On directed versions of the Corrádi-Hajnal Corollary

On directed versions of the Corrádi-Hajnal Corollary Andrzej Czygrinow H. A. Kierstead heodore Molla October 4, 01 Abstract For k N, Corrádi and Hajnal proved that every graph G on k vertices with minimum degree δ(g) k has a C -factor, i.e., a partitioning of the vertex set so that each part induces the -cycle C. Wang proved that every directed graph G on k vertices with minimum total degree δ t ( G) := min v V (deg (v)+deg + (v)) (k 1)/ has a C -factor, where C is the directed -cycle. The degree bound in Wang s result is tight. However, our main result implies that for all integers a 1 and b 0 with a + b = k, every directed graph G on k vertices with minimum total degree δ t ( G) 4k 1 has a factor consisting of a copies of T and b copies of C, where T is the transitive tournament on three vertices. In particular, using b = 0, there is a T -factor of G, and using a = 1, it is possible to obtain a C -factor of G by reversing just one edge of G. All these results are phrased and proved more generally in terms of undirected multigraphs. We conjecture that every directed graph G on k vertices with minimum semidegree δ 0 ( G) := min v V min(deg (v), deg + (v)) k has a C -factor, and prove that this is asymptotically correct. 1 Introduction For a graph G = (V, E) set G := V and G := E. Let d(v) denote the degree of a vertex v, δ(g) := min{d(v) : v V } denote the minimum degree of G, and σ := min vw / E(G) d(v) + d(w) denote the minimum Ore-degree of G. Two subgraphs of G are independent if their vertex sets are disjoint. In 196 Corrádi and Hajnal [] proved: Theorem 1. Every graph G with G k and δ(g) k contains k independent cycles. School of Mathematical Sciences and Statistics, Arizona State University, Tempe, AZ 8587, USA. E-mail address: aczygri@asu.edu. School of Mathematical Sciences and Statistics, Arizona State University, Tempe, AZ 8587, USA. E-mail address: kierstead@asu.edu. Research of this author is supported in part by NSA grant H980-1-1-01. School of Mathematical Sciences and Statistics, Arizona State University, Tempe, AZ 8587, USA. E-mail address: tmolla@asu.edu. Research of this author is supported in part by NSA grant H980-1-1-01. 1

In 1998, Enomoto [4] proved an Ore-type version of Theorem 1: Theorem. Every graph G with G k and σ (G) 4k 1 contains k independent cycles. A tiling of a graph G is a set of independent subgraphs, called tiles. A tiling is a factor, if its union spans G. For a subgraph H G, an H-tiling is a tiling whose tiles are all isomorphic to H. The -cycle C is called a triangle. Theorem 1 has the following corollary, whose complementary version is a precursor and special case of the 1970 Hajnal-Szemerédi [7] theorem on equitable coloring. Corollary. Every graph G with G divisible by and δ(g) G has a C -factor. This paper is motivated by the problem of proving versions of Corollary for directed graphs G := (V, E ). Our directed graphs are simple, in the sense that they have no loops and for all x, y V there are at most two edges, one of form xy and one of form yx, whose ends are in {x, y}. The in- and out-degree of a vertex v are denoted by deg (v) and deg + (v); the total degree of v is the sum deg t (v) := deg (v) + deg + (v), and the semidegree of v is deg 0 (v) := min{deg (v), deg + (v)}. The minimum total degree of G is δ t ( G) := min{deg t (v) : v V }, and the minimum semidegree of G is δ 0 ( G) := min{deg 0 (v) : v V }. Let K G be a subgraph on three vertices x, y, z. If { xy, yz, xz} E( K) then K is a transitive triangle, denoted by T ; if { xy, yz, zx} E( K) then K is a cyclic triangle, denoted by C. Wang [1] proved the following directed version of Corollary. Theorem 4. Every directed graph G with δ t ( G) G has a C -tiling of size G. The degree condition of Theorem 4 is tight: Example 5. Suppose n = k + 1 is odd and divisible by. Let G be the directed graph with V ( G) = V 1 V, where V 1 V =, k = V 1, V = k + 1, and E( G) = F := { xy : x, y V 1 x, y V (x V 1 y V )}. Then δ t ( G) = k 1 = G 1, and no C contains vertices from both parts V 1 and V. So no tiling of G contains n cyclic triangles. While Theorem 4 is tight, we can significantly relax the minimum degree condition for the cost of at most one incorrectly oriented edge. Our main result implies: Corollary 6 (to Theorem 10). Suppose G is a directed graph with δ t ( G) 4 G, and c 0 and t 1 are integers with c + t = G. Then G has a tiling consisting of c cyclic triangles and t transitive triangles.

