On problems about judicious bipartitions of graphs Yuliang Ji∗ Jie Ma† Juan Yan‡ Xingxing Yu§ 7 1 0 2 n Abstract: Bollobás and Scott [5]conjectured that every graph Ghas a balanced bipartite a spanningsubgraphH suchthatforeachv V(G),d (v) (d (v) 1)/2. Inthispaper,we J ∈ H ≥ G − show that every graphic sequence has a realization for which this Bollobás-Scott conjecture 5 2 holds, confirming a conjecture of Hartke and Seacrest [10]. On the other hand, we give an infinite family of counterexamples to this Bollobás-Scott conjecture, which indicates ] O that (dG(v) 1)/2 (rather than (dG(v) 1)/2) is probably the correct lower bound. We ⌊ − ⌋ − C also study bipartitions V1,V2 of graphs with a fixed number of edges. We provide a (best . possible) upper bound on e(V1)λ +e(V2)λ for any real λ 1 (the case λ = 2 is a question h ≥ of Scott [13]) and answer a question of Scott [13] on max e(V ),e(V ) . t 1 2 a { } m Keywords: bipartition; bisection; degree sequence; complete k-partite graph; ℓ -norm λ [ 1 AMS classification: 05C07, 05C70 v 2 6 1 7 0 . 1 0 7 1 : v i X r a ∗School of Mathematical Sciences, Universityof Science and Technology of China, Hefei, Anhui230026, China. Email: jiyl@mail.ustc.edu.cn. †School of Mathematical Sciences, Universityof Science and Technology of China, Hefei, Anhui230026, China. Email: jiema@ustc.edu.cn. Partially supported byNSFC projects 11501539 and 11622110. ‡College of Mathematics and Systems Science, Xinjiang University, Urumqi, Xinjiang 830046, China. Email: yanjuan207@163.com. Partially supported by NSFCproject 11501486. §School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332, USA. Email: yu@math.gatech.edu. Partially supported by NSFgrants DMS–1265564 and DMS-1600387. 1 1 Introduction For any positive integer k, let [k] := 1,...,k . Let G be a graph and V ,...,V be a 1 k { } partition ofV(G). Whenk = 2,suchapartition issaidtobeabipartitionofG. Asubgraph H of a graph G is said to bea bisection of G if H is a bipartite spanningsubgraphof G and the partition sets of H differ in size by at most one. For i,j [k], we use e(V ) to denote i ∈ the number of edges of G with both ends in V and use e(V ,V ) to denote the number of i i j edges between V and V . Judicious partitioning problems for graphs ask for partitions of i j graphs that bound a number of quantities simultaneously, such as all e(V ) and e(V ,V ). i i j There has been extensive research on this type of problems over the past two decades. As an attempt to better understand how edges of a graph are distributed, we study several judicious bipartitioning problems. Specifically, we study a conjecture of Bollobás and Scott [5] and its degree sequence version conjectured by Hartke and Seacrest [10]. We also study two questions of Scott [13] on bipartitions V ,V of a graph with m edges, 1 2 bounding e(V )2+e(V )2 and max e(V ),e(V ) in terms of m. 1 2 1 2 { } For a graph G and for any v V(G), we use d (v) to denote the degree of the vertex v G ∈ in G. It is well known that if H is a maximum bipartite spanning subgraph of a graph G, then d (v) d (v)/2 for each v V(G). This, however, may not be true if one requires H G ≥ ∈ H to be a bisection, as observed by Bollobás and Scott [5] by considering the complete bipartite graphs K for m 2ℓ + 3. In an attempt to obtain a similar result for 2ℓ+1,m ≥ bisections, Bollobás and Scott [5] conjectured that every graph G has a bisection H such that d (v) (d (v) 1)/2 for all v V(G). (1) H G ≥ − ∈ This conjecture for regular graphs was made by Häggkvist [8] in 1978, and variations of this