Bounds for pairs in partitions of graphs

Transcription

1 Bounds for pairs in partitions of graphs Jie Ma Xingxing Yu School of Mathematics Georgia Institute of Technology Atlanta, GA , USA Abstract In this paper we study the following problem of Bollobás and Scott: What is the smallest fk, m such that for any integer k 2 and any graph G with m edges, there is a partition V G = k i=1 V i such that for 1 i j k, ev i V j fk, m? We show that fk, m < 1.6m/k + om, and fk, m < 1.5m/k + om for k 23. While the graph K 1,n shows that fk, m m/k 1, which is 1.5m/k when k = 3. We also show that f4, m m/3+om and f5, m 4m/15+om, providing evidence to a conjecture of Bollobás and Scott. For dense graphs, we improve the bound to 4m/k 2 + om, which, for large graphs, answers in the affirmative a related question of Bollobás and Scott. AMS Subject Classification: 05C35, 05D40 Keywords: Graph partition; Judicious partition; Azuma-Hoeffding inequality Partially supported by NSA and by NSFC Project

2 1 Introduction For a graph G, we use V G and EG to denote the vertex set and edge set of G, respectively. We use δg to denote the minimum degree of G. For subsets S,T of V G, we use es,t to denote the number of edges of G with one end in S and the other in T; es to denote the number of edges with both ends in S; and ds to denote the number of edges with at least one end in S. Classical graph partition problems often ask for partitions of a graph that optimize a single quantity. For example, the Maximum Bipartite Subgraph Problem asks for a partition V 1,V 2 of the vertices of a graph that maximizes ev 1,V 2. This problem is NP-hard, see [11]. However, it is easy to prove that any graph with m edges has a partition V 1,V 2 with ev 1,V 2 m/2. Edwards [8,9] improved this lower bound to m/ m + 1/4 1/2, which is best possible for complete graphs K 2n+1. A different type of partition problems ask for a partition of a given graph that optimizes several quantities simultaneously. Such problems are called Judicious Partition Problems by Bollobás and Scott [3]. The Bottleneck Bipartition Problem, raised by Entringer see, for example, [13, 15] is a judicious partition problem: Find a partition V 1,V 2 of the vertex set of a graph G that minimizes max{ev 1,eV 2 }. Shahrokhi and Székely [16] showed that this problem is also NP-hard. Porter [13] proved that any graph with m edges has a partition V 1,V 2 with ev i m/4 + O m, establishing a conjecture of Erdös. A matrix version of this Erdös conjecture was formulated by Entringer, and was solved by Porter and Székely [14]. Bollobás and Scott [5] improved this to ev i m/ m + 1/4 1/2, and showed that K 2n+1 are the only extremal graphs. We note that in [1] a connection is given between the Maximum Bipartite Subgraph Problem and the Bottleneck Bipartition Problem. Bollobás and Scott [5] proved that for any integer k 1 and any graph G of size m, V G can be partitioned into V 1,...,V k such that ev i m k + k 1 2 2k 2m + 1/4 1/2 for 2 i {1, 2,..., k}. The complete graphs of order kn + 1 are the only extremal graphs modulo isolated vertices. In this paper we study the following judicious partition problem of Bollobás and Scott [7]. Problem 1.1 What is the smallest fk,m such that for any integer k 2 and any graph G with m edges, there is a partition V G = k i=1 V i such that for 1 i j k, ev i V j fk,m? Note that the case k = 2 for Problem 1.1 is trivial. For k = 3, we note that for each permutation ijk of {1,2,3}, dv i = m ev j V k ; so Problem 1.1 asks for a lower bound on min{dv i : i = 1,2,3} which is studied in [12]. For k 4, bounding max{ev i V j : 1 i j k} is much more difficult than bounding max{ev i : 1 i k}: In the former case one needs to bound k 2 quantities resulted from a k-partition, while in the latter case one only needs to bound k quantities. In Section 2, we use probabilistic method to show that fk,m < 1.6m/k +om, and that fk,m < 1.5m/k + om for k 23. The following example shows that fk,m m/k 1, which is close to 1.6m/k when k = 3. For k 3, take the graph K 1,n with n k 1, and let x be the vertex of degree n. Let V 1,...,V k be a k-partition of V G, with x V 1. Without loss of generality, we may assume that V 2 n + 1 V 1 /k 1. Now ev 1 V 2 n + 1 V 1 /k 1 + V 1 1 = 2