The degree condition of Corollary 6 is also tight. Example 7. There exists a directed graph G such that δ t ( G) 4 G 1, but G cannot be tiled with any combination of cyclic and transitive triangles: Suppose G = k. Set V ( G) = V 1 V, where V 1 V =, V 1 = k + 1, and V = k 1, and let E( G) = { xy : x / V 1 y / V 1 }. Then δ t ( G) = 4k = 4n 1, but G does not have any triangle factor, since every vertex in V 1 would need to be paired with two vertices from V, and there are too few vertices in V. Our methods are obscured by the elementary proofs of Theorem 10 and Corollary 6. We used stability techniques (regularity-blow-up, absorbing structures) to discover what should be true, and only then were able to concentrate our energy on a successful elementary argument of the optimal result. Sometimes the process works in the other direction. The striking gap between Wang s Theorem and Corollary 6, suggests that there is a better theorem. An important characteristic of Example 5 is that while every vertex has large total degree, it also has small semidegree. This led us to the following conjecture. Conjecture 8. Every directed graph G with G = k and δ 0 ( G) k has a C -factor. In support of this conjecture, we use stability techniques to prove the following asymptotic version. We expect that with more effort this approach can be improved to a proof of the conjecture for sufficiently large graphs by using the techniques similar to those of Levitt, Sárközy and Semerédi [11]. Corollary 9 (to Theorem 1). For every ε > 0 there exists n 0 N such that every directed graph G with G n 0 and minimum semidegree δ 0 ( G) ( + ε) G has a C -tiling of size. G It turns out that our results can be phased more generally, and proved more easily, in terms of multigraphs. Suppose that M = (V, E) is a multigraph. For two vertices x, y V let µ(x, y) be the number of edges with ends x and y. In particular, if xy / E then µ(x, y) = 0. Let µ(m) := max x,y V µ(x, y). The degree of x is d(x) := y V µ(x, y). The minimum degree of M is δ(m) := min{d(v) : v V }. A k-triangle is a multigraph T k such that C T k, T k = and T k = k. The underlying multigraph M of a simple directed graph G is obtained by removing the orientation of all edges of G. In particular, if xy and yx are both edges of G then µ M (x, y) =. By our definitions, δ t ( G) = δ(m). If M contains a 4-triangle with vertices x, y, z and µ M (x, y) = then G contains a transitive triangle with the same vertex set, since regardless of the orientation of xz and yz, one of the orientations xy or yx completes a transitive triangle on {x, y, z}. Thus if M contains t independent 4-triangles then G contains t independent transitive triangles. Notice that the converse is not true. For instance, if G is an orientation then µ(m) = 1, and so M contains no 4-triangle. Similarly, if M contains a 5-triangle with vertices x, y, z and µ(xy) = = µ(yz) then G contains a cyclic triangle (and

also a transitive triangle) on the same vertex set, since any orientation of the edge xz can be extended to a cyclic triangle by choosing the orientations of the other two edges carefully. It is convenient to introduce the following terminology and notation. A multigraph M := (V, E) is standard if µ(m). For a fixed standard multigraph M we use the following default notation. Two simple graphs G := G M := (V, E G ) and H := H M := (V, E H ) are defined by E G := {xy : µ(x, y) 1} and E H := {xy : µ(x, y) = }. Edges xy E H are said to be heavy, and y is said to be a heavy neighbor of x; edges in E G E H are light. We also set n := M. The following is our main theorem. Theorem 10. Every standard multigraph M with δ(m) 4n has a tiling of size n, where one tile is a 4-triangle and the remaining n 1 tiles are 5-triangles. Other authors have studied different degree conditions for directed graphs. For example Ghouila-Houri [6] and Woodall [14] proved analogs of Dirac s and Ore s theorems for directed graphs. Orientations of graphs (directed graphs with neither multiple edges nor -cycles) lead to another group of results. For example, Keevash, Kuhn, and Osthus [9] proved that if G is an oriented graph then δ 0 ( G) ( G 4)/8 guarantees the existence of a Hamilton cycle. Recently, Keevash and Sudakov [8] showed that there is some ɛ > 0 such that for sufficiently large n if G is an oriented graph on at least n vertices with δ 0 ( G) (1/ ɛ) G then G contains a packing of directed triangles covering all but at most three vertices. The paper is organized as follows. In the remainder of this section we review some additional notation. In Section we warm up by giving a short self-contained proof of the generalization of Wang s Theorem to standard multigraphs. This generalization in needed in Section 4. In Section we prove our main result, and in Section 4 we prove stability results related to Conjecture 8. 1.1 Additional notation Fix a multigraph (or graph) M = (V, E). Set M := V and M := E. For a subset U V, let U := M[U] = 1 e E(U) µ(e); and let U := V U. For any vertex v V (M), let v, U := u U µ(v, u). For U, U V (G), let U, U := v U v, U, and let E(U, U ) be the set of edges with one end in U and the other end in U. Then U, U = E(U, U ) + U U. Viewing an edge or nonedge xy as a set, these definitions imply µ(xy) = xy. If T is a triangle and V (T ) = {x, y, z} we may identify T by listing the vertices as xyz or by listing an edge and a vertex, that is, if e = yz then we may refer to T as ex or xe. Set [n] := {1,..., n}. If i, j [n] is clear from the context, we may write i j for i + j (mod n) with out explicitly mentioning n. 4

Warm-up In this section we warm up by proving the multigraph generalizations of Theorem 4 and the case c = 0 of Corollary 6. For completeness, and to illustrate the origins of our methods, we begin with a short proof of Corollary based on Enomoto s proof of Theorem. Proof of Corollary. Let G = (V, E) be an edge-maximal counterexample. Then n = k, δ(g) k, G does not contain a C -factor (so G K k ), but the graph G + obtained by adding a new edge a 1 a does have a C -factor. So G has a near triangle factor T, i.e., a factor such that A := a 1 a a T is a path and every H T A is a triangle. Claim. Suppose T is a near triangle factor of G with path A := a 1 a a and triangle B := b 1 b b. If {a 1, a }, B 5 then a, B = 0. Proof. Choose notation so that a 1, B = and a, B. Suppose b i N(a ). Then either {a 1 b i 1 b i, a a b i } or {a 1 a b i, a b i 1 b i } is a C -factor of G[A B], depending on whether b i N(a ). Regardless, this contradicts the minimality of G. Since {a 1, a }, G 4k, but {a 1, a }, A = < 4, there is a triangle B := b 1 b b T with {a 1, a }, B 5. Choose notation so that b 1, b, b N(a 1 ) and b, b N(a ). Applying the claim to A yields a, B = 0. Thus {b 1, a }, A B + {a 1, a }, A B (4 + ) + (1 + ) = 0 < 4 = 6. Since {b 1, a }, G + {a 1, a }, G 1k, some triangle C := c 1 c c T satisfies: {b 1, a }, C + {a 1, a }, C 1. Then a, C, {a 1, a }, C > 0. By Claim, {a 1, a }, C 4; so {b 1, a }, C 5. Claim applied to T {b 1 a 1 a, a b b } {A, B} yields a 1, C = 0. So a, C > 0 and either {b 1, a }, C = 6 or a, C =. Thus some i [] satisfies c i a, c i a, c i 1 b 1, c i b 1 E(G). So T {c i a a, b 1 c i 1 c i, a 1 b b } {A, B, C} is a C -factor of G. Next we use Corollary to prove Theorem 11. Theorem 11. Every standard multigraph M on n vertices with δ(m) 4n contains n independent 4-triangles. Proof. We consider three cases depending on n (mod ). Case 0 : n 0 (mod ). Since δ(g M ) 1δ(M) n, Corollary implies M has a triangle factor T. Choose T having the maximum number of 4-triangles. We are done, unless A = for some A = a 1 a a T. Since A, M ( ) 4n, ( ) n A, M A 4n A, A = 4n 9 > 1. 5

