Decomposition of random graphs into complete bipartite graphs

Transcription

1 Decomposition of random graphs into complete bipartite graphs Fan Chung Xing Peng Abstract We consider the problem of partitioning the edge set of a graph G into the minimum number τg) of edge-disjoint complete bipartite subgraphs. We show that for a random graph G in Gn, p), where p is a constant no greater than 1/2, almost surely τg) is between n clog n) 3+ɛ and n 2 log 1/p n for any positive constants c and ɛ. 1 Introduction For a graph G, the bipartition number, denoted by τg), is the minimum number of complete bipartite subgraphs that are edge-disjoint and whose union is the edge set of G. In 1971, Graham and Pollak [5] proved that τk n ) = n 1. 1) In particular, they showed that for a graph G on n vertices, the bipartition number τg) is bounded below as follows: τg) max {n +, n } 2) where n + is the number of positive eigenvalues and n is the number of negative eigenvalues of the adjacency matrix of G. Then, 1) follows from 2). Since then, there have been a number of alternative proofs for 1) by using linear algebra [9, 10, 11] or by using matrix enumeration [12, 13]. Let αg) denote the independence number of G, which is the maximum number of vertices so that there are no edges among some set of αg) vertices in G. A star is a special bipartite graph in which all edges share a common vertex which we call the center of the star. For a graph G on n vertices, the edge set of G can obviously be decomposed into n αg) stars. It follows immediately that τg) n αg). 3) It was mentioned in a 1988 paper [8], by Kratzke, Reznick and West, that Erdős conjectured the equality in 3) holds for almost all graphs G Gn, 1/2), although we could not find this conjecture in any other publication on Erdős problems. Let βg) be the size of the largest induced complete bipartite graph in G. Another possible upper bound for τg), as pointed out by Alon [1], is τg) n βg) + 1. Department of Mathematics, University of California, San Diego, La Jolla, CA 92093, fan@ucsd.edu, x2peng@ucsd.edu). Research is supported in part by ONR MURI N , and AFSOR AF/SUB

2 This follows from the fact that edges in G can be partitioned into n βg) stars and a largest induced complete bipartite graph. Therefore we get τg) min{n αg), n βg) + 1}. A random graph almost surely has an independent set of order c log n and therefore τg) n c log n. For the lower bound, for a random graph G, it is well known that almost surely the number of positive and negative eigenvalues of the adjacency matrix of G is bounded above by n/2 + c n. Consequently, the inequality in 2) yields a rather weak lower bound of τg). We will prove the following theorem. Theorem 1 For a random graph G in Gn, 1/2), almost surely the bipartition number τg) of G satisfies n clog 2 n) 3+ɛ τg) n 2 log 2 n for any positive constants c and ɛ. Theorem 1 is a special case of the following theorem. Theorem 2 For a random graph G in Gn, p), where p is a constant no greater than 1/2, almost surely the bipartition number τg) of G satisfies n clog b n) 3+ɛ τg) n 2 log b n for b = 1 p and any positive constants c and ɛ. For sparser random graphs, Alon [1] recently proved that there exists some small) constant c such that for 2 n p ) c, the bipartition number for a random graph G in Gn, p) satisfies τg) = n Θ. log np p We remark that the difficulty for computing τg) is closely related to the intractability of computing αg). In general, the problem of determining αg) is an NP-complete problem, as one of the original 21 NP-complete problems in Karp [7]. If G does not contain a 4-cycle, then τg) = n αg). Schrijver showed that the problem of determining αg) for the family of C 4 -free graphs G remains NP-complete [8]. Therefore the problem of determining τg) is also NP-complete. Nevertheless, Theorem 2 implies that for almost all graphs G, we can bound τg) within a relatively small range. We also consider a variation of the bipartition number by requiring an additional condition that no complete bipartite graph in the partition is a star. We define the strong bipartition number, denoted by τ G), to be the minimum number of complete bipartite graphs which are not stars) needed to partition the edge set of G. If there is no such a partition then we define τ G) as ; if V G) 2 then we define τ G) to be zero. We will show that for a random graph G Gn, p), the strong bipartition number satisfies τ G) n if p is a constant and p 1 2. The paper is organized as follows: In the next section, we state some definitions and basic facts that we will use later. In Section 3, we establish upper bounds for the number of edges incident to several specified families of complete bipartite subgraphs. In Section 4, we consider the remaining uncovered edges and give corresponding lower bounds that our main theorem needs. In Section 5, we show that almost surely the strong bipartition number is at least n for a random graph on n vertices. In Section 6, we use the lemmas and the strong bipartition theorem to prove Theorem 2. A number of problems and remarks are mentioned in Section 7. 2