3 n + k 2 V 1 1/k 1 n/k 1 = m/k 1, where m is the number of edges in K 1,n. On the other hand, the complete graph K k+2 has m = k+2 2 edges, and any k-partition V 1,...,V k of K k+2 has two sets, say V 1,V 2, such that V 1 V 2 = 4. So ev 1 V 2 = 6 = 12m 12m k+2k+1. This shows that fk,m k+2k+1. For general complete graphs K n, a simple counting shows that for any k-partition V 1,...,V k of K n, k 2, there exist V i,v j such that V i + V j 2n/k; and hence ev i V j 2n/k 2. From this, we can deduce that fk,m 4m/k 2 + On, and this bound is achieved by taking a balanced k-partition of V K n i.e., any two partition sets differ in size by at most one. Note that K 1,n is sparse, i.e. the number of edges is On. The consideration of K 1,n and K k+2 led Bollobás and Scott [7] to the following conjecture. Conjecture 1.2 fk, m 12m k+2k+1 + On. The case k = 2 for Conjecture 1.2 is trivial as the bound becomes m + On. For k = 3, Conjecture 1.2 is equivalent to the following problem: Find a partition V G = V 1 V 2 V 3 so that dv i 2m/5 + On. It is shown in [12] that if G is a graph with m edges then there is a partition V 1,...,V k of V G such that dv i m/k 1 + Om 4/5 establishing a conjecture of Bollobás and Scott [6, 7], for large graphs. This result implies f3,m < m/2 + om 4/5 ; so Conjecture 1.2 holds for k = 3. In Section 3, we prove the bound 4m/k 2 + om for dense graphs, which implies that Conjecture 1.2 holds for dense graphs. As a consequence, we establish the following conjecture of Bollobás and Scott [7] for large graphs. Conjecture 1.3 For each k 2 there is a constant c k > 0 such that if G is a graph with m edges, n vertices, and δg c k n, then there is a partition V 1,...,V k of V G such that for 1 i j k, 12m ev i,v j k + 2k + 1. In Section 4, we show f4,m m/3 + om and f5,m 4m/15 + 0m, which implies Conjecture 1.2 for k = 4 and k = 5. In Section 5, we study partitions V 1,...,V k of graphs that optimize both max{ev i : 1 i k} and max{ev i V j : 1 i j k}. Bollobás and Scott [7] asked whether it is possible to find a partition V 1,...,V k such that ev i m + k 1 2m + 1/4 1/2 for 1 i k, and k 2 2k 12m 2 ev i V j k+2k+1 + On for 1 i j k. We show that for k = 3 and k = 4 one can find a partition satisfying these bounds asymptotically. 2 A bound for k-partitions In this section, we prove a bound on fk,m in Problem 1.1. First, we state the Azuma- Hoeffding inequality [2,10], which will be used to bound deviations. We use the version given in [4]. 3

4 Lemma 2.1 Let Z 1,...,Z n be independent random variables taking values in {1,...,k}, let Z := Z 1,...,Z n, and let f : {1,...,k} n N such that fy fy c i for any Y,Y {1,...,k} n that differ only in the i th coordinate. Then for any z > 0, z 2 P fz EfZ + z exp P fz EfZ z exp 2 k i=1 c2 i z 2 2 k i=1 c2 i We need a simple lemma which will also be used in Section 4 for finding probabilities when finding 4-partitions. Lemma 2.2 Let a j 0 for j {1,2,3,4} such that α := 4 j=1 a j > 0, and let f ij x i,x j = a i + a j x i + x j for 1 i j 4. Then there exist p i [0,1/2], 1 i 4, such that 4 i=1 p i = 1 and, for 1 i j 4, f ij p i,p j α/3.,. Proof. First, assume a i α/2 for all 1 i 4. Then p i := 1/2 a i /α [0, 1 2 ], and f ij p i,p j = a i + a j 1 a i + a j = 1 α α a i + a j α α 4 α 4. So we may assume without loss of generality that a 4 > α/2. Then a i + a j α/2 for all 1 i j 3. Let p 1 = p 2 = p 3 = 1/3 and p 4 = 0. Then for 1 i 3, f i4 = a i +a 4 /3 α/3; and for 1 i j 3, f ij = a i + a j 2/3 α/22/3 = α/3. Remark. From the proof of Lemma 2.2, we may choose p i = 0 when a i > α/2, and p i max{1/2 a i /α,1/3} when a i α/2. We need another lemma. Lemma 2.3 Let h 4 = 1/3. There exist t k,h k for k 5 such that h k = 2 2t k, and k 2t k 2 2t k = k 3 k 2t k k h k 1 + hk 1 k + 4 2t k. kk 1 Moreover, h k < 1.6/k, and h k < 1.5/k for k 23. Proof. We first show that there exist t k 0,1/2 and h k 1/k 1,2/k, k 5, such that h k = 2 2t k, and k 2t k 2 2t k = k 3 k 2t k k h k 1 + hk 1 k + 4 2t k. kk 1 Suppose k 5. Let f k t = 2 2t k 2t 4