Thus A, B 1 for some B = b 1 b b T. Suppose a 1, B a, B a, B. Then 5 a 1, B 6 and {a, a }, B 7. Hence, {a, a }, b i for some i []; so T {a a b i, a 1 b i 1 b i } {A, B} is a 4-triangle factor of M. Case 1 : n 1 (mod ). Pick v V, and set M := M v. Then M 0 (mod ), and 4n 9 δ(m ) δ(m) µ(m) = 4(n 1) 4 M. By Case 0, M, and also M, contains M = n independent 4-triangles. Case : n (mod ). Form M + M by adding a new vertex x and heavy edges xv for all v V (M). Then M + 0 (mod ) and δ(m + ) 4 M +. By Case 0, M + contains M + independent 4-triangles. So M = M + x contains M + 1 = n of them. Now we consider 5-triangle tilings. First we prove Proposition 1, which is also needed in the next section. Then we strengthen Wang s Theorem to standard multigraphs. Proposition 1. Let T = v 1 v v M be a 5-triangle, and x V (M T ). If x, T 4 then xe is a ( x, T + 1)-triangle for some e E(T ). Proof. Suppose v 1 v, v 1 v E H. If N(x) {v v } then xv v is a ( x, T +1)-triangle. Else, x, v 1 v i x, T 1 for some i {, }. So xv 1 v i is a ( x, T + 1)-triangle. Theorem 1. Every standard multigraph M with δ(m) n contains n independent 5-triangles. Proof. Consider two cases depending on whether n (mod ). Case 1 : n (mod ). By Theorem 11, M has a tiling T consisting of n independent 4- and 5-triangles. Over all such tilings, select T with the maximum number of 5-triangles. We are done, unless there exists A = a 1 a a T such that A = 4. Assume a 1 a is the heavy edge of A. By the case, L := V T has at most one vertex. If L then let a L; otherwise set a := a. Also set A := A + a. Then a, A + ( A ) = A 4, since otherwise G[A ] contains a 5-triangle. So A, A A 4( A ). For B T, define f(b) := A, B + a, B. Then f(a) = 8 + a, A 4 + A. So B T f(b) = d(a 1 ) + d(a ) + d(a ) + d(a ) A, A A 4 n 6n 6 4( A ) = 6(n A ) + (4 + A ) + > 18 ( T 1) + f(a). A, A A Thus f(b) 19 for some B T A. If B is a 4-triangle then set B := B + e, where e is parallel to some e E(B) with µ(e) = 1, and set M := M + e. Otherwise, set B := B and M := M. It suffices to prove that M [A B ] contains two independent 5-triangles, since in either case another 5-triangle can be added to T, a contradiction. Label the vertices of B as b 1, b, b so that b 1 b and b 1 b are heavy edges. Since a 1 a is a heavy edge, if {a 1, a }, b then a 1 a b is a 5-triangle for all b V (B). Consider 6

three cases based on k := max{ a, B, a, B }. Let a {a, a } satisfy a, B = k. Since f(b) 19, we have 4 a, B 6. If a, B = 4 then {a 1, a }, B 11. By Proposition 1, there exists i [] such that ab i b i 1 is a 5-triangle, and a 1 a b i is another disjoint 5-triangle. If a, B = 5 then {a 1, a }, B 9. So there exists i {, } such that a 1 a b i is a 5-triangle; and ab 1 b 5 i is another 5-triangle. Finally, if a, B = 6 then {a 1, a }, B 7. So there exists i [] such that a 1 a b i is a 5-triangle; and ab i 1 b i is another 5-triangle. Case : n (mod ). Form M M by adding a vertex x and heavy edges xv for all v V. Then M 0 (mod ) and δ(m ) M. By Case 1, M contains M + independent 5-triangles. So M = M x contains M 1 = n of them. Main Theorem In this section we prove our main result, Theorem 10. Let M be a standard multigraph with δ(m) 4n. We start with three Propositions used in the proof. Proposition 14. Suppose T = v 1 v v M is a 5-triangle, and x 1, x V (M T ) are distinct vertices with {x 1, x }, T 9. Then M[{x 1, x } V (T )] has a factor containing a 5-triangle and an edge e such that e is heavy if min i [] { x i, T } 4. Proof. Label so that v 1 v, v 1 v E H and x 1, T x, T. First suppose x, T 4. If x v i E H for some i {, } then {x 1 v 1 v 5 i, x v i } works. Else V (T ) N(x ). Also x 1 v j E H for some j {, }. So {x 1 v j, x v 1 v 5 j } works. Otherwise, x, T = and x 1, T = 6. So {x 1 v i 1 v i, x v i } works for some i []. Proposition 15. Suppose T = v 1 v v M is a 5-triangle, and e 1, e E(M T ) are independent heavy edges with e 1, T 9 and e, T 7. Then M[e 1 e V (T )] contains two independent 5-triangles. Proof. Choose notation so that e 1, v i for both i []. There exists j [] so that e, v j. Pick i [] j. Then e 1 v i and e v j are disjoint 5-triangles. Proposition 16. Suppose T M is a 5-triangle, and xyz is a path in H M T. If xz, T 9 and y, T 1 then M[{x, y, z} V (T )] has a factor containing a 5- and a 4-triangle. Proof. Choose notation so that x, T z, T, and T = v 1 v v with v 1 N(y). We identify a 4-triangle A and a 5-triangle B depending on several cases. Suppose x, T = 6 and z, T. If zv 1 E then set A := yzv 1 and B := xv v ; else set A := zv v and B := xyv 1. Otherwise x, T = 5 and z, T 4. If zv 1 / E then set A := xyv 1 and B := zv v. Otherwise zv 1 E. If zv 1 is heavy then set A := xv v and B := zyv 1 ; if xv 1 is light then set A := zyv 1 and B := xv v. Otherwise zv 1 is light and xv 1 is heavy. Set A := zv v and B := xyv 1. 7

