# Solutions to problems about potentially Ks,t-bigraphic pair

• Jian-Hua Yin and Liang Zhang
From the journal Open Mathematics

## Abstract

Let S = ( a 1 , , a m ; b 1 , , b n ) , where a 1 , , a m and b 1 , , b n are two nonincreasing sequences of nonnegative integers. The pair S = ( a 1 , , a m ; b 1 , , b n ) is said to be a bigraphic pair if there is a simple bipartite graph G = ( X Y , E ) such that a 1 , , a m and b 1 , , b n are the degrees of the vertices in X and Y , respectively. In this case, G is referred to as a realization of S . Given a bigraphic pair S , and a complete bipartite graph K s , t , we say that S is a potentially K s , t -bigraphic pair if some realization of S contains K s , t as a subgraph (with s vertices in the part of size m and t in the part of size n ). Ferrara et al. (Potentially H-bigraphic sequences, Discuss. Math. Graph Theory 29 (2009), 583–596) defined σ ( K s , t , m , n ) to be the minimum integer k such that every bigraphic pair S = ( a 1 , , a m ; b 1 , , b n ) with σ ( S ) = a 1 + + a m k is a potentially K s , t -bigraphic pair. This problem can be viewed as a “potential” degree sequence relaxation of the (forcible) Turán problem. Ferrara et al. determined σ ( K s , t , m , n ) for n m 9 s 4 t 4 . In this paper, we further determine σ ( K s , t , m , n ) for n m s and n + m 2 t 2 + t + s . As two corollaries, if n m t 2 + t + s 2 or if n m s and n 2 t 2 + t , the values σ ( K s , t , m , n ) are determined completely. These results give a solution to a problem due to Ferrara et al. and a solution to a problem due to Yin and Wang.

MSC 2010: 05C07; 05C35

## 1 Introduction

The study of vertex degrees in graphs has a long history, often asking when an n -tuple of nonnegative integers is realizable as the vertex degrees of a simple n -vertex graph with specified properties. Analogous problems are also studied for bipartite graphs. Let S = ( a 1 , , a m ; b 1 , , b n ) , where a 1 , , a m and b 1 , , b n are two sequences of nonnegative integers with a 1 a m and b 1 b n . We say that S is a bigraphic pair if there is a simple bipartite graph G with partite sets { x 1 , , x m } and { y 1 , , y n } such that the degree of x i is a i and the degree of y j is b j . In this case, we say that G is a realization of S . Two methods to determine if S is a bigraphic pair are the Gale-Ryser criteria [1,2] and the Havel-Hakimi-type algorithm [3].

## Theorem 1.1

[1,2] S is a bigraphic pair if and only if i = 1 m a i = i = 1 n b i and i = 1 k a i i = 1 n min { k , b i } for k = 1 , , m (or i = 1 k b i i = 1 m min { k , a i } for k = 1 , , n ).

For 1 p m and 1 q n , let S ( a p ) = ( a 1 , , a p 1 , a p + 1 , , a m ; b 1 , , b n ) and S ( b q ) = ( a 1 , , a m ; b 1 , , b q 1 , b q + 1 , , b n ) , where b 1 b n is a rearrangement in nonincreasing order of b 1 1 , , b a p 1 , b a p + 1 , , b n and a 1 a m is a rearrangement in nonincreasing order of a 1 1 , , a b q 1 , a b q + 1 , , a m . We say that S ( a p ) (resp. S ( b q ) ) is the residual pair obtained from S by laying off a p (resp. b q ).

## Theorem 1.2

[3] S is a bigraphic pair if and only if S ( a p ) (or S ( b q ) ) is a bigraphic pair.

We can also ask whether there is a realization satisfying a particular property. Let S = ( a 1 , , a m ; b 1 , , b n ) be a bigraphic pair, and let K s , t be the complete bipartite graph with partite sets of size s and t . We say that S is a potentially K s , t -bigraphic pair if some realization of S contains K s , t (with s vertices in the part of size m and t in the part of size n ). If some realization of S contains K s , t on those vertices having degree a 1 , , a s , b 1 , , b t , we say that S is a potentially A s , t -bigraphic pair. Ferrara et al. [4] proved that S is a potentially A s , t -bigraphic pair if and only if it is a potentially K s , t -bigraphic pair. Yin and Wang [5] developed a Havel-Hakimi-type algorithm to determine if S is a potentially K s , t -bigraphic pair. This algorithm can also be used to construct a graph with degree sequence pair S and containing K s , t on those vertices having degree a 1 , , a s , b 1 , , b t .

