) = 0 with u, E V(Q’). We claim that

with equality only if d(u,-,,zw) = 2(d(u,+l,zw) = 2) whenever d(u , , zw) = 0 with

If d ( u 1 u 2 , z w ) 5 3, then (3.3) follows directly from (3.1) and (3.2). Suppose that d(ulu-2, zw) = 4. Then d(u.3, zw) = 0 or d(u, , zw) = 0, for otherwise z or w is joined to four consecutive vertices or one of u, u1 zwu2u3, u,u1wzu2uy is a square-path. Without loss of generality, suppose that d ( u 3 , zw) = 0. Then, d(zw , Q) = d(zw, Q - us) 5 IV(Q)I - 1, with equality only if d ( ~ , - ~ , ~ w ) = 2 whenever d(u , , zw) = 0, where u, E V(Q - u3).

ut E V(Q)(uz E V(Q‘1).


This together with (3.2) gives (3.3). Let R = C - (Q u Q’ u {ulu2}) . By Lemma 2.l(a),

2 2 3 3 d ( z , R) 5 -(RI + 1 and d(w, R) 5 -IRI + 1. (3.4)

Combining with (3.3),

J

Substituting IRI = c - IV(Q)I - IV(Q’)I - 2 ,

(a) Since c > 2p + 3, the proofs of (3.1) and (3.2) also implies, by the maximality of IV(Q)I + IV(Q’)I, that IV(Q)I 2 p and IV(Q‘)I L p . If both have strict inequalities, then IV(Q)l+lV(Q’)I L 2p+2 and the result follows from (3.6). Thus, without loss of generality, we assume that IV(Q)I = p , noting that IV(Q)I + IV(Q’)I 2 2p. If equality does not hold in (3.31, then the result follows easily from (3.5) and (3.6). Thus we may also assume that equality holds in (3.3). If IRI f 0 (mod 31, then d ( z , R) 5 $IRI +$ and d(w, R) 5 $ IRI + $; if IRI = 0 (mod 3) and there is one strict inequality in (3.4), then d(zw, R) 5 $ IRI + 1. In either case, d ( z w , R) I 4 IRI + 4 , which implies, as seen in (3.5) and (3.61, that d(zw, C ) 5 $c - $ p + g, as required. Thus we may further assume that both equalities hold in (3.4) (so \RI = 0 (mod 3)). By definition, R = uq+luq+2 . . . ugr-l. Since both equalities hold in (3.4) and by Lemma 2.l(a), we have d(zw, uq+luq+2) = 4. Let P be an almost-cube-path connecting xy to zw or wz, say to zw. Now, C‘ = u1u2Puq+luq+2. ‘ . u,ul is a square- cycle. By the maximality of c, IV(P)I = p = IV(Q)1 and C’ has the same length as C. We shall arrive at a contradiction by showing that C’ has more 3-chords then C does.

If d(zw,u,) # 0, say d(z,u,) = 1, then d ( z , R U {u,}) = ( $ ( R ( + 1) + 1 > $(IRI + 1) + 1, contradicting Lemma 2.l(a). Thus, d ( z w , u,,) = 0, and since equality holds in (3.3), d(zw, uq-l) = 2 . If d ( z w , uq-2) # 0, then replacing uq by zw or wz yields a longer square- cycle. Thus ~ ( z w , uq-2) = 0, and again since equality holds in (3.3), d(zw, uq-3) = 2. (We note that IV(Q)I = p 2 4.) We show now that there is no 3-chord incident with any vertex of { U ~ - ~ , U ~ - ~ , U ~ } in C. If E E ( G ) , then u1u2Puq-luq+luquq+2uq+3 . . . u,ul is a square-cycle of length c + 2. If uquq-3 E E(G), then u1u2Puq--3uq-luquq+1 ...u,u1 is a square-cycle of length c + 3. If uq--2uq+l E E(G) , then u1u2Puq--3uQ-l uq-2uquq+1 . . ’ucu l is a square-cycle of length c + 4. If uq-2uq-5 E E(G) , then uq--5 uq-4uq-2uq-3uq-122vuq+l~q+2 . . . uq--(iuq-5 is a square-cycle of length c+ 1. If u,,-1u,+2 E E(G) , then u ~ u ~ P u q ~ ~ u q + ~ u q+2~~~ucu1 is a square-cycle of length c + 1. Finally, if uq-] uq-4 E E(G), then D = u q - 4 u q - 3 u q - 1 z ~ u q + l u q + ~ ~ ~ ~ u q--5uq-4 is a square-cycle of length c in which zuq+2 and wuqP3 are 3-chords. Since we have just showed that C has no 3-chord incident with uqP2 or uq and also uq-1uq+2 $ E(G) , the only 3-chord lost is uq-Iuq-4. However, we gain two 3-chords in D: wuq-3 and zuqf2 . So D has more 3-chords than C, contradicting the optimality of C. This proves that there is no 3-chord incident with any vertex of {uq-2,uq--1,uq} in C. Since P is an almost-cube-path and since uly and zuqf2 are 3-chords in C’, it follows that C’ has more 3-chords than C. This contradiction proves (a).