Proof of Theorem 10. We consider three cases depending on n (mod ). Case 0 : n 0 (mod ). Let n =: k, and let M be a maximal counterexample. Let T be a maximum T 5 -tiling of M and U = T T V (T ). Claim 1. T = k 1. Proof. Let e E. By the maximality of M, M + e has a factor T consisting of 5-triangles and one 4-triangle A 1. If e A 1 then the 5-triangles are contained in M, and so we are done. Otherwise, e E(A + ) for some 5-triangle A + T. Set A := A + e, and A := A 1 A. Then A satisfies: (i) A = 6, (ii) M[A] contains two independent heavy edges, and (iii) M[V A] has a T 5 -factor. Over all vertex sets satisfying (i iii), select A and independent heavy edges e 1, e M[A] so that z 1 z is maximized, where {z 1, z } := A (e 1 e ). Let T be a T 5 -factor of M[V A]. Set A i := e i + z i, for i []. If M[A] contains a 5-triangle we are done. Otherwise x, A 4 for all x V (A 1 ), and so A = A 1 + A + A 1, A 0. Thus ( ) 4 A, V A 6 n 1 40 > 4(k ). So A, B 5 for some B = b 1 b b T. It suffices to show that M[A B] contains two independent T 5. Suppose {z 1, z }, B 9. If there exists h [] such that z h, B = 6 then choose i [] with b i, A 9. There exists j [] with b i e j 5; also z h b i 1 b i 5. So we are done. Otherwise, z h, B 4 and z h, B 5 for some h []. By Proposition 14, M[V (B) + z 1 + z ] has a factor consisting of a heavy edge and a T 5, implying, by the maximality of z 1 z, that z 1 z =. Set e := z 1 z. Choose distinct i, j [] so that e i, B 9 and e j, B 7. By Proposition 15, there are two T 5 in M[e i e j V (B)]. By Claim 1, W := V U satisfies W =. Choose T with W H maximum. Claim. W 4. Proof. Suppose not. Then W, U ( 4 n 1) > 1(k 1). So W, T 1 for some T T. Thus there exist w, w W with w, T 4 and w, T 5. By Proposition 14, M[V (T ) {w, w }] has a factor containing a T 5 and a heavy edge. By the choice of W this implies W H = 1. Set W =: {x, y, z} where xy is heavy. Since ( ) 4n ( n ) z, U + xy, U 4 1 5 > 16 1 = 16(k 1), some T = v 1 v v T satisfies z, T + xy, T 17. Suppose v 1 v, v 1 v E H. To contradict the maximality of W, it suffices to find i [] so that {M[{x, y, v i }], M[{z, v i 1, v i }]} contains a T 5, and a graph with at least four edges. If z, T = then xy, T 11. Choose i {, } so that z, v 1 v i 1. If z, T = 4 then xy, T 9. Choose i {, } so that xyv i is a 5-triangle. If z, T = 5 then xy, T 7. Choose i {, } so that xy, v i. Otherwise, z, T = 6 and xy, T 5. Choose i [] so that xy, v i. 8

Since M is a counterexample and W 4, we have M[W ] =: xyz is a path in M H. Claim. There exists A T and a labeling {a 1, a, a } of V (A) such that (a) x is adjacent to a 1 ; (b) one of xa a and za a is a 5-triangle and the other is at least a 4-triangle; (c) if xa 1 is light then both xa a and za a are 5-triangles; and (d) y, A = 0. Proof. There exists A = a 1 a a T such that xz, A 9, since ( ) 4n xz, U 1 xz, W 8n ( n ) 4 > 8 1 = 8(k 1). Say x, A z, A. Since M is a counterexample, Proposition 16 implies (d) y, A = 0. If z, A = then, by Proposition 1, za a is a 4-triangle for some a, a A. In this case x, A = 6, so xa a is a 5-triangle and xa 1 is a heavy edge. So (a c) hold. If z, A 4 then, by Proposition 1, za a is a 5-triangle for some a a A. In this case x, A 5, so xa a is a 4-triangle and x is adjacent to a 1. Furthermore, if xa 1 is light then xa a is a 5-triangle. Again (a c) hold. Claim 4. There exists B T A such that a 1 y, B + xz, B 5. Proof. Set U := U V (A). Since xz / E and y, A = 0, ( ) 4 a 1 y, U + xz, U 6 n 1 a 1 y, W V (A) xz, W A 4n 6 (8 + 4) (8 + 8) > 4 ( n ) = 4(k ). So there exists B T A with a 1 y, B + xz, B 5. Let W := W {a 1 } V (B). For any edge e {a 1 x, xy, yz} define Q(e) := {u V (B) : e, u } and for any vertex v {a 1, x, y, z} and k {4, 5} define P k (v) := {u B : T k M[B u + v]}. Claim 5. If v / e and there exists u P k (v) Q(e) then M[(V (B) e) + v] can be factored into a ( + e )-triangle and a k-triangle. Moreover: e, B 6 (a) Q(e) (b) P 5 (v) v, B (c) P 4 (v) = if v, B 5 and P 4 (v) ( v, B ) otherwise. (1) Proof. For the first sentence apply definitions; for (1) check each argument value. 9