Let S = ( a 1 , , a m ; b 1 , , b n ) , where a 1 , , a m and b 1 , , b n are two nonincreasing sequences of nonnegative integers. Let 1 s m , 1 t n , a s t and b t s . We first define pairs S 0 , , S s as follows. Let S 0 = S . Let

S 1 = ( a 2 , , a m ; b 1 1 , , b t 1 , b t + 1 ( 1 ) , , b n ( 1 ) ) ,

where b t + 1 ( 1 ) b n ( 1 ) is a rearrangement in nonincreasing order of b t + 1 1 , , b a 1 1 , b a 1 + 1 , , b n . For 2 i s , given S i 1 = ( a i , , a m ; b 1 i + 1 , , b t i + 1 , b t + 1 ( i 1 ) , , b n ( i 1 ) ) , let

S i = ( a i + 1 , , a m ; b 1 i , , b t i , b t + 1 ( i ) , , b n ( i ) ) ,

where b t + 1 ( i ) b n ( i ) is a rearrangement in nonincreasing order of b t + 1 ( i 1 ) 1 , , b a i ( i 1 ) 1 , b a i + 1 ( i 1 ) , , b n ( i 1 ) .

We now define pairs S 0 , , S t as follows. Let S 0 = S . Let

S 1 = ( a 1 1 , , a s 1 , a s + 1 ( 1 ) , , a m ( 1 ) ; b 2 , , b n ) ,

where a s + 1 ( 1 ) a m ( 1 ) is a rearrangement in nonincreasing order of a s + 1 1 , , a b 1 1 , a b 1 + 1 , , a m . For 2 i t , given S i 1 = ( a 1 i + 1 , , a s i + 1 , a s + 1 ( i 1 ) , , a m ( i 1 ) ; b i , , b n ) , let

S i = ( a 1 i , , a s i , a s + 1 ( i ) , , a m ( i ) ; b i + 1 , , b n ) ,

where a s + 1 ( i ) a m ( i ) is a rearrangement in nonincreasing order of a s + 1 ( i 1 ) 1 , , a b i ( i 1 ) 1 , a b i + 1 ( i 1 ) , , a m ( i 1 ) .

## Theorem 1.3

[5] S is a potentially A s , t -bigraphic pair if and only if S s (or S t ) is a bigraphic pair.

Motivated by the problem due to Erdős et al. [6] of finding the minimum integer k such that every realizable n -tuple with a sum of at least k is potentially K r -graphic, Ferrara et al. [4] investigated analogous problem for bipartite graphs. They defined σ ( K s , t , m , n ) to be the minimum integer k such that every bigraphic pair S = ( a 1 , , a m ; b 1 , , b n ) with σ ( S ) = a 1 + + a m k is a potentially K s , t -bigraphic pair. They determined σ ( K s , t , m , n ) when m and n are sufficiently large in terms of s and t . This problem can be viewed as a “potential” degree sequence relaxation of the (forcible) Turán problem.

## Theorem 1.4

[4] If t s 1 and n m 9 s 4 t 4 , then σ ( K s , t , m , n ) = n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 .

Ferrara et al. proposed a problem as follows.

## Problem 1.1

[4] This would be useful if one were interested in finding smaller bounds on the n and m necessary to assure Theorem 1.4.

Yin and Wang proved a new result as follows.

## Theorem 1.5

[5] If t s 1 , n m s and n ( s + 1 ) t 2 ( 2 s 1 ) t + s 1 , then σ ( K s , t , m , n ) = n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 .

Yin and Wang also proposed a problem as follows.

## Problem 1.2

[5] It would be meaningful to investigate a lower bound on n + m necessary to assure Theorem 1.5.

The purpose of this paper is to improve Theorem 1.5 and determine σ ( K s , t , m , n ) for n m s and n + m 2 t 2 + t + s , that is, a solution to Problems 1.2. As two corollaries, if n m t 2 + t + s 2 or if n m s and n 2 t 2 + t , the values σ ( K s , t , m , n ) are determined completely, that is, a solution to Problem 1.1.

## Theorem 1.6

If t s 1 , n m s and n + m 2 t 2 + t + s , then σ ( K s , t , m , n ) = n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 .

## Corollary 1.1

If t s 1 and n m t 2 + t + s 2 , then σ ( K s , t , m , n ) = n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 .

## Corollary 1.2

If t s 1 , n m s and n 2 t 2 + t , then σ ( K s , t , m , n ) = n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 .

## 2 Proof of Theorem 1.6

In order to prove Theorem 1.6, we need some lemmas.

## Lemma 2.1

[7] Theorem 1.1 remains valid if i = 1 k a i i = 1 n min { k , b i } is assumed only for those k for which a k > a k + 1 or k = m (or i = 1 k b i i = 1 m min { k , a i } is assumed only for those k for which b k > b k + 1 or k = n ).

## Lemma 2.2

[5] Let S = ( a 1 , , a m ; b 1 , , b n ) be a bigraphic pair with a s t , b t s , m 1 b 1 b t = = b a 1 + 1 b a 1 + 2 b n and n 1 a 1 a s = = a b 1 + 1 a b 1 + 2 a m . For each S i = ( a i + 1 , , a m ; b 1 i , , b t i , b t + 1 ( i ) , , b n ( i ) ) with 0 i s , let t i = max { j b t + 1 ( i ) b t + j ( i ) 1 } . Then

1. t s t s 1 t 0 a 1 + 1 t .

2. For each i with 1 i s , we have b t + k ( i ) = b t + k ( i 1 ) for k > t i . Consequently, b t + k ( s ) = b t + k for k > t s .

## Lemma 2.3

Let S = ( a 1 , , a m ; b 1 , , b n ) be a bigraphic pair with a s t , b t s , m 1 b 1 b t = = b a 1 + 1 b a 1 + 2 b n and n 1 a 1 a s = = a b 1 + 1 a b 1 + 2 a m . If i = 1 t ( b i b t ) + i = a s + 1 + 1 n b i t s , then S is a potentially A s , t -bigraphic pair.

## Proof

It is trivial for s = 1 . Assume s 2 . By Theorem 1.3, we only need to check that S s = ( a s + 1 , , a m ; b 1 s , , b t s , b t + 1 ( s ) , , b n ( s ) ) is a bigraphic pair. Clearly, a s + 1 + + a m = ( b 1 s ) + + ( b t s ) + b t + 1 ( s ) + + b n ( s ) . Denote = b t and p = max { i a s + i = a s } . Then s + p b 1 + 1 , i.e., p b 1 + 1 s . By Lemma 2.1, it is enough to check that i = 1 k a s + i i = 1 t min { k , b i s } + i = t + 1 n min { k , b i ( s ) } for p k m s . Denote x = b t + 1 ( s ) . By b t + 1 ( s ) b t + 1 = , we have x . If k x , by k p b 1 + 1 s > b i s for 1 i t , then i = 1 t min { k , b i s } + i = t + 1 n min { k , b i ( s ) } = i = 1 t ( b i s ) + i = t + 1 n b i ( s ) = a s + 1 + + a m i = 1 k a s + i . Assume p k x 1 . If t s a s + 1 , by b a s + 1 + t b t + t s x 1 k , then i = 1 t min { k , b i s } + i = t + 1 n min { k , b i ( s ) } i = t + 1 a s + 1 + t min { k , b i ( s ) } = k a s + 1 i = 1 k a s + i . Assume t s < a s + 1 . Then by Lemma 2.2, b t + j ( s ) = b t + j for j a s + 1 . If k b a s + 1 + t , then i = 1 t min { k , b i s } + i = t + 1 n min { k , b i ( s ) } i = t + 1 a s + 1 + t min { k , b i ( s ) } = k a s + 1 i = 1 k a s + i . Assume k > b a s + 1 + t . For each i with a s + 1 + 1 i t + t s , we have min { k , b i ( s ) } = k = ( k ) b i ( k ) . Also, for each i with t + t s + 1 i a s + 1 + t , by Lemma 2.2, we have min { k , b i ( s ) } = min { k , b i } = min { ( k ) , b i } min { b i ( k ) , b i } = b i ( k ) . Therefore, i = 1 t min { k , b i s } + i = t + 1 n min { k , b i ( s ) } = i = 1 t ( b i s ) + i = t + 1 a s + 1 min { k , b i ( s ) } + i = a s + 1 + 1 a s + 1 + t min { k , b i ( s ) } + i = a s + 1 + t + 1 n min { k , b i ( s ) } i = 1 t ( ( b i ) + ( s ) ) + k ( a s + 1 t ) + i = a s + 1 + 1 a s + 1 + t ( b i ( k ) ) + i = a s + 1 + t + 1 n b i = i = 1 t ( b i ) + i = a s + 1 + 1 n b i + ( s ) t + k ( a s + 1 t ) ( k ) t t s + ( s ) t + k ( a s + 1 t ) ( k ) t = k a s + 1 i = 1 k a s + i .□

## Lemma 2.4

Let S = ( a 1 , , a m ; b 1 , , b n ) be a bigraphic pair with a s t , b t s , m 1 b 1 b t = = b a 1 + 1 b a 1 + 2 b n and n 1 a 1 a s = = a b 1 + 1 a b 1 + 2 a m . If i = 1 s ( a i a s ) + i = b t + 1 + 1 m a i t s , then S is a potentially A s , t -bigraphic pair.

## Proof

By the symmetry, the proof of Lemma 2.4 is similar to that of Lemma 2.3.□

## Lemma 2.5

[4] Suppose that S = ( a 1 , , a m ; b 1 , , b n ) is not a potentially A s , t -bigraphic pair. Let G be a realization of S with partite sets X and Y , with X = m and Y = n . Let X s be the set of s highest degree vertices of X , and Y t be the set of t highest degree vertices of Y . Assume that G is a realization of S that maximizes the number of edges between X s and Y t . Let x and y be nonadjacent members of X s and Y t , and let A = N G ( x ) Y t and B = N G ( y ) X s . Then both A and B contain at most ( s 1 ) ( t 1 ) vertices.

## Lemma 2.6

[7] If S = ( a 1 , , a m ; b 1 , , b n ) is a bigraphic pair with a s 2 t 1 and b t 2 s 1 , then S is a potentially A s , t -bigraphic pair.

## Lemma 2.7

Let S = ( a 1 , , a m ; b 1 , , b n ) be a bigraphic pair with m 1 b 1 b t = = b a 1 + 1 b a 1 + 2 b n and n 1 a 1 a s = = a b 1 + 1 a b 1 + 2 a m . If n ( s 1 ) + m ( t 1 ) max { 2 s t 2 2 t 2 + t s , 2 t s 2 2 s 2 + s t } and σ ( S ) n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 , then S is a potentially A s , t -bigraphic pair.

## Proof

By σ ( S ) n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 , it is straightforward to show that a s t and b t s . On the contrary, we assume that S is not a potentially A s , t -bigraphic pair. Let G be a realization of S with partite sets X and Y , with X = m and Y = n . Let X s be the set of s highest degree vertices of X , and Y t be the set of t highest degree vertices of Y . Assume that G is a realization of S that maximizes the number of edges between X s and Y t . Let x and y be nonadjacent members of X s and Y t , and let A = N G ( x ) Y t and B = N G ( y ) X s . By Lemma 2.5, both A and B contain at most ( s 1 ) ( t 1 ) vertices. This implies a s d G ( x ) A + Y t 1 ( s 1 ) ( t 1 ) + t 1 = s t s and b t d G ( y ) B + X s 1 ( s 1 ) ( t 1 ) + s 1 = s t t . By Lemma 2.6, we have a s 2 t 2 or b t 2 s 2 , and so we may consider the following two cases.

Case 1. a s 2 t 2 .

It follows from Lemma 2.3 that σ ( S ) = i = 1 t b i + i = t + 1 a s + 1 b i + i = a s + 1 + 1 n b i = i = 1 t ( b i b t + b t ) + i = t + 1 a s + 1 b i +  i = a s + 1 + 1 n b i i = 1 t ( b i b t ) + i = a s + 1 + 1 n b i  +  t b t + ( a s + 1 t ) b t t s 1 + a s + 1 b t t s 1 + ( 2 t 2 ) ( s t t ) < ( 2 s t 2 2 t 2 + t s ) ( t 1 ) ( s 1 ) + 1 n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 , a contradiction.

Case 2. b t 2 s 2 .

It follows from Lemma 2.4 that σ ( S ) = i = 1 s a i + i = s + 1 b t + 1 a i + i = b t + 1 + 1 m a i = i = 1 s ( a i a s + a s ) + i = s + 1 b t + 1 a i + i = b t + 1 + 1 m a i i = 1 s ( a i a s ) + i = b t + 1 + 1 m a i + s a s + ( b t + 1 s ) a s t s 1 + b t + 1 a s t s 1 + ( 2 s 2 ) ( s t s ) < ( 2 t s 2 2 s 2 + s t ) ( t 1 ) ( s 1 ) + 1 n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 , a contradiction.□

## Lemma 2.8

Let S = ( a 1 , , a m ; b 1 , , b n ) be a bigraphic pair. If n + m 2 max { t 2 , s 2 } + t + s and σ ( S ) n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 , then S is a potentially A s , t -bigraphic pair.

## Proof

It is straightforward to show that a s t and b t s . We use induction on s + t . It is trivial for s = 1 or t = 1 . Assume s 2 and t 2 . If a 1 = n or there exists an integer k with t k a 1 such that b k > b k + 1 , then the residual pair S ( a 1 ) = ( a 2 , , a m ; b 1 , , b n ) obtained from S by laying off a 1 satisfies n + ( m 1 ) 2 max { t 2 , s 2 } + t + ( s 1 ) 2 max { t 2 , ( s 1 ) 2 } + t + ( s 1 ) , σ ( S ( a 1 ) ) = σ ( S ) a 1 n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 n = n ( s 2 ) + ( m 1 ) ( t 1 ) ( t 1 ) ( s 2 ) + 1 and b 1 = b 1 1 , , b t = b t 1 . By Theorem 1.2 and the induction hypothesis, S ( a 1 ) is a potentially A s 1 , t -bigraphic pair, and hence S is a potentially A s , t -bigraphic pair. So we may assume a 1 n 1 and b 1 b t = = b a 1 + 1 b a 1 + 2 b n . If b 1 = m or there exists an integer k with s k b 1 such that a k > a k + 1 , then the residual pair S ( b 1 ) = ( a 1 , , a m ; b 2 , , b n ) obtained from S by laying off b 1 satisfies ( n 1 ) + m 2 max { t 2 , s 2 } + ( t 1 ) + s 2 max { ( t 1 ) 2 , s 2 } + ( t 1 ) + s , σ ( S ( b 1 ) ) = σ ( S ) b 1 n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 m = ( n 1 ) ( s 1 ) + m ( t 2 ) ( t 2 ) ( s 1 ) + 1 and a 1 = a 1 1 , , a s = a s 1 . By Theorem 1.2 and the induction hypothesis, S ( b 1 ) is a potentially A s , t 1 -bigraphic pair, and hence S is a potentially A s , t -bigraphic pair. So we may further assume b 1 m 1 and a 1 a s = = a b 1 + 1 a b 1 + 2 a m . If s t , then 2 s t 2 2 t 2 + t s 2 t s 2 2 s 2 + s t and n ( s 1 ) + m ( t 1 ) ( n + m ) ( s 1 ) ( 2 t 2 + t + s ) ( s 1 ) = 2 s t 2 2 t 2 + ( t + s ) ( s 1 ) 2 s t 2 2 t 2 + t + s , implying that n ( s 1 ) + m ( t 1 ) max { 2 s t 2 2 t 2 + t s , 2 t s 2 2 s 2 + s t } . Similarly, if t s , then 2 t s 2 2 s 2 + s t 2 s t 2 2 t 2 + t s and n ( s 1 ) + m ( t 1 ) ( n + m ) ( t 1 ) ( 2 s 2 + t + s ) ( t 1 ) = 2 t s 2 2 s 2 + ( t + s ) ( t 1 ) 2 t s 2 2 s 2 + t + s , implying that n ( s 1 ) + m ( t 1 ) max { 2 s t 2 2 t 2 + t s , 2 t s 2 2 s 2 + s t } . Thus by Lemma 2.7, S is a potentially A s , t -bigraphic pair.□

## Proof of Theorem 1.6

To show the lower bound, Ferrara et al. [4] considered the bigraphic pair S = ( n s 1 , ( t 1 ) m s + 1 ; m s 1 , ( t 1 ) m s + 1 , ( s 1 ) n m ) , where the symbol x y stands for y consecutive terms, each equal to x . Clearly, neither partite set in any realization of S has s vertices of degree t . Hence, S is not a potentially K s , t -bigraphic pair. Thus, σ ( K s , t , m , n ) σ ( S ) + 1 = n ( s 1 ) + m ( t 1 ) ( t 1 ) ( s 1 ) + 1 . The upper bound directly follows from Lemma 2.8.□

## Acknowledgements

The authors would like to thank the referees for their helpful suggestions.

1. Funding information: This research was supported by Hainan Provincial Natural Science Foundation of China (Nos. 122RC545, 2019RC085), National Natural Science Foundation of China (No. 11961019), and Key Laboratory of Engineering Modeling and Statistical Computation of Hainan Province.

2. Conflict of interest: The authors state no conflict of interest.

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