3 2 Preliminaries Let G = V, E) be a graph. For a vertex v V G), the neighborhood N G v) of v is the set {u: u V G) and u, v) EG)} and the degree d G v) of v is N G v). For a hypergraph H = V, E) and v V H), we define the degree d H v) to be {F : v F and F EH)}. For U V G), let eu) be the set of edges of G with both endpoints in U and G[U] be the subgraph induced by U. Furthermore, 2 U denotes the power set of U. For two subsets A and B of V, we define EA, B) = {u, v) E : u A and v B}. We say A and B form a complete bipartite graph if A B = and u, v) EG) for all u A and v B. We will use the following versions of Chernoff s inequality and Azuma s inequality. Theorem 3 [3] Let X 1,..., X n be independent random variables with PrX i = 1) = p i, PrX i = 0) = 1 p i. We consider the sum X = n i=1 X i with expectation EX) = n i=1 p i. Then we have Lower tail) PrX EX) λ) e λ2 /2EX), Upper tail) PrX EX) + λ) e λ 2 2EX)+λ/3). Theorem 4 [2] Let X be a random variable determined by m trials T 1,..., T m, such that for each i, and any two possible sequences of outcomes t 1,..., t i and t 1,..., t i 1, t i : E X T 1 = t 1,..., T i = t i ) E X T 1 = t 1,..., T i 1 = t i 1, T i = t i) c i then Pr X EX) λ) 2e λ2 /2 m i=1 c2 i. The following lemma on edge density will be useful later. Lemma 1 Almost surely a random graph G in Gn, p) satisfies, for all U V G) with U log n, eu) p 2 U 2 C U 3/2 log 1/2 n where C is some positive constant. The lemma follows from Theorem 3. The following lemma is along the lines of a classical result of Erdős for random graphs [4]. We include the statement and a short proof here for the sake of completeness. Lemma 2 For G Gn, p), where p is a constant no greater than 1 2, almost surely all complete bipartite graphs K A,B in G with A B satisfy A 2 log b n where b = 1/p. Proof: For two subsets A and B of V G), with A = B = k, the probability that A and B form a complete bipartite graph in Gn, p) is at most p k2. There are at most n n k) k) choices for A and B. For k 2 log b n, we have ) 2 n p k2 = o1) k as p is a constant. The lemma then follows. We note that Erdős result on Ramsey s theorem [4] states that every 2-coloring of the edges of the complete graph K n contains a monochromatic clique of order 1 2 log 2 n. It is not 3

4 hard to show along the same lines that Gn, 1/2) contains a complete bipartite graph K A,B with A = B = 1 4 log 2 n and the bound in Lemma 2 is tight up to a constant factor. The upper bound in Theorem 2 is an immediate consequence of 3) and the classical results on the independence number αg) of a random graph. An upper bound for αg) can be found in [4]. The problem of determining the independence number for a random graph has been extensively studied in the literature. The asymptotic order of αg) for G in Gn, p) was determined in [6]. Theorem 5 [6] If p is a constant, p < 1 c, and G Gn, p), then almost surely αg) is of order αg) = 2 log b n + olog n) where b = 1/p and c is a positive constant. 3 Edges covered by a given family of subsets For a graph G = V, E) and A V, we define V G, A) = {v : v V G) \ A and {u, v} E for all u A}. It immediately follows that A and B form a complete bipartite graph if B is contained in V G, A), namely, B V G, A). We say an edge u, v) E is covered by A if either u A and v V G, A) or v A and u V G, A). For A = {A 1, A 2,..., A k } 2 V and σ, a linear ordering of [k], we define a function l as follows. For notational convenience, we use i to denote the i-th element under the ordering σ. For each 1 i k, we define G i and li) recursively. We let G 1 = G and let l1) be an arbitrary subset of V G 1, A 1 ). Given G i 1, we let G i be a new graph with the vertex set V G) and the edge set EG i 1 ) \ EA i 1, li 1)). We set li) to be an arbitrary subset of V G i, A i ). We define fg, A) = max σ max l k EA i, li)). Basically, for given A i s, we wish to choose B i s so that the complete bipartite graphs K A1,B 1,..., K Ak,B k cover as many edges in G as possible. An example is illustrated in Figure 1 for A = {A 1, A 2, A 3 } with A 1 = {a, b}, A 2 = {b, c}, and A 3 = {c, d}. Here fg, A) = 4 is achieved by σ = identity, l1) = {e}, l2) =, and l3) = {e}, or σ = 213), l1) = and l2) = l3) = {e}. We observe fg, A) k i=1 EA i, V G, A i )) as li) V G, A i ). When e i=1 a b c d Figure 1: An illustration of fg, A). U V G) and A U for each A A, we use fg, U, A) to denote fg[u], A). Note that G[U] denotes the induced subgraph of G on a subset U of V G). 4