To obtain a contradiction, it suffices to find two independent triangles C, D M[W w] for some w {a 1, x, y} so that {C, D, wa a } is a factor of M[W V (A) V (B)] consisting of two 5-triangles and one 4-triangle. We further refine this notation by setting D := vb i 1 b i and C := b i e, where v {a 1, x, y, z}, b i B and e E({a 1, x, y, z} v). Then w is defined by w ({a 1, x, y, z} e) v; set W := wa a. Claim 6. None of the following statements is true: (a) P 4 (v) Q(e) for some e {xy, yz} and v {x, z} e. (b) P 5 (v) Q(e) for some e {a 1 x, xy, yz} and v {a 1, x, y, z} e such that y e+v. (c) P 4 (a 1 ) Q(xy) and P 4 (a 1 ) Q(yz). (d) There exists b i P 5 (a 1 ) such that x, y, z N(b i ). Proof. By Claim, each case implies M is not a counterexample, a contradiction: (a) Then w = a 1, and so W 5, C 5 and D 4. (b) Then D 5. If e = then C 5 and W 4; otherwise e = a 1 x and, by Claim (c), W 5 and C 4. (c) By Claim (b), wa a 5 for some w {x, z}. Set v := a 1 and e := {x, y, z} w. Then W 5, C 5 and D 4. (d) Set v := a 1, choose w {x, z} so that W 5, and set e := {x, y, z} w. Then D 5 and C 4. Claim 7. a 1, B < 5. Proof. Suppose not. Let {x, z } = {x, z}, where x, B z, B. For k {4, 5}, define s k (e, v) := Q(e) + P k (v). Then s k (e, v) > implies Q(e) P k (v). We use Claim 5 to calculate s k (e, v). Observe 5 a 1, B x y, B + yz, B, and so x y, B 1 a 1, B. If a 1, B = 6 then s 5 (x y, a 1 ) 1+, contradicting Claim 6 (b). Otherwise, a 1, B = 5. Either x y, B 9 or z y, B 7. In the first case, s 5 (x y, a 1 ) +, contradicting Claim 6 (b). In the second case, s 4 (x y, a 1 ), s 4 (z y, a 1 ) > 1+, contradicting Claim 6 (c). Claim 8. {a 1, y}, B < 9. Proof. Suppose {a 1, y}, B 9. We consider several cases. Case 1 : a 1, B = 4 and y, B = 6. By Proposition 1 there are distinct b, b, b V (B) with b P 5 (a 1 ) and 1 a 1, b a 1, b =. Claim 6 (b) implies b / Q(xy) Q(yz); so x, z / N(b), since b, y =. By Claim 5 (b), P 5 (y) = B; so Q(a 1, x) = by Claim 6 (b). Thus x, b a 1, b and x, b = 0. By the case {x, z}, B 5; thus x, b = 1 and z, {b, b } = 4; so a 1, b = 1 = a 1, b. Thus b P 4 (a 1 ) Q(xy) Q(yz), contradicting Claim 6 (c). Case : a 1, B = and y, B = 6. Then {x, z}, B 7. For {u, v} = {x, y}, we have u, B 1. So Claim 5 (a) implies Q(uy) 1. Thus Claim 6 (a) implies P 4 (v). So Claim 5 (c) implies v, B 4. Thus x, B, z, B 4. 10

By Proposition 1, b P 4 (a 1 ) for some b B. So b / Q(xy) Q(yz) by Proposition 6 (b). Thus b / N(x) N(z). Also P 5 (y) = B by y, B = 6. By Claim 6 (b), Q(a 1 x) =. Thus x, B b. So xb E and z, B b = z, B. Thus b P 4 (z) Q(xy), contradicting Claim 6 (a). Case : a 1, B = 4 and y, B = 5. Then (i) {x, z}, B 7; let {x, z } := {x, z}, where x, B z, B. Claim 5 (b) implies P 5 (y) ; Proposition 1 implies b i P 5 (a 1 ) for some i []. So by Claim 6 (b,d), (ii) Q(a 1 x) 1, (iii) b i / Q(xy) Q(yz), and (iv) xyz N(b i ). Thus by (iii) and Claim 5 (a), x, B 5, and so by (i), z, B. By Claim 5 (a), Q(x y) and Q(yz ) 1. Claim 6 (a) then implies P 4 (x ) meaning (v) 4 x, B z, B and, therefore, by Claim 5 (a), (vi) Q(a 1 x) 1. By Claim 6 (b,c) P 5 (a 1 ) 1 and P 4 (a 1 ). This implies (vii) b i / N(a 1 ): Otherwise, since b i P 5 (a 1 ) implies b i, a 1 1, there exist h, j [] i with b h, a 1 = and b i, a 1 = 1 = a 1, B. If b i b j = 1 then P 5 (a 1 ) = ; if b i b j = then P 4 (a 1 ) =. Either is a contradiction. By (ii), (vi) and (vii), Q(a 1 x) = {b j } for some j [] i, and a 1, b h b j = 4, where h = 6 i j. Thus N(x) B = {b i, b j }. By (iv), z / N(b i ), and so N(z) B = {b h, b j }. So yzb h = z, B + y, zb h zb j + = 4 xb i b j = x, B + b i b j + 1 = 4 yzb h + xb i b j = {x, z}, B + y, zb h zb j + b i b j {x, z}, B + 9 by (v) by (v) by (i) Thus {yzb h, xb i b j, A} is a factor of M(A B W ) with two T 5 and a T 4, a contradiction. Claim 9. {a 1, y}, B 9. Proof. Suppose {a 1, y}, B 8. Then {x, z}, B 9 and y, B 1. Proposition 16 implies there exist independent 4- and 5-triangles in M[W V (B)], a contradiction. Observing that Claim 8 contradicts Claim 9, completes the proof of Case 0. Case 1 : n 1 (mod ). Choose any vertex v V, and set M = M v. By Case 0, M, and so M, contains n 4 = n independent 5-triangles and a 4-triangle. Case : n (mod ). Add a new vertex x together with all edges of the form xv, v V to M to get M +. By Case 0, M + contains n = n + 1 independent 5-triangles and a 4-triangle, at most one of them contains x. So M contains n independent 5-triangles and a 4-triangle. 4 Relaxed degree conditions In this section we attack Conjecture 8 and prove Corollary 9. We rely on ideas from Levitt, Sárközy and Semerédi [11]. 11

