Solutions to problems about potentially K_s,t-bigraphic pair

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 ).

Theorem 1.2

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

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.

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 .

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.

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 .

Problem 1.2

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

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 .

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 ℓ = 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 = ℓ , we have x ≤ ℓ . 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 = ℓ − ( ℓ − k ) ≥ b i − ( ℓ − 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 { ℓ − ( ℓ − k ) , b i } ≥ min { b i − ( ℓ − k ) , b i } = b i − ( ℓ − 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 − ℓ ) + ( ℓ − s ) ) + k ( a s + 1 − t ) + ∑ i = a s + 1 + 1 a s + 1 + t ( b i − ( ℓ − k ) ) + ∑ i = a s + 1 + t + 1 n b i = ∑ i = 1 t ( b i − ℓ ) + ∑ i = a s + 1 + 1 n b i + ( ℓ − s ) t + k ( a s + 1 − t ) − ( ℓ − k ) t ≥ t s + ( ℓ − s ) t + k ( a s + 1 − t ) − ( ℓ − 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.□