aa r X i v : . [ m a t h . C O ] F e b GROWING BALANCED COVERING SETS
TUNG H. NGUYEN
Abstract.
Given a bipartite graph with bipartition (
A, B ) where B is equipartitioned into k ≥ A be picked one by one so that at every step, the picked vertices cover roughly the samenumber of vertices in each of these blocks? We show that, if each block has cardinality m say, the vertices in B have the same degree, and each vertex in A has at most cm neighbors in every block where c > v , . . . , v n of the vertices in A such that for every j ∈ { , . . . , n } , thenumbers of vertices with a neighbor in { v , . . . , v j } in every two blocks differ by at most p k − c / m .This is related to a well-known lemma of Steinitz, and partially answers an unpublished question of Scottand Seymour. Introduction
For every integer n ≥
1, let [ n ] := { , . . . , n } . The motivation of this note is an unpublished question ofAlex Scott and Paul Seymour [2] on balanced covers of bipartite graphs related to the opening questionin the abstract. Here, bipartite graphs have no multiple edges. Question 1.
For every integer k ≥ , does there exist f ( k ) > with the following property? Let m ≥ bean integer and c ∈ (0 , be a constant independent of m . Consider a bipartite graph with bipartition ( A, B ) where B is partitioned into k ≥ blocks B , . . . , B k each of size m , the vertices in B have the same degree,and each vertex in A has at most cm neighbors in each B i . For every i ∈ [ k ] and X ⊆ A , let N ( X, B i ) be the set of vertices in B i with a neighbor in X . Then, there is a chain of sets ∅ ( A ( . . . ( A n = A where n = | A | such that max i ,i ∈ [ k ] || N ( A j , B i ) | − | N ( A j , B i ) || ≤ f ( k ) cm for all j ∈ [ n ] . Our focus in this note is a more convenient formulation of Question 1. Let us introduce some notation.We denote the set of nonnegative real numbers by R + . For a finite set S and an integer r ≥
1, let S ( r ) bethe family of all subsets of cardinality r of S ; we identify S (1) with S . For integers n ≥ r ≥
1, a weighted r -uniform hypergraph on [ n ] is a function w : [ n ] ( r ) → R + satisfying P R ∈ [ n ] ( r ) w ( R ) = 1. For every S ⊆ [ n ],let w ( S ) := P R ∈ S ( r ) w ( R ). Thus w ( S ) = 0 when | S | < r . We can think of w ( S ) as the “mass” of S .Now, Question 1 can be rephrased as follows. Question 2.
For every integer k ≥ , does there exist f ( k ) > with the following property? Let c ∈ (0 , .Let w , . . . , w k be weighted r -uniform hypergraphs on [ n ] for some n ≥ r ≥ . If w i ([ n ] \ { j } ) ≥ − c foreach i ∈ [ k ] and j ∈ [ n ] , then there is a chain ∅ ( S ( . . . ( S n = [ n ] such that | w i ( S j ) − w i ( S j ) | ≤ f ( k ) c for all i , i ∈ [ k ] and j ∈ [ n ] . To see that Question 2 extends Question 1, we identify A with [ n ], let r be the common degree of B ,and let w i ( R ) := 1 − | N ( A \ R, B i ) | /m for every i ∈ [ k ] and R ∈ A ( r ) ; then w i ( S ) = 1 − | N ( A \ S, B i ) | /m for every i ∈ [ k ] and S ⊆ A = [ n ], in particular w i ([ n ] \ { j } ) = 1 − | N ( j, B i ) | /m ≥ − c for every j ∈ [ n ].To see that Question 1 extends Question 2, observe that it suffices to consider when each p i assumesrational values, in which case there exists m such that m · w i ( R ) ∈ Z for all i ∈ [ k ] and R ∈ [ n ] ( r ) ; then welet A := [ n ], and let each B i have size m and have exactly m · w i ( R ) vertices whose neighborhood is R foreach R ∈ [ n ] ( r ) . The desired chain ∅ ( A ( . . . ( A n = A of Question 1 corresponds to the desired chain ∅ ( S ( . . . ( S n = [ n ] of Question 2 via the relation A j = A \ S n − j = [ n ] \ S n − j for every j ∈ [ n − In Questions 1 and 2, one can easily choose f (2) = 1, but it is still open whether f (3) exists. It wouldalso be helpful to know whether f ( k ) exists when r = 2. When r = 1, one can choose f ( k ) = k , and thisfollows from a well-known lemma of Steinitz (see [1] for its history and related results), which we state here. Theorem 1.
Let B be the unit ball of a given norm in R d where d ≥ . For every finite subset V of B with P v ∈ V v = , there is an ordering v , . . . , v n of the vectors in V for which v + · · · + v j ∈ d B for all j ∈ [ n ] . To see how f ( k ) can be chosen as k when r = 1 in Question 2, put d := k and let γ : [ d ] → [ k ] × [ k ]be a bijection; for every j ∈ [ n ], let v j be the vector in R d whose ℓ -th component is w i ( j ) − w i ( j ) where γ ( ℓ ) = ( i , i ) for each ℓ ∈ [ d ]. We then apply Theorem 1 for the ∞ -norm to V = { v , . . . , v n } , notingthat the ∞ -norm of v j is at most c for every j ∈ [ n ] and w i ( S ) = P j ∈ S w i ( j ) for all i ∈ [ k ] and S ⊆ [ n ].The main result of this note provides a partial answer to Question 2. More precisely, it says that f ( k )can be chosen as p k −
1) if c is replaced by √ c . Theorem 2.
Let k ≥ be an integer, and let c ∈ (0 , . Let w , . . . , w k be weighted r -uniform hypergraphson [ n ] for some n ≥ r ≥ . If w i ([ n ] \{ j } ) ≥ − c for every i ∈ [ k ] and j ∈ [ n ] , then there is a chain of sets ∅ ( S ( . . . ( S n = [ n ] such that | w i ( S j ) − w i ( S j ) | ≤ p k − c for all i , i ∈ [ k ] and j ∈ [ n ] . Theorem 2 does not imply Theorem 1 as far as we know. By the previous discussion on the equivalenceof Questions 1 and 2, Theorem 2 is equivalent to a result that partially answers Question 1; we state ithere for completeness.
Theorem 3.
Let k ≥ , m ≥ be integers, and let c ∈ (0 , be a constant independent of m . Considera bipartite graph with bipartition ( A, B ) where B is partitioned into k blocks B , . . . , B k each of size m ,the vertices in B have the same degree, and each vertex in A has at most cm neighbors in each B i . Forevery i ∈ [ k ] and X ⊆ A , let N ( X, B i ) be the set of vertices in B i with a neighbor in X . Then, there isa chain ∅ ( A ( . . . ( A n = A where n = | A | such that || N ( A j , B i ) | − | N ( A j , B i ) || ≤ p k − c / m for all i , i ∈ [ k ] and j ∈ [ n ] . Proof of Theorem 2
For every i ∈ [ k ], S ⊆ [ n ], and j ∈ S , let δ i ( j, S ) := w i ( S ) − w i ( S \ { j } ) = X R ∈ S ( r ) ,j ∈ R w i ( R ) . Then 0 ≤ δ i ( j, S ) ≤ δ i ( j, [ n ]) ≤ c obviously, and the crucial point is that, for all i ∈ [ k ] and S ⊆ [ n ], X j ∈ S δ i ( j, S ) = r · w i ( S ) . (1)For S ⊆ [ n ], let the unbalance of S be the quantity max i ,i ∈ [ k ] | w i ( S ) − w i ( S ) | . To prove Theorem 2,we shall build the desired chain ∅ ( S ( . . . ( S n = [ n ] in reverse so that S j has unbalance at most p k − c for all j ∈ [ n ]. So, given a nonempty subset S of [ n ] whose unbalance is reasonably small, itmight be helpful to see how we can remove some j ∈ S while maintaining reasonably small unbalance.This is easy when k = 2, because (1) implies that there is some j ∈ S for which δ ( j, S ) − δ ( j, S ) has thesame sign as w ( S ) − w ( S ) has, and thus S \ { j } has unbalance at most c if S has unbalance at most c .(This explains why one can choose f (2) = 1 in Questions 1 and 2.) Still, when k ≥
3, it seems difficult tosimultaneously control | w i ( S ) − w i ( S ) | for all i , i ∈ [ k ]. A trick to overcome this difficulty is to note that( w i ( S ) − w i ( S )) ≤ w i ( S ) − w k ( S )) + ( w i ( S ) − w k ( S )) ] ≤ k − X i =1 ( w i ( S ) − w k ( S )) , (2)which leads us to pay attention to how the more observable k ϕ ( S ) k = k − X i =1 ( w i ( S ) − w k ( S )) changes when we delete an element from S . Here, ϕ ( S ) is the vector in R k − whose i -th component is w i ( S ) − w k ( S ), and k . k is the Euclidean norm in R k − . The following lemma is motivated by this idea,showing that if k ϕ ( S ) k is reasonably small and j is a uniformly random element of S , then k ϕ ( S \ { j } ) k is reasonably small in expectation. Lemma 4. If S is a nonempty subset of [ n ] , then | S | X j ∈ S k ϕ ( S \ { j } ) k ≤ k ϕ ( S ) k − r | S | ( k ϕ ( S ) k − ( k − c ) . Proof.
For every j ∈ S , let x j := ϕ ( S ) − ϕ ( S \ { j } ) = ( δ ( j, S ) − δ k ( j, S ) , . . . , δ k − ( j, S ) − δ k ( j, S )) , then (1) yields X j ∈ S x j = r ( w ( S ) − w k ( S ) , . . . , w k − ( S ) − w k ( S )) = r ϕ ( S ) . Write h ., . i for the standard inner product in R k − . It follows that X j ∈ S k ϕ ( S \ { j } ) k = | S | · k ϕ ( S ) k − X j ∈ S h ϕ ( S ) , x j i + X j ∈ S k x j k = | S | · k ϕ ( S ) k − r k ϕ ( S ) k + X j ∈ S k x j k . To conclude the proof of the lemma, it suffices to show that P j ∈ S k x j k ≤ r ( k − c . To this end, let x ij := δ i ( j, S ) − δ k ( j, S ) for every i ∈ [ k −
1] and j ∈ S . Observe that, for each i ∈ [ k − X j ∈ S x ij ≤ X j ∈ S ( δ i ( j, S ) + δ k ( j, S ) ) ≤ c X j ∈ S ( δ i ( j, S ) + δ k ( j, S )) (1) = rc ( w i ( S ) + w k ( S )) ≤ rc. Therefore P j ∈ S k x j k = P j ∈ S P k − i =1 x ij = P k − i =1 P j ∈ S x ij ≤ r ( k − c, as claimed. (cid:3) The rest of the proof of is now easy. If n ≤ r , we order the elements of [ n ] arbitrarily, obtaining thechain ∅ ( S ( . . . ( S n = [ n ]. For every j ∈ [ n ], w i ( S j ) ≤ c for each i ∈ [ k ] by (1), so S j has unbalance atmost 2 c ≤ p k − c where we assumed k ≥
3. If n > r , we first construct S r ( . . . ( S n − ( S n = [ n ]by backward induction where | S j | = j and k ϕ ( S j ) k ≤ ( k − c for all j with 2 r < j ≤ n ; by (2), it followsthat S j has unbalance at most p k − c for such j . Initially k ϕ ( S n ) k = 0 as S n = [ n ]. For 2 r < j ≤ n ,suppose that we have constructed S j , . . . , S n − , S n . By Lemma 4 with S = S j , there is some j ∈ S j suchthat k ϕ ( S j \ { j } ) k ≤ ( k − c , and we let S j − := S j \ { j } . This finishes the construction of S r , . . . ,S n − , S n . We then order the elements of S r arbitrarily, obtaining the chain ∅ ( S ( . . . ( S r − ( S r ,and arguing similarly as in the case n ≤ r . This completes the construction and the proof of Theorem 2. Acknowledgement.
The author would like to thank Paul Seymour for introducing him to Question 1,for encouragement, and for helpful comments.
References [1] I. B´ar´any, On the power of linear dependencies,
Building Bridges , Bolyai Soc. Math. Stud., vol. 19, Springer, Berlin,2008, pp. 31–45.[2] P. Seymour, personal communication.
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