5 Lemma 3 Suppose that for G Gn, p) and U V G), A is a family of 2-sets of U with A U. Then almost surely we have fg, U, A) 2p 2 A U + 8 U U log n for p 1 2. Proof: We list edges with both endpoints in U as e 1, e 2,..., e m where m = ) U 2. For each e i = u i, v i ) for 1 i m, we consider T i {H, T} where T i = H means e i is an edge and T i = T means e i is not an edge. To simplify the notation we use X to denote the random variable fg, U, A) and notice that X is determined by T 1,..., T m. Given the outcome t j of T j for each 1 j i 1 we wish to establish an upper bound for EX T 1 = t 1,..., T i 1 = t i 1, T i = H) EX T 1 = t 1,..., T i 1 = t i 1, T i = T). 4) Let K 1 be the set of graphs over U such that e j is given by t j for each 1 j i 1 and e i is a non-edge. Similarly, let K 2 be the set of graphs over U such that e j is given by t j for each 1 j i 1 and e i is an edge. We have K 1 = K 2. Thus we get EX T 1 = t 1,..., T i 1 = t i 1, T i = H) = K K 1 fk, A)PrK K 1 ) and EX T 1 = t 1,..., T i 1 = t i 1, T i = T) = K K 2 fk, A)PrK K 2 ). Define a mapping µ : K 1 K 2 such that EK) and EµK)) differ only by e i for each K K 1. We get µ is a bijection and PrK K 1 ) = PrµK) K 2 ). Therefore the expression 4) can be bounded from above by K K 1 fk, A) fµk), A) PrK K 1 ). Notice that each edge can be covered by at most once. We observe fk, A) fµk), A) 2 because e i and the other edge sharing one endpoint with e i could be covered by A in µk) but not in K. Therefore 4) is bounded above by 2. Now we apply Theorem 4 for λ = 8 U U log n and c i = 2. Then we have Pr X EX) 8 U ) U log n 2e 64 U 3 log n/2 m i=1 c2 i 2e 4 U log n, 5) using the fact m U 2 2. To estimate EX), we note that EfG, U, A)) 2p2 U for a fixed A A. Therefore, EX) A A E fg, U, A)) 2p 2 A U. Thus 5) implies Pr X 2p 2 A U + 8 U ) U log n Pr X EX) 8 U ) U log n 2e 4 U log n. Recall the assumptions A U and A = 2 for each A A. For a fixed size of U, the number of choices for U and A is at most n U U 2 U which is less than n 3 U. Therefore the probability that there are some U and A which violate the assertion in the lemma is at most ne 4 U log n n 3 U < 1 n for sufficiently large n as U 2. The lemma is proved. The following lemmas for other families of sets A have proofs which are quite similar to the proof of Lemma 3. We will sketch proofs here. 5

6 Lemma 4 Suppose that for G Gn, p), A is a family of subsets of U V G) satisfying A U and 2 A 2 log 2 n for each A A. Then almost surely we have for p 1 2. fg, U, A) 2p 2 A U + 8 U log 2 n U log 2 n log U Proof: We will use the Azuma s inequality. For e i = u i, v i ), we define K 1, K 2, and a bijection µ similarly. The only difference is that fk, A) fµk), A) 2 log 2 n for each K K 1. This is because e i and at most other 2 log 2 n 1 edges sharing the same endpoint with e i could by covered by A in µk) but not in K. Therefore the corresponding expression for 4) can be upper bounded by 2 log 2 n. We can then estimate X = fg, U, A) by applying Theorem 4 with λ = 8 U log 2 n U log 2 n log U and c i = 2 log 2 n. This leads to Pr X EX) 8 U log 2 n ) U log 2 n log U 2e 4 U log 2 n log U. 6) Since EX) A A EfG, U, A)) 2p2 A U and the number of choices for U and A can be bounded from above by n 1+ U U 2 U log 2 n, we can bound the probability in 6) by 2e 4 U log 2 n log U n 1+ U U 2 U log 2 n 1 n as U 2. This completes the proof of the lemma. Lemma 5 Suppose that for G Gn, p), A is a family of subsets of U V G) satisfying A U and 3 A 2 log 2 n for each A A. Then almost surely we have for p 1 2. fg, U, A) 3p 3 A U + 8 U log 2 n U log U log 2 n Proof: The proof is similar to that of Lemma 4. The only difference is that we assume A 3 and therefore EfG, U, A)) A A EfG, U, A)) 3p 3 A U. We use Theorem 4 in a similar way as in the proof of Lemma 4 to complete the proof of Lemma 5. Lemma 6 Suppose that for G Gn, p), A is a family of subsets of U V G) satisfying A U 1+δ and δ log b U A 2 log 2 n for each A A, where b = 1 p and δ is some positive constant. Then almost surely we have provided p 1 2. fg, U, A) δ A U 1 δ log b U + 8 U 3+δ)/2 log 2 n log U log 2 n Proof: We use the assumptions on A to derive EfG, U, A)) A A EfG, U, A)) δ log b U p δ log b U A U δ A U 1 δ log b U. Then we bound the number of choices for U and A from above by n 1+ U U 2 U 1+δ log 2 n. Applying Theorem 4 for λ = 8 U 3+δ)/2 log 2 n log U log 2 n and c i = 2 log 2 n, the lemma then follows. 6

