On problems about judicious bipartitions of graphs

TL;DR

This paper investigates bipartitions of graphs, confirming a conjecture for some realizations, providing counterexamples to refine bounds, and studying edge distributions with new bounds and answers to open questions.

Contribution

It proves the Bollobás-Scott conjecture for certain graph realizations, presents counterexamples suggesting a revised bound, and establishes optimal bounds on bipartition edge distributions.

Findings

Confirmed the Bollobás-Scott conjecture for some graph realizations.

Provided infinite counterexamples indicating the need for a revised bound.

Established optimal bounds on bipartition edge sums and answered related open questions.

Abstract

Bollob\'{a}s and Scott [5] conjectured that every graph $G$ has a balanced bipartite spanning subgraph $H$ such that for each $v\in V(G)$, $d_H(v)\ge (d_G(v)-1)/2$. In this paper, we show that every graphic sequence has a realization for which this Bollob\'{a}s-Scott conjecture holds, confirming a conjecture of Hartke and Seacrest [10]. On the other hand, we give an infinite family of counterexamples to this Bollob\'{a}s-Scott conjecture, which indicates that $\lfloor (d_G(v)-1)/2\rfloor$ (rather than $(d_G(v)-1)/2$) is probably the correct lower bound. We also study bipartitions $V_1, V_2$ of graphs with a fixed number of edges. We provide a (best possible) upper bound on $e(V_1)^{\lambda}+e(V_2)^{\lambda}$ for any real $\lambda\geq 1$ (the case $\lambda=2$ is a question of Scott [13]) and answer a question of Scott [13] on $\max\{e(V_1),e(V_2)\}$.

Equations

$$d_H(v)\ge (d_G(v)-1)/2\text{ for all }v\in V(G).$$

$$t(m):=\sqrt{m/2+1/16}-1/4.$$

$$\underset{V(G)=V_1\cup V_2}{\min }e(V_1{)}^2+e(V_2{)}^2$$

$$\underset{V(G)=V_1\cup V_2}{\min }e(V_1{)}^{\lambda }+e(V_2{)}^{\lambda }\le {\left(\genfrac{}{}{0ex}{}{t(m)}{2}\right)}^{\lambda }+{\left(\genfrac{}{}{0ex}{}{t(m)+1}{2}\right)}^{\lambda }.$$

$$d_{\ell }=d_{\ell +1}=k.$$

$$X_1=\{i\in [n-2]:f_i^{\prime \prime }=f_i-1\},X_2=\{i\in [n-2]:f_i^{\prime \prime }=f_i-2\}$$

$$K=d_{\ell}^{\prime}=\left\{\begin{array}[]{ll}k-1&\text{ if }d^{\prime}_{\ell}=d_{\ell}-1,\\ k&\text{ if }d^{\prime}_{\ell}=d_{\ell}.\end{array}\right.$$

$$|X_1|+2|X_2|=2K.$$

$$\begin{array}[]{ll}A=\{i\in[n-2]:f_{i}\geqslant s+2\},{}{}{}{}{}&D=\{i\in[n-2]:f_{i}=s,f_{i}^{\prime}=f_{i}-1\},\\ B=\{i\in[n-2]:f_{i}=s+1\},{}{}{}{}{}&E=\{i\in[n-2]:f_{i}=s-1\},\\ C=\{i\in[n-2]:f_{i}=s,f_{i}^{\prime}=f_{i}\},{}{}{}{}{}&F=\{i\in[n-2]:f_{i}\leqslant s-2\}.\end{array}$$

$$\begin{array}[]{ll}\forall\ i\in A,&~{}f_{i}^{\prime}=f_{i}-1\geqslant s+1,\\ \forall\ i\in B,&~{}f_{i}^{\prime}=f_{i}-1=s,\\ \forall\ i\in C,&~{}f_{i}^{\prime}=f_{i}=s,\\ \forall\ i\in D,&~{}f_{i}^{\prime}=f_{i}-1=s-1,\\ \forall\ i\in E,&~{}f_{i}^{\prime}=f_{i}=s-1,\\ \forall\ i\in F,&~{}f_{i}^{\prime}=f_{i}\leqslant s-2.\end{array}$$

$$A\subseteq Y\text{ and }|Y|=K=|A|+|B|+|D|.$$

$$d_J(w_i)\ge (d_F(w_i)-1)/2\text{ for all }i\in [n-2].$$

$$\epsilon=\left\{\begin{array}[]{ll}1,&\text{ if }K=k-1,\\ 0,&\text{ if }K=k.\end{array}\right.$$

$$V_1=W_1\cup \{a\}\text{ and }{V}_2=W_2\cup \{b\}.$$

$$2d_H(a)-d_G(a)=\left(|X_2\cap I_2|-|X_2\cap I_1|\right)+|X_1\cap I_2|+ϵ\ge -1.$$

$$2d_H(b)-d_G(b)=\left(|X_2\cap I_1|-|X_2\cap I_2|\right)+|X_1\cap I_1|+ϵ\ge -1.$$

$$-2\le |X_2\cap I_2|-|X_2\cap I_1|\le -1.$$

$$2d_H(a)-d_G(a)=(|X_2\cap I_2|-|X_2\cap I_1|)+|X_1\cap I_2|-1+ϵ\ge -1.$$

$$2d_H(b)-d_G(b)=(|X_2\cap I_1|-|X_2\cap I_2|)+|X_1\cap I_1|-1+ϵ\ge 0.$$

$$0\le |X_2\cap I_2|-|X_2\cap I_1|\le 1.$$

$$2d_H(a)-d_G(a)=(|X_2\cap I_2|-|X_2\cap I_1|)+|X_1\cap I_2|-1+ϵ\ge -1.$$

$$2d_H(b)-d_G(b)=(|X_2\cap I_1|-|X_2\cap I_2|)+|X_1\cap I_1|-1+ϵ\ge 0.$$

$${\cal X}_{0}^{\prime}=\{X_{i}:i\in[k],X_{i}\mbox{ crosses $H$, }|X_{i}|\equiv 0\mod 2,\mbox{ and }||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||=2\}$$

$${\cal X}_{1}=\{X_{i}:i\in[k],\mbox{ $X_{i}$ crosses $H$, and }|X_{i}|\equiv 1\mod 2\}.$$

$$s({\mathcal{A}}^{\prime })=s(\mathcal{A})/2+(m+2n-1)/2,$$

$$|W_2\cap V_2|-|W_2\cap V_1|=|(W_0^{\prime }\cup W_2)\cap V_2|-|(W_0^{\prime }\cup W_2)\cap V_1|=t+2t^{\prime }-1.$$

$$|V_1|-|V_2|=2|{\mathcal{S}}_0^{\prime }|+|{\mathcal{S}}_1\setminus A|+s(\mathcal{A})-2s({\mathcal{A}}^{\prime })=1.$$

$$|\overline{X_i}\cap V_1|-|\overline{X_i}\cap V_2|=(|V_1|-|V_2|)-(|X_i\cap V_1|-|X_i\cap V_2|)=1-(|X_i\cap V_1|-|X_i\cap V_2|).$$

$$s({\mathcal{A}}^{\prime })=s(\mathcal{A})/2-|\mathcal{A}|/2+1.$$

$$(d_{G^{\prime }}(v)-1)/2\le d_{H^{\prime }}(v)\le d_H(v)<(d_G(v)-1)/2\le d_{G^{\prime }}(v)/2$$

Full text

On problems about judicious bipartitions of graphs

Yuliang Ji       Jie Ma       Juan Yan       Xingxing Yu School of Mathematical Sciences, University of Science and Technology of China, Hefei, Anhui 230026, China. Email: jiyl@mail.ustc.edu.cn.School of Mathematical Sciences, University of Science and Technology of China, Hefei, Anhui 230026, China. Email: jiema@ustc.edu.cn. Partially supported by NSFC projects 11501539 and 11622110.College of Mathematics and Systems Science, Xinjiang University, Urumqi, Xinjiang 830046, China. Email: yanjuan207@163.com. Partially supported by NSFC project 11501486.School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332, USA. Email: yu@math.gatech.edu. Partially supported by NSF grants DMS–1265564 and DMS-1600387.

Abstract:

Bollobás and Scott [5] conjectured that every graph $G$ has a balanced bipartite spanning subgraph $H$ such that for each $v\in V(G)$, $d_{H}(v)\geq(d_{G}(v)-1)/2$. In this paper, we show that every graphic sequence has a realization for which this Bollobás-Scott conjecture 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 that $\lfloor(d_{G}(v)-1)/2\rfloor$ (rather than $(d_{G}(v)-1)/2$) is probably the correct lower bound. We also study bipartitions $V_{1},V_{2}$ of graphs with a fixed number of edges. We provide a (best possible) upper bound on $e(V_{1})^{\lambda}+e(V_{2})^{\lambda}$ for any real $\lambda\geq 1$ (the case $\lambda=2$ is a question of Scott [13]) and answer a question of Scott [13] on $\max\{e(V_{1}),e(V_{2})\}$.