5 and g k t = k 3 k h hk 1 4 k t. k kk 1 1 It is easy to see that f k t is decreasing, and g k t is increasing. Now assume that h k 1 < 2 k 1 for some k 5. Note that and g k 0 = k 3 k h k 1 < k 3 2 k k 1 < 2 k = f k0, g k 1/2 = k 2 k h 4 k 1 + kk 1 k 2 kk kk 1 > 1 k 1 = f k1/2. k 1 Therefore, since f k t is decreasing and g k t is increasing, there exists t k 0,1/2, for each k 5, such that f k t k = g k t k. Let h k := f k t k = 2 2t k k 2t k. Then since t k 0,1/2, 1/k 1 < h k < 2/k for k 5. Next, we show that h k < 1.6/k, and h k < 1.5/k for k 23. Let h k = c k /k, and it suffices to show c k < 1.6, and c k < 1.5 = 3/2 for k 23. Since h k 1/k 1,2/k, c k 1,2. Note that c k = 2 2t k k = k 3h k 1 + h k t k = k 3 k 2t k k 1 k 1 c k c k 1 k 1 2t k. From c k = 2 2t k k 2t k k we deduce t k = 2k kc k 2k 2c k ; and so c k = k 3 k 1 c k c k 12k kc k. k 1k c k With h 4 = 1/3 and hence c 4 = 4/3 and using MATLAB, we have c k < 1.6 for k = 5,...,22, and c < 3/2. Now assume k 24 and c k 1 < 3/2. Then c k < k /22k kc k, k 1 2 k 1k c k and so Hence, since c k 1,2, 2k 1c k < 3k c k c k c k /k c k. 2k + 9c k < 3k c kc k k c k = 3k c k 2 k c k < 3k /k 2. Therefore, c k < 3k k k + 9k 2 3/2. The last inequality holds since we assume k 24. We can now prove the main lemma for k-partitions. 5

6 Lemma 2.4 Let k 4 be an integer, let a j 0 for j {1,...,k} such that α := k j=1 a j > 0, and let f ij x i,x j = a i + a j x i + x j for 1 i j k. Let h k be defined as in Lemma 2.3. Then there exist p i [0,2/k], 1 i k, such that k i=1 p i = 1 and, for 1 i j k, f ij p i,p j 1.6α/k, and f ij p i,p j 1.5α/k for k 23. Proof. We apply induction on k; the case k = 4 follows from Lemma 2.2 as h 4 = 1/3. Suppose k 5. First, assume that there exists some l {1,...,k} such that a l tα, say l = k. Let p i = x for 1 i < k, with 0 x 1 k 1, and let p k = 1 k 1x. Then k i=1 p i = 1; for 1 i k 1, and for 1 i j k 1, f ik p i,p k 1 k 2xα; f ij p i,p j 2xa i + a j 2xα a k 1 t2xα. We wish to minimize max{1 k 2x,1 t2x}. Setting 1 k 2x = 1 t2x, we have and, for 1 i j k, x = 1 k 2t f ij p i,p j 2 2t k 2t α. Since 0 x 1 k 1 and x = 1/k 2t, we have 0 t 1 2. Now let us assume that a i tα for all 1 i k. By induction hypothesis, for any l {1,...,k} there exist p l i [0,2/k 1], i {1,...,k} \ {l}, such that i {1,...,k}\{l} pl i = 1 and for any {i,j} {1,...,k} \ {l}, a i + a j p l i + p l j h k 1 α a l. For 1 i k, let p i = 1 k p l i. l {1,...,k}\{i} Since p l i 2/k 1 for i {1,...,k} \ {l}, we have p i [0,2/k] for 1 i k. Also, k p i = 1 k i=1 k i=1 l {1,...,k}\{i} p l i = 1 k k l=1 i {1,...,k}\{l} p l i = 1 k k 1 = 1. l=1 6