7 4 Bounding uncovered edges In order to prove the bipartite decomposition theorem, we also need to establish lower bounds for the number of uncovered edges for a given family A of subsets. First, we will derive a lower bound on the number of uncovered edges for a collection A of 2-sets of V G). Let S 0 be the set of u V G) such that u is in only one A A. For u in S 0, we denote the only 2-set containing u by A u. Our goal is to give a lower bound on the number of uncovered edges with both endpoints in S 0. To simplify the estimate, we impose some technical restrictions and work on a subset S of S 0. To do so, we will lose at most a factor of 2 in the lower bound estimate which is tolerable). To form S, for each A u = {u, v} with u, v S 0, we delete one of u and v arbitrarily from S 0. Let T = u S A u \ u). Clearly S and T are disjoint. Furthermore, S T. We define E to be the set of edges u, v) with u S and v T and E to be the set of edges u, v) with u, v S. We assume E = p. For a fixed ordering σ of edges in E and an ordering τ of edges in E, we define sets E i and E i for each 0 i p recursively. Let E 0 = E and E 0 = E. Given E i 1 and E i 1 for each 1 i p, we assume the first edge in E i 1 is e = u, v) with u S and v T. Let u, v ) E i 1 be the first edge such that u = u and A v = {v, v }. We note edges u, v) and u, v ) are covered by {v, v }. We define E i = E i 1 \ u, v) and E i = E i 1 \ u, v ). If there is no such an edge u, v ) then we define E i = E i 1 \ u, v) and E i = E i 1. Finally we define gg, S, T ) = min σ min τ E p, where σ and τ range over all orderings of edges in ES, T ) and ES), respectively. Lemma 7 Suppose that a graph G and each edge of G is contained in at most one of the complete bipartite graphs K Ai,B i with A i = 2 B i. For A = {A : A = A i for some i}, let S and T denote two disjoint subsets of V G) satisfying the properties that i) each u S is in a unique A A, ii) S does not contain any A in A, iii) T = {v : {u, v} A for some u S}. Then gg, S, T ) is a lower bound for the number of uncovered edges by A with both endpoints in S. Proof: We can choose an arbitrary order σ on edges in ES, T ). Suppose an edge e = u, v), with u S and v T, is uniquely covered by A in A. If A = {w, v} for some w in S, we associate with e the edge u, w) ES, S). In addition, we choose the order τ on ES, S) to be consistent with σ in the sense that the associated edges in ES, S) maintain the same order. If A = {w, v} for some w not in S or e is not covered by any A, then we do not associate any edge to e. We note that if each edge e = u, v ) ES, S) is contained in a unique bipartite graph K A,B, then A is in A from the assumptions i) iii). From the definition of g, e will be removed in the process. Thus, gg, S, T ) is a lower bound for the number of uncovered edges by A. Lemma 8 For G Gn, p) and disjoint subsets S and T in V G), almost surely we have ) S gg, S, T ) p 3 5 S S log n 2 for all choices of S and T, provided p 1 2 and S 4. Proof: We sketch the proof here which is similar to that of Lemma 3. We list edges with endpoints in S T as e 1,..., e m, where m = ) S T 2. For each ei = u i, v i ) for 1 i m, we consider T i {H, T} where T i = H means e i is an edge and T i = T means e i is not an edge. 7

8 Let X denote the random variable gg, S, T ) for G Gn, p). We note that X is determined by T 1,..., T m. For the fixed outcome t j of T j for 1 j i 1, we consider EX T 1 = t 1,..., T i 1 = t i 1, T i = H) EX T 1 = t 1,..., T i 1 = t i 1, T i = T). 7) For e i = u i, v i ), if u i, v i T then the outcome of T i does not contribute to 7). If u i, v i S then the outcome of T i can change 7) by at most one depending on whether e i is covered or not. If u i S and v i T then the outcome of T i could effect 7) by at most one. This is because e i could make another edge u i, w) covered by the 2-set {v i, w}. Thus 7) is bounded above by one. Applying Azuma s theorem as stated in Theorem 4 with λ = 5 S S log n and c i = 1, we have Pr X EX) 5 S ) S log n 2e 6 S log n, 8) using m 2 S 2. To estimate EX), we define X u,v, for u, v S, to be the event u, v) EG), u, A v \ {v}) EG), and v, A u \ {u}) EG). Here G Gn, p) and A v resp.a u ) is the only 2-set containing vresp.u). Let I u,v denote the random indicator variable for X u,v. We note PrI u,v = 1) = p 3 and EX) ) S PrI u,v = 1) = p 3. 2 Thus ) S Pr X p 3 2 u,v S 5 S ) S log n Pr X EX) 5 S ) S log n 2e 6 S log n. For a given size of S, there are at most n 2 S choices for S and T and S 2 S choices for σ and τ. By 8), the probability that there are some S and T which violate the lemma is at most 2n 3 S +1 e 6 S log n < 1 n provided n is sufficiently large and S 4. This proves the lemma. Next, we wish to establish a lower bound on the number of uncovered edges for general cases of A. For W U V G), we consider L: W 2 U\W with the property Lw) Lw ) = for w, w W. We define hg, U, W, L) to be the number of edges w, w ) in G such that w, w W, w, z) EG) for each z Lw ), and w, z ) EG) for each z Lw). We will use the following Lemma late we will show that hg, U, W, L) gives a lower bound for the number of uncovered edges). Lemma 9 For G Gn, p), where p is a constant no greater than 1/2, b = 1 p, and U 4, suppose W U satisfies W = U / log 2 b U and L as defined above satisfies 1 Lw) c log b U for some positive constant c. Then almost surely we have hg, U, W, L) c U 2 2c / log 4 b U 2 U 3/2 log n for all choices of U, W and L, where c is some positive constant. Proof: For u, v W, let X u,v denote the event that u, v) EG), u, w) EG) for each w Lv), and v, z) EG) for each z Lu). Here G Gn, p). The random indicator variable for X u,v is written as I u,v. From the definition of h, we have hg, U, W, L) = u,v W I u,v = Y for G Gn, p). Since PrI u,v = 1) b 1 2c log b U, we have ) EY ) b 1 2c log b W U c U 2 2c / log 4 b U, 2 8