Keywords: bipartition; bisection; degree sequence; complete $k$-partite graph; $\ell_{\lambda}$-norm

AMS classification: 05C07, 05C70

1 Introduction

For any positive integer $k$, let $[k]:=\{1,\ldots,k\}$. Let $G$ be a graph and $V_{1},\ldots,V_{k}$ be a partition of $V(G)$. When $k=2$, such a partition is said to be a bipartition of $G$. A subgraph $H$ of a graph $G$ is said to be a bisection of $G$ if $H$ is a bipartite spanning subgraph of $G$ and the partition sets of $H$ differ in size by at most one. For $i,j\in[k]$, we use $e(V_{i})$ to denote the number of edges of $G$ with both ends in $V_{i}$ and use $e(V_{i},V_{j})$ to denote the number of edges between $V_{i}$ and $V_{j}$. Judicious partitioning problems for graphs ask for partitions of graphs that bound a number of quantities simultaneously, such as all $e(V_{i})$ and $e(V_{i},V_{j})$. 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 $V_{1},V_{2}$ of a graph with $m$ edges, bounding $e(V_{1})^{2}+e(V_{2})^{2}$ and $\max\{e(V_{1}),e(V_{2})\}$ in terms of $m$.

For a graph $G$ and for any $v\in V(G)$, we use $d_{G}(v)$ to denote the degree of the vertex $v$ in $G$. It is well known that if $H$ is a maximum bipartite spanning subgraph of a graph $G$, then $d_{H}(v)\geq d_{G}(v)/2$ for each $v\in V(G)$. This, however, may not be true if one requires $H$ to be a bisection, as observed by Bollobás and Scott [5] by considering the complete bipartite graphs $K_{2\ell+1,m}$ for $m\geq 2\ell+3$. In an attempt to obtain a similar result for bisections, Bollobás and Scott [5] conjectured that every graph $G$ has a bisection $H$ such that

[TABLE]

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 $\pi$ (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 $\pi$. Hartke and Seacrest [10] proved that for any graphic sequence $\pi$ with even length, $\pi$ has a realization $G$ which admits a bisection $H$ such that for all $v\in V(G)$, $d_{H}(v)\geq\lfloor(d_{G}(v)-1)/2\rfloor$. They further conjectured that for any graphic sequence $\pi$ with even length, $\pi$ 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 $V(G)=\{v_{1},\dots,v_{n}\}$, we define the parity bisection of $G$ to be the bisection with partition sets $V_{1}$ and $V_{2}$, where $V_{i}=\{v_{j}\in V(G):j\equiv i\mod 2\}$ for each $i\in[2]$, and $E(H)=\{uv\in E(G):u\in V_{1}\mbox{ and }v\in V_{2}\}$.

Theorem 1.1**.**

Let $\pi=(d_{1},\dots,d_{n})$ be any graphic sequence with $d_{1}\geq\cdots\geq d_{n}$. Then there exists a realization $G$ of $\pi$ with $V(G)=\{v_{1},\dots,v_{n}\}$ and $d_{G}(v_{i})=d_{i}$ for $i\in[n]$, such that if $H$ denotes the parity bisection of $G$ then $d_{H}(v_{i})\geq(d_{G}(v_{i})-1)/2$ for $i\in[n]$.

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 $\pi=(d_{1},\dots,d_{n})$ with $d_{1}=\cdots=d_{k}=n-1$ and $d_{k+1}=\cdots=d_{n}=k$. Let $H$ be an arbitrary bisection of $G$ with parts $A,B$ and, without loss of generality, assume that $|A\cap V(K)|\leq k/2$. Since $k<n/2$, there must exist a vertex $v\in B\cap I$. So $d_{H}(v)=|A\cap V(K)|\leq\lfloor k/2\rfloor$. Since $d_{G}(v)=k$ and $k$ is odd, we see that $d_{H}(v)\leq(d_{G}(v)-1)/2$.

The second result in this paper gives indication that perhaps the lower bound in the original Bollobás-Scott conjecture was meant to be $\lfloor(d_{G}(v)-1)/2\rfloor$ (rather than $(d_{G}(v)-1)/2$).

Proposition 1.2**.**

Let $r_{1},r_{2},r_{3}$ be pairwise distinct odd integers such that for every $i\in[3]$, $r_{i}\notin\left\{1,\lfloor(r_{1}+r_{2}+r_{3})/2\rfloor,\lceil(r_{1}+r_{2}+r_{3})/2\rceil\right\}$. Then for any bisection $H$ of the complete 3-partite graph $G:=K_{r_{1},r_{2},r_{3}}$, there always exists a vertex $v$ with $d_{H}(v)<(d_{G}(v)-1)/2$.

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 $d_{H}(v)\geq\lfloor(d_{G}(v)-1)/2\rfloor$ for all $v\in V(G)$. 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 $d_{H}(v)\geq(d_{G}(v)-c)/2$ for all $v\in V(G)$.

We now turn our discussion to problems on general bipartitions. Answering a question of Erdős, Edwards [6] showed in 1973 that every graph with $m$ edges admits a bipartition $V_{1},V_{2}$ such that $e(V_{1},V_{2})\geq m/2+t(m)/2$, where

[TABLE]

This bound is best possible for the complete graphs of odd order. Bollobás and Scott [3] extended Edwards’ bound by showing that every graph $G$ with $m$ edges has a bipartition $V_{1},V_{2}$ simultaneously satisfying $e(V_{1},V_{2})\geq m/2+t(m)/2$ and $\max\{e(V_{1}),e(V_{2})\}\leq m/4+t(m)/4$, 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 $\lambda>0$ and a bipartition $V_{1},V_{2}$ of a graph $G$, define the $\ell_{\lambda}$-norm of $(V_{1},V_{2})$ to be $\left(e(V_{1})^{\lambda}+e(V_{2})^{\lambda}\right)^{1/\lambda}$. Then to maximize $e(V_{1},V_{2})$ is equivalent to minimize the $\ell_{1}$-norm of $(V_{1},V_{2})$, while minimizing $\max\{e(V_{1}),e(V_{2})\}$ is the same as minimizing the $\ell_{\infty}$-norm of $(V_{1},V_{2})$. It is natural to consider other norms. In particular, Scott asked for the maximum of

[TABLE]

over graphs $G$ with $m$ edges, see Problem 3.18 in [13]. We provide an answer to this question by proving the following general result.

Theorem 1.4**.**

Let $m$ be any positive integer and $\lambda\geq 1$ be any real number. Then, for any graph $G$ with $m$ edges,

[TABLE]

Moreover, the equality holds if and only if $G$ is a complete graph of odd order.

We also consider analogous questions for $k$-partitions in Section 4.

Though Edward’s bound is tight for all integers $m=\binom{n}{2}$, Erdős [7] conjectured that the difference between Edwards’ bound and the truth can still be arbitrarily large for other $m$. This was confirmed by Alon [1]: every graph with $m=n^{2}/2$ edges admits a bipartition $V_{1},V_{2}$ such that $e(V_{1},V_{2})\geq m/2+t(m)/2+\Omega(m^{1/4})$. Bollobás and Scott [5, 13] made a similar conjecture for $\max\{e(V_{1}),e(V_{2})\}$: for certain $m$, $\max\{e(V_{1}),e(V_{2})\}$ can be arbitrary far from $m/4+t(m)/4$. Ma and Yu [12] proved that every graph with $m=n^{2}/2$ edges admits a bipartition $V_{1},V_{2}$ such that $\max\{e(V_{1}),e(V_{2})\}\leq m/4+t(m)/4-\Omega(m^{1/4})$. Another result in the same spirit was given by Hofmeister and Lefmann [9] that any graph with $\binom{kn}{2}$ edges has a $k$-partition $V_{1},...,V_{k}$ with $\sum_{i=1}^{k}e(V_{i})\leq k\binom{n}{2}$, which beats the trivial upper bound $\frac{1}{k}\binom{nk}{2}$.

Motivated by these results, Scott asked the following question: does every graph $G$ with $\binom{kn}{2}$ edges have a vertex partition into $k$ sets, each of which contains at most $\binom{n}{2}$ edges? ( See Problem 3.9 in [13].) We show that the answer to this question is negative for $k=2$.

Theorem 1.5**.**

There exist infinitely many positive integers $n$ and for each such $n$ there is a graph $G$ with $\binom{2n}{2}$ edges, such that, for every bipartition $V_{1},V_{2}$ of $G$, $\max\{e(V_{1}),e(V_{2})\}\geq\binom{n}{2}+5n/48.$

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 non-negative integers. Let $\pi=(d_{1},\dots,d_{n})$ with $d_{1}\geq\cdots\geq d_{n}$. By removing $d_{i}$ from $\pi$ and subtracting $1$ from the $d_{i}$ remaining elements of $\pi$ with lowest indices, we obtain a new sequence $\pi^{\prime}=(d^{\prime}_{1},\dots,d^{\prime}_{i-1},d^{\prime}_{i+1},\dots,d^{\prime}_{n})$, and we say that $\pi^{\prime}$ is obtained from $\pi$ by laying off $d_{i}$. This operation was introduced by Kleitman and Wang [11], and they proved the following.

Lemma 2.1** (Kleitman-Wang [11]).**

For any $i\in[n]$, the sequence $\pi=(d_{1},\dots,d_{n})$ with $d_{1}\geq\ldots\geq d_{n}$ is graphic if and only if the sequence $\pi^{\prime}$ obtained from $\pi$ by laying off $d_{i}$ is graphic.

It is easy to see that the sequence $\pi^{\prime}$ obtained from $\pi$ by laying off $d_{i}$ need not be non-increasing. To avoid this issue, Hartke and Seacrest [10] introduced a variation of the above operation. Choose a fixed $i\in[n]$. Let $s$ be the smallest value among the $d_{i}$ largest elements in $\pi$, not including the $i$th element of $\pi$ (namely, $d_{i}$ itself). Let $S=\{j\in[n]-\{i\}:d_{j}>s\}$. Note that $|S|<d_{i}$. Let $T$ be the set of $d_{i}-|S|$ largest indices $j$ with $j\not=i$ and $d_{j}=s$. Then by laying off $d_{i}$ with order, we remove $d_{i}$ from $\pi$ and subtract $1$ from $d_{j}$ for all $j\in S\cup T$. If $\pi^{\prime}=(d^{\prime}_{1},\dots d^{\prime}_{i-1},d^{\prime}_{i+1},\dots,d^{\prime}_{n})$ denotes the new sequence, then it has the monotone property $d^{\prime}_{1}\geq\cdots\geq d^{\prime}_{n}$.