(b) Since c 5 2 p + 3, by the maximality of IV(Q)I + IV(Q’)I we have JRI 5 1, and it follows from (3.3) that d ( z w , C) 1. IV(Q)I + IV(Q’)I + 3 + 21RI = c + 1 + IRI 5 c + 2, as

Let H = G - C and h = 1 V ( H ) 1. Let K be a maximum complete subgraph in H with IV(K)I = k . By Lemma 2.7, we have k 2 4. By the maximality of K,d(K,u) < k - l f o r a l l u ~ V ( H - K ) a n d h e n c e d ( K , H ) 5 h(k- l ) . I fd(K,u iu i+l ) 5 k + 1 for every edge U , U ~ + ~ E E(C) , then summing these inequalities over all i , 1 5 i L c, yields that d(K, G) 5 (k + 1)/2c, and consequently,

required. I

Proof of Theorem 3.2.

(3.7)

which gives c 5 i n . If equality holds, then d(v , G) = i n for all v E V ( K ) . So d ( T , G ) = 2n for any triangle T in K. Therefore we assume that there is some edge U , U , + ~ E E(C) such that d ( K , u , ~ , + ~ ) 2 k + 2, which implies that d ( ~ , u , + ~ , e ) = 4 for some edge e E E ( K ) . Let P = ~ ~ 5 ~ 5 : ~ . . I G ~ - ~ ~ ~ be a longest cube-path in H starting at some edge zlx2 with d(x lz2 , u2u,+1) = 4 for some i, where p 2 k 2 4. By relabeling, we may assume i = 1. That is, d(zlz2, u1u2) = 4.

p = 4. So V ( P ) induces a K4. Clearly, c 2 Ic 2 4. By Lemma 2.5(b), noting that d(z1x2 ,u1u2) = 4, we have that d(xlz2,G) < i c . Without loss of generality we assume that d(z2, C) < $c. If c > 2 p + 3 , then, by Lemma 3.3(a), d(53x4,C) 2 $ c - 1 and hence d(x2z3z4, C) < 2c-1. By the maximality of P, d(z2x3xq, H - P ) 5 2(h-p) = 2h-8, and since V ( P ) induces a K4, we have that d(z3z4, P ) = 9, and consequently,

Case 1.

2n 1. d ( z 2 ~ 3 ~ 4 , G) < (2c - 1) + (2h - 8) + 9 = 2n, (3.8)

a contradiction. Therefore we suppose that c 5 2p + 3. By Lemma 3.3(b), d(x3z4, C) 5 c + 2. Using 4 ~ ~ x 4 , H ) 5 2(h - l ) ,

4 3 -n 5 ~ ( Z Q Z ~ , G) 5 c + 2 + 2(h - 1) = 2n - c, (3.9)

which gives that c 5 in. If equality, c = i n , holds, then d(x3,G) = d(xq,G) = i n , and moreover, d(a3z4, H ) = 2(h - l ) , which implies, by the maximality of P, that d(x2 , H - P ) = 0. Using d(z2 ,P) = 3, we obtain that 4 x 2 , G) < $ c + 3 I $ n + $, and hence d(z2 , G) = $n. Let T be the triangle on ( 5 2 , z3, xq}. Then d ( T , G) = 2n, as required.