9 for some constant c. From the definition of L, we have X u,v are independent of one another. By applying the Chernoff s bound for the lower tail in Theorem 3 with λ = 2 U 3/2 log n, we have PrY c U 2 2c / log 4 b U 2 U 3/2 log n) PrY EY ) 2 U 3/2 log n) e 4 U 3 log n 2EY ) e 2 U log n, using the fact that EY ) U 2. For a given size U of U, it is straightforward to bound the number of choices for U, W and L from above by n U U U / log2 b U U c U / log b U < n U U 2c U / log b U n 1.5 U, when n is sufficiently large. The probability that there is some U, W and L which violate the lemma is at most ne 2 U log n n 1.5 U < 1 n if n is large enough. This completes the proof of the lemma. 5 A theorem on strong bipartition decompositions Recall the strong bipartition number τ G) is the minimum number of complete bipartite graphs whose edges partition the edge set of G and none of them is a star. We will prove the following theorem. Theorem 6 Suppose G Gn, p), U V G) is a vertex subset with U clog b n) 3+ɛ where b = 1/p, c and ɛ are positive constants. For p being a positive constant no greater than 1/2, almost surely we have τ G[U]) U. The proof of Theorem 6 is based on several lemmas which we will first prove. In this section, we may assume that G Gn, p) satisfies the statements in all lemmas in the preceding sections. By Lemma 1, the number of edges in G[U] satisfies eg[u]) = p 2 + o1))u2, here U = u. We will prove Theorem 6 by contradiction. Suppose EG[U]) = m i=1ek Ai,B i ) and m < ) u where denotes the disjoint union. We assume further Ai B i for each 1 i m. Lemma 2 implies A i 2 log 2 n for each 1 i m. We define L = {A i : 1 i m} and note that L could be a multi-set. We consider three subsets of L defined as follows: L 1 = {A i L: A i < δ 1 log b u} L 2 = {A i L: A i < δ 2 log b u} L 3 = {A i L: A i = 2}, where We have the following lemma. { } ɛ δ 1 = min 43 + ɛ), and δ 2 = δ Lemma 10 If L ) u, then we have L ) 1 u. 9

10 Proof: We will first prove the following claim: Claim 1: L ) 1 u. Proof of Claim 1: Suppose the contrary. By Lemma 3, the number of edges covered by L 3 is at most 2p 2 L 3 u + 8u u log n. By Lemma 5, the number of edges covered by L 2 \ L 3 i.e., A i 3) is at most 3p 3 L 2 \ L 3 u + 8u log 2 n u log u log 2 n. Therefore the total number of edges covered by L 2 is at most 2p 2 L 3 u + 3p 3 L 2 \ L 3 u + 8u u log n + 8u log 2 n u log u log 2 n. Thus the number of edges which are not in any K Ai,B i with A i L 2 is at least p ) 2 + o1) u 2 2p 2 L 3 u 3p 3 L 2 \ L 3 u 8u u log n 8u log 2 n u log u log 2 n. Observe that the expression above is a decreasing function if we view L 3 as the variable. From the assumptions L 3 < 1 2 ) u and L ) 1 u, the number of edges which are not contained in any K Ai,B i with A i L 2 is at least p 1 2 u2 2 1 ) 2p 2 u 2 7 ) p p3 u 2 + ou 2 2 ) 125 7p o1) u 2 when n is large enough. Here we note that u log 2 n u log u log 2 n = ou 2 ) as we assume u clog b n) 3+ɛ. Since p 1 p2 2 and p is a constant, we get that 125 7p3 500 is a positive constant. Applying Lemma 6 with δ = δ 2, the number of edges covered by L \ L 2 i.e., A i δ 2 log b n) is at most δ 2 L \ L 2 u 1 δ2 log b u + 8u 3+δ2)/2 log 2 n log u log 2 n. Since u 3+δ2)/2 log 2 n log u log 2 n = ou 2 ) by the choice of δ 2, in order to cover the remaining edges, we need at least C 1 u 1+δ2/2 extra complete bipartite graphs for some positive constant C 1, i.e., L \ L 2 C 1 u 1+δ2/2. Since C 1 u 1+δ2/2 > ) 1 u for sufficiently large n, this leads to a contradiction. Thus we have L ) and the claim is proved. Now we proceed to prove the lemma using the fact that L u). We consider a auxiliary graph U whose vertex set is U and edge set is L 3. It is possible that U has multiedges. We partition the vertex set of U into three sets U 1, U 2, and U 3, where U 1 = {v V U ): d U v) = 0}, U 2 = {v V U ): d U v) = 1}, and U 3 = {v V U ): d U v) 2}. We will prove the following. Claim 2: The number of edges not contained in any K Ai,B i for A i L 3 is at least p U 1 ) 2 + p 3 U 2 /2) 2 + ou 2 ). Proof of Claim 2: The first part of the sum follows from Lemma 1. For the second part of the sum, we let U 2 U 2 such that for each v U 2, the neighbor of v in U is not in U 2. We have U 2 U 2 /2. Then we apply Lemma 8 with S = U 2 and T consisting of neighbors of S in U. To finish the proof of Claim 2, we use the fact S S log n = ou 2 ) as u = clog b n) 3+ɛ. We will prove Lemma 10 by contradiction. Suppose L ) 1 u. This implies that the average degree of U is at most We consider the following cases. 10