Clearly, the sequence obtained from $\pi$ by laying off $d_{i}$ with order is just a permutation of the sequence obtained from $\pi$ by laying off $d_{i}$. So the following is true.

Lemma 2.2** (Hartke and Seacrest [10]).**

For any $i\in[n]$, the sequence $\pi=(d_{1},\dots,d_{n})$ with $d_{1}\geq\ldots\geq d_{n}$ is graphic if and only if the sequence obtained from $\pi$ by laying off $d_{i}$ with order is graphic.

We give a brief outline of our proof of Theorem 1.1. We choose two consecutive elements $d_{\ell}$ and $d_{\ell+1}$ of $\pi$. Using Lemma 2.2 we obtain a new graphic sequence $\pi^{\prime\prime}$ of length $n-2$ by first laying off $d_{\ell+1}$ with order and then laying off $d_{\ell}$ with order. By induction, $\pi^{\prime\prime}$ has an $(n-2)$-vertex realization $F$ whose parity bisection $J$ has the desired property. We then show that one can form $G$ from $F$ by adding two new vertices (for $d_{\ell}$ and $d_{\ell+1}$) and choosing 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 $\pi=(d_{1},\ldots,d_{n})$ with $d_{1}\geq\ldots\geq d_{n}$. The assertion is trivial when $n=1,2$. So we may assume that $n\geq 3$ and the assertion holds for all graphic sequences with length less than $n$. Then there exist two consecutive elements of $\pi$ that are identical; so let $\ell\in[n-1]$ be fixed such that

[TABLE]

Let $\pi^{\prime}=(d_{1}^{\prime},...,d_{\ell}^{\prime},d_{\ell+2}^{\prime},...,d_{n}^{\prime})$ be the sequence obtained from $\pi$ by laying off $d_{\ell+1}$ with order. Let $\pi^{\prime\prime}=(d_{1}^{\prime\prime},...,d_{\ell-1}^{\prime\prime},d_{\ell+2}^{\prime\prime},...,d_{n}^{\prime\prime})$ be the sequence obtained from $\pi^{\prime}$ by laying off $d_{l}^{\prime}$ with order. By Lemma 2.2, $\pi^{\prime}$ and $\pi^{\prime\prime}$ both are graphic sequences.

Let $\omega=(f_{1},...,f_{n-2})$ be the sequence obtained from $\pi$ with $d_{\ell}$ and $d_{\ell+1}$ removed, and re-indexed so that the indices are consecutive, i.e., $f_{i}=d_{i}$ for $i\in[\ell-1]$ and $f_{i}=d_{i+2}$ for $i\in[n-2]\setminus[\ell-1]$. Let $\omega^{\prime}=(f_{1}^{\prime},...,f_{n-2}^{\prime})$ be the sequence obtained from $\pi^{\prime}$ with $d^{\prime}_{\ell}$ removed, and re-indexed so that the indices are consecutive. Also, let $\omega^{\prime\prime}=(f_{1}^{\prime\prime},...,f_{n-2}^{\prime\prime})$ be the sequence obtained from $\pi^{\prime\prime}$ by re-indexing so that the indices are consecutive. Note that $\omega^{\prime\prime}$ is a graphic sequence.

To turn a realization of $\omega^{\prime\prime}$ to a realization of $\pi$, we need to track the changes between $f_{i}$ and $f_{i}^{\prime\prime}$ for all $i\in[n-2]$. Note that $0\leq f_{i}-f_{i}^{\prime\prime}\leq 2$. Let

[TABLE]

and

[TABLE]

So $K=\sum_{i\in[n-2]}|f_{i}-f_{i}^{\prime}|=\sum_{i\in[n-2]}|f_{i}^{\prime}-f_{i}^{\prime\prime}|$; hence

[TABLE]

We now prove two claims asserting certain properties on $X_{1}$ and $X_{2}$. For convenience, 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, $A_{1},...,A_{t}$, of integers is said to be consecutive if $A_{1}\cup...\cup A_{t}$ is consecutive and, for any $i,j\in[t]$ with $i<j$ and $A_{i}$ and $A_{j}$ nonempty, we have $A_{i}<A_{j}$.

Claim 1. There exist consecutive sets $R_{1},R_{2},R_{1}^{\prime},R_{2}^{\prime},Q$ such that $X_{1}=R_{1}^{\prime}\cup R_{2}^{\prime}$ and $X_{2}=R_{1}\cup R_{2}$ such that

• (a)

the sequence $R_{1},R_{1}^{\prime},Q,R_{2}^{\prime}$ is consecutive,

• (b)

either $R_{2}=\emptyset$ or $R_{2}=Q$, and

• (c)

$f_{i}^{\prime\prime}=f_{j}^{\prime\prime}+1$ for all $i\in R_{1}^{\prime},j\in R_{2}^{\prime}$.

Proof of Claim 1..

Let $s$ be the minimum of the largest $K$ numbers in $\omega=(f_{1},...,f_{n-2})$. (Note that this $s$ is the same as the $s$ in the definition of laying off $d_{\ell+1}$ with order from $\pi$.) In order to keep track whether $f_{i}^{\prime}=f_{i}$ or $f_{i}^{\prime}=f_{i}-1$ and whether $f_{i}^{\prime\prime}=f_{i}^{\prime}$ or $f_{i}^{\prime\prime}=f_{i}^{\prime}-1$, we divide $[n-2]$ into six pairwise disjoint sets:

[TABLE]

By the definitions of $\pi^{\prime}$ and $\omega^{\prime}$, we see that $A,B,C,D,E,F$ is consecutive and

[TABLE]

Thus, it is easy to see that $A\cup B\cup D=\{i\in[n-2]:f^{\prime}_{i}=f_{i}-1\}$; so $|A|+|B|+|D|=K.$

Let $Y=\{i\in[n-2]:f^{\prime\prime}_{i}=f_{i}^{\prime}-1\}$. Then it follows that

[TABLE]

To complete our proof of Claim 1, we distinguish four cases based on relations among the sizes of $B,C,D,E$.

First, suppose $|C|\geq|B|+|D|$. Let $C^{\prime\prime}$ consist of the last $|B|+|D|$ integers in $C$, and $C^{\prime}:=C\setminus C^{\prime\prime}$. Then we see that $Y=A\cup C^{\prime\prime}$. Let $R_{1}=A$, $R_{2}=\emptyset$, $R_{1}^{\prime}=B$, $R_{2}^{\prime}=C^{\prime\prime}\cup D$ and $Q=C^{\prime}$. It is easy to check that $X_{1}=R_{1}^{\prime}\cup R_{2}^{\prime}$ and $X_{2}=R_{1}\cup R_{2}$, and (a) and (b) holds. Note that $f_{i}^{\prime\prime}=s$ for $i\in R_{1}^{\prime}$, and $f_{j}^{\prime\prime}=s-1$ for $j\in R_{2}^{\prime}$; so (c) holds.

Next, suppose $|D|\leq|C|<|B|+|D|$. Let $B^{\prime\prime}$ consist of the last $|B|+|D|-|C|$ integers in $B$, and $B^{\prime}=B\setminus B^{\prime\prime}$. Then $Y=A\cup B^{\prime\prime}\cup C$. Let $R_{1}=A$, $R_{2}=Q=B^{\prime\prime}$, $R_{1}^{\prime}=B^{\prime}$ and $R_{2}^{\prime}=C\cup D$. It is easy to check that $X_{1}=R_{1}^{\prime}\cup R_{2}^{\prime}$ and $X_{2}=R_{1}\cup R_{2}$, and that (a) and (b) holds. Note that $f_{i}^{\prime\prime}=s$ for $i\in R_{1}^{\prime}$, and $f_{j}^{\prime\prime}=s-1$ for $j\in R_{2}^{\prime}$; so (c) holds.

Now assume $|C|<|D|\leq|C|+|E|$. Let $E^{\prime\prime}$ consist of the last $|D|-|C|$ integers in $E$, and $E^{\prime}=E\setminus E^{\prime\prime}$. Then $Y=A\cup B\cup C\cup E^{\prime\prime}$. Let $R_{1}=A\cup B$, $R_{2}=\emptyset$, $R_{1}^{\prime}=C\cup D$, $R_{2}^{\prime}=E^{\prime\prime}$, and $Q=E^{\prime}$. It is easy to check that $X_{1}=R_{1}^{\prime}\cup R_{2}^{\prime}$ and $X_{2}=R_{1}\cup R_{2}$, and (a) and (b) holds. Note that $f_{i}^{\prime\prime}=s-1$ for $i\in R_{1}^{\prime}$ and $f_{j}^{\prime\prime}=s-2$ for $j\in R_{2}^{\prime}$; so (c) holds.

Finally we consider the case when $|D|>|C|+|E|$. Let $D^{\prime\prime}$ consist of the last $|D|-|C|-|E|$ integers in $D$, and $D^{\prime}=D\setminus D^{\prime\prime}$. Then $Y=A\cup B\cup C\cup D^{\prime\prime}\cup E$. Let $R_{1}=A\cup B$, $R_{2}=Q=D^{\prime\prime}$, $R_{1}^{\prime}=C\cup D^{\prime}$ and $R_{2}^{\prime}=E$. It is easy to check that $X_{1}=R_{1}^{\prime}\cup R_{2}^{\prime}$ and $X_{2}=R_{1}\cup R_{2}$, and (a) and (b) holds. Note that $f_{i}^{\prime\prime}=s-1$ for $i\in R_{1}^{\prime}$ and $f_{j}^{\prime\prime}=s-2$ for $j\in R_{2}^{\prime}$; so (c) holds.

Let $I_{1}=\{i\in[n-2]:i\equiv 1\mod 2\}$ and $I_{2}=\{i\in[n-2]:i\equiv 0\mod 2\}$.

Claim 2. $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|\in\{-2,0,2\}$. Moreover, $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|=0$ implies $|X_{2}\cap I_{1}|-|X_{2}\cap I_{2}|\in\{-1,0,1\}.$

Proof of Claim 2..

By (2), we see $|X_{1}|$ must be even. So $|R^{\prime}_{1}|$ and $|R^{\prime}_{2}|$ are of the same parity. Since both $R_{1}^{\prime}$ and $R_{2}^{\prime}$ are consecutive, $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|\in\{-2,0,2\}$.

Now suppose $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|=0$. If $R_{2}=\emptyset$, then $X_{2}=R_{1}$ is a consecutive set and thus $|X_{2}\cap I_{1}|$ and $|X_{2}\cap I_{2}|$ differ by at most one. So we may assume $R_{2}\neq\emptyset$. Then $R_{2}=Q$ by Claim 1. As the sequence $R_{1},R_{1}^{\prime},Q,R_{2}^{\prime}$ is consecutive, we see that $X_{1}\cup X_{2}$ is a consecutive set; so $|(X_{1}\cup X_{2})\cap I_{1}|$ and $|(X_{1}\cup X_{2})\cap I_{2}|$ differ by at most one. Hence, since $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|=0$, $||X_{2}\cap I_{1}|-|X_{2}\cap I_{2}||\leq 1$.

We are ready to construct a realization of $\pi=(d_{1},...,d_{n})$. Recall that $\omega^{\prime\prime}=(f_{1}^{\prime\prime},...,f_{n-2}^{\prime\prime})$ is a graphic sequence. By induction hypothesis, there exists a realization $F$ of $\omega^{\prime\prime}$ with $V(F)=\{w_{1},...,w_{n-2}\}$ and $d_{F}(w_{i})=f_{i}^{\prime\prime}$ for $i\in[n-2]$, such that the parity bisection $J$ of $F$ satisfies

[TABLE]

Let $W_{j}=\{w_{i}:i\equiv j\mod 2\}$ for $j\in[2]$.

In what follows, we will construct a graph $G$ as the realization of $\pi$ such that its parity bisection $H$ of $G$ satisfies $d_{H}(v)\geq(d_{G}(v)-1)/2$ for all $v\in V(G)$, by adding two new vertices $a,b$ (so $V(G)=V(F)\cup\{a,b\}$) 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

[TABLE]

We write $V(G)=\{v_{1},...,v_{n}\}$ such that $v_{i}=w_{i}$ for $i<\ell$, $\{v_{\ell},v_{\ell+1}\}=\{a,b\}$, and $v_{i}=w_{i-2}$ for $\ell+1<i\leq n$.

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 $\{v_{\ell},v_{\ell+1}\}$ with odd index. So the parity partition of $V(G)$ is

[TABLE]

Case $1$. $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|=0$.

We know $F\subseteq G$ and $V(G)=V(F)\cup\{a,b\}$, and we need to add edges at $a$ and $b$ to form $G$, a realization of $\pi$. Add $ab$ if $\epsilon=1$, $av_{i}$ for all $i\in X_{2}\cup(X_{1}\cap I_{2})$, and $bv_{j}$ for all $j\in X_{2}\cup(X_{1}\cap I_{1})$. Since $|X_{1}\cap I_{1}|=|X_{1}\cap I_{2}|$, $G$ is a realization of $\pi$. Let $H$ denote the parity bisection of $G$; so $V_{1},V_{2}$ are the partition sets of $H$. We need to show that $d_{H}(v)\geq(d_{G}(v)-1)/2$ for all $v\in V(G)$.

For each $w_{i}$ with $i\notin X_{1}\cup X_{2}$, its neighborhoods in $F,G$ are the same; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})\geq(d_{F}(w_{i})-1)/2=(d_{G}(w_{i})-1)/2$.

For vertices $w_{i}$ with $i\in X_{2}$, we have $d_{G}(w_{i})=d_{F}(w_{i})+2$ and $d_{H}(w_{i})=d_{J}(w_{i})+1$; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})+1\geq(d_{F}(w_{i})-1)/2+1=(d_{G}(w_{i})-1)/2$.

For vertices $w_{i}$ with $i\in X_{1}$, we have $d_{G}(w_{i})=d_{F}(w_{i})+1$ and $d_{H}(w_{i})=d_{J}(w_{i})+1$; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})+1\geq(d_{F}(w_{i})-1)/2+1>(d_{G}(w_{i})-1)/2$.

For the vertex $a$, we have $d_{G}(a)=|X_{2}|+|X_{1}\cap I_{2}|+\epsilon$ and $d_{H}(a)=|X_{2}\cap I_{2}|+|X_{1}\cap I_{2}|+\epsilon$. Note that in this case, by Claim 2, we have $||X_{2}\cap I_{1}|-|X_{2}\cap I_{2}||\leq 1$, which implies that

[TABLE]

Hence, $d_{H}(a)\geq(d_{G}(a)-1)/2$.

Similarly, for the vertex $b$, we have $d_{G}(b)=|X_{2}|+|X_{1}\cap I_{1}|+\epsilon$ and $d_{H}(b)=|X_{2}\cap I_{1}|+|X_{1}\cap I_{1}|+\epsilon$. Note, by Claim 2, $||X_{2}\cap I_{1}|-|X_{2}\cap I_{2}||\leq 1$; so

[TABLE]

Hence, $d_{H}(b)\geq(d_{G}(b)-1)/2$.

Case $2$. $|X_{1}\cap I_{2}|-|X_{1}\cap I_{1}|=2$.

Recall that $X_{1}=R^{\prime}_{1}\cup R^{\prime}_{2}$, where each $R^{\prime}_{i}$ is consecutive. Thus it follows that $|R^{\prime}_{i}\cap I_{2}|=|R^{\prime}_{i}\cap I_{1}|+1$ for $i\in[2]$. Since the sequence $R_{1},R^{\prime}_{1},Q,R^{\prime}_{2}$ is consecutive and starts from the integer $1$, we see that $|R_{1}\cap I_{2}|=|R_{1}\cap I_{1}|-1$ and $|Q\cap I_{2}|=|Q\cap I_{1}|-1$. Therefore, since $R_{2}=\emptyset$ or $R_{2}=Q$ (by (b) of Claim 1), we have

[TABLE]

We claim that there exists some $z\in X_{1}\cap I_{2}$ with $d_{J}(w_{z})\geq d_{F}(w_{z})/2$. To see this, choose $x\in R^{\prime}_{1}\cap I_{2}$ and $y\in R^{\prime}_{2}\cap I_{2}$. By (3) , we have $d_{J}(w_{x})\geq(d_{F}(w_{x})-1)/2$ and $d_{J}(w_{y})\geq(d_{F}(w_{y})-1)/2$. By (c) of Claim 1, $d_{F}(w_{x})=d_{F}(w_{y})+1$. Observe that for any vertex $u$ of $F$, $d_{F}(u)$ and $2d_{J}(u)-d_{F}(u)$ are of the same parity; so $2d_{J}(w_{x})-d_{F}(w_{x})$ and $2d_{J}(w_{y})-d_{F}(w_{y})$ must have different parities. Therefore there exists $z\in\{x,y\}$ such that $d_{J}(w_{z})\geq d_{F}(w_{z})/2$, proving the claim.

We now add edges at $a$ and $b$ to form $G$ from $F$: add $ab$ if $\epsilon=1$, $av_{i}$ for all $i\in X_{2}\cup(X_{1}\cap I_{2})\setminus\{z\}$, and $bv_{j}$ for all $j\in X_{2}\cup(X_{1}\cap I_{1})\cup\{z\}$. Since $|X_{1}\cap I_{2}|=|X_{1}\cap I_{1}|+2$, $G$ is a realization of $\pi$. Next we show that the parity bisection $H$ of $G$ satisfies the property that $d_{H}(v)\geq(d_{G}(v)-1)/2$ for all $v\in V(G)$.

For each $w_{i}$ with $i\notin X_{1}\cup X_{2}$, its neighborhoods in $F,G$ are the same; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})\geq(d_{F}(w_{i})-1)/2=(d_{G}(w_{i})-1)/2$.

For each $w_{i}$ with $i\in X_{2}$, $d_{G}(w_{i})=d_{F}(w_{i})+2$ and $d_{H}(w_{i})=d_{J}(w_{i})+1$. Hence by (3) and the way we choose $z$, $d_{H}(w_{i})=d_{J}(w_{i})+1\geq(d_{F}(w_{i})-1)/2+1=(d_{G}(w_{i})-1)/2$.

For $w_{i}$ with $i\in X_{1}\setminus\{z\}$, we have $d_{G}(w_{i})=d_{F}(w_{i})+1$ and $d_{H}(w_{i})=d_{J}(w_{i})+1$; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})+1\geq(d_{F}(w_{i})-1)/2+1>(d_{G}(w_{i})-1)/2$.

The vertex $w_{z}$ satisfies $d_{G}(w_{z})=d_{F}(w_{z})+1$ and $d_{H}(w_{z})=d_{J}(w_{z})$. Hence by (3), $d_{H}(w_{z})=d_{J}(w_{z})\geq d_{F}(w_{z})/2=(d_{G}(w_{z})-1)/2$.

For the vertex $a$, by definition we have $d_{G}(a)=|X_{2}|+|X_{1}\cap I_{2}|-1+\epsilon$ and $d_{H}(a)=|X_{2}\cap I_{2}|+|X_{1}\cap I_{2}|-1+\epsilon$. By (4) and the fact that $|X_{1}\cap I_{2}|\geq 2$,

[TABLE]

Hence, $d_{H}(a)\geq(d_{G}(a)-1)/2$.

For the vertex $b$, we have $d_{G}(b)=|X_{2}|+|X_{1}\cap I_{1}|+1+\epsilon$ and $d_{H}(b)=|X_{2}\cap I_{1}|+|X_{1}\cap I_{1}|+\epsilon$. This, together with (4), imply that

[TABLE]

Hence, $d_{H}(b)\geq(d_{G}(b)-1)/2$.

Case $3$. $|X_{1}\cap I_{1}|-|X_{1}\cap I_{2}|=2$.

In this case, we have $|R^{\prime}_{i}\cap I_{1}|=|R^{\prime}_{i}\cap I_{2}|+1$ for $i\in[2]$ (as $R_{1}^{\prime}$ and $R_{2}^{\prime}$ are consecutive). Because $R_{1},R_{1}^{\prime},Q,R_{2}^{\prime}$ is consecutive, it follows that $|R_{1}\cap I_{1}|=|R_{1}\cap I_{2}|$ and $|Q\cap I_{1}|=|Q\cap I_{2}|-1$. Since $R_{2}=\emptyset$ or $R_{2}=Q$ (by (b) of Claim 1),

[TABLE]

Since $|X_{1}|$ is even and $|R^{\prime}_{i}\cap I_{1}|=|R^{\prime}_{i}\cap I_{2}|+1$ for $i\in[2]$, there exist $x\in R_{1}^{\prime}\cap I_{1}$ and $y\in R_{2}^{\prime}\cap I_{1}$. By (3) and (c) of Claim 1, we have $d_{J}(w_{x})\geq(d_{F}(w_{x})-1)/2$, $d_{J}(w_{y})\geq(d_{F}(w_{y})-1)/2$, and $d_{F}(w_{x})=d_{F}(w_{y})+1$. Since for any vertex $u$ of $F$, $d_{F}(u)$ and $2d_{J}(u)-d_{F}(u)$ are of the same parity, $2d_{J}(w_{x})-d_{F}(w_{x})$ and $2d_{J}(w_{y})-d_{F}(w_{y})$ must have different parities. Therefore there exists $z\in\{x,y\}$ such that $d_{J}(w_{z})\geq d_{F}(w_{z})/2$.

We now add edges at $a$ and $b$ to form the graph $G$: add $ab$ if $\epsilon=1$, $av_{i}$ for all $i\in X_{2}\cup(X_{1}\cap I_{2})\cup\{z\}$, and $bv_{j}$ for all $j\in X_{2}\cup(X_{1}\cap I_{1})\setminus\{z\}$. Since $|X_{1}\cap I_{1}|=|X_{1}\cap I_{2}|+2$, $G$ is a realization of $\pi$. We need to verify that $d_{H}(v)\geq(d_{G}(v)-1)/2$ for all $v\in V(G)$.

If $v=w_{i}$ for some $i\notin X_{1}\cup X_{2}$, then $d_{G}(w_{i})=d_{F}(w_{i})$ and $d_{H}(w_{i})=d_{J}(w_{i})$; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})\geq(d_{F}(w_{i})-1)/2=(d_{G}(w_{i})-1)/2$.

If $v=w_{i}$ for some $i\in X_{2}$, then $d_{G}(w_{i})=d_{F}(w_{i})+2$ and $d_{H}(w_{i})=d_{J}(w_{i})+1$; again by (3), $d_{H}(w_{i})=d_{J}(w_{i})+1\geq(d_{F}(w_{i})-1)/2+1=(d_{G}(w_{i})-1)/2$.

If $v=w_{i}$ for some $i\in X_{1}\setminus\{z\}$, then $d_{G}(w_{i})=d_{F}(w_{i})+1$ and $d_{H}(w_{i})=d_{J}(w_{i})+1$; so by (3), $d_{H}(w_{i})=d_{J}(w_{i})+1\geq(d_{F}(w_{i})-1)/2+1>(d_{G}(w_{i})-1)/2$.

Now suppose $v=w_{z}$. Note that $d_{G}(w_{z})=d_{F}(w_{z})+1$ and $d_{H}(w_{z})=d_{J}(w_{z})$. So $d_{H}(w_{z})=d_{J}(w_{z})\geq d_{F}(w_{z})/2=(d_{G}(w_{z})-1)/2$.

Suppose $v=a$. Note that $d_{G}(a)=|X_{2}|+|X_{1}\cap I_{2}|+1+\epsilon$ and $d_{H}(a)=|X_{2}\cap I_{2}|+|X_{1}\cap I_{2}|+\epsilon$. By (5),

[TABLE]

So $d_{H}(a)\geq(d_{G}(a)-1)/2$.

Finally, suppose $v=b$. We have $d_{G}(b)=|X_{2}|+|X_{1}\cap I_{1}|-1+\epsilon$ and $d_{H}(b)=|X_{2}\cap I_{1}|+|X_{1}\cap I_{1}|-1+\epsilon$. By (5) and the fact that $|X_{1}\cap I_{1}|\geq 2$,

[TABLE]

So $d_{H}(b)\geq(d_{G}(b)-1)/2$.

3 Complete multipartite graphs

For convenience, we say that a bisection $H$ of a graph $G$ is good if for each $v\in V(G)$, $2d_{H}(v)\geq d_{G}(v)-1$. Thus the Bollobás-Scott conjecture says that every graph contains a good bisection. Here we discuss which complete multipartite graphs have good bisections. Throughout the rest of this section, let $G:=K_{r_{1},\dots,r_{k}}$, and let $X_{1},\ldots,X_{k}$ denote the partition sets of $G$ with $|X_{i}|=r_{i}$ for all $i\in[k]$.

First, we note that whenever $|V(G)|$ is even, $G$ has a good bisection. Since $|V(G)|$ is even, $V(G)$ has a partition $V_{1},V_{2}$ such that $|V_{1}|=|V_{2}|$, $||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||=1$ if $|X_{i}|$ is odd, and $|X_{i}\cap V_{1}|=|X_{i}\cap V_{2}|$ if $|X_{i}|$ is even. Let $H$ denote the maximum bisection of $G$ with partition sets $V_{1}$ and $V_{2}$. For any $v\in V(G)$, $v\in X_{i}$ for some $i\in[k]$. Note that $d_{G}(v)=|V(G)|-|X_{i}|$ and $d_{H}(v)\geq(|V(G)|-|X_{i}|-1)/2=(d_{G}(v)-1)/2$.

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 $v\in V(G)$, $d_{H}(v)\geq\lfloor(d_{G}(v)-1)/2\rfloor$.)

For a bisection $H$ of $G$ with partition sets $V_{1}$ and $V_{2}$, we say that $X_{i}$ crosses $H$ if $X_{i}\cap V_{j}\not=\emptyset$ for $j\in[2]$. For a subset $W\subseteq V(G)$, let $\overline{W}=V(G)\setminus W$. We need two easy lemmas.

Lemma 3.1**.**

Let $G=K_{r_{1},\dots,r_{k}}$ and let $X_{i}$, $i\in[k]$, be the partition sets of $G$. Suppose $H$ is a good bisection of $G$ with partition sets $V_{1}$ and $V_{2}$. Then the following statements hold for $i\in[k]$.

• $(i)$

If $X_{i}$ crosses $H$ and $|\overline{X_{i}}|$ is even then $|\overline{X_{i}}\cap V_{1}|=|\overline{X_{i}}\cap V_{2}|$ and $||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||\leq 1$.

• $(ii)$

If $X_{i}$ crosses $H$ and $|\overline{X_{i}}|$ is odd then $||\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}||=1$ and $||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||\leq 2$.

Proof.

Suppose $X_{i}$ crosses $H$. Then there exist $v_{j}\in X_{i}\cap V_{j}$ for $j\in[2]$. Note that $d_{G}(v_{1})=d_{G}(v_{2})=|\overline{X_{i}}|$. Thus, since $H$ is a good bisection, $d_{H}(v_{j})=|\overline{X_{i}}\cap V_{3-j}|\geq\lfloor|\overline{X_{i}}|/2\rfloor$ for $j\in[2]$. Notice that $|\overline{X_{i}}\cap V_{1}|+|\overline{X_{i}}\cap V_{2}|=|\overline{X_{i}}|.$ So $||\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}||\leq 1$.

If $|\overline{X_{i}}|$ is even then $|\overline{X_{i}}\cap V_{1}|=|\overline{X_{i}}\cap V_{2}|$, and $(i)$ holds. So assume $|\overline{X_{i}}|$ is odd. Since $||V_{1}|-|V_{2}||\leq 1$, $||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||\leq 2$, and $(ii)$ holds.

Lemma 3.2**.**

Let $G=K_{r_{1},\dots,r_{k}}$ with $|V(G)|$ odd, let $X_{1},\dots,X_{k}$ be the partition sets of $G$, and let $H$ be a good bisection of $G$ with partition sets $V_{1},V_{2}$ such that $|V_{1}|=|V_{2}|+1$. Let

[TABLE]

and

[TABLE]

Let $W_{1}=\bigcup_{X_{i}\in{\cal X}_{1}}X_{i}$, $W_{0}=\bigcup_{X_{i}\in{\cal X}_{0}^{\prime}}X_{i}$, $|{\cal X}_{1}|=t$, and $|{\cal X}_{0}^{\prime}|=t^{\prime}$. Then

• $(i)$

$|W_{1}\cap V_{1}|-|W_{1}\cap V_{2}|=t$,

• $(ii)$

$|W_{0}\cap V_{1}|-|W_{0}\cap V_{2}|=2t^{\prime}$, and

• $(iii)$

$|\overline{W_{1}\cup W_{0}}\cap V_{1}|-|\overline{W_{1}\cup W_{0}}\cap V_{2}|=-(t+2t^{\prime}-1)$.

Proof.

First, we prove $(i)$. If $W_{1}=\emptyset$, then $|W_{1}\cap V_{1}|-|W_{1}\cap V_{2}|=0=t$. So assume $W_{1}\neq\emptyset$ and let $X_{i}\in{\cal X}_{1}$. Then $|X_{i}|$ is odd and $X_{i}$ crosses $H$. Hence, since $|V(G)|$ is odd, $|\overline{X_{i}}|$ is even. By Lemma 3.1, $|\overline{X_{i}}\cap V_{1}|=|\overline{X_{i}}\cap V_{2}|$. Because $|V_{1}|=|V_{2}|+1$, $|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|=1$. Hence, $|W_{1}\cap V_{1}|-|W_{1}\cap V_{2}|=\sum_{X_{i}\in{\cal X}_{1}}(|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|)=t$.

We now prove $(ii)$. If $W_{0}=\emptyset$ then $t^{\prime}=0$ and the result holds trivially. So assume $W_{0}\neq\emptyset$ and let $X_{i}\in{\cal X}_{0}^{\prime}$. Then $|X_{i}|$ is even and $X_{i}$ crosses $H$. Since $|V(G)|$ is odd, $|\overline{X_{i}}|$ is odd. By Lemma 3.1, $||\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}||=1$, and by the definition of ${\cal X}_{0}^{\prime}$, $||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||=2$. Therefore, because $|V_{1}|=|V_{2}|+1$, $|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|=2$ (as well as $|\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}|=-1$). Hence $|W_{0}\cap V_{1}|-|W_{0}\cap V_{2}|=\sum_{X_{i}\in{\cal X}^{\prime}_{0}}(|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|)=2t^{\prime}$.

It is easy to see that $(iii)$ follows from $(i)$, $(ii)$ and the assumption $|V_{1}|=|V_{2}|+1$.

We now give a necessary and sufficient condition for a complete multipartite graph with odd order to admit a good bisection. Let $G=K_{r_{1},\dots,r_{k}}$ with partition sets $X_{i}$, $i\in[k]$, such that $|X_{i}|=r_{i}$. Let ${\cal X}=\{X_{i}:i\in[k]\}$, ${\cal S}_{1}=\{X_{i}:i\in[k]\mbox{ and }|X_{i}|\equiv 1\mod 2\}$ and ${\cal S}_{0}=\{X_{i}:i\in[k]\mbox{ and }|X_{i}|\equiv 0\mod 2\}$. For any ${\cal A}\subseteq{\cal X}$, let $s({\cal A})=\sum_{X_{i}\in{\cal A}}|X_{i}|$. We say that ${\cal A}$ is good if there exists ${\cal A}^{\prime}\subseteq{\cal A}$ such that

[TABLE]

where $m=|{\cal S}_{1}\setminus{\cal A}|$ and $n$ is a nonnegative integer with $n\leq|{\cal S}_{0}\setminus{\cal A}|$.

Proposition 3.3**.**

Let $G=K_{r_{1},\dots,r_{k}}$ with partition sets $X_{1},\ldots,X_{k}$, and assume $|V(G)|$ is odd. Let ${\cal X}=\{X_{i}:i\in[k]\}$. Then $G$ has a good bisection if and only if ${\cal X}$ has a good subset.

Proof.

First, we prove that if $G$ has a good bisection, then ${\cal X}$ has a good subset. Let $H$ be a good bisection of $G$ and let $V_{1},V_{2}$ be the corresponding partition sets of $H$. Since $|V(G)|$ is odd, we may assume that $|V_{1}|=|V_{2}|+1$. Let ${\cal S}=\{X_{i}:i\in[k]\mbox{ and }X_{i}\mbox{ crosses }H\}$, {\cal X}_{1}=\{X_{i}:i\in[k],X_{i}\mbox{ crosses H, and }|X_{i}|\equiv 1\mod 2\}, {\cal X}^{\prime}_{0}=\{X_{i}:i\in[k],X_{i}\mbox{ crosses H, }|X_{i}|\equiv 0\mod 2,\mbox{ and }||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||=2\}, and {\cal X}_{0}^{\prime\prime}=\{X_{i}:i\in[k],X_{i}\mbox{ crosses H, }|X_{i}|\equiv 0\mod 2,\mbox{ and }|X_{i}\cap V_{1}|=|X_{i}\cap V_{2}|\}.

By Lemma 3.1 $(ii)$ and the assumption $|V_{1}|=|V_{2}|+1$, for every $X_{i}\in{\cal X}$ crossing $H$ and $|X_{i}|$ even, $||X_{i}\cap V_{1}|-|X_{i}\cap V_{2}||=2$ or $|X_{i}\cap V_{1}|=|X_{i}\cap V_{2}|$. Hence, ${\cal S}={\cal X}_{1}\cup{\cal X}_{0}^{\prime}\cup{\cal X}_{0}^{\prime\prime}$. Let ${\cal A}={\cal X}\setminus{\cal S}$.

Moreover, let $W_{1}=\bigcup_{X_{i}\in{\cal X}_{1}}X_{i}$, $W_{0}=\bigcup_{X_{i}\in{\cal X}^{\prime}_{0}}X_{i}$, $W_{0}^{\prime}=\bigcup_{X_{i}\in{\cal X}^{\prime\prime}_{0}}X_{i}$ and $W_{2}=\bigcup_{X_{i}\in{\cal A}}X_{i}$. Then $V(H)=W_{1}\cup W_{0}\cup W_{0}^{\prime}\cup W_{2}$. By the definition of ${\cal X}^{\prime\prime}_{0}$, $|W_{0}^{\prime}\cap V_{1}|=|W_{0}^{\prime}\cap V_{2}|$. Let $|{\cal X}_{1}|=t$ and $|{\cal X}^{\prime}_{0}|=t^{\prime}$; then by Lemma 3.2 $(iii)$, $|(W_{0}^{\prime}\cup W_{2})\cap V_{2}|-|(W_{0}^{\prime}\cup W_{2})\cap V_{1}|=t+2t^{\prime}-1$. Combining these two equalities, we get

[TABLE]

Since $X_{i}$ does not cross $H$ for any $X_{i}\subseteq{\cal A}$, there exists ${\cal A}^{\prime}\subseteq{\cal A}$ such that $W_{2}\cap V_{2}=\bigcup_{X_{i}\in{\cal A^{\prime}}}X_{i}$ and $W_{2}\cap V_{1}=\bigcup_{X_{i}\in{\cal A}\setminus{\cal A}^{\prime}}X_{i}$. Now, $|W_{2}\cap V_{2}|=s({\cal A}^{\prime})$ and $|W_{2}\cap V_{1}|=s({\cal A})-s({\cal A}^{\prime})$. Thus $s({\cal A}^{\prime})=s({\cal A})/2+(t+2t^{\prime}-1)/2$ by (6). Note that $t=|{\cal X}_{1}|=|{\cal S}_{1}\setminus{\cal A}|$ and $t^{\prime}=|{\cal X}^{\prime}_{0}|\leq|{\cal X}_{0}^{\prime}\cup{\cal X}_{0}^{\prime\prime}|=|{\cal S}_{0}\setminus{\cal A}|$. So ${\cal A}$ is a good subset of ${\cal X}$.

Now, we prove that if ${\cal X}$ has a good subset, then $G$ has a good bisection. Let ${\cal A}$ be a good subset of ${\cal X}$. Then there exists ${\cal A}^{\prime}\subseteq{\cal A}$ such that $s({\cal A}^{\prime})=s({\cal A})/2+(m+2n-1)/2$, where $m=|{\cal S}_{1}\setminus{\cal A}|$ and $n\leq|{\cal S}_{0}\setminus{\cal A}|$. Let ${\cal S}_{0}^{\prime}\subseteq{\cal S}_{0}\setminus{\cal A}$ with $|{\cal S}_{0}^{\prime}|=n$, and let ${\cal S}_{0}^{\prime\prime}=({\cal S}_{0}\setminus{\cal A})\setminus{\cal S}_{0}^{\prime}$. We partition $V(G)$ into $V_{1}$ and $V_{2}$ such that

• •

$|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|=1$ if $X_{i}\in{\cal S}_{1}\setminus{\cal A}$,

• •

$|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|=2$ if $X_{i}\in{\cal S}_{0}^{\prime}$,

• •

$|X_{i}\cap V_{1}|-|X_{i}\cap V_{2}|=0$ if $X_{i}\subseteq{\cal S}_{0}^{\prime\prime}$, and

• •

$X_{i}\subseteq V_{1}$ if $X_{i}\in{\cal A}\setminus{\cal A}^{\prime}$, and $X_{i}\subseteq V_{2}$ if $X_{i}\in{\cal A}^{\prime}$.

Then

[TABLE]

Let $H$ be the bisection of $G$ with partition sets $V_{1}$ and $V_{2}$ and edge set $E(H)=\{uv\in E(G):u\in V_{1}\mbox{ and }v\in V_{2}\}$. Next, we show that $H$ is a good bisection of $G$. Note that, for each $v\in X_{i}\subseteq V(G)$, $d_{G}(v)=|\overline{X_{i}}|$, $d_{H}(v)=|\overline{X_{i}}\cap V_{1}|$ if $v\in V_{2}$, and $d_{H}(v)=|\overline{X_{i}}\cap V_{2}|$ if $v\in V_{1}$. Also note that

[TABLE]

If $v\in X_{i}$ for some $X_{i}\in{\cal S}_{1}\setminus{\cal A}$, then $|\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}|=1-1=0$; so $d_{H}(v)=d_{G}(v)/2$.

If $v\in X_{i}$ for some $X_{i}\in{\cal S}_{0}^{\prime}$ then $|\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}|=1-2=-1$; so $d_{H}(v)\geq(d_{G}(v)-1)/2$.

If $v\in X_{i}$ for some $X_{i}\in{\cal S}_{0}^{\prime\prime}$ then $|\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}|=1-0=1$; so $d_{H}(v)\geq(d_{G}(v)-1)/2$.

If $v\in X_{i}\cap V_{2}$ for some $X_{i}\in{\cal A}$ then $|\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}|=1+|X_{i}|>0$; so $d_{H}(v)=|\overline{X_{i}}\cap V_{1}|\geq d_{G}(v)/2$.

Finally, suppose $v\in X_{i}\cap V_{1}$ for some $X_{i}\in{\cal A}$. Then $|\overline{X_{i}}\cap V_{1}|-|\overline{X_{i}}\cap V_{2}|=1-|X_{i}|$, i.e., $|\overline{X_{i}}\cap V_{2}|-|\overline{X_{i}}\cap V_{1}|=|X_{i}|-1\geq 0$. This implies that $d_{H}(v)=|\overline{X_{i}}\cap V_{2}|\geq d_{G}(v)/2$.

Proof of Proposition 1.2.

Let $G=K_{r_{1},r_{2},r_{3}}$ and ${\cal X}=\{X_{1},X_{2},X_{3}\}$ such that $X_{1},X_{2},X_{3}$ are the partition sets of $G$ and $|X_{i}|=r_{i}$ for $i\in[3]$. Let ${\cal S}_{0}=\{X_{i}:i\in[3]\mbox{ and }|X_{i}|\equiv 0\mod 2\}$ and ${\cal S}_{1}=\{X_{i}:i\in[3]\mbox{ and }|X_{i}|\equiv 1\mod 2\}$. Then ${\cal S}_{0}=\emptyset$ and ${\cal S}_{1}={\cal X}$.

If ${\cal X}$ has no good subset then the assertion follows from Proposition 3.3. So assume that ${\cal A}$ is a good subset of ${\cal X}$ with ${\cal A}^{\prime}\subseteq{\cal A}$ such that $s({\cal A}^{\prime})=s({\cal A})/2+(m+2n-1)/2$, where $m=|{\cal S}_{1}\setminus{\cal A}|=3-|{\cal A}|$ and $n\leq|{\cal S}_{0}\setminus{\cal A}|=0$. So

[TABLE]

It is easy to see that ${\cal A}\neq\emptyset$. Since $r_{i}\geq 3\geq|{\cal A}|$, ${\cal A}^{\prime}\neq\emptyset$ and, hence, $|{\cal A}|\neq 1$. Since $r_{1},r_{2},r_{3}$ are all distinct, $|{\cal A}|\neq 2$. So $|{\cal A}|=3$. Now a straightforward analysis shows that for some $i\in[3]$, $r_{i}\in\{\lfloor(r_{1}+r_{2}+r_{3})/2\rfloor,\lceil(r_{1}+r_{2}+r_{3})/2\rceil\}$. This is a contradiction.

Proposition 3.3 characterizes those complete multipartite graphs which do not have a good bisection. The next result says that there are more such examples.

Proposition 3.4**.**

Let $G=K_{r_{1},\dots,r_{k}}$ where $r_{i}\geq 7$ for every $i\in[k]$. Suppose $G$ does not have a good bisection. Then for any edge $e\in E(G)$, $G-e$ does not have a good bisection.

Proof.

Assume, to the contrary, that $G^{\prime}=G-e$ has a good bisection $H^{\prime}$ with partition sets $V_{1},V_{2}$. We may assume that $E(H^{\prime})=\{xy\in E(G^{\prime}):x\in V_{1}\mbox{ and }y\in V_{2}\}$. Then for every vertex $v\in V(G^{\prime})$, $d_{H^{\prime}}(v)\geq(d_{G^{\prime}}(v)-1)/2$. Let $H$ be the bisection of $G$ with partition sets $V_{1}$ and $V_{2}$ such that $E(H)=\{xy\in E(G):x\in V_{1}\mbox{ and }y\in V_{2}\}$. Let $e=uw$.

Then $d_{G}(v)=d_{G^{\prime}}(v)$ for all $v\in V(G)\setminus\{u,w\}$, and $d_{G}(v)=d_{G^{\prime}}(v)+1$ for $v\in\{u,w\}$. Also, we have $d_{H}(v)=d_{H^{\prime}}(v)$ for all $v\in V(G)\setminus\{u,w\}$, and $d_{H^{\prime}}(v)\leq d_{H}(v)\leq d_{H^{\prime}}(v)+1$ for $v\in\{u,w\}$.

Since $H$ is not a good bisection of $G$, there exists a vertex $v\in V(G)$ such that $d_{H}(v)<(d_{G}(v)-1)/2$. So we have

[TABLE]

which implies that $d_{G^{\prime}}(v)$ is odd (since $d_{H}(v)$ is an integer), $d_{G}(v)=d_{G^{\prime}}(v)+1$, and $(d_{G^{\prime}}(v)-1)/2=d_{H^{\prime}}(v)=d_{H}(v)$. Since $d_{G}(v)=d_{G^{\prime}}(v)+1$, $v\in\{u,w\}$.

Assume, without loss of generality, that $v=u\in X_{i}\cap V_{1}$, where $X_{1},\ldots,X_{k}$ are the partition sets of $G$. (Then $w\notin X_{i}$ as $uw\in E(G)$.) So $d_{G^{\prime}}(u)=|\overline{X_{i}}|-1$, $|\overline{X_{i}}|$ is even, and $w\in V_{1}$. Thus, $|V_{2}\cap\overline{X_{i}}|=d_{H^{\prime}}(u)=(d_{G^{\prime}}(u)-1)/2=|\overline{X_{i}}|/2-1.$ Therefore, $|V_{1}\cap\overline{X_{i}}|=|\overline{X_{i}}|-|V_{2}\cap\overline{X_{i}}|=|\overline{X_{i}}|/2+1.$ So $|V_{1}\cap\overline{X_{i}}|-|V_{2}\cap\overline{X_{i}}|=2$. Because $||V_{1}|-|V_{2}||=1$, $||V_{1}\cap X_{i}|-|V_{2}\cap X_{i}||\leq 3$. Since $|X_{i}|\geq 7$, $|V_{1}\cap X_{i}|\geq 2$. Therefore, there exists a vertex $v_{1}\in V_{1}\cap X_{i}$ such that $v_{1}\not=u$. Also, $v_{1}\neq w$ since $w\notin X_{i}$. Then $d_{G^{\prime}}(v_{1})=d_{G}(v_{1})=|\overline{X_{i}}|$ is even. Thus $d_{H^{\prime}}(v_{1})=|V_{2}\cap\overline{X_{i}}|=|\overline{X_{i}}|/2-1<(d_{G^{\prime}}(v_{1})-1)/2$. This contradicts the assumption that $H^{\prime}$ is a good bisection of $G^{\prime}$.

4 Scott’s questions on bipartitions

In this section, we address two questions of Scott [13] on bipartitions of graphs. First, we prove Theorem 1.4 on $\ell_{\lambda}$-norm of bipartitions (with $\lambda\geq 1$), for which we need a result of Bollobás and Scott [3] on judicious bipartitions. Recall the definition of $t(m)=\sqrt{m/2+1/16}-1/4$.

Lemma 4.1** (Bollobás and Scott).**

Let $G$ be a graph with $m$ edges. Then there exists a bipartition $V(G)=V_{1}\cup V_{2}$ such that $e(V_{1},V_{2})\geq m/2+t(m)/2$ and $\max\{e(V_{1}),e(V_{2})\}\leq m/4+t(m)/4$. Moreover, if for every such bipartition $V(G)=V_{1}\cup V_{2}$ it always holds that $\max\{e(V_{1}),e(V_{2})\}=m/4+t(m)/4$, then $G$ must be a complete graph of odd order.

Proof of Theorem 1.4..

Let $\lambda\geq 1$ and let $G$ be a graph with $m$ edges. By Lemma 4.1, there is a bipartition $V_{1},V_{2}$ of $V(G)$ such that $e(V_{1},V_{2})\geq m/2+t(m)/2$ and, for $i\in[2]$, $e(V_{i})\leq m/4+t(m)/4=\binom{t(m)+1}{2}$. Note that

[TABLE]

Without loss of generality, we assume that $e(V_{1})\geq e(V_{2})$. Then $e(V_{2})\leq t(m)^{2}/2$.

We claim that $e(V_{1})^{\lambda}+e(V_{2})^{\lambda}\leq{t(m)\choose 2}^{\lambda}+{t(m)+1\choose 2}^{\lambda}.$ This is true if $e(V_{2})\leq\binom{t(m)}{2}$. So we may assume $\binom{t(m)}{2}\leq e(V_{2})\leq t(m)^{2}/2$. For $\lambda\geq 1$, the function $f(x)=(t(m)^{2}-x)^{\lambda}+x^{\lambda}$ is strictly decreasing when $\binom{t(m)}{2}\leq x\leq t(m)^{2}/2.$ Therefore,

[TABLE]

Now assume that for every bipartition $V_{1},V_{2}$ of $V(G)$, we have $e(V_{1})^{\lambda}+e(V_{2})^{\lambda}={t(m)\choose 2}^{\lambda}+{t(m)+1\choose 2}^{\lambda}.$ Then it follows from the above arguments, $e(V_{2})=\binom{t(m)}{2}$ and $e(V_{1})+e(V_{2})=t(m)^{2}$. So $\max\{e(V_{1}),e(V_{2})\}=e(V_{1})=t(m)^{2}-\binom{t(m)}{2}=m/4+t(m)/4$. By Lemma 4.1, $G$ is a complete graph of odd order.

Remark. From the above proof, we see that actually $V(G)$ has a bipartition $V_{1},V_{2}$ such that $e(V_{1})^{\lambda}+e(V_{2})^{\lambda}\leq{t(m)\choose 2}^{\lambda}+{t(m)+1\choose 2}^{\lambda}$ for all $\lambda\geq 1$.

To extend Theorem 1.4 to $k$-partitions for $k\geq 3$, we need the following result of Xu and Yu [14] on $k$-partitions.

Lemma 4.2** (Xu and Yu).**

Let $G$ be a graph with $m$ edges and let $k\geq 3$ be a positive integer. Then there exists a $k$-partition $V(G)=V_{1}\cup...\cup V_{k}$ such that

[TABLE]

and for $i\in[k]$,

[TABLE]

We now determine the $\ell_{\lambda}$-norm (where $\lambda\geq 1$) for $k$-partitions, up to an additive term $O(m^{\lambda-1})$. The proof is similar to the bipartition case.

Theorem 4.3**.**

Let $k\geq 3$ be an integer and $\lambda\geq 1$ be a real number. Then any graph $G$ with $m$ edges has a $k$-partition $V(G)=V_{1}\cup...\cup V_{k}$ such that

[TABLE]

Proof.

Let $G$ be a graph with $m$ edges. By Lemma 4.2, there exists a $k$-partition $V(G)=V_{1}\cup...\cup V_{k}$ such that

[TABLE]

and for $i\in[k]$,

[TABLE]

Without loss of generality, let $e(V_{1})\geq e(V_{2})\geq\ldots\geq e(V_{k})$ and let $\alpha:=e(V_{1})-m/k^{2}$.

If $\alpha\leq-\frac{k-1}{k^{2}}t(m)$ then

[TABLE]

So we may assume that

[TABLE]

Note that we may assume $\sum_{i=1}^{k}e(V_{i})=\frac{m}{k}-\frac{k-1}{k}t(m)+\frac{17k}{8}$. Also note that $\sum_{i=1}^{k}e(V_{i})^{\lambda}$ increases if we replace $e(V_{k})$ by $e(V_{k})-1$ and replaces $e(V_{i})$ by $e(V_{i})+1$, for any $i\in[k-1]$. Therefore, we may further assume that $e(V_{1})=...=e(V_{k-1})=\frac{m}{k^{2}}+\alpha$ and $e(V_{k})=\frac{m}{k^{2}}-\frac{k-1}{k}t(m)+\frac{17k}{8}-(k-1)\alpha$. So by (7), we have

[TABLE]

where the second inequality holds because the expression in the second line is an increasing function of $\alpha$, for $-\frac{k-1}{k^{2}}t(m)\leq\alpha\leq\frac{k-1}{k^{2}}t(m)$.

We remark that the bound in the above theorem is tight up to the term $O(m^{\lambda-1})$, by considering the complete graph $K_{ks}$ which has $m=\binom{ks}{2}$ edges. Thus $s=(2t(m)+1)/k$. The minimum $\sum_{i=1}^{k}e(V_{i})^{\lambda}$ over all $k$-partitions $V_{1},\ldots,V_{k}$ of $V(K_{ks})$ is attained when $|V_{i}|=s$ for $i\in[k]$, and this minimum value equals

[TABLE]

Using $2t(m)^{2}+t(m)=m$ and $t(m)=\Theta(\sqrt{m})$, we see that

[TABLE]

It would be interesting to find the optimal upper bound in Theorem 4.3. We believe that the extremal graphs for $\ell_{\lambda}$-norms of $k$-partitions (where $\lambda\geq 1$) should be the complete graphs $K_{kn+\lfloor k/2\rfloor}$. We formulate the following question. For fixed $\lambda\geq 1$ and integer $k\geq 2$, let $s:=s(m)$ be such that $m=\binom{ks+\lfloor k/2\rfloor}{2}$ and let

[TABLE]

Question 4.4**.**

Fix any real $\lambda\geq 1$ and integer $k\geq 2$. For any positive integer $m$, is it true that

[TABLE]

for all graphs $G$ with $m$ edges, with equality if and only if $m=\binom{ks+\lfloor k/2\rfloor}{2}$ for some integer $s$? Does the equality hold only for $K_{ks+\lfloor k/2\rfloor}$ (modulo some isolated vertices)?

A result of Bollobás and Scott [4] shows that this is true for $\lambda=1$ and any $k$. Theorem 1.4 provides an affirmative answer for the case $k=2$.

We now turn to the following question of Scott [13].

Question 4.5**.**

Does every graph $G$ with $\binom{kn}{2}$ edges have a vertex partition into $k$ sets, each of which contains at most $\binom{n}{2}$ edges?

We give a negative answer to this question in the case $k=2$. For this we need to show that there exist an infinite sequence of pairs of integers with certain properties.

Lemma 4.6**.**

There are pairs $(a_{i},b_{i})$ of integers for all $i\geq 0$ such that

• $(i)$

$a_{i}\geq 36$* and $a_{i}$ is even, and $b_{i}\geq 21$ and $b_{i}$ is odd,*

• $(ii)$

$3b_{i}(b_{i}-1)=a_{i}(a_{i}-1)$, and

• $(iii)$

$b_{i}\leq 7a_{i}/12$.

Proof.

We recursively define integer pairs $(n_{i},t_{i})$ as follows, such that the desired sequence $\{(a_{i},b_{i})\}_{i\geq 0}$ will be a subsequence of $\{(n_{i},t_{i})\}_{i\geq 0}$.

Let

[TABLE]

and, for $i\geq 1$, let

[TABLE]

For convenience, we write

[TABLE]

for $i\geq 0$, and

[TABLE]

for $i\geq 1$.

We claim that $\alpha_{i}=0$ for $i\geq 0$ and that $\beta_{i}=0$ for $i\geq 1$. By a direct calculation, we see that $\alpha_{0}=0$, $\alpha_{1}=0$ and $\beta_{1}=0$. Now assume for some $i\geq 1$, we have $\alpha_{j}=0$ for $j\in[i]\cup\{0\}$, and $\beta_{j}=0$ for $j\in[i]$. Using (8) and the definition of $\alpha_{i+1}$ and $\beta_{i+1}$, we have

[TABLE]

Thus, the claim follows from induction.

From (8), we see that both $\{n_{i}\}_{i\geq 0}$ and $\{t_{i}\}_{i\geq 0}$ are increasing sequences; so $n_{i}\geq 36$ and $t_{i}\geq 21$ for $i\geq 0$. Moreover, $t_{i}\leq 7n_{i}/12$ for $i\geq 0$. For otherwise, $t_{i}>7n_{i}/12$ for some $i$. Then $i\geq 1$ and

[TABLE]

which is larger than $n_{i}(n_{i}-1)$ (since $n_{i}\geq 36$), a contradiction.

Using (8), it is easy to observe that $n_{i}$ is even if and only if $i\equiv 0,3\mod 4$, and that $t_{i}$ is odd if and only if $i\equiv 0,1\mod 4$. Therefore letting $a_{i}=n_{4i}$ and $b_{i}=t_{4i}$ for $i\geq 0$, we see that the pairs $(a_{i},b_{i})$ satisfy all requirements $(i)$, $(ii)$ and $(iii)$.

We now give the proof of Theorem 1.5.

Proof of Theorem 1.5..

By Lemma 4.6, there exist infinitely many pairs $(2n,t)$ of positive integers such that $t$ is odd, $t\leq 7n/6$, and $3t(t-1)=2n(2n-1)$.

Let $G$ be the union of three pairwise disjoint copies of the clique $K_{t}$. Then $|V(G)|=3t$ and

[TABLE]

Let $V_{1},V_{2}$ be a bipartition of $V(G)$. Without loss of generality, we assume that $|V_{1}|\geq|V_{2}|$.

Then $G[V_{1}]$ is the disjoint union of three cliques, say $K_{a},K_{b},$ and $K_{c}$. (We set $K_{0}=\emptyset$.) Hence, $G[V_{2}]$ is the disjoint union of cliques $K_{t-a},K_{t-b}$ and $K_{t-c}$. As $t$ is odd, we have

[TABLE]

Choose integers $a^{\prime},b^{\prime},c^{\prime}$ such that $a^{\prime}\leq a,b^{\prime}\leq b,c^{\prime}\leq c$ and

[TABLE]

We also need an easy property of binomial coefficients that for any integers $m-n\geq 2$,

[TABLE]

Then we have

[TABLE]

This completes the proof of Theorem 1.5.

It seems likely that similar result holds for general $k$-partitions, though we are not able to construct such graphs due to difficulties in proving a more general version of Lemma 4.6.