Example 17. Let M be a underlying multigraph of the directed graph G from Example 5. Then M = n = k + 1, V = V 1 V, V 1 V =, V 1 = k, V = k + 1, every pair xy is an edge, and xy is heavy if and only if x, y V 1 x, y V. Since no 5-triangle contains vertices of both V 1 and V, and V 1 is not divisible by, M does not have a 5-triangle factor. Moreover, δ(m) = ( V 1 1) + V = k 1 = n 1. Finally, notice that E H (V 1, V ) =. Example 17 shows that Theorem 1 is tight but it also suggests that requiring H M to be connected may allow us to relax the degree condition. Conjecture 18. If M is a standard multigraph on n vertices where n is divisible by, δ(m) 4 n 1 and H M is connected then M has a T 5 -factor. As a side note, Conjecture 18 implies both both Corollary and the following Theorem of Enomoto, Kaneko and Tuza [5]. Theorem 19. If G is a connected graph on k vertices and δ(g) k then G has a k independent paths on vertices. Definition 0. For α 0 call a graph G on n vertices α-splittable if there exists a partition {A, B} of V (G) such that A, B n and A, B G αn. In this section, we will prove the following theorem, which supports Conjecture 18, and yields Corollary 9. Theorem 1. For every ε, α > 0 there exists n 0 := n 0 (ε, α) such that for every standard multigraph M = (V, E) on n n 0 vertices where n is divisible by the following holds. If δ(m) ( 4 + ε) n and H M is not α-splittable then M has a 5-triangle factor. Before the proof, we will first collect a few simple facts and definitions that will be used throughout, and show that Theorem 1 implies Corollary 9. Let ε > 0 and let M = (V, E) be standard multigraph on n vertices such that δ(m) ( 4 + ε) n. Note that for every u V N(u) ( + ε ) ( ) 1 n and N H (u) + ε n. For any U V and k 1 define Q k (U) := {v V : v, U k}. For any e E, ( ) 4 + ε n e, V Q 4 (e) + Q (e) + Q (e) = Q 4 (e) + Q (e) + n. Therefore, Q 4 (e) + Q (e) ( ) + ε n, () 1

and since Q 4 (e) Q (e), Q (e) For any u V, let F (u) := {e E H : u Q (e)}. Note that F (u) N(u) N H (v), v N H (u) and for every v N H (u), N(u) N H (v) ( + ε F (u) 1 ( ) 1 + ε n. () ) ( n + 1 + ε) n n = ε n. So ( ) 1 ε + ε n > ε 4 n. (4) We are now ready to prove Corollary 9. At the same time we will prove the following corollary showing that Conjecture 18 implies Conjecture 8. Corollary. If every standard multigraph M with M = k, δ(m) 4k 1, and H M connected has a T 5 -factor, then every directed graph G with G = k and δ 0 ( G) k has a C -factor. Proof of Corollaries 9 and. For Corollary 9 we are given ε and set α := ε ; for Corollary 4 set α, ε := 0. Let M be the underlying multigraph of G. Note that δ(m) ( 4 + ε) n, and δ(h M ) ( + ε)n. If H := H M is not α -splittable then Corollary 9 follows from Theorem 1. Moreover, if α = 0 then H M is connected any component has size at least 1 + δ(h) and so Conjecture 18 implies Conjecture 8. So assume H is α -splittable, but G does not have a C -factor. Partition V := V (H) as {A, B} so that A, B n and e H(A, B) α n. Set Z := {v V : E H (v) E H (A, B) > αn} and note that Z αn. For every z Z, N H (z) ( 1 + 4α) n so we can find a matching K in H such that Z K. By () each e M satisfies Q (e) ( 1 + ε) n; so there exists x e (V K) Q (e), and ex e is a 5-triangle. Let Y := e K ex e, V := V Y, A := A Y and B := B Y. Then Y Z αn and E H (v) E H (A, B ) αn for every v V. So each C {A, B } (i) δ 0 ( ( ) ( ) 1 G[V ]) + α n, (ii) δ(h[v ]) + α n, and (5) (iii) δ 0 ( G[C ]) 1 n for C {A, B }. In particular, (5iii) implies n < A, B < n. 1

If A B 0 (mod ) then let A := A and B := B. If not, since A B is divisible by, without loss of generality we can assume that A 1 (mod ) and B (mod ). Let v B. Since B < n there exists u N + (v) A. By (5i,iii) ( ) N + (u) (N H (v) B ) + α n + 1 n (n 1) > 0. So there exists w B v with T := uvw = C. Let A := A V (T ) = A v and B := B V (T ) = B u v. In either case, ( A B B + 4) > n > B 1, so ( G[B ) δ t ] δ 0 ( G[V ]) ( v, A GM + v, A HM ) v, B B M ( ) + α n ( A + αn) B B (by (5i)) = 4 (n A B B 1) + 1 ( A B B + 4) > ( B 1). A similarly calculation gives that δ t ( G[A ]) ( A 1). G[A ] and G[B ] have cyclic triangle factors. Hence, by Theorem 4, both Proof of Theorem 1. This proof uses the probabilistic absorbing method as in [11]. Let 0 < σ < min{ ε, α σ45 } and τ := and assume throughout that n is sufficiently large. Let 1 16 4 H := H M. Definition. For any disjoint X, Y V, we will say that Y absorbs X if M[Y ] and M[Y X] both have 5-triangle factors. For any X ( V ), call (y1,..., y 45 ) V 45 an X- sponge when Y := {y 1,..., y 45 }, Y = 45 and Y absorbs X, and let A X be the set of X-sponges. Definition 4. For k > 0 the tuple (z 1,..., z k 1 ) V k 1 is a k-chain if (a) z 1,..., z k 1 are distinct vertices, (b) z i z i 1 is a heavy edge for 1 i k, and (c) z i Q (z i z i 1 ) Q (z i+1 z i+ ) for 1 i k 1. For u, v V if u Q (z 1 z ) and v Q (z i z i 1 ) for some 1 i k and u, v / {z 1,..., z k 1 } then we say that the k-chain joins u and v (see Figure 1). For k > 0, if there are at least (σn) k 1 k-chains that join u and v we say that u is k-joined with v. Note that for 1 i < k 5 if u is i-joined with v then u is k-joined with v. Indeed, using () and (4), we can extend any i-chain that joins u and v by iteratively picking a vertex z j Q (z j, z j 1 ) and then a heavy edge z j+1 z j+1 F (z j ) that avoids the vertices {u, v, z 1,..., z j 1 } for j from i + 1 to k in at least (σn) j ways. For any u V define L k (u) := {v V : v is k-joined with u for some 1 i k}. 14