problem were studied by Ban and Linial [2]. Hartke and Seacrest [10] studied a degree sequence version of this Bollobás-Scott conjecture. A nondecreasing sequence π (of nonnegative integers) is said to be graphic if it is the degree sequence of some finite simple graph G; and such G is called a realization of the sequence π. Hartke and Seacrest [10] proved that for any graphic sequence π with even length, π has a realization G which admits a bisection H such that for all v V(G), ∈ d (v) (d (v) 1)/2 . They further conjectured that for any graphic sequence π with H G ≥ ⌊ − ⌋ even length, π has a realization G for which (1) holds. We prove this Hartke-Seacrest conjecture for all graphic sequences. For a graph G and a labeling of its vertices V(G) = v ,...,v , we define the parity 1 n { } bisection of G to be the bisection with partition sets V and V , where V = v V(G) : 1 2 i j { ∈ j i mod 2 for each i [2], and E(H) = uv E(G) : u V and v V . 1 2 ≡ } ∈ { ∈ ∈ ∈ } Theorem 1.1. Let π = (d ,...,d ) be any graphic sequence with d d . Then 1 n 1 n ≥ ··· ≥ there exists a realization G of π with V(G) = v ,...,v and d (v )= d for i [n], such 1 n G i i { } ∈ that if H denotes the parity bisection of G then d (v ) (d (v ) 1)/2 for i [n]. H i G i ≥ − ∈ The bound in Theorem 1.1 is best possible, as shown by the following example given by Hartke and Seacrest [10]. Let G be the join of a clique K on k vertices and an independent set I on n k vertices, where n is even and k < n/2 is odd. It is not hard to show that G − in fact is the unique realization of the sequence π = (d ,...,d ) with d = = d = n 1 1 n 1 k ··· − and d = = d = k. Let H be an arbitrary bisection of G with parts A,B and, k+1 n ··· 2 without loss of generality, assume that A V(K) k/2. Since k < n/2, there must exist | ∩ |≤ a vertex v B I. So d (v) = A V(K) k/2 . Since d (v) = k and k is odd, we see H G ∈ ∩ | ∩ | ≤ ⌊ ⌋ that d (v) (d (v) 1)/2. H G ≤ − The second result in this paper gives indication that perhaps the lower bound in the original Bollobás-Scott conjecture was meant to be (d (v) 1)/2 (rather than (d (v) G G ⌊ − ⌋ − 1)/2). Proposition 1.2. Let r ,r ,r be pairwise distinct odd integers such that for every i [3], 1 2 3 ∈ r / 1, (r +r +r )/2 , (r +r +r )/2 . Then for any bisection H of the complete i 1 2 3 1 2 3 ∈ { ⌊ ⌋ ⌈ ⌉} 3-partite graph G := K , there always exists a vertex v with d (v) < (d (v) 1)/2. r1,r2,r3 H G − This result will follow from a more general result, Proposition 3.3, on all complete multipartite graphs. We remark here that, for each complete multipartite graph G, it is easy (as we will see in Section 3) to find a bisection H of G such that d (v) (d (v) 1)/2 H G ≥ ⌊ − ⌋ for all v V(G). However, for general graphs, even the following weaker version of the ∈ Bollobaás-Scott conjecture seems quite difficult to prove (or disprove). Conjecture 1.3. There exists some absolute constant c > 0 such that every graph G has a bisection H with d (v) (d (v) c)/2 for all v V(G). H G ≥ − ∈ We now turn our discussion to problems on general bipartitions. Answering a question of Erd˝os, Edwards [6] showed in 1973 that every graph with m edges admits a bipartition V ,V such that e(V ,V ) m/2+t(m)/2, where 1 2 1 2 ≥ t(m) := m/2+1/16 1/4. − p This bound is best possible for the complete graphs of odd order. Bolloba´s and Scott [3] extended Edwards’ bound by showing that every graph G with m edges has a bipartition V ,V simultaneously satisfying e(V ,V ) m/2+t(m)/2 and max e(V ),e(V ) m/4+ 1 2 1 2 1 2 ≥ { } ≤ t(m)/4, where both bounds are tight for the