11 Case 1: U u. By considering the total sum of degrees of U, we have 2 U 3 + u U 1 U 3 ) ) u Thus, U u implies U u. Claim 2 together with this lower bound on U 1 implies that the number of edges not in any K Ai,B i with A i L 3 is at least p 72 + o1)) u 2. By Claim 1 we have L ) 1 u, so the number of additional 7 complete bipartite graphs K Ai,B i with A i L 2 \ L 3 is at most 1500u using the assumption L ) 1 u. These complete bipartite graphs can cover at most 7p o1))u2 edges by Lemma 5. Thus we conclude that the number of edges not covered by any of K Ai,B i with A i L 2 is at least ) p 72 7p o1) u 2. Note that p 72 7p3 500 is a positive constant when p is constant and p 1 2. By applying Lemma 6 with δ = δ 2, the bipartite graphs K Ai,B i with A i L \ L 2 i.e., A i δ 2 log b n) can cover at most δ 2 L \ L 2 u 1 δ2 log b u + 8u 3+δ2)/2 log 2 n log u log 2 n edges. We note u 3+δ2)/2 log 2 n log u log 2 n = ou 2 ) because of the choice of δ 2. To cover the remaining edges, we need at least C 1u 1+δ2/2 extra complete bipartite graphs K Ai,B i with A i L\L 2 for some positive constant C 1, i.e., L\L 2 C 1u 1+δ2/2. Since C 1u 1+δ2/2 > ) 1 u for n large enough, we get a contradiction to the assumption L Case 2: U 3 < 1 5 u ) u. In this case we have U 1 + U 2 4 5u. Note that the lower bound given by Claim 2 is minimized when U 2 = 4 5 u, i.e., the number of edges not contained in any K A i,b i with A i L 3 is at least 2 25 p3 + o1) ) u 2. By the same argument as in Case 1 we can show the number of edges in K Ai,B i with A i L 2 \ L 3 is at most 7p o1))u2. Now there are at least ) p3 + o1) u 2 edges which is not in any K Ai,B i with A i L 2. We note that p3 is a positive constant as we assume p is a constant. By using Lemma 6 with δ = δ 2, the bipartite graphs K Ai,B i with A i L \ L 2 i.e., A i δ 2 log b n) can cover at most δ 2 L \ L 2 u 1 δ2 log b u + ou 2 ) edges. As in Case 1, we consider the number of extra bipartite graphs K Ai,B i with A i L \ L 2 needed to cover the reaming edges, leading to the same contradiction to the assumption on L. Therefore we have proved L 3 > ). Lemma 11 Let H be a hypergraph with the vertex set U and the edge set L 1. There is some positive constant C 2 such that there are C 2 u vertices of H with degree less than δ ) δ1 logb u. 11