Case 2. p 2 5. Let T be the triangle on ( ~ ~ - 2 , zp-l , z p } . If c > 2p + 3, by Lemmas 3.2 and 3.3(a),

4 2 5 3 3 3

3 3 3 3 3 3

5 -c - T p + -, d ( x p - 2 5 p , C )

5 -c 4 2 5 - - p + -, and ~ ( Z ~ - - ~ Z + I , C) L -c 4 2 - - (P - 1) + -. 5

Adding these inequalities yields that 2d(T, C) 5 4c-2p+5+$, which gives, since d ( T , C) is an integer, that d(T, C) 5 2c - p + 2. By the maximality of P, d ( T , H - P ) 5 2(h - p ) , and using d ( T , P ) 1. 3 ( p - l ) ,

2n 5 d ( T , G) 5 2c - p + 2 + 2(h - p ) + 3 ( p - 1) = 2n - 1. (3.10)


This contradiction shows that c i 2 p + 3 . By Lemmas 3.2 and 3.3(b), and as seen in (3.91, we have that

C) 5 c t 2 ,

4 -n 5 d(~:,-lzp, G) 5 c + 2 + 2(h - 1) = 2n, - C, (3.1 1) 3

which gives that c 5 $n with equality only if ~ ( Z , - ~ , G ) = d(z,,G) = $n. If equality holds, then, by applying Lemmas 3.2 and 3.3(b) to the edge xp-2xp, we have that d ( ~ ~ , - ~ , G) = d(x,, G) = jn, and hence d ( T , G) = 2n. This proves Theorem 3.1.

Theorem 3.2. Let C = ulu2 . . . u,u1 be a chord-maximum square-cycle in a graph G with n vertices. Suppose that G - C # 8 and d(v, G) 2 i n + r for all v E V ( G - C ) , where r is a nonnegative real number. Then c L: $n - 2r.

Let H = G - C and h = IV(H)I. As in the proof of Theorem 3.1, let K be a maximum complete subgraph in H with IV(K)\ = lc. If d(K,u,ui+l) 5 k + 1 for every edge uiu,L+l E E(C) , then, instead of (3.7), we have that

We may modify the proof of Theorem 3.1 to obtain the following generalization.

Proof.

J

which gives c I $n - 2 r k / ( k - 3 ) 5 f n - 2r, as required. Thus, as in the proof of Theorem 3.1, we may let P = ~ 1 ~ 2 x 3 . . . xp-lxp be a longest cube-path in H starting at some edge 51x2 E E ( H ) with d(zlz2,ulu2) = 4. Clearly, we still have contradictions in (3.8) and (3.10). Note that (3.9) can be included in (3.11) by allowing p = 4. Instead of (3.11), we now have that $n + 2r 5 ~(J: , -~J: , ,G) 5 2n - c, which gives that c 5 i n - 2r. This

I completes the proof of Theorem 3.2.

4. A LOWER BOUND FOR LONGEST SQUARE-CYCLES

Theorem 4.1. If G is a graph on n vertices with S(G) 2 $n, then G has a square-cycle of length at least n/2.

We need the following three lemmas in the proof of Theorem 4.1.

Lemma 4.2 [4, Theorem 2.11. Let G be a graph on n vertices. If 6(G) > $n, then any two independent ordered edges are connected by a square-path in G.

Lemma 4.3. Let G be a graph on n vertices with S(G) 2 $n. Let C = ~ 1 ~ 2 . . . u,u1 be a chord-maximum square-cycle in G. Set H = G - C and h = IV(H)I. If c < h, then the following hold.

(a) d(z,C) I $c for all J: E v ( H ) . (b) d(e , C ) L: (4c - 1)/3 for all e E E ( H ) .