z z 5 z 8 z 11 v z 14 u z z 6 z 9 z 1 z 1 z 4 z 7 z 10 z 1 Figure 1: The 5-chain (z 1,..., z 14 ) joins u and v. Note that, by the previous comment, L 1 (u) L 5 (u). Let {x 1, x, x } := X ( V ), Y := (z1,..., z 45 ) V 45, and recall m(i) := 15(i 1). It is not hard to see that Y A X if Y satisfies the following for i []: the vertices z 1,..., z 45 are distinct, M[{z m(1)+1, z m()+1, z m()+1 }] is a 5-triangle, and (z m(i)+,..., z m(i)+15 ) is a 5-chain that joins z m(i)+1 and x i. Our plan is to show (i) there is a small set A of disjoint sponges such that for all -sets X ( V ) there exists an X-sponge Y A, and (ii) there exists a -set X V A such that M (X A) has a 5-triangle factor C. Since there exists an X-sponge in A, this will imply that M has a 5- triangle factor. To prove (i) we first show that every -set is absorbed by a positive fraction of all 45- tuples, and then use Chernoff s inequality to find A. The following claim is our main tool. Claim 10. L 5 (x) = V for every vertex x V. Proof. We will first show that, for every u V, ( 1 L 1 (u) + ε ) n and u L 1 (u). Since F (u) (αn), u L 1 (u). Let t := e F (u) Q (e). By (), t F (u) ( 1 + ε) n. If v / L 1 (u) then there are less than (σn) < ε(αn) ε F (u) edges e F (u) for which v Q (e). Therefore, ( ) 1 F (u) + ε n t < F (u) L 1 (u) + ε F (u) L 1 (u) ε F (u) n + (1 ε) F (u) L 1 (u), and L 1 (u) > n (1 ε) 1 > ( 1 + ε ) n. Note that for any u, v V if L i (u) L j (v) σn and 1 i, j then v L i+j (u). Indeed, we can pick w L i (u) L j (v) in one of σn ways and we can then pick an i-chain (u 1,..., u i 1 ) that joins u and w and a j-chain (v 1,..., v i 1 ) that joins v and w so that u, u 1,..., u i, w, v j,..., v 1 and v are all distinct in 1 (σn)(i+j) ways. Since (u 1,..., u i, w, v j,..., v 1 ) is a (i + j)-chain that joins u and v and there are (σn) (i+j) 1 such -chains, v L (u). 15

Let x V and suppose, by way of contradiction, that there exists y V such that y / L 5 (x). If there exists z / L (x) L (y), from the preceding argument, we have L 1 (u) L 1 (v) < σn for any distinct u, v {x, y, z}. But this is a contradiction, because ( 1 + ε ) n (σ)n > n. Therefore, if we let X := L (x) and Y := L (y) L (x), {X, Y } is a partition of V. We have that X L 1 (x) ( 1 + ε ) n and, since y / L4 (x), L (y) L (x) < σn so Y L 1 (y) σn ( 1 + ε 6) n Call a 4-tuple (v 1, v, v, v 4 ) connecting if v 1 X and v 4 Y, v v E H and v 1, v 4 Q (v v ). Since M is not α-splittable, E H (X, Y ) αn. Pick some e := x y E H (X, Y ) where x X and y Y. We will show that there are at least (σn) connecting 4-tuples which contain x and y. Since M[v 1, v, v, v 4 ] can contain at most 4 edges from E H (X, Y ), this will imply that there are at least 1 4 αn (σn) 4(σn) 4 connecting 4-tuples and this will prove that y L 5 (x), a contradiction. Indeed, select a connecting 4-tuple (v 1, v, v, v 4 ) in 4(σn) 4 ways. Since v 1 is -joined with x there are at least 1 (σn)5 -chains that join x and v 1 and avoid {v 1, v, v, v 4 }. Similarly, there are 1 (σn)5 -chains that join v 4 and y and avoid all previously selected vertices. Therefore, there are at least (σn) 14 5-chains that join x and y. So, by way of contradiction, assume there are less than (σn) connecting 4-tuples containing e. Suppose Q 4 (e) σn and pick z Q 4 (e) and let T := {x, y, z}. Note that M[T ] is a 6-triangle and that ( ) 4 + ε n T, V 6 Q 5 (T ) + 4 Q 5 (T ) = Q 5 (T ) + 4n so Q 5 (T ) εn. Pick w Q 5(T ). Note that there are at least σn εn (σn) choices for the pair (z, w) and that if w X then (w, x, z, y ) is a connecting 4-tuple and if w Y then (x, z, y, w) is a connecting 4-tuple. Therefore, we can assume Q 4 (e) < σn which, by (), implies that Q (e) ( + ε) n. For any v 1 Q (e) X and v 4 Q (e) Y, (v 1, x, y, v 4 ) is a connecting 4-tuple. Therefore, we cannot have Q (e) X σn and Q (e) Y σn. So suppose Q (e) X < σn. Then Y Q (e) Y > n which contradicts the fact that X > n. Since a similar argument holds when Q (e) Y < σn, the proof is complete. Claim 11. For every X ( V ), AX 4τn 45. Proof. Define m(i) := 15(i 1) and let {x 1, x, x } := X. Pick v m(1)+1 v m()+1 := e from one of the at least 1 n edges in H X. By (), we can pick v m()+1 from one of the more than 1 n vertices in Q (e) X. We have that M[{v m(1)+1, v m()+1, v m()+1 ] is a 5-triangle. For i [], pick a 5-chain (v m(i)+,..., v m(i)+15 ) that joins v m(i)+1 and x i in one of 1 (σn)14 ways. Note that (v 1,..., v 45 ) A X and there are at least 1 7 σ4 n 45 4τn 45 such tuples. The next part of the proof is probabilistic. The tools we require are the union bound, the linearity of expectation, Markov s inequality and the following theorem of Chernoff: Theorem 5. Let X be a random variable with binomial distribution. Then the following hold for any t 0: 16