12 Proof: We consider several cases. Case a: L 2 > ) 1 u. The sum of degrees in H is less than 1 δ 2 L 2 log b u + δ 1 L 1 \ L 2 log b u δ ) 1500 δ1 2 δ u log b u + δ ) u log 1500 b u ) u log b u. Here we used the assumption L 1 L = m < ) u and the choice of δ2. Case b: L ) 1 u. By Lemma 10, L L 3 + δ 1 L 1 \ L 3 log b u ) u. The sum of degrees is at most ) 1 u + δ δ1 2 δ ) 1 u log 2000 b u. ) u log b u We have proved that the sum of degrees of H is less than δ ) δ1 u logb u. Let U be the set of vertices with degree at least δ ) δ1 logb u. We consider U δ1 2 δ ) 1 δ1 log 3000 b u 2 δ ) 1 u log 2000 b u, which yields U 1 C 2 )u for some positive constant C 2. Each vertex in U \U has degree less than δ ) δ1 logb u and U \ U C 2 u. The lemma is proved. We recall that G[U] is the subgraph of G induced by U. We have the following lemma. Lemma 12 The number of edges in G[U] which are not contained in any K Ai,B i L 1 is at least C 3 u 2 δ1+δ1/2000 for some positive constant C 3. with A i Proof: We consider the hypergraph H with the vertex set U and the edge set L 1 as defined in Lemma 11. Let W be the set of vertices with degree less than δ ) δ1 logb u in H; we have W C 2 u for some positive constant C 2 by Lemma 11. We will use Lemma 9 to prove Lemma 12. In order to apply Lemma 9, we will first find a subset W of W such that for any u, v W there is no A i L 1 containing u and v. Also we will associate each w W with a set Lw) U \ W satisfying the property that Lw) Lw ) = for each w w W. To do so, we consider an arbitrary linear ordering of vertices in W. Let q = W / log 2 b u, W 0 = W, Z 0 = and H 0 = H. For each 1 i q, we recursively define a vertex v i, a set W i, a set Z i and a hypergraph H i as follows: For given W i 1 and H i 1, we let v i be the first vertex in W i 1 and define F v i ) = {A EH i 1 ): v i A}. By the assumption on the size of sets in L 1 and the degree upper bound for vertices in W, we have A F vi) A log 2 b u/2. We define Z i = {A EH i 1 ): A \ A F v i)a ) = 1}. Then A Zi A \ A F v i)a ) log 2 b u/2 since each A F v i ) can contribute at most δ 1 log b u to the sum and F v i ) δ1 2 log b u because of the degree upper bound for vertices in W. We define W i = W i 1 \ ) A Zi F v i)a and H i to be the new hypergraph with the vertex set V H i 1 ) \ ) A Zi F v i)a. If A EH i 1 ) then A \ ) A Zi F v i)a EH i ). Thus V H i ) = W i W i 1 log 2 b u and W q 1. Therefore, v i is well-defined for 1 i q. We write W = {v 1, v 2,..., v q }. 12

13 For each A F v i ) and A F v j ) with i < j we have A A = as we delete the set A Zi F v i)a in step i. For each v i W and each A F v i ), we let fa) be an arbitrary vertex other than v i from A and F v i ) = A F vi){fa)}. It follows from the preceding definitions that F v i ) F v j ) = for 1 i j q. Furthermore, for each v i and each A L 1 containing v i, either A is in F v i ) or a subset of A with size at least two is in F v i ). Hence, A F v i ). For an edge v i, v j ), if v i, z) is a non-edge for each z F v j ) and v j, z ) is a non-edge for each z F v i ), then the edge v i, v j ) is uncovered by the family of sets L 1. Suppose v i, v j ) is in K A,B for some A L 1. We have either v i A or v j A. In the former case we get A Lv i ) by the definition of Lv i ). Let z A Lv i ). Then A and B does not form a complete bipartite graph since v j, z) is not an edge by the assumption. We get a contradiction and we have a similar argument for the later case. Therefore the function hg, U, W, L) gives a lower bound for the number of uncovered edges in G for the given family of sets L 1. Now we apply Lemma 9 with U = V H), W = W, Lv i ) = F v i ) for each v i and c = δ1 2 δ The lemma then follows. We are ready to prove Theorem 6. Proof of Theorem 6: Suppose that EG[U]) = m i=1ek Ai,B i ). If m > ) u, then we are done. Otherwise, Lemma 12 implies that there are at least C 3 u 2 δ1+δ1/2000 edges uncovered after we delete the edges in K Ai,B i for each A i L 1. We then apply Lemma 6 with δ = δ 1 which gives an upper bound for the number of edges covered by L \ L 1 i.e., A i δ 1 log b u) : δ 1 L \ L 1 u 1 δ1 log b u + 8u 3+δ1)/2 log 2 n log u log 2 n. Here we note that u 3+δ1)/2 log 2 n log u log 2 n = ou 2 δ1+δ1/2000 ) because of the choice of δ 1. Therefore we need at least C 4 u 1+δ1/2500 additional complete bipartite graphs K Ai,B i with A i L \ L 1 to cover the remaining edges, where C 4 is some positive constant. Since C 4 u 1+δ1/2500 > u when n is sufficiently large and we get a contradiction. Theorem 6 is proved. 6 Proof of Theorem 2 Before proving Theorem 2, we first prove the following lemma. Lemma 13 Suppose that edges of G can be decomposed into k 1 complete bipartite graphs, of which k 2 complete bipartite graphs are stars for some k 2 k 1. Then G has an edge decomposition EG) = k i=1 EK A i,b i ) with k k 1 such that for i k 2, K Ai,B i are stars and for j > k 2, we have A j, B j V G) \ k1 i=1 A i. Proof: For an edge decomposition B = {K A1,B 1,..., K Ak1,B k1 }, we can modify B by the following algorithm. Algorithm A Input G and B. Step 1: Set G = G, V = and B =. Step 2: If none of K Ai,B i B is a star, then stop and output B. Otherwise go to Step 3. 13