Proof. By Lemma 2.4, c 2 5. Let UZI E E(G) with u E V ( C ) and ZI E V ( H ) . Clearly, d ( u , N ) 2 $n-(c-l) = (2h-c)/3+1. By Lemma2.l(b), d(w, H ) 2 (2h-1)/3. Therefore,

h - c - 1 d(u, H ) + d(v , H ) 2 h + 1 + and so l N ~ ( u v ) 1 2 1, (4.1)


with equality only if d ( u , C) = c - 1 and d(v, C) = (2c + 1)/3. (a) If the statement is not true, let z E V ( H ) with d(z , C) 2 (2c + 1)/3. By Lemma

2.l(b), d ( z , C ) = (2c + 1)/3 (so c 2 7). Let H’ = H - z. We first claim that there is y f V ( H ’ ) such that d ( y , ~ ~ u , + ~ ) = 2 for some j , 1 5 j I c. If this is false, then d ( y , C) 5 c /2 for all y E V ( H ’ ) and so

(4.2)

Furthermore, since d ( y , ~ p , + ~ ) 5 1 for ally E V ( H ’ ) , we have that 3n 5 d(u,u,+l,G) 5 h + 1 + 2(c - l), which yields, using n = h + c, that c 2 ( h + 3)/2. Applying this to (4.2) yields that S(H’) 2 $ ( h - 1) 2 $ ( h - 1). By the result in [3], H’ has a hamiltonian square-cycle. This implies that H has a square-cycle of length at least h - 1, and it follows from Lemma 2.8 that H has a hamiltonian square-cycle, contradicting c < h. This shows that there is y E V ( H ’ ) such that d(y , ujuj+l) = 2 for some j, 1 I j 5 c , as claimed. Since d(z , C) = (2c+1)/3 and c 2 7, there is uiulLi+l E E ( C - U ~ U , ~ + ~ ) such that d(z , ~ , & u i + ~ ) = 2. Choose i and j such that the two edges ~ ~ u , + ~ and u,7uj+l are as close as possible. By relabeling, we may assume that i = 1, that is, d(z,u1u2) = 2. By Lemma 2.l(b) and by the choice of i and j, 3 5 j L 5. By (4.11, we may let a E N ~ ( z u 2 ) and b E N ~ ( y u , , ) .

a # y and b $ { a , z } . By Lemma 2.5, d ( a z , H ) _> $n. - $ c = i h . Since h 2 c + 1 2 8, we see that I N ~ f ( a z ) l 2 3. Let f E N ~ ( u z ) \ {b,y). Now consider the subgraph H’ = H - z. For any v E V ( H ’ ) , by Lemma 2.3, if d(v, C) = (2c + 1)/3, then v z 4 E(G). This implies that S(H’) > $lV(H’)l , and it follows from Lemma 4.2 that a f is connected to by by a square-path P in H‘. Then u1u2xPufufII . . . ucuI is a square cycle of length longer than C.

a f y and b = 2. By Lemma 2.3, d(y,C) # (2c + 1)/3, and by (4.1), INH(yu3)( 2 2. Let z E NH(yuj)\{z). If z # a , we havecase la; if z = a,ulu2sayuju,+l . . . u,ul is a longer square-cycle.

a # y and b = a. If I N ~ ( y u j ) l 2 2 or I N ~ ( z u ~ ) l 2 2, then we have either Case l a or Case l b above. Suppose that this is not the case. By (4.1), d(z , C) = d(y , C) = (2c + 1)/3, and by Lemma 2.4, N c ( z ) = Nc(y) and so z is joined to both u j and uj+l. Since we have equality in (4.11, d(u2, C) = c- 1 and so uLL,-1ucu2u1 is a square-path, which means that we may use xu1 as xu2 in the above arguments and derive that NH(zu l ) = {a>. Now ulu2azuju,,+1 ..’ucul is a square-cycle. Thus j = 5, and by the choice of i and j , no vertex in H is joined to both ug and u4. So d(ugu4,H) I h. Let w E {u3,u4} with d ( w , H ) 5 i h . Then d(w ,C) 2 i n - i h = (3c + n)/6. Using n 2 2 c + 1, we derive that d(w, C) 2 (5c + 1)/6. Let P = u 4 u g u g . . . ucu1u2. It follows from Lemma 2.9 that there is a square-path P’ starting at u4u5 and ending at ulu2 with V(P’) = V ( P ) u {w}. Then P‘u2azu4 is a square-cycle of length c + 1.