complete graphs of odd order. Scott [13] provided an interesting viewpoint by introducing norm for partitions. For a real number λ > 0 and a bipartition V ,V of a graph G, define the ℓ -norm of (V ,V ) to 1 2 λ 1 2 be e(V )λ+e(V )λ 1/λ. Then to maximize e(V ,V ) is equivalent to minimize the ℓ -norm 1 2 1 2 1 of ((cid:0)V ,V ), while mi(cid:1)nimizing max e(V ),e(V ) is the same as minimizing the ℓ -norm of 1 2 1 2 ∞ { } (V ,V ). It is natural to consider other norms. In particular, Scott asked for the maximum 1 2 of min e(V )2+e(V )2 1 2 V(G)=V1∪V2 overgraphsGwithmedges,seeProblem3.18in[13]. Weprovideananswertothisquestion by proving the following general result. Theorem 1.4. Let m be any positive integer and λ 1 be any real number. Then, for any ≥ graph G with m edges, λ λ t(m) t(m)+1 min e(V )λ+e(V )λ + . 1 2 V(G)=V1∪V2 ≤ (cid:18) 2 (cid:19) (cid:18) 2 (cid:19) Moreover, the equality holds if and only if G is a complete graph of odd order. 3 We also consider analogous questions for k-partitions in Section 4. Though Edward’s bound is tight for all integers m = n , Erd˝os [7] conjectured that 2 the difference between Edwards’ bound and the truth can st(cid:0)ill(cid:1)be arbitrarily large for other m. This was confirmed by Alon [1]: every graph with m = n2/2 edges admits a bipartition V ,V such that e(V ,V ) m/2+t(m)/2+Ω(m1/4). Bollobás and Scott [5,13] made a 1 2 1 2 ≥ similar conjecture for max e(V ),e(V ) : for certain m, max e(V ),e(V ) can be arbitrary 1 2 1 2 { } { } far from m/4 +t(m)/4. Ma and Yu [12] proved that every graph with m = n2/2 edges admitsabipartitionV ,V suchthatmax e(V ),e(V ) m/4+t(m)/4 Ω(m1/4). Another 1 2 1 2 { } ≤ − result in the same spirit was given by Hofmeister and Lefmann [9] that any graph with kn 2 edges has a k-partition V ,...,V with k e(V ) k n , which beats the trivial up(cid:0)pe(cid:1)r 1 k i=1 i ≤ 2 bound 1 nk . P (cid:0) (cid:1) k 2 Motiv(cid:0)ate(cid:1)d by theseresults, Scott asked the following question: doesevery graph Gwith kn edges have a vertex partition into k sets, each of which contains at most n edges? ( 2 2 S(cid:0)ee(cid:1)Problem 3.9 in [13].) We show that the answer to this question is negative(cid:0)f(cid:1)or k = 2. Theorem 1.5. There exist infinitely many positive integers n and for each such n there is a graph G with 2n edges, such that, for every bipartition V ,V of G, max e(V ),e(V ) 2 1 2 { 1 2 } ≥ n +5n/48. (cid:0) (cid:1) 2 (cid:0) (cid:1) This paper is organized as follows. We prove Theorem 1.1 in Section 2, and then investigate complete multipartite graphs for the Bollobás-Scott conjecture in Section 3. In Section 4, we discuss the questions of Scott and complete the proofs of Theorems 1.4 and 1.5. 2 Hartke-Seacrest conjecture In this section, we prove Theorem 1.1. We need two operations on a sequence of nonnegative integers. Let π = (d ,...,d ) with d d . By removing d from π and 1 n 1 n i ≥ ··· ≥ subtracting 1 from the d remaining elements of π with lowest indices, we obtain a new i sequence π′ = (d′,...,d′ ,d′ ,...,d′ ), and we say that π′ is obtained from π by laying 1 i−1 i+1 n off d . This operation was introduced by Kleitman and Wang [11], and they proved the i following. Lemma 2.1 (Kleitman-Wang [11]). For any i [n], the sequence π = (d ,...,d ) with 1 n ∈ d ... d is graphic if and only if the sequence π′ obtained from π by laying off d is 1 n i ≥ ≥ graphic. It is easy to see that the sequence π′ obtained from π by laying off d need not be non- i increasing. To avoid this issue, Hartke andSeacrest [10]introduceda variation of theabove operation. Choosea fixedi [n]. Let sbethe smallest value among thed largest elements i ∈ in π, not including the ith element of π (namely, d itself). Let S = j [n] i : d > s . i j { ∈ −{ } } Note that S < d . LetT bethesetof d S largest indices j withj = iand d = s. Then i i j | | −| | 6 by laying off d with order, we remove d from π and subtract 1 from d for all j S T. If i i j ∈ ∪ π′ = (d′,...d′ ,d′ ,...,d′ )denotesthenewsequence,thenithasthemonotoneproperty 1 i−1 i+1 n d′ d′ . 