7 Moreover, for 1 i j k, f ij p i,p j = a i + a j p i + p j = 1 k a i + a j = 1 k h k 1 k h k 1 k l {1,...,k}\{i,j} l {1,...,k}\{i,j} k 3 k h k 1α + l {1,...,k}\{i} p l i + l {1,...,k}\{j} a i + a j p l i + pl j + 1 k a i + a j p j i + pi j p l j α a l + 1 k a i + a j p j i + pi j 4 k 3α + a i + a j + kk 1 a i + a j hk 1 + 2tα. k 4 kk 1 By Lemma 2.3 and since h 4 = 1/3, there exist t k,h k for k 5 such that h k = 2 2t k k 2t k = k 3 k h k 1 + hk 1 k + 4 2t k, kk 1 h k < 1.6/k, and h k < 1.5/k for k 23. This completes the proof of the lemma. Theorem 2.5 Let k 4 be an integer. Then fk,m h k m + Om 4/5, where h k < 1.6/k, and h k < 1.5/k for k 23. Proof. Let G be a graph with m edges, and we may assume that G is connected as otherwise we simply consider individual components. Let V G = {v 1,...,v n } such that dv 1 dv 2... dv n. Let V 1 = {v 1,...,v t } with t = m α, where 0 < α < 1/2 and will be optimized later. Then t < n since m < n 2 /2. Moreover, ev 1 < t 2 /2 1 2 m2α and dv t+1 2m 1 α since t + 1dv t+1 t+1 i=1 dv i 2m. Label the vertices in V 2 := V G\V 1 as u 1,...,u n t such that eu i,v 1 {u 1,...,u i 1 } > 0 for i = 1,...,n t. Note that this can be done since G is connected. Fix a random k-partition V 1 = k i=1 Y i, and assign each member of Y i the color i, 1 i k. Extend this coloring to V G such that each vertex u i V 2 is independently assigned the color j with probability p i j, where k j=1 pi j = 1 and pi j will be determined later. Let Z i denote the indicator random variable of the event of coloring u i. Hence Z i = j iff u i is assigned the color j. Let G i = G[V 1 {u 1,,u i }] for i = 1,...,n t, and let G 0 = G[V 1 ]. Let Xj 0 = Y j for 1 j k, and x 0 jl = ex0 j X0 l for 1 j l k. For i = 1,...,n t and 1 j,l k, define X i j := {vertices of G i with color j}, x i jl := exi j Xi l, x i jl := xi jl xi 1 jl, b i j := eu i,x i 1 j. 7

8 Note that b i j depends on Z 1,...,Z i 1 only. Hence for 1 i n t and 1 j l k, and so where here E x i jl Z 1,...,Z i 1 = b i j + bi l pi j + pi l, E x i jl = ai j + ai l pi j + pi l, a i j = Z 1,...,Z i 1 PZ 1,...,Z i 1 b i j. Since b i j is determined by Z 1,...,Z i 1, a i j is determined by ps j, 1 j k and 1 s k i 1. Note that b i j = eu i,g i 1 > 0, and that eu i,g i 1 is independent of Z 1,...,Z n t. Moreover, j=1 k k a i j = j=1 = = j=1 Z 1,...,Z i 1 Z 1,...,Z i 1 Z 1,...,Z i 1 = eu i,g i 1 > 0. PZ 1,...,Z i 1 b i j PZ 1,...,Z i 1 k j=1 b i j PZ 1,...,Z i 1 eu i,g i 1 So by Lemma 2.4, there exist p i j [0,1], 1 j k, such that k and 1 j l k, k E x i jl h k a i j = h k eu i,g i 1. j=1 j=1 pi j = 1 and, for 1 i n t Note that p i j is determined by ai j, 1 i k; and hence pi j is recursively determined by ps j, 1 j k and 1 s i 1. Also note that m = eg 0 + n t i=1 eu i,g i 1. Now Ex n t jl = n t E x i jl + Ex0 jl i=1 eu i,g i 1 + x 0 jl h k n t i=1 h k m + ev 1 h k m m2α. 8

9 Clearly, changing the color of u i i.e., changing Z i affects x jl := x n t jl by at most du i. So by Lemma 2.1, P x jl > Ex jl + z exp z 2 2 n t i=1 du i 2 z 2 exp 2 n t i=1 du idv t+1 z 2 < exp 4m2m 1 α exp z2 8m 2 α. Let z = 8lnkk 1/2 1 2 m 1 α 2. Then for 1 j l k, P x jl > Ex jl + z < exp lnkk 1/2 = 2 kk 1. So there exists a partition V G = k i=1 X i such that for 1 j l k, ex j X l Ex jl + z h k m m2α + z h k m + om, where the om term in the expression is 1 2 m2α + 8lnkk 1/2 1 2 m 1 α 2. Choosing α = 2 5 to minimize max{2α,1 α/2}, the om term becomes Om Dense graphs We now prove Conjecture 1.2 for graphs with large minimum degree. The approach is similar to that for proving Theorem 2.5, but simpler because the large minimum degree condition helps to bound ev 1,V 2. Note that the term 4m/k 2 in the theorem below is best possible by simply taking a random k-partition. Theorem 3.1 Let k 2 be an integer and let ǫ > 0. If G is a graph with m edges and δg ǫn, then there is a partition X 1,...,X k of V G such that for 1 i j k, ex i X j 4 2/ǫ k 2m + kk 1 + 8ln m 5/6. 2 Proof. We may assume that G is connected otherwise it suffices to consider individual components. Let V G = {v 1,...,v n } such that dv 1 dv 2... dv n. Let V 1 = {v 1,...,v t } with t = m α, where 0 < α < 1/2. Then t < n, ev 1 m 2α /2, and dv t+1 2m 1 α. Let V 2 = V G\V 1 = {u 1,...,u n t } such that eu i,v 1 {u 1,...,u i 1 } > 0 for i = 1,...,n t. 9