a = y. If (N~(zu~)l 2 2, then it is one of the above cases. We thus assume that N ~ ( z u 2 ) = {y}. So d(u2,C) = c - 1. As in Case lc, we have that Nli(zul) = {v}. Since y cannot be joined to four consecutive vertices of C, we have j 2 4. No matter whether j = 4 or j = 5, by the choice of i , j and by Lemma 2.l(b), we always have xu3 E E(G) . So ulu~zyujuj+l .. .u,:u1 is a square-cycle, which implies that j = 5. As in Case lc, using d(u3u4, H ) 5 h, we obtain a longer square-cycle containing w E {us, u4} . In any case, we have a contradiction to the maximality of c. This proves (a).

(b) By statement (a), d ( e , C ) 5 $c for all e E E ( H ) . If (b) is false, let zy E E ( H ) with d(zy,C) = :c (so c = 0 (mod 3) and c 2 6). By Lemma 2.5, with a proper labeling we have that z E N ~ ( u l u 2 ) and y E N H ( u ~ u ~ + I ) , where 3 I t 5 5. By (4.11,

Case la.

Case lb.

Case Ic.

Case 2.


/ N I J ( Z U ~ ) ( 2 2 and I N ~ ( y u t ) ( 2 2. If there is z E V ( H ) such that N ~ ( z u ~ ) = {z,y}, then u I u 2 z ~ y u t u t + l . . . uCul is a square-cycle of length at least c + 1. Otherwise, there are distinct a , b E V ( H ) \ {z, y} such that a E Nfi(zu2) and b E N H ( y u t ) . By (a), d(az , H ) 2 3h, 3 and since h 2 c 2 7,(N~(az)l 2 3. Similarly, l N ~ ( b y ) I 2 3. If d ( a z , b y ) = 4, then ulu2zabyutut+l . . . ucul is a longer square-cycle. Thus d(az, by) 5 3 and it follows that there are distinct u’,b’ E V ( H ) \ { a , b,z ,y} such that u’ E N ~ ( a z ) and b’ E N ~ ( b y ) . Let H’ = H - z - y. For any v E V ( H ) , by Lemma 2.6, if d ( v , C) = gc, then d(v, zy) 5 1. This means, since c = 0 (mod 31, that S(H’ ) 2 i n - i c - 1 > :(h - 2). By Lemma 4.2, aa‘ is connected to b’b by a square-path Q in HI. Then u1 u2xQyutut+l . . . u,ul is a longer

I

Lemma 4.4. Let G be a graph on n vertices with S(G) 2 $ 7 ~ and let C = u1u2 . . . u,u1 be a chord-maximum cycle in G. Set H = G - C and h = IV(H)I . Let z E V ( H ) with d ( z , C) = max{d(v, C ) : u E V ( H ) } . If c < h, then the following hold.

square-cycle. This contradiction proves (b) and so the lemma.

(a) For distinct u, v E V ( C ) and y E V ( H ) \{x}, if uz, yv E E(G), then uz is connected to yv through a square-path Q in H with IV(Q)I 2 1.

(b) z E NH(uL,_lu,) for some i, 1 I i 5 c, and if N ~ ( u ~ u ~ + ~ ) \ { z } # 0, where3 2 i+2 , then j - i 2 6 with equality only if d(u,+l, H ) > h/2.

Proof. By applying Lemma 4.3(a) to (4.1), we have that INH(uz)) 2 2 and INH(vy)I 2 2. If N H ( U Z ) = {y, z } and N ~ ( y v ) = (5 , z } for some z E V ( H ) , then uzzyv is a square- path and the result holds with V ( Q ) = {.}. Otherwise, there are distinct wl , w2 E V ( H ) \ {z, y} such that w1 c N ~ ( u z ) and wz E N ~ ( v y ) . If d ( z , C) < Zc, then, by the choice of z , S ( W ) > $h, and by Lemma 4.2, xwl is connected to way by a square-path in H and so uz is connected to yw through a square-path Q in H with {w1,w2} C V ( Q ) . Thus we assume that d(z , C) 2 $c, and by Lemma 4.3(a), d(z , C) = $ c (hence c 2 6 by Lemma 2.4). Using Lemma 4.3(b), we derive that d(zwl, H ) 2 (4h + 1)/3. Since h 2 c + 1 2 7 , it follows that \ N ~ ( z w ~ ) \ 2 3 and there is z E NH(zwl) \ {w2,y}. Let H’ = H - z. For any w E V ( H ’ ) , by Lemma 4.3(b), if d(v ,C) = gc, then vuz $! E(G). This implies that S(H’) > $ ( h - l), and by Lemma 4.2, wlz is connected to w2y by a square-path in HI, which gives a square-path from uz to yu with the required property. This proves (a).