1 ≥ ··· ≥ n 4 Clearly, the sequence obtained from π by laying off d with order is just a permutation i of the sequence obtained from π by laying off d . So the following is true. i Lemma 2.2 (Hartke and Seacrest [10]). For any i [n], the sequence π = (d ,...,d ) 1 n ∈ with d ... d is graphic if and only if the sequence obtained from π by laying off d 1 n i ≥ ≥ with order is graphic. Wegive abriefoutlineofourproofofTheorem1.1. Wechoosetwoconsecutive elements d and d of π. Using Lemma 2.2 we obtain a new graphic sequence π′′ of length n 2 ℓ ℓ+1 − by first laying off d with order and then laying off d with order. By induction, π′′ has ℓ+1 ℓ an (n 2)-vertex realization F whose parity bisection J has the desired property. We then − showthatonecanformGfromF byaddingtwonewvertices (ford andd )andchoosing ℓ ℓ+1 their neighbors, so that the parity bisection of G satisfies Theorem 1.1. Proof of Theorem 1.1. We apply induction on the length n of the graphic sequence π = (d ,...,d ) with d ... d . The assertion is trivial when n = 1,2. So we may assume 1 n 1 n ≥ ≥ that n 3 and the assertion holds for all graphic sequences with length less than n. Then ≥ there exist two consecutive elements of π that are identical; so let ℓ [n 1] be fixed such ∈ − that d = d = k. ℓ ℓ+1 Let π′ = (d′,...,d′,d′ ,...,d′ ) be the sequence obtained from π by laying off d with 1 ℓ ℓ+2 n ℓ+1 order. Let π′′ = (d′′,...,d′′ ,d′′ ,...,d′′) be the sequence obtained from π′ by laying off d′ 1 ℓ−1 ℓ+2 n l with order. By Lemma 2.2, π′ and π′′ both are graphic sequences. Let ω = (f ,...,f ) be the sequence obtained from π with d and d removed, and 1 n−2 ℓ ℓ+1 re-indexed so that the indices are consecutive, i.e., f = d for i [ℓ 1] and f = d i i i i+2 ∈ − for i [n 2] [ℓ 1]. Let ω′ = (f′,...,f′ ) be the sequence obtained from π′ with d′ ∈ − \ − 1 n−2 ℓ removed, and re-indexed so that the indices are consecutive. Also, let ω′′ = (f′′,...,f′′ ) 1 n−2 be the sequence obtained from π′′ by re-indexing so that the indices are consecutive. Note that ω′′ is a graphic sequence. To turn a realization of ω′′ to a realization of π, we need to track the changes between f and f′′ for all i [n 2]. Note that 0 f f′′ 2. Let i i ∈ − ≤ i− i ≤ X = i [n 2] :f′′ = f 1 , X = i [n 2] : f′′ = f 2 1 { ∈ − i i− } 2 { ∈ − i i− } and k 1 if d′ = d 1, K = d′ = − ℓ ℓ − ℓ (cid:26) k if d′ = d . ℓ ℓ So K = f f′ = f′ f′′ ; hence i∈[n−2]| i− i| i∈[n−2]| i − i | P P X +2X = 2K. (2) 1 2 | | | | We now prove two claims asserting certain properties on X and X . For convenience, 1 2 we introduce some notation. For nonempty sets A and B of integers, we write A < B if the maximum integer in A is less than the minimum integer in B. A set S of integers is consecutive if it consists of consecutive integers. A sequence of pairwise disjoint sets, A ,...,A , of integers is said to be consecutive if A ... A is consecutive and, for any 1 t 1 t ∪ ∪ i,j [t] with i< j and A and A nonempty, we have A < A . i j i j ∈ 5 Claim 1. There exist consecutive sets R ,R ,R′,R′,Q such that X = R′ R′ and 1 2 1 2 1 1 ∪ 2 X = R R such that 2 1 2 ∪ (a) the sequence R ,R′,Q,R′ is consecutive, 1 1 2 (b) either R = or R = Q, and 2 2 ∅ (c) f′′ = f′′+1 for all i R′,j R′. i j ∈ 1 ∈ 2 Proof of Claim 1. Let s be the minimum of the largest K numbers in ω = (f ,...,f ). 