14 Step 3: For i = 1,..., k 1, if K Ai,B i is a star, add the center of the star to V and add K Ai,B i to B. Step 4: For each K Ai,B i B \ B, replace it by K A i,b i where A i = A i \ V and B i \ V. For each star S = K Ai,B i in B with A i = {v i }, define S i to be the star centered at v i containing all edges incident to v i in G \ {v 1,..., v i 1 }. Then we replace S = K Ai,B i by S in B. Step5: Output B. We note that the cardinality of B does not increase throughout Algorithm A, although it is possible that some member of B might have no edge left. In that case, B decreases. Lemma 14 For a graph G, there is a subset T V G) such that τg) = n T + τ G[T ]). Proof: We use the notation in the proof of Lemma 13. Note that for k 1 = τg), we have k k 1 so that the output of Algorithm A have size k = k 1. We choose T = V \ V. Then all non-stars in the decomposition in B are an edge decomposition for G[T ]. Therefore, we have τg) = k = V + τ G[T ]). We are ready to prove Theorem 2. Proof of Theorem 2: The upper bound follows from the well known fact see Theorem 5) that almost surely a random graph G Gn, p) has an independent set I with size 2 log b n where b = 1/p and p is constant. We consider vertices v 1,..., v m with m = n 2 log b n, which are not contained in I. For each 1 i m we define a star K Ai,B i with A i = {v i } and B i = {v j : j > i and v i, v j ) EG)}. We have EG) = m i=1ek Ai,B i ). Therefore we have τg) n 2 log b n. For the lower bound, we may assume that G Gn, p) satisfies all statements in the lemmas in the preceding sections. Suppose G has an edge decomposition: EG) = k i=1ek Ai,B i ), with k = τg) and assume that for some l k, we have A i = {v i } for 1 i l. Let W = {v 1,..., v l }. If W = then Theorem 2 follows from Theorem 6 directly. We need only to consider the case W. By Algorithm A we can assume EG ) = k i=l+1 EK A i,b i ) where G is the subgraph induced by T = V G) \ W. We get τg) = W + τ G ). 9) We will prove l > n clog b n) 3+ɛ for any positive constants c and ɛ. Suppose l n clog b n) 3+ɛ for some c and ɛ. Thus, T clog b n) 1+ɛ. By Theorem 6 we have τ G[T ]) ) T. Therefore τg) = W + τ G ) W ) T n, which is a contradiction. Theorem 2 is proved. 14

15 7 Problems and remarks The results on the bipartite decompositions in this paper lead to many questions, several of which we mention here. Conjecture 1: For a random graph G Gn, p), where p is a constant no greater than 1/2, almost surely τg) = min{n αg), n βg) + 1}, where βg) is the size of largest induced complete bipartite graph in G. A somewhat weaker conjecture is the following: Conjecture 2: For a random graph G Gn, p), where p is a constant no greater than 1/2, almost surely τg) = n 2 + o1)) log b n, where b = 1/p and log denotes the natural logarithm. For sparser random graphs, Alon [1] showed that there exists some small) constant ) c such that for 2 n p c, a random graph G in Gn, p) satisfies τg) = n Θ log np p. It will be of interest to further sharpen the lower bound. Conjecture 3: For a random graph G Gn, p), with p = o1), almost surely τg) = n 1 + o1)) 2 p log np. Conjecture 4: For a random graph G Gn, p), where p is a constant no greater than 1/2, suppose that an edge decomposition EG) = k i=1 EK A i,b i ) achieves τg) = k. Then almost surely at least n on) of the bipartite subgraphs K Ai,B i are stars. In this paper, we have given rather crude estimates for the constants involved. In particular, for the strong bipartition number τ G), a consequence of Theorem 6 states that for G Gn, p), where p is a constant no greater than 1/2, almost surely τ G) n. For the case of p c for some small c, Alon [1] showed that almost surely τ G) 2n for G Gn, p). A natural question is to improve the lower bound for τ G). In the other direction, it is of interest to characterize graphs with specified upper bounds for τ. Problem 5: Characterize graphs G which satisfy References τ G) n. [1] N. Alon, Bipartite decomposition of random graphs, arxiv: [math.CO]. [2] K. Azuma, Weighted sums of certain dependent random variables, Tohuku Math. Journal, 193) 1967), [3] H. Chernoff, A note on an inequality involving the normal distribution, Ann. Probab., ), [4] P. Erdős, Some remarks on the theory of graphs, Bull. Amer. Math. Soc., ), [5] R. L. Graham and H. O. Pollak, On the addressing problem for loop switching, Bell Syst. Tech. J., 50 8) 1971),

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