Induced bisecting families for hypergraphs
Niranjan Balachandran, Rogers Mathew, Tapas Kumar Mishra, Sudebkumar Prasant Pal
IInduced-bisecting families of bicolorings for hypergraphs
Niranjan Balachandran , Rogers Mathew , Tapas Kumar Mishra , andSudebkumar Prasant Pal Department of Mathematics, Indian Institute of Technology, Bombay 400076, India [email protected] Department of Computer Science and Engineering, Indian Institute of Technology, Kharagpur721302, India {rogers,tkmishra,spp}@cse.iitkgp.ernet.in
Abstract.
Two n -dimensional vectors A and B , A, B ∈ R n , are said to be trivially or-thogonal if in every coordinate i ∈ [ n ] , at least one of A ( i ) or B ( i ) is zero. Given the n -dimensional Hamming cube { , } n , we study the minimum cardinality of a set V of n -dimensional {− , , } vectors, each containing exactly d non-zero entries, such that every‘possible’ point A ∈ { , } n in the Hamming cube has some V ∈ V which is orthogonal,but not trivially orthogonal, to A . We give asymptotically tight lower and (constructive)upper bounds for such a set V except for the even values of d ∈ Ω ( n . (cid:15) ) , for any (cid:15) , < (cid:15) ≤ . . Keywords: Hamming cube, Covering, Hyperplane, Bicoloring
Two n -dimensional vectors A and B , A, B ∈ R n , are said to be trivially orthogonal if inevery coordinate i ∈ [ n ] , at least one of A ( i ) or B ( i ) is zero. The vectors A and B are non-trivially orthogonal if they are orthogonal, but not trivially orthogonal. Consider the followingproblem: "Given the n -dimensional Hamming cube { , } n , what is the minimum cardinalityof a subset V of n -dimensional {− , , } vectors, each containing exactly d non-zero entries,such that every point A ∈ { , } n in the Hamming cube has some V ∈ V which is non-trivially orthogonal to A ?". It is not hard to see that the all-zero vector and the unit vectors { (1 , , . . . , , (0 , , . . . , , . . . , (0 , , . . . , } can never have any non-trivially orthogonal vectorin {− , , } n . Additionally, the all-ones vector (1 , . . . , cannot be non-trivially orthogonal toany vector in {− , , } n consisting of exactly d non-zero entries, when d is odd. We callthe vectors (0 , . . . , , (1 , , . . . , , . . . , (0 , , . . . , (and additionally, (1 , . . . , when d is odd)as trivial . Since no n -dimensional {− , , } vector with exactly one non-zero entry is non-trivially orthogonal to any non-trivial point of the Hamming cube, we assume that d ≥ inthe rest of the paper. Definition 1.
Let ≤ d ≤ n , where d and n are integers. We define β d ( n ) as the minimumcardinality of a subset V of n -dimensional {− , , } vectors, each containing exactly d non-zero entries, such that every non-trivial point in the Hamming cube { , } n has a non-triviallyorthogonal vector V ∈ V . In this paper, we study the problem of estimation of bounds for β d ( n ) .We now define a general version of the aforementioned problem in terms of bicoloringsof a hypergraph. Let G be a hypergraph on the vertex set [ n ] . Corresponding to the trivialvectors/points of the Hamming cube, the singleton sets and the empty set (and additionally,the set [ n ] when d is odd) are the trivial hyperedges or trivial subsets of [ n ] . Let X S denote a ± bicoloring of vertices of S ⊆ [ n ] , i.e. X S : S → { +1 , − } , for some S ⊆ [ n ] . We abuse thenotation to denote the subset of vertices colored with +1 (-1) with respect to bicoloring X S as X S (+1) (resp., X S ( − ). a r X i v : . [ m a t h . C O ] J a n Definition 2.
Given a hypergraph G , a hyperedge A ∈ E ( G ) is said to be induced-bisected bya bicoloring X S of a subset S ⊆ V ( G ) , if | A ∩ X S (+1) | = | A ∩ X S ( − | 6 = 0 . A set X = { X S , . . . , X S t } of t bicolorings is called an induced-bisecting family of order d for G if1. each S i ⊆ [ n ] has exactly d vertices, ≤ i ≤ t , ≤ d ≤ n , and2. every non-trivial hyperedge A ∈ E ( G ) is induced-bisected by at least one X S i , ≤ i ≤ t .Let β d ( G ) denote the minimum cardinality of an induced-bisecting family of order d for hyper-graph G . From Definitions 1 and 2, it is clear that the maximum of β d ( G ) over all hypergraphs G on [ n ] is β d ( n ) . Example 1.