1 n−2 (Note that this s is the same as the s in the definition of laying off d with order from π.) ℓ+1 In order to keep track whether f′ = f or f′ = f 1 and whether f′′ = f′ or f′′ = f′ 1, i i i i− i i i i − we divide [n 2] into six pairwise disjoint sets: − A = i [n 2] :f > s+2 , D = i [n 2] :f = s,f′ = f 1 , { ∈ − i } { ∈ − i i i− } B = i [n 2] :f = s+1 , E = i [n 2] : f = s 1 , i i { ∈ − } { ∈ − − } C = i [n 2] :f = s,f′ = f , F = i [n 2] : f 6 s 2 . { ∈ − i i i} { ∈ − i − } By the definitions of π′ and ω′, we see that A,B,C,D,E,F is consecutive and i A, f′ =f 1 > s+1, ∀ ∈ i i− i B, f′ =f 1 = s, ∀ ∈ i i− i C, f′ =f = s, ∀ ∈ i i i D, f′ =f 1 = s 1, ∀ ∈ i i− − i E, f′ =f = s 1, ∀ ∈ i i − i F, f′ =f 6 s 2. ∀ ∈ i i − Thus, it is easy to see that A B D = i [n 2] :f′ = f 1 ; so A + B + D = K. ∪ ∪ { ∈ − i i− } | | | | | | Let Y = i [n 2] : f′′ = f′ 1 . Then it follows that { ∈ − i i − } A Y and Y = K = A + B + D . ⊆ | | | | | | | | To complete our proof of Claim 1, we distinguish four cases based on relations among the sizes of B,C,D,E. First, suppose C B + D . Let C′′ consist of the last B + D integers in C, and | | ≥ | | | | | | | | C′ := C C′′. Then we see that Y = A C′′. Let R = A, R = , R′ = B, R′ = C′′ D \ ∪ 1 2 ∅ 1 2 ∪ and Q = C′. It is easy to check that X = R′ R′ and X = R R , and (a) and (b) 1 1 ∪ 2 2 1 ∪ 2 holds. Note that f′′ = s for i R′, and f′′ = s 1 for j R′; so (c) holds. i ∈ 1 j − ∈ 2 Next, suppose D C < B + D . Let B′′ consist of the last B + D C integers | |≤ | | | | | | | | | |−| | in B, and B′ = B B′′. Then Y = A B′′ C. Let R = A, R = Q = B′′, R′ = B′ and \ ∪ ∪ 1 2 1 R′ = C D. It is easy to check that X = R′ R′ and X = R R , and that (a) and 2 ∪ 1 1 ∪ 2 2 1 ∪ 2 (b) holds. Note that f′′ = s for i R′, and f′′ = s 1 for j R′; so (c) holds. i ∈ 1 j − ∈ 2 Now assume C < D C + E . Let E′′ consist of the last D C integers in E, | | | | ≤ | | | | | |−| | and E′ = E E′′. Then Y = A B C E′′. Let R = A B, R = , R′ = C D, \ ∪ ∪ ∪ 1 ∪ 2 ∅ 1 ∪ R′ = E′′, and Q = E′. It is easy to check that X = R′ R′ and X = R R , and (a) 2 1 1 ∪ 2 2 1 ∪ 2 and (b) holds. Note that f′′ = s 1 for i R′ and f′′ = s 2 for j R′; so (c) holds. i − ∈ 1 j − ∈ 2 Finally we consider the case when D > C + E . Let D′′ consist of the last D | | | | | | | |− C E integers in D, and D′ = D D′′. Then Y = A B C D′′ E. Let R = A B, 1 | |−| | \ ∪ ∪ ∪ ∪ ∪ 6 R = Q = D′′, R′ = C D′ and R′ = E. It is easy to check that X = R′ R′ and 2 1 ∪ 2 1 1 ∪ 2 X = R R , and (a) and (b) holds. Note that f′′ = s 1 for i R′ and f′′ = s 2 for 2 1 ∪ 2 i − ∈ 1 j − j R′; so (c) holds. ∈ 2 Let I = i [n 2] :i 1 mod 2 and I = i [n 2] :i 0 mod 2 . 1 2 { ∈ − ≡ } { ∈ − ≡ } Claim 2. X I X I 2,0,2 . Moreover, X I X I = 0 implies 1 1 1 2 1 1 1 2 | ∩ |−| ∩ | ∈ {− } | ∩ |−| ∩ | X I X I 1,0,1 . 2 1 2 2 | ∩ |−| ∩ |∈ {− } Proof of Claim 2. By (2), we see X must be even. So R′ and R′ are of the same | 1| | 1| | 2| parity. Since both R′ and R′ are consecutive, X I X I 2,0,2 . 1 2 | 1∩ 1|−| 1 ∩ 2|∈ {− } Now suppose X I X I = 0. If R = , then X = R is a consecutive set 1 1 1 2 2 2 1 | ∩ |−| ∩ | ∅ and thus X I and X I differ by at most one. So we may assume R = . Then 2 1 2 2 2 | ∩ | | ∩ | 6 ∅ R = Q by Claim 1. As the sequence R ,R′,Q,R′ is consecutive, we see that X X is a 2 1 1 2 1∪ 2 consecutive set; so (X X ) I and (X X ) I differ by at most one. Hence, since 1 2 1 1 2 2 | ∪ ∩ | | ∪ ∩ | X I X I = 0, X I X I 1. 1 1 1 2 2 1 2 2 | ∩ |−| ∩ | || ∩ |−| ∩ || ≤ Wearereadytoconstructarealizationofπ = (d ,...,d ). Recallthatω′′ =(f′′,...,f′′ ) 1 n 1 n−2 is a graphic sequence. By induction hypothesis, there exists a realization F of ω′′ with V(F) = w ,...,w and d (w ) = f′′ for i [n 2], such that the parity bisection J of { 1 n−2} F i i ∈ − F satisfies d (w ) (d (w ) 1)/2 for all i [n 2]. (3) J i F i ≥ − ∈ − Let W = w :i j mod 2 for j [2]. j i { ≡ } ∈ In what follows, we will construct a graph G as the realization of π such that its parity bisection H of G satisfies d (v) (d (v) 1)/2 for all v V(G), by adding two new H G ≥ − ∈ vertices a,b (so V(G) = V(F) a,b ) and some edges from these two vertices to F (which ∪{ } we will describe in three separate cases). Notice that if K = k 1, then we would add the − edge ab as well; so for convenience, let 1, if K = k 1, ǫ = − (cid:26) 0, if K = k. We write V(G) = v ,...,v such that v = w for i< ℓ, v ,v = a,b , and v = w 1 n i i ℓ ℓ+1 i i−2 { } { } { } for ℓ+1 < i n. ≤ In view of Claim 2, we consider the following three cases. In each of these three cases, we use a to represent the vertex in v ,v with odd index. So the parity partition of ℓ ℓ+1 { } V(G) is V = W a and V = W b . 1 1 2 2 ∪{ } ∪{ } Case 1. X I X I = 0. 1 1 1 2 | ∩ |−| ∩ | We know F G and V(G) = V(F) a,b , and we need to add edges at a and b to ⊆ ∪{ } form G, a realization of π. Add ab if ǫ = 1, av for all i X (X I ), and bv for all i 2 1 2 j ∈ ∪ ∩ j X (X I ). Since X I = X I , G is a realization of π. Let H denote 2 1 1 1 1 1 2 ∈ ∪ ∩ | ∩ | | ∩ | the parity bisection of G; so V ,V are the partition sets of H. We need to show that 1 2 d (v) (d (v) 1)/2 for all v V(G). H G ≥ − ∈ 7 For each w with i / X X , its neighborhoods in F,G are the same; so by (3), i 1 2 ∈ ∪ d (w ) = d (w ) (d (w ) 1)/2 = (d (w ) 1)/2. H i J i F i G i ≥ − − For vertices w with i X , we have d (w )= d (w )+2 and d (w )= d (w )+1; so i 2 G i F i H i J i ∈ by (3), d (w )= d (w )+1 (d (w ) 1)/2+1 = (d (w ) 1)/2. H i J i F i G i ≥ − − For vertices w with i X , we have d (w )= d (w )+1 and d (w )= d (w )+1; so i 1 G i F i H i J i ∈ by (3), d (w )= d (w )+1 (d (w ) 1)/2+1 > (d (w ) 1)/2. H i J i F i G i ≥ − − For thevertex a,wehaved (a) = X + X I +ǫandd (a) = X I + X I +ǫ. G 2 1 2 H 2 2 1 2 | | | ∩ | | ∩ | | ∩ | Note that in this case, by Claim 2, we have X I X I 1, which implies that 2 1 2 2 || ∩ |−| ∩ || ≤ 2d (a) d (a) = (X I X I )+ X I +ǫ 1. H G 2 2 2 1 1 2 − | ∩ |−| ∩ | | ∩ | ≥ − Hence, d (a) (d (a) 1)/2. H G ≥ − Similarly, for the vertex b, we have d (b) = X + X I +ǫ and d (b) = X I + G 2 1 1 H 2 1 | | | ∩ | | ∩ | X I +ǫ. Note, by Claim 2, X I X I 1; so 1 1 2 1 2 2 | ∩ | || ∩ |−| ∩ || ≤ 2d (b) d (b) = (X I X I )+ X I +ǫ 1. H G 2 1 2 2 1 1 − | ∩ |−| ∩ | | ∩ | ≥ − Hence, d (b) (d (b) 1)/2. H G ≥ − Case 2. X I X I = 2. 1 2 1 1 | ∩ |−| ∩ | Recall that X =R′ R′, where each R′ is consecutive. Thus it follows that R′ I = 1 1∪ 2 i | i∩ 2| R′ I +1 for i [2]. Since the sequence R ,R′,Q,R′ is consecutive and starts from the | i∩ 1| ∈ 1 1 2 integer 1, we see that R I = R I 1 and Q I = Q I 1. Therefore, since 1 2 1 1 2 1 | ∩ | | ∩ |− | ∩ | | ∩ |− R = or R = Q (by (b) of Claim 1), we have 2 2 ∅ 2 X I X I 1. (4) 2 2 2 1 − ≤ | ∩ |−| ∩ | ≤ − We claim that there exists some z X I with d (w ) d (w )/2. To see this, 1 2 J z F z ∈ ∩ ≥ choose x R′ I and y R′ I . By (3) , we have d (w ) (d (w ) 1)/2 and ∈ 1 ∩ 2 ∈ 2 ∩ 2 J x ≥ F x − d (w ) (d (w ) 1)/2. By (c) of Claim 1, d (w ) = d (w )+1. Observe that for any J y F y F x F y ≥ − vertex u of F, d (u) and 2d (u) d (u) are of the same parity; so 2d (w ) d (w ) and F J F J x F x − − 2d (w ) d (w ) must have different parities. Therefore there exists z x,y such that J y F y − ∈ { } d (w ) d (w )/2, proving the claim. J z F z ≥ We now add edges at a and b to form G from F: add ab if ǫ = 1, av for all i i ∈ X (X I ) z , and bv for all j X (X I ) z . Since X I = X I +2, 2 1 2 j 2 1 1 1 2 1 1 ∪ ∩ \{ } ∈ ∪ ∩ ∪{ } | ∩ | | ∩ | G is a realization of π. Next we show that the parity bisection H of G satisfies the property that d (v) (d (v) 1)/2 for all v V(G). H G ≥ − ∈ For each w with i / X X , its neighborhoods in F,G are the same; so by (3), i 1 2 ∈ ∪ d (w ) = d (w ) (d (w ) 1)/2 = (d (w ) 1)/2. H i J i F i G i ≥ − − For each w with i X , d (w ) = d (w )+2 and d (w ) = d (w )+1. Hence by (3) i 2 G i F i H i J i ∈ and the way we choose z, d (w ) = d (w )+1 (d (w ) 1)/2+1 = (d (w ) 1)/2. H i J i F i G i ≥ − − For w with i X z , we have d (w )= d (w )+1 and d (w )= d (w )+1; so by i 1 G i F i H i J i ∈ \{ } (3), d (w ) = d (w )+1 (d (w ) 1)/2+1 > (d (w ) 1)/2. H i J i F i G i ≥ − − The vertex w satisfies d (w ) = d (w ) + 1 and d (w ) = d (w ). Hence by (3), z G z F z H z J z d (w ) = d (w ) d (w )/2 = (d (w ) 1)/2. H z J z F z G z ≥ − 8 For the vertex a, by definition we have d (a) = X + X I 1+ǫ and d (a) = G 2 1 2 H | | | ∩ |− X I + X I 1+ǫ. By (4) and the fact that X I 2, 2 2 1 2 1 2 | ∩ | | ∩ |− | ∩ | ≥ 2d (a) d (a) = (X I X I )+ X I 1+ǫ 1. H G 2 2 2 1 1 2 − | ∩ |−| ∩ | | ∩ |− ≥ − Hence, d (a) (d (a) 1)/2. H G ≥ − Forthevertexb,wehaved (b) = X + X I +1+ǫandd (b) = X I + X I +ǫ. G 2 1 1 H 2 1 1 1 | | | ∩ | | ∩ | | ∩ | This, together with (4), imply that 2d (b) d (b) = (X I X I )+ X I 1+ǫ 0. H G 2 1 2 2 1 1 − | ∩ |−| ∩ | | ∩ |− ≥ Hence, d (b) (d (b) 1)/2. H G ≥ − Case 3. X I X I = 2. 1 1 1 2 | ∩ |−| ∩ | In this case, we have R′ I = R′ I +1 for i [2] (as R′ and R′ are consecutive). | i∩ 1| | i∩ 2| ∈ 1 2 Because R ,R′,Q,R′ is consecutive, it follows that R I = R I and Q I = 1 1 2 | 1 ∩ 1| | 1 ∩ 2| | ∩ 1| Q I 1. Since R = or R = Q (by (b) of Claim 1), 2 2 2 | ∩ |− ∅ 0 X I X I 1. (5) 2 2 2 1 ≤ | ∩ |−| ∩ | ≤ Since X is even and R′ I = R′ I + 1 for i [2], there exist x R′ | 1| | i ∩ 1| | i ∩ 2| ∈ ∈ 1 ∩ I and y R′ I . By (3) and (c) of Claim 1, we have d (w ) (d (w ) 1)/2, 1 ∈ 2 ∩ 1 J x ≥ F x − d (w ) (d (w ) 1)/2, and d (w ) = d (w )+1. Since for any vertex u of F, d (u) J y F y F x F y F ≥ − and 2d (u) d (u) are of the same parity, 2d (w ) d (w ) and 2d (w ) d (w ) must J F J x F x J y F y − − − have different parities. Therefore there exists z x,y such that d (w ) d (w )/2. J z F z ∈ { } ≥ We now add edges at a and b to form the graph G: add ab if ǫ = 1, av for all i i X (X I ) z , andbv forallj X (X I ) z . Since X I = X I +2, 2 1 2 j 2 1 1 1 1 1 2 ∈ ∪ ∩ ∪{ } ∈ ∪ ∩ \{ } | ∩ | | ∩ | G is a realization of π. We need to verify that d (v) (d (v) 1)/2 for all v V(G). H G ≥ − ∈ If v = w for some i / X X , then d (w ) = d (w ) and d (w )= d (w ); so by (3), i 1 2 G i F i H i J i ∈ ∪ d (w ) = d (w ) (d (w ) 1)/2 = (d (w ) 1)/2. H i J i F i G i ≥ − − If v = w for some i X , then d (w ) = d (w )+2 and d (w ) = d (w )+1; again i 2 G i F i H i J i ∈ by (3), d (w )= d (w )+1 (d (w ) 1)/2+1 = (d (w ) 1)/2. H i J i F i G i ≥ − − If v = w for some i X z , then d (w ) = d (w )+1 and d (w ) = d (w )+1; so i 1 G i F i H i J i ∈ \{ } by (3), d (w )= d (w )+1 (d (w ) 1)/2+1 > (d (w ) 1)/2. H i J i F i G i ≥ − − Now suppose v = w . Note that d (w ) = d (w ) + 1 and d (w ) = d (w ). So z G z F z H z J z d (w ) = d (w ) d (w )/2 = (d (w ) 1)/2. H z J z F z G z ≥ − Suppose v = a. Note that d (a) = X + X I +1+ǫ and d (a) = X I + G 2 1 2 H 2 2 | | | ∩ | | ∩ | X I +ǫ. By (5), 1 2 | ∩ | 2d (a) d (a) = (X I X I )+ X I 1+ǫ 1. H G 2 2 2 1 1 2 − | ∩ |−| ∩ | | ∩ |− ≥ − So d (a) (d (a) 1)/2. H G ≥ − Finally, suppose v = b. We have d (b) = X + X I 1 + ǫ and d (b) = G 2 1 1 H | | | ∩ | − X I + X I 1+ǫ. By (5) and the fact that X I 2, 2 1 1 1 1 1 | ∩ | | ∩ |− | ∩ | ≥ 2d (b) d (b) = (X I X I )+ X I 1+ǫ 0. H G 2 1 2 2 1 1 − | ∩ |−| ∩ | | ∩ |− ≥ So d (b) (d (b) 1)/2. H G ≥ − 9 3 Complete multipartite graphs For convenience, we say that a bisection H of a graph G is good if for each v V(G), ∈ 2d (v) d (v) 1. Thus the Bollobás-Scott conjecture says that every graph contains a H G ≥ − good bisection. Here we discuss which complete multipartite graphs have good bisections. Throughout the rest of this section, let G := K , and let X ,...,X denote the r1,...,rk 1 k partition sets of G with X = r for all i [k]. i i | | ∈ First, we note that whenever V(G) is even, G has a good bisection. Since V(G) is | | | | even, V(G) has a partition V ,V such that V = V , X V X V = 1 if X 1 2 1 2 i 1 i 2 i | | | | || ∩ |−| ∩ || | | is odd, and X V = X V if X is even. Let H denote the maximum bisection of i 1 i 2 i | ∩ | | ∩ | | | G with partition sets V and V . For any v V(G), v X for some i [k]. Note that 1 2 i ∈ ∈ ∈ d (v) = V(G) X and d (v) (V(G) X 1)/2 = (d (v) 1)/2. G i H i G | |−| | ≥ | |−| |− − We will see that this is not always the case when V(G) is odd. The main result of | | this section is a necessary and sufficient condition for a complete multipartite graph with odd order to contain a good bisection. As a consequence, we show that for many complete multipartite graphs G, G (and even G minus an edge) does not have a good bisection. (However, it is not hard to see that such G does have a bisection H such that for each v V(G), d (v) (d (v) 1)/2 .) H G ∈ ≥ ⌊ − ⌋ For a bisection H of G with partition sets V and V , we say that X crosses H if 1 2 i X V = for j [2]. For a subset W V(G), let W = V(G) W. We need two easy i j ∩ 6 ∅ ∈ ⊆ \ lemmas. Lemma 3.1. Let G = K and let X , i [k], be the partition sets of G. Suppose H r1,...,rk i ∈ is a good bisection of G with partition sets V and V . Then the following statements hold 1 2 for i [k]. ∈ (i) If X crosses H and X iseventhen X V = X V and X V X V 1. i i i 1 i 2 i 1 i 2 | | | ∩ | | ∩ | || ∩ |−| ∩ || ≤ (ii) IfX crosses H and X isodd then X V X V = 1and X V X V i i i 1 i 2 i 1 i 2 | | || ∩ |−| ∩ || || ∩ |−| ∩ ||≤ 2. Proof. Suppose X crosses H. Then there exist v X V for j [2]. Note that i j i j ∈ ∩ ∈ d (v )= d (v )= X . Thus, since H is a good bisection, d (v )= X V X /2 G 1 G 2 i H j i 3−j i | | | ∩ |≥ ⌊| | ⌋ for j [2]. Notice that X V + X V = X . So X V X V 1. i 1 i 2 i i 1 i 2 ∈ | ∩ | | ∩ | | | || ∩ |−| ∩ || ≤ If X is even then X V = X V , and (i) holds. So assume X is odd. Since i i 1 i 2 i | | | ∩ | | ∩ | | | V V 1, X V X V 2, and (ii) holds. 1 2 i 1 i 2 || |−| || ≤ || ∩ |−| ∩ || ≤ Lemma 3.2. Let G = K with V(G) odd, let X ,...,X be the partition sets of G, r1,...,rk | | 1 k and let H be a good bisection of G with partition sets V ,V such that V = V +1. Let 1 2 1 2 | | | | ′ = X : i [k],X crosses H, X 0 mod 2, and X V X V = 2 X0 { i ∈ i | i| ≡ || i ∩ 1|−| i ∩ 2|| } and = X : i [k], X crosses H, and X 1 mod 2 . 1 i i i X { ∈ | | ≡ } Let W = X , W = X , = t, and ′ =t′. Then 1 Xi∈X1 i 0 Xi∈X0′ i |X1| |X0| S S (i) W V W V =t, 1 1 1 2 | ∩ |−| ∩ | 10

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