If x y! NH(u,-~u,) for all i, 1 5 i 5 c, then d(z , C) 5 i c . By the choice of z,

2 1 3 h + n 2 6 ’

S ( H ) 2 3n - -c = ~ (4.3)

and moreover, d ( H , C ) 5 ach, which implies that gnc 5 d(C,G) 5 +ch + c(c - l), and hence, n 2 ;(h+2). Applying this to (4.3) gives that 6 ( H ) 2 (3h+2)/4 > qh. By the result in [3], H has a hamiltonian square-cycle, contradicting c < h. Therefore, z E NH(u,-lu2) for some i. Now let y E N ~ ( u ~ u ~ + ~ ) \ {z} and choose j such that j - i is as small as possible. By statement (a), there is a square-path Q in H , where IV(Q)l 2 1, such that u,-1u2zQyu,uj+1 . . . uLL,_zu,-l is a square-cycle of length at least c - ( j - i - 1) + 3. Thus j - i 2 4. For convenience, let L = a1a2 . . . at be the segment of C from u,+] to uJ- l , where a1 = U , + ~ , U ~ = u,+2,. . . , and t = j - i - 1. If t 2 6, the proof of part (b) is completed, and thus we assume that t 5 5. By the minimality of j - i, NH(ulul+l) = 0 for all 1 ,1 I 1 I t - 1, and then,

d ( a l a l + l , H ) l h f o r d l l , l < l 5 t - l . (4.4)


Let A = { u E V ( L ) : d ( a , H ) 5 h/2}. By (4.4), ]A1 2 ( t - l ) / 2 with equality only if al $! A . We note that, for each a E A , d(u, C ) >_ $n - f h = (3c + n)/6 2 (5c + 1)/6. By applying Lemma 2.9 to the graph induced by V ( C ) with P = uJu,+1u3+2.. . u,-~u, (by the minimality of j - 2, IV(P)I 2 t + 4 > 2t - 5), we obtain a square-path P’ starting at u7uJ+1 and ending at u,-1uL2 with V(P’) = V ( P ) u A. Then P’u,xQyuZLJ is a square-cycle of length at least c - t + IA( + 3, which implies that t > 5 with equality only if a1 $ A.

Let C = u1u2 . . . u,ul be a chord-maximum square-cycle in G. Assume, contrary to assertion, that c < n/2. By Lemma 2.4, c > 5. Set H = G - C and h = IV(H)I. So c < h. Let x E V ( H ) with d ( x , C ) = max{d(v,C) : ’u E V ( H ) } . By Lemma 4.4(b), x E N ( U , ~ , + ~ ) for some 2. By relabeling, we assume that z E N(ucul ) , and again by Lemma 4.4(b), both N ~ ( u ~ u ~ ) \ {x} = 0 and N H ( U ~ U ? ) \ {x} = 0. If we also have N ~ ( u 2 u ~ ) \ {z} = 0, then d(u2u3u4, H ) 5 h + 2 and so

This completes the proof of part (b), and so the lemma. a Proof of Theorem 4.1.

2n 5 d ( ~ 2 ~ 3 ~ 4 , G) 5 h + 2 + 3 ( ~ - I ) = + 2~ - 1,

which gives that c 2 (n + 1)/2, a contradiction. Therefore N , Z I ( U ~ U ~ ) \ (z} # 0. Let y E N H ( u z u ~ ) \ { ~ } . By Lemma 4.4(a), u1x is connected to yu2 and to yu4 through square- paths Q and &‘ in H , respectively, where lV(Q)l > 1 and lV(Q’)I > 1. If u2u5 E E(G), then u,ulxQyu2u4u5. . . u,ul is a square-cycle of length at least c+2. Thus u2u5 @ E(G). Since x cannot be joined to four consecutive vertices on C, x $ N(u2u3) and so d(u2u3, H ) 5 h. Consequently,

which gives that c 2 n / 3 + 3. Using c 5 (n - 1)/2, we obtain that n > 21, and therefore, c > 10. Now, if u3u6 E E(G) , then u , u l ~ Q y u 2 u 4 u ~ ~ u ~ u ~ ~ ~ ~ u , u 1 is a square-cycle of length at least c + 3, which is impossible. So u3u6 $ E(G). Then, instead of (4.3, we have that

4 -n 5 d ( ~ 2 ~ 3 , G) 5 h + 2 ( ~ - 1) - 2 = 72 + c - 4, 3

and as above, c 2 13. Using Lemma 4.4(b), we derive that

d ( u , ~ , + ~ , H ) 5 h + 1

with equality j = 7 only if d(u2 , H ) > h/2.

(= C - u2u3uq) and let D = C - u 4 (so u2u3 is the single edge of D - P). By (4.6),

for all j, 2 5 j 5 7, (4.6)

Case la. d (u3 ,H) 5 (h + 1)/2 and d ( u 5 , H ) 5 ( h + 1)/2. Let P = u5u6 . . . u,ul

8 2 3 3

d ( ~ 2 ~ 3 , D ) + d ( ~ g ~ g , D ) > - T X - 2(h + 1) - 4 = 2~ + -n - 6.

Using n 2 2c + 1 and IV(D)I = c - 1,

and furthermore, since d(u3, H ) 5 (h + 1)/2 and d(u5, H ) 5 (h + 1)/2,


We have showed that u2u5 $ E(G) and so u5 $ N(u2u3), it follows from Lemma 2.10 that the subgraph induced by V ( D ) has a hamiltonian square-path P’ starting at u2u3 and ending at u,u1. Then uCu1~Q’yu4u2u3P’ is a square-cycle of length at least c + 3.

Case lb. d(u3, H ) 5 ( h + 1 ) / 2 and d(u5, H ) > ( h + 1) /2 . Then by (4.61, d(u6, H ) 5 h/2. Let L = U g u 7 . . . U,UI and F = C - ~ 4 ~ 5 . We have that

8 10lV(F)I - 8 3 3

d ( ~ 2 ~ 3 , F ) + d ( ~ g ~ 7 , F ) 2 -n - 2(h + 1 ) - 8 2

and

5 d ( ~ y , F ) + d ( ~ g , F ) 2

In the proof of c 2 13, we showed that u3u6 $ E(G) , and so u6 f N(u2u3). It follows from Lemma 2.10 that the subgraph induced by V ( F ) has a hamiltonian square-path L’ starting at ~ 2 ~ 3 and ending at u,ul. Then u,u12Q’yu4u2u3P’ is a square-cycle of length at least c + 2.

Case 2. d(u3,H) > ( h + 1 ) / 2 . By (4.6), d ( u 2 , H ) 5 h/2 and d ( u q , H ) 5 h/2. Since d ( u 2 , H ) 5 h/2, we do not have equality j = 7 in (4.6) and so d(u7u8,H) 5 h + 1. As before, we have z E NH (u3u5)\{~} and u1z is connected to zu3 and to zug by square-paths in H . Using d(u2, H ) 5 h/2 and n 2 2c + 1 ,

2 h 3 c + n 5 c + 1 3 2 6 6

d(u2, C ) 2 -n - - = __ 2 -.

Let W = u7ulg.. . u,u1 and S = ~ 2 ~ 3 . . . u6. Since c 2 13, we have that IV(W)l = c - 5 2 8 > 2(V(S)I - 5. It follows from Lemma 2.9 that there is a square-path W’ starting at u7u8 and ending at u,ul with V(W’) = V ( W ) U {u2}, and the proof is completed by the same arguments as those used in Cases l a and lb.

5. PARTITION INTO TWO SQUARE-CYCLES

Theorem 5.1. Let G be a graph on n vertices with S(G) 2 $n. Then G has two vertex- disjoint square-cycles C1 and C2 such that V ( C l ) u V(C2) = V ( G ) and C1 is chord- maximum.

Proof. If n 5 5 , G is complete and the theorem holds with C1 = G and C2 = 0. Assume that n 2 6 and let C1 be a chord-maximum square-cycle of length c1 in G. By Lemma 2.4, c1 2 5. If c1 = n, the theorem holds with C2 = 0. We thus assume that c1 < n. Let H = G - C and h = IV(H)I. By Theorem 4.1, c1 2 h. Let C2 be a chord-maximum square-cycle of length c2 in H . If c2 = h, we are done, and thus we assume that c2 < h. Set R = H - C2 and T = IR1. By Lemma 2.8, T 2 4.

If 27- 5 cl, let D = G - C2. Clearly C1 is also a chord-maximum square-cycle in D. If d(v,C2) 5 $c2 for all v E V ( R ) , then d(v, D ) 2 $(n - c2) for all v E V ( R ) , and it follows from Theorem 3.1 that 27- 2 cl, and so 27- = c1 and there is a triangle T in R such that d(T, D ) = 2(n - c2). By Lemma 2.6, d ( T , C2) 5 2c2 - 1. It follows that


This contradiction shows that there is z E V ( R ) such that d(z ,C2) 2 (2c2 + 1)/3, which means, by Lemma 2.l(b), that d(z,C2) = (2c2 + 1)/3. Let D’ = G - C2 - z. By Lemma 2.3(a), for each w E V ( R - z), if d(v, C2) = (2c2 + 1)/3, then wz $! E(G) , which implies that d(w, 0’) > $lV(D‘l for each w E V ( R - z). As before, 2 ( ~ - 1) 2 el, a contradiction.

If 2r 2 c1 + 1, then T 2 ( h + 1)/2 and c2 5 ( h - 1)/2. Since C2 is chord-maximum in H and by Theorem 4.1, we have that S ( H ) < $h. Thus there is x E V ( H ) such that d(z,C1) = (2c1+1)/3 (so c1 = 1 (mod 3)). Let H’ = H - x . As before, S(H’) > $lV(H’)I, and it follows from Theorem 4.1 that H’ has a square-cycle of length at least ( h - 1)/2. Therefore, c2 = ( h - 1)/2,r = ( h + 1)/2, and c1 = h. Then n = 2 (mod 3) and so S(G) 2 (an + 1)/3, which implies that S ( H ) 2 (2n + 1)/3 - (2cl + 1)/3 = $h, a contradiction.

I

Note added in proofi Komlos, Siirkozy, and SzemerCdi have just used the Regularity Lemma and a new lemma they call the Blow-up Lemma to prove that P6sa’s conjecture is true for sufficiently large n, where “sufficiently large” means much more than Z1Oo.

This completes the proof of the theorem.

References

[ 11

[2]

[3]

[4]

[5]

(61

[7]

G. A. Dirac, Some theorems on abstract graphs. Proc. London Math Soc. 2 (1952), 68-81.

P. Erdos, Problem 9, Theory of Graphs and its Applications (M. Fieldler ed.), Czech. Acad. Sci. Publ., Prague ( 1 964), 159. G. Fan and R. Haggkvist, The square of a hamiltonian cycle, SIAM J. Discrete Math., 203-212.

G. Fan and H. A. Kierstead, The square of paths and cycles, J. Combinat. Theory Ser: B 63 (199% 5544 .

G. Fan and H. A. Kierstead, Hamiltonian square-paths, preprint, .I. Combinat. Theory Sex B, in press.

A. Hajnal and E. Szemeredi, Proof of a conjecture of P. Erdos, Combinatorid Theory and its Application (P. Erdos, A. Renyi, and Vera T. Sos eds), North-Holland, London (1970), 601423.

P. Seymour, Problem section, Combinatorics: Proceedings of the British Combinatorid Confer- ence 1973 (T. P. McDonough and V. C. Mavron eds), Cambridge University Press, Cambridge (1974), 201-202.

Received April 18, I995

partitioning a graph into two square-cycles

Documents