( (a) Pr[X E[X] + t] exp ( (b) Pr[X E[X] t] exp t (E[X]+t/) t E[X] ). ) ; The next lemma completes step (i) of the proof. Lemma 6. There exists a set A of disjoint sponges such that A ε n and for every -set 45 X there exists an X-sponge in A. Proof. Let F be a set of 45-tuples chosen by picking each 45-tuple Y V 45 randomly and independently with probability ρn 44 where ρ := τ. Let ε := ε/45 and note that ρ < ε. 4 10 Clearly E[ F ] = ρn, and, by Claim 11, E[ F A X ] 4τρn for each X ( V ). Let O be the set of pairs of overlapping tuples in V 45, that is { ( ) } V 45 O = {(x 1,..., x 45 ), (y 1,..., y 45 )} : x i = y j for some i, j {1,..., 45} ; note that O 1(n 45 45 n88 ) < 000n 89. For any P O, Pr[P F] = (ρn 44 ), so if O F := {P O : P F} then, by the linearity of expectation, E[ O F ] Pr[P F] < 000n 89 ρ n 88 = τρ n. P O By Theorem 5, and, for every X ( V ), Pr[ F ε n] = Pr[ F E[ F ] + (ε ρ)n] exp Pr[ A X F τρn] = Pr[ A X F E[X] τρn] exp ( ) (ε ρ) n (ρ + ε ) ( ) (τρ) n 8τρ Since by Markov s inequality we have that Pr[ O F τρn] 1, if n is large enough Pr[ F ε n] + Pr[ A X F τρn] + Pr[ O F τρn] < 1. X ( V ) Therefore, by the union bound, we can fix F V 45 so that, F < ε n, A X F > τρn for every X ( V ), and OF < τρn. Note that O F O F τρn and let { ( ) ( )} V A = T F OF : T A X for some X and let A = {{v 1,..., v 45 } : (v 1,..., v 45 ) F}. Every T A is an absorbing tuple for some X ( V ), so M[T ] has a 5-triangle factor. Also, note that for distinct T, S A, {S, T } / O, that is, no pair of tuples in F overlap. Furthermore, A X A A X F O F τρn 1. 17

Let A be the set of 45-tuples quaranteed by Lemma 6 and let A be the set of vertices that are contained in some tuple of A. Let M = M A. By Theorem 10, because A 45ε n εn, there is a 5-triangle packing of M X where X is a -set of V A. There exists (y 1,..., y 45 ) A X A. Let Y := {y 1,..., y 45 }. By the definition of an X-sponge there is a 5-triangle factor of M[X Y ] and, since every tuple in A is a sponge, there is a 5-triangle factor of M[A Y ]. This completes the proof. References [1] H. Chernoff, A measure of asymptotic efficiency for tests of a hypothesis based on the sum of observations, The Annals of Mathematical Statistics (195), no. 4, 49 507. [] K. Corrádi and A. Hajnal, On the maximal number of independent circuits in a graph, Acta Mathematica Hungarica 14 (196), no., 4 49. 1 [] G.A. Dirac, Some theorems on abstract graphs, Proceedings of the London Mathematical Society (195), no. 1, 69 81. [4] Enomoto, H., On the existence of disjoint cycles in a graph, Combinatorica, 18, (1998), no. 4, 487 49. 1 [5] Enomoto, H., Kaneko, A. and Tuza, Z., P -factors and covering cycles in graphs of minimum degree n/, Combinatorics, 5 (1987), 1 0. 4 [6] A. Ghouila-Houri, Une condition suffisante dõexistence dõun circuit hamiltonien, C. R. Math. Acad. Sci. Paris, 5, (1960), 495 497. [7] A. Hajnal and E. Szemerédi, Proof of a conjecture of P. Erdős, Combinatorial Theory and Its Application (1970), 601 6. [8] P. Keevash, B. Sudakov, Triangle packings and 1-factors in oriented graphs, Journal Combinatorial Theory B 99, (009), 709 77. [9] P. Keevash, D. Kuhn and D. Osthus, An exact minimum degree condition for Hamilton cycles in oriented graphs, J. London Math. Soc. 79, (009), 144 166. 1 [10] J. Komlós, G.N. Sárközy, and E. Szemerédi, On the square of a hamiltonian cycle in dense graphs, Random Structures & Algorithms 9 (1996), no. 1-, 19 11. 1 [11] I. Levitt, G.N. Sárközy, and E. Szemerédi, How to avoid using the regularity lemma: Pósa s conjecture revisited, Discrete Mathematics 10 (010), no., 60 641. 1 [1] C.St.J.A. Nash-Williams, Edge-disjoint hamiltonian circuits in graphs with vertices of large valency, Studies in Pure Mathematics (Presented to Richard Rado), 1971, pp. 157 18. 1 18

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