10 Now assume δg ǫn. Then 2m = v V G dv ǫn2. So n 2m/ǫ. Thus, ev 1,V 2 + 2eV 1 = t dv i < tn m α 2m/ǫ = 2/ǫm 1/2+α. i=1 Fix a random partition V 1 = Y 1 Y 2... Y k and, for each i {1,...,k}, assign the color i to all vertices in Y i. We extend this coloring to V G by independently assigning the color j for each j {1,...,k} to each vertex u i V 2 with probability 1/k. Let Z i denote the indicator random variable of the event of coloring u i. Let X i be the set of vertices of G with color i. Then Y i X i for 1 i k; and for 1 i j k, EeX i X j = EeX i X j V 2 + EeX i X j V 2,Y i Y j + ey i Y j 2/k 2 ev 2 + ev 1,V 2 + ev 1 4 k 2m + 2/ǫm 1/2+α. Clearly, changing the color of u i i.e., changing Z i affects ex i X j by at most du i. Then as in the proof of Theorem 2.5, we apply Lemma 2.1 to conclude that for any 1 i j k, z 2 P ex i X j > EeX i X j + z exp 2 n t i=1 du exp z2 i 2 8m 2 α. Let z = 8lnkk 1/2m 1 α/2. Then for 1 i j k, kk 1 P ex i X j > EeX i X j + z < exp ln = 2 2 kk 1. So there exists a partition V G = X 1 X 2... X k such that, for 1 i j k, ex i X j 4 k 2m + 2/ǫm 1/2+α + z 4 k 2m + 2/ǫm 1/2+α + 8lnkk 1/2m 1 α/2 Picking α = 1/3 to minimize max{1/2 + α,1 α/2}, we have the desired bound. As a corollary, Conjecture 1.3 holds for graphs with Ωk 12 ln k 3 edges. Hence Conjecture 1.2 holds for all graphs G with δg ǫn, for any fixed k 2 and ǫ > 0. 4 Bounds for 4-partitions and 5-partitions In this section, we prove Conjecture 1.2 for 4-partitions and 5-partitions. We use Lemma 2.2 for 4-partitions. For 5-partitions, we need the following lemma. Lemma 4.1 Let a j 0 for j {1,...,5} such that α := 5 j=1 a j > 0, and let f ij x i,x j = a i + a j x i + x j for 1 i j 5. Then there exist p i [0,2/5], 1 i 5, such that 5 i=1 p i = 1 and, for 1 i j 5, f ij p i,p j 4α/15. 10

11 Proof. If there exists some l {1,...,5} such that a l 5α/11, then a i + a j 6α/11 for {i,j} {1,...,5} \ {l}. Let p l = 1/45 and let p i = 11/45 for i {1,...,5} \ {l}. Then for i {1,...,5} \ {l}, 11 f il p i,p l = a i + a l p i + p l α = α; and for {i,j} {1,...,5} \ {l}, f ij = a i + a j p i + p j 6α = α. Therefore, we may assume that a i < 5α/11 for all 1 i 5. By Lemma 2.2, for any 1 l 5 there exist p l i [0,1/2], i {1,...,5} \ {l}, such that i {1,...,5}\{l} pl i = 1 and, for {i,j} {1,...,5} \ {l}, a i + a j p l i + pl j 1 3 α a l. Indeed, by the remark following Lemma 2.2, we may choose p l i, i {1,...,5} \ {l}, such that p l i = 0 when a i > α a l /2, and p l i max{1/2 a i/α a l,1/3} when a i α a l /2. For 1 i 5, let p i = 1 5 l {1,...,5}\{i} pl i. Then p i [0,2/5], and 5 p i = 1 5 i=1 5 i=1 l {1,...,5}\{i} p l i = l=1 i {1,...,5}\{l} p l i = = 1. l=1 So for 1 i j 5, f ij p i,p j = a i + a j p i + p j = 1 5 a i + a j = l {1,...,5}\{i,j} l {1,...,5}\{i,j} l {1,...,5}\{i} p l i + l {1,...,5}\{j} a i + a j p l i + pl j a i + a j p j i + pi j p l j α a l a i + a j p j i + pi j = α + a i + a j a i + a j p j i + pi j = α + a i + a j pj i + pi j. We need to show that f ij p i,p j 4 15α for 1 i j 5. If a i > α a j /2 and a j > α a i /2, then p j i = pi j = 0, and hence f ijp i,p j 3 15 α < 4 15 α. Now assume a i > α a j /2 and a j α a i /2. Then p j i = 0 and p i j max{1/2 a j /α a i,1/3}. Suppose 1/2 a j /α a i > 1/3. Then a j < α a i /6; and hence, since 11

12 a i > α a j /2, we have a i > α α/6 + a i /6/2. Solving this inequality for a i, we have a i > 5α/11 which is in conflict with our assumption. Therefore, 1/2 a j /α a i 1/3, and so p i j 1/3. Hence f ij p i,p j α + a i + a j α. By symmetry, if a j > α a i /2 and a i α a j /2, then f ij p i,p j 4 15 α. So we are left with the case when a i α a j /2 and a j α a i /2. Then a i + a j α a i + a j /2, and so a i + a j 2α/3. Moreover, p j i max{1/2 a i /α a j,1/3} and p i j max{1/2 a j/α a i,1/3}. If 1/2 a i /α a j > 1/3 and 1/2 a j /α a i > 1/3, then 6a i +a j < α and 6a j +a i < α. Hence a i + a j < 2α/7, and so since p j i 1/2 and pi j 1/2, f ij p i,p j α + a i + a j < 2 15 α α < 4 15 α. If 1/2 a i /α a j > 1/3 and 1/2 a j /α a i 1/3, then 6a i + a j α and p i j 1/3. Since a j α a i /2, a i + 2a j α. So 11a i + a j = 6a i + a j + 5a i + 2a j 6α, and hence a i + a j 6α/11. Then f ij p i,p j α + a i + a j α α < 4 15 α. The case when 1/2 a i /α a j 1/3 and 1/2 a j /α a i > 1/3 is symmetric. Therefore, we may assume that 1/2 a i /α a j 1/3 and 1/2 a j /α a i 1/3. Then p j i 1/3 and pi j 1/3. Recall that a i + a j 2α/3. Hence f ij p i,p j α + a i + a j α α = 4 15 α. Using the same proof of Theorem 2.5, with Lemma 2.2 and Lemma 4.1 in place of Lemma 2.4, we have the following results on 4-partitions and 5-partitions. Theorem 4.2 f4,m m/3 + Om 4/5. Theorem 4.3 f5,m 4m/15 + Om 4/5. Recall that the graphs K 1,n give f4,m m/3 and f5,m m/4. When k = 4, 12m/k + 2k + 1 = 3/5 > 1/3. So as a consequence of Theorem 4.2, Conjecture 1.2 holds for k = 4. When k = 5, 12m/k + 2k + 1 = 2/7 > 4/15. Hence, Theorem 4.3 establishes Conjecture 1.2 for k = 5. 12

13 5 Simultaneous bounds for 3-partitions and 4-partitions In this section, we study the following problem suggested by Bollobás and Scott [7]. Problem 5.1 For any integer k 2 and for any graph G with m edges and n vertices, is it possible to find a partition V 1,...,V k of V G such that for 1 i k, ev i m k 2 + k 1 2k 2 2m , 2 and for 1 i j k, ev i V j 12m k + 2k On? Recall that Bollobás and Scott [5] showed the existence of a k-partition satisfying the above bound on ev i, and K kn+1 are the only extremal graphs. Also recall that the bound on ev i V j is best possible for K k+2. We show that for k = 3 and k = 4, one can find partitions that satisfy these bounds asymptotically. For large k, a similar approach as in the proofs of Lemma 2.4 and Theorem 2.5 may be used to give some bounds. Note that in the proofs to follow, we will use the fact that the maximum of xa x, a > 0, is a 2 /4. Lemma 5.2 Let a j 0 for j = 1,2,3 such that α := a 1 + a 2 + a 3 > 0, let f ij x i,x j = a i + a j x i + x j for 1 i j 3, and let g i x i = a i x i for 1 i 3. Then there exist p i [0,2/3], 1 i 3, such that 3 i=1 p i = 1, f ij p i,p j 5α/9 for 1 i j 3, and g i p i α/9 for 1 i 3. Proof. First, assume that a i < 2α/3 for all i = 1,2,3. Let p i = 2/3 a i /α. Then p i [0,2/3], i = 1,2,3, and p 1 + p 2 + p 3 = 1. Moreover, for 1 i j 3, f ij p i,p j = a i + a j 4 α 3 a i + a j α 4 α 9 α < 5 9 α and, for i = 1,2,3, g i p i = a i 2 α 3 a i α 1 α 9 α. Next assume that some a i > 5α/6, say a 3 > 5α/6. So a 1 + a 2 α/6. We choose p 1 = p 2 = 4/9 and p 3 = 1/9. Then f 12 p 1,p 2 < α/6 < 5α/9; f i3 p i,p 3 5α/9 for i = 1,2; g 3 p 3 α/9; and g i p i α/64/9 = 2α/27 < α/9 for i = 1,2. Therefore, we may assume that there exists some a i, say a 3, such that 2α/3 a 3 5α/6. Then α/6 a 1 + a 2 α/3. Let p 3 = 0 and p i = 2/3 a i /3a 1 + a 2 for i = 1,2. Then p i [0,2/3] and p 1 + p 2 + p 3 = 1. Clearly, g 3 p 3 = 0 and, for i = 1,2, g i p i = a i 3a 1 + a a i 3a 1 + a a 1 + a a 1 + a α.

14 Note that f 12 p 1,p 2 a 1 + a 2 α/3 < 5α/9. So it remains to show that f 13 p 1,p 3 5α/9 and f 23 p 2,p 3 5α/9. By symmetry we only need to prove f 13 p 1,p 3 5α/9. Note that f 13 p 1,p 3 = a 1 +a 3 2/3 a 1 /3α a 3, which may be viewed as a function of a 1,a 3 while fixing α. We look for the maximal value of ha 1,a 3 := f 13 p 1,p 3 subject to 2α/3 a 1 + a 3 α and 2α/3 a 3 5α/6. Taking partial derivatives and setting them to 0, we have h = 2 a 1 3 a 1 3α a 3 a 1 + a 3 3α a 3 = 0, and h = 2 a 3 3 a 1 3α a a a 1 + a 3 1 α a 3 2 = 0. Then a 1 /α a 3 = 1 from h a 1 = h a 3, and hence a 3 = 0 from h a 1 = 0, a contradiction. So the maximal value of h occurs on the boundary of the region defined by 2α/3 a 1 + a 3 α and 2α/3 a 3 5α/6. When a 1 +a 3 = 2α/3, then a 1 = 0 and a 3 = 2α/3, and hence h = 4α/9. When a 1 +a 3 = α then h = α/3. When a 3 = 2α/3 then h = a 1 + 2α/32/3 a 1 /α = 2/3 + a 1 /α2/3 a 1 /αα 4α/9. When a 3 = 5α/6, then h a 1 + 5α/62/3 2a 1 /α = 5/6 + a 1 /α2/3 2a 1 /αα 5α/9. Hence f 13 p 1,p 3 5α/9. The next lemma is for 4-partitions. Lemma 5.3 Let a j 0 for j = 1,2,3,4 such that α := a 1 + a 2 + a 3 + a 4 > 0, let f ij x i,x j = a i + a j x i + x j for 1 i j 4, and let g i x i = a i x i for 1 i 4. Then there exist p i [0,1/2], 1 i 4, such that 4 i=1 p i = 1, f ij p i,p j 2α/5 for 1 i j 4, and g i p i α/16 for 1 i 4. Proof. First, suppose a i < α/2 for all 1 i 4. Let p i = 1/2 a i /α. Then p i [0,1/2] for 1 i 4, and 4 i=1 p i = 1. Moreover, for 1 i j 4, and for 1 i 4, f ij p i,p j = a i + a j α 1 a i + a j α g i p i = a i 1 α 2 a i α 1 α 16 α. α 1 4 α < 2 5 α, Now assume that some a i > 4α/5, say a 4 > 4α/5. Then a 1 +a 2 +a 3 α/5. Let p 1 = p 2 = p 3 = 5/16 and p 4 = 1/16. Then for i = 1,2,3, f i4 p i,p 4 6α/16 < 2α/5; for 1 i j 3, f ij p i,p j α/5 < 2α/5; g 4 p 4 α/16; and for i = 1,2,3, g i p i α/55/16 = α/16. So we may assume that there exists some a i, say a 4, such that α/2 a 4 4α/5. Then α/5 a 1 + a 2 + a 3 α/2. Let p 4 = 0 and p i = 1/2 a i /2α a 4. Then p i [0,1/2] and 4 i=1 p i = 1. Clearly, g 4 p 4 = 0. Note that α a 4 α/2. So for i = 1,2,3 a i 1 g i p i = 2α a 4 2 a i 2α a 4 1 2α a 4 16 α; 14

15 and for 1 i j 3, f ij p i,p j = a i + a j 2α a 4 1 a i + a j 2α a 4 1 2α a 4 4 α < 2 5 α. Thus it remains to prove f i4 p i,p 4 2α/5 for i = 1,2,3. By symmetry, we only prove f 14 p 1,p 4 2α/5. Note that ha 1,a 4 := f 14 p 1,p 4 = a 1 + a 4 1/2 a 1 /2α a 4 may be viewed as a function of a 1,a 4 while fixing α, and we look for its maximal value subject to α/2 a 1 + a 4 α and α/2 a 4 4α/5. Taking partial derivatives and setting them to 0, we have and h = 1 a 1 2 a 1 2α a 4 1 a 1 + a 4 = 0, 2 α a 4 h = 1 a 4 2 a 1 2α a a a 1 + a 4 1 α a 4 2 = 0. Then a 1 /α a 4 = 1 from h a 1 = h a 4, and so a 4 < 0 from h a 1 = 0, a contradiction. Thus, the maximal value of h occurs when a 1 + a 4 {α/2,α} or a 4 {α/2,4α/5}. When a 1 +a 4 = α/2, we have a 1 = 0 and a 4 = α/2, and hence h = α/4. When a 1 +a 4 = α, then h = 0. When a 4 = α/2 then h = α1/2 + a 1 /α1/2 a 1 /α α/4. When a 4 = 4α/5, then h = α4/5 + a 1 /α1/2 5a 1 /2α 2α/5. Hence f 14 a 1,a 4 2α/5. Now we use Lemma 5.2 and essentially the same proof of Theorem 2.5 to prove the following. Theorem 5.4 Let G be a graph with m edges. Then there is a partition X 1,X 2,X 3 of V G such that for 1 i 3, and for 1 i j 3, ex i 1 9 m + Om4/5, ex i X j 5 9 m + Om4/5. Proof. We may assume that G is connected. Let V G = {v 1,...,v n } such that dv 1 dv 2... dv n. Let V 1 = {v 1,...,v t } with t = m α, where 0 < α < 1/2. Then t < n, ev m2α, and dv t+1 2m 1 α. Let V 2 := V G \ V 1 = {u 1,...,u n t } such that eu i,v 1 {u 1,...,u i 1 } > 0 for i = 1,...,n t. Fix a random 3-partition V 1 = Y 1 Y 2 Y 3, and assign each member of Y i the color i, 1 i 3. Extend this coloring to V G such that each vertex u i V 2 is independently assigned the color j with probability p i j, where 3 j=1 pi j = 1 and pi j will be determined later. Let Z i denote the indicator random variable of the event of coloring u i. Let G i = G[V 1 {u 1,,u i }] for i = 1,...,n t, and let G 0 = G[V 1 ]. Let Xj 0 = Y j and x 0 jl = ex0 j X0 l for 1 j,l 3. For i = 1,...,n t and 1 j,l 3, define X i j := {vertices of G i with color j}, x i jl := exi j Xi l, x i jl := xi jl xi 1 jl, b i j := eu i,x i 1 j. 15

16 When j = l, let x i j := xi jl and xi j = xi jl. Note that bi j depends on Z 1,...,Z i 1 only and 3 j=1 bi j = eu i,g i 1 is in dependent of Z 1,...,Z i 1. Let a i j = Z 1,...,Z PZ i 1 1,...,Z i 1 b i j, which is determined by p s j, 1 j 3 and 1 s i 1. As in the proof of Theorem 2.5, for 1 i n t and 1 j l 3 we have E x i jl = ai j + ai l pi j + pi l, and for 1 i n t we have E x i j = a i jp i j. By Lemma 5.2, there exist p i j [0,2/3], 1 j 3, such that 3 and 1 j l 3, E x i jl 5 3 a i j 9 = 5 3 b i j 9 = 5 9 eu i,g i 1 ; and for 1 i n t, j=1 j=1 j=1 pi j = 1; for 1 i n t E x i j a i j = 1 9 j=1 3 b i j = 1 9 eu i,g i 1. Note that p i j is determined by ai j, 1 j 3; and hence pi j is recursively defined by ps j, 1 j 3 and 1 s i 1. Now i=1 j=1 Ex n t jl = 5 n t eu i,g i 1 + x 0 jl m + ev 1, and Exj n t 1 n t eu i,g i 1 + x 0 j m + ev 1. i=1 Clearly, changing the color of u i i.e., changing Z i affects x jl := x n t jl at most du i. So by Lemma 2.1, P x jl > Ex jl + z exp z2 8m 2 α, and x j := x n t j by and P x j > Ex j + z exp z2 8m 2 α. Let z = 8ln m 1 α 2. Then for 1 j l 3, P x jl > Ex jl + z < 1 6, and for 1 j 3, P x j > Ex j + z <

17 So there exists a partition V G = X 1 X 2 X 3 such that for 1 j l 3, ex j X l Ex jl + z 5 9 m + om, and for 1 j 3, ex j Ex j + z 1 9 m + om. The om term in both expressions is 1 2 m2α + 8ln m 1 α 2. Picking α = 2 5 to minimize max{2α,1 α/2}, the om term becomes Om4 5. By the same argument as in the proof of Theorem 5.4, using Lemma 5.3 instead of Lemma 5.2, we have the following result. Theorem 5.5 Let G be a graph with m edges. Then there is a partition X 1,X 2,X 3,X 4 of V G such that for 1 i 4, and for 1 i j 4, ex i 1 16 m + Om4/5, ex i X j 2 5 m + Om4/5. 17

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