Let H be the hypergraph with all the n − n − non-trivial subsets of [ n ] ashyperedges and let d = 2 . For any S ∈ (cid:0) [ n ]2 (cid:1) , let X S color one point in S with color +1 and theother with -1. Observe that X = { X S | S ∈ (cid:0) [ n ]2 (cid:1) } forms an induced-bisecting family of order2 for H . β ( H ) ≤ (cid:0) n (cid:1) . Moreover, this upper bound is also tight: if X { a,b }
6∈ X , { a, b } ∈ (cid:0) [ n ]2 (cid:1) ,then the hyperedge { a, b } ∈ H cannot be induced-bisected. The problem addressed in this paper can be viewed as a generalization of the problem of bisecting families [3]. Let n ∈ N and let A be a family of subsets of [ n ] . Another family B ofsubsets of [ n ] is called a bisecting family for A , if for each A ∈ A , there exists a B ∈ B suchthat | A ∩ B | ∈ {d | A | e , b | A | c} . In the bicoloring terminology, letting S = [ n ] , X S (+1) = B , X S ( −
1) = [ n ] \ B , the bisecting family B maps to a collection X of bicolorings such that foreach A ∈ A , there exists a bicoloring X ∈ X such that | A ∩ X (+1) | − | A ∩ X ( − | ∈ {− , , } .In [3], the authors define β [ ± ( n ) as the minimum cardinality of a bisecting family for thefamily consisting of all the non-empty subsets of [ n ] , and they prove that β [ ± ( n ) = d n e [3]. Note that when d = n and A e denotes the family of non-trivial even subsets of [ n ] , anyinduced bisecting family of order d for A e is a bisecting family for the family consisting of allthe non-empty subsets of [ n ] . In other words, β n ( A e ) = β [ ± ( n ) . However, observe that when d = n , i.e. S = [ n ] , no odd subset of [ n ] can be induced-bisected: this follows from the factthat for any odd subset A , | A ∩ X S (+1) | − | A ∩ X S ( − | is odd.An affine hyperplane is a set of vectors H ( a, b ) = { x ∈ R n : h a, x i = b } , where a ∈ R n , b ∈ R . Covering the { , } n Hamming cube with the minimum number of affine hyperplaneshas been well studied - a point x ∈ { , } n is said to be covered by a hyperplane H ( a, b ) if h a, x i = b . Without any further restriction, note that H ( e , and H ( e , covers every pointon the { , } n Hamming cube, where e = (1 , , . . . , is the first unit vector. Alon and Füredi[1] show that the covering-by-hyperplanes problem becomes substantially nontrivial underthe restriction that only the nonzero vectors are covered. They demonstrated, using the notionof Combinatorial Nullstellensatz [2], that we need at least n affine hyperplanes when the zerovector remains uncovered. This can be achieved by the set of hyperplanes { H ( e i , } , where e i is the i th unit vector, ≤ i ≤ n . Many other extensions of this covering problem involvingother restrictions have been studied in detail (see [4,7,6]). The problem of bisecting families[3] imposes the following constraints on the minimum cardinality set of covering hyperplanes { H i ( a i , b i ) } : (i) b i ∈ {− , , } ; (ii) a i ∈ {− , } n . The problem of induced-bisecting familiesputs stronger restrictions not just on the hyperplanes, but also on the definition of ‘covering’by a hyperplane { H i ( a i , b i ) } : (i) b i = 0 ; (ii) a i consists of exactly d non-zero coordinates, a i ∈ {− , , } n and d ∈ [ n ] ; (iii) we say a point x is covered by a hyperplane H ( a, b ) when a is nontrivially orthogonal to x . Main result
In this paper, we establish the following theorem.
Theorem 1.
Let ≤ d ≤ n , where d and n are integers. Then, n ( n − d ≤ β d ( n ) ≤ (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) . Moreover, β d ( n ) ≥ n − , when d is odd. This establishes asymptotically tight bounds on β d ( n ) for all values of n , when d is odd.Moreover, the bound is asymptotically tight when d ∈ O ( √ n ) , even if d is even. However,when d ∈ Ω ( n . (cid:15) ) and d is even, the above lower bound may not be asymptotically tight,for any (cid:15) , < (cid:15) ≤ . . Let H denote the hypergraph consisting of all the non-trivial subsets of [ n ] . Let the set X = { X S , . . . , X S t } of bicolorings be any optimal induced-bisecting family of order d for H , where t ∈ N .Considering only the two sized subsets of [ n ] , we note that every two element hyperedge { a, b } , a , b ∈ [ n ] , must lie in some S i , S i ∈ { S , . . . , S t } ; otherwise, no bicoloring in X caninduced-bisect { a, b } . So, it follows that P X S ∈X (cid:0) d (cid:1) ≥ (cid:0) n (cid:1) , i.e., β d ( n ) ≥ n ( n − d ( d − . A constantfactor improvement in the lower bound can be obtained by the following observation: themaximum number of two element subsets { a, b } that can be induced-bisected by any X S ∈ X , | S | = d , is d . So, we have the following proposition. Proposition 1. β d ( n ) ≥ n ( n − d . Observe that when d is large, say d ∈ Ω ( n . (cid:15) ) , where < (cid:15) ≤ . , Proposition onlyyields a sublinear lower bound. When d is odd, we can prove a general lower bound of n − on β d ( n ) using the following version of Cayley-Bacharach theorem by Riehl and Graham [5]on the maximum number of common zeros between n quadratics and any polynomial P ofsmaller degree. Theorem 2. [5] Given the n quadratics in n variables x ( x − , . . . , x n ( x n − with n com-mon zeros, the maximum number of those common zeros a polynomial P of degree k can gothrough without going through them all is n − n − k . Lemma 1. β d ( n ) ≥ n − , when d is odd.Proof. Let B be a minimum-cardinality induced-bisecting family for all the non-trivial subsets A ⊆ [ n ] . Let R B denote the n -dimensional vector representing the bicoloring B ∈ B , i.e. R B ∈ {− , , } n and R B contains exactly d nonzero entries. Consider the polynomials M ( X ) , N ( X ) , and P ( X ) , X ∈ { , } n . M ( X = ( x , . . . , x n )) = Y B ∈B < R B , X > . (1) N ( X = ( x , . . . , x n )) = n X i =1 x i − . (2) P ( X ) = M ( X ) N ( X ) . (3)Let X A denote the 0-1 n -dimensional incidence vector corresponding to A ⊆ [ n ] . Note that M ( X A ) vanishes for each A ⊆ [ n ] except (i) the all 1’s vector, (1 , . . . , , since d is odd, and(ii) possibly the singleton sets. Since N ( X A ) vanishes for all singleton sets, P ( X A ) vanisheson all subsets A ⊆ [ n ] except for the set [ n ] (corresponding to the the all 1’s vector). Since thedegree of P is |B| + 1 and P in non-zero only at X A = (1 , . . . , , using Theorem 2, we have |B| ≥ n − . (cid:50) However, when d is even, the above lower bounding technique does not work since thepolynomial M may vanish at every point of the Hamming cube { , } n . In this case, we canobtain a lower bound of Ω ( √ d ) by considering the maximum number of hyperedges that canbe induced-bisected by a single bicoloring. n is d + 1 In what follows, we consider the hypergraph H consisting of all the non-trivial hyperedges of [ n ] , where n = d + 1 and demonstrate a construction of an induced-bisecting family of order d of cardinality d + 1 . Theorem 3.
Let d be an integer greater than 1. Then, d ≤ β d ( d + 1) ≤ d + 1 . Moreover, β d ( d + 1) = d + 1 , when d is even.Proof. We consider the cases when d is even and d is odd separately. We start our analysiswith the case when d is even. Let v , . . . , v d +1 denote the d + 1 vertices. Consider a circularclockwise arrangement of d + 1 slots, namely P , . . . , P d +1 in that order. The slots P to P d are colored with +1, slots P d +2 to P d +1 are colored with -1, and only slot P d +1 remainsuncolored. Each slot can contain exactly one vertex and each vertex takes the color of the slotit resides in. As for the initial configuration, let v i ∈ P i , for ≤ i ≤ d + 1 . This configurationgives the coloring X , where (i) X (+1) = { v , . . . , v d } , (ii) X ( −
1) = { v d +2 , . . . , v d +1 } ,and, (iii) the vertex v d +1 remains uncolored. We obtain the second coloring X from X by one clockwise rotation of the vertices in the circular arrangement. Therefore, we have, X (+1) = { v d +1 , v , . . . , v d − } , X ( −
1) = { v d +1 , . . . , v d } ; the vertex v d remains uncolored.See Figure 1 for an illustration. Similarly, repeating the process d times, we obtain the set X = { X , . . . , X d +1 } of bicolorings. We have the following observations. P P P P P P P P P P P P P P P P P P P P P P P P P v v v v v v v v v v v v v v v v v v v v v v v v v X X X X X Fig. 1.
Vertices in (i) P and P are colored with +1, (ii) P and P are colored with -1; the vertex in P remains uncolored. X = { X , . . . , X } is an induced bisecting family when n = d + 1 = 5 . Observation 1 If X induced-bisects every non-trivial odd subset of [ d +1] , then X induced-bisectsevery non-trivial even subset of [ d + 1] as well. To prove the observation, consider an even hyperedge A e ⊂ [ d + 1] , and let X ∈ X be thebicoloring that induced-bisects the odd hyperedge ¯ A e = [ d + 1] \ A e . Note that one vertex in ¯ A e remains uncolored under X . Otherwise, ¯ A e cannot get induced-bisected under X . Since | X (+1) | = d and | ¯ A e ∩ X (+1) | = | ¯ A e |− , it follows that | A e ∩ X (+1) | = | X (+1) \ ( ¯ A e ∩ X (+1)) | = d − | ¯ A e |− . Similarly, | A e ∩ X ( − | = d − | ¯ A e |− . So, A e is induced-bisected under X . This completes the proof of Observation .Therefore, it suffices to prove that X induced-bisects every non-trivial odd subset of [ d +1] .For the sake of contradiction, assume that A is an odd hyperedge not induced-bisected by X .Let c i = | A ∩ X i +1 (+1) | − | A ∩ X i +1 ( − | , for ≤ i ≤ d . All additions/subtractions in thesubscript of c are modulo d + 1 . Our assumption implies that c i = 0 for all ≤ i ≤ d . Observation 2 | c i − c i +1 | ≤ , for ≤ i ≤ d . Furthermore, if c i > c i +1 and c i is odd, then c i − c i +1 = 1 . The first part of Observation follows from the construction and we omit the details forbrevity. Note that when c i is odd, the element in P d +1 cannot belong to the odd hyperedge A . This takes care of the second part of Observation .Observe that a bicoloring X j ∈ X , ≤ j ≤ d + 1 , induced-bisects the odd hyperedge A ifand only if c j is 0. We know that bicoloring X ( X i +1 ) is obtained from X ( X i , respectively)by one clockwise rotation of vertices in the circular arrangement. Thus, during the construc-tion of bicolorings X through X d +1 , we perform a full rotation of the vertices with respect to their starting arrangement in X . So, it follows that there exist i and j such that c i is positiveand c i + j is negative. Combined with the second part of Observation , this implies the exis-tence of an index p such that c p = 0 . This is a contradiction to the assumption that A is notinduced-bisected by X . Therefore, every odd subset of [ d + 1] is induced-bisected by X , andusing Observation , the upper bound on β d ( d + 1) follows.To see that the upper bound is tight, observe that exactly one d -sized hyperedge can getinduced bisected under a single bicoloring - the hyperedge missing the uncolored vertex. Thiscompletes the proof of Theorem 3 for even values of d .Recall that along with the empty set and the singleton sets, the set [ d + 1] becomes trivialwhen d is odd. When d is odd, the slots P to P d +12 − are colored with +1, slots P d +12 +1 to P d +1 are colored with -1, and only slot P d +12 remains uncolored. If we generate the bicolorings { X , . . . , X d +1 } as in the proof for the even values of d , by similar arguments, it can be shownthat { X , . . . , X d +1 } is indeed an induced-bisecting family for the hypergraph consisting of allthe non-trivial hyperedges (see Appendix A for a proