On problems about judicious bipartitions of graphs

0.2 MB·

by  Yuliang Ji

On problems about judicious bipartitions of graphs
Yuliang Ji∗ Jie Ma† Juan Yan‡ Xingxing Yu§
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n Abstract: Bollob´as 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 Bollob´as-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 Bollob´as-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
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1 AMS classification: 05C07, 05C70
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∗School of Mathematical Sciences, Universityof Science and Technology of China, Hefei, Anhui230026,
China. Email: [email protected].
†School of Mathematical Sciences, Universityof Science and Technology of China, Hefei, Anhui230026,
China. Email: [email protected]. Partially supported byNSFC projects 11501539 and 11622110.
‡College of Mathematics and Systems Science, Xinjiang University, Urumqi, Xinjiang 830046, China.
Email: [email protected]. Partially supported by NSFCproject 11501486.
§School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332, USA. Email:
[email protected]. 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 Bollob´as
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 Bollob´as 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, Bollob´as 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 H¨aggkvist [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 Bollob´as-Scott con-
jecture. 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 Bollob´as-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 multi-
partite 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
∈
Bolloba´as-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). Bollob´as 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 Bollob´as-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 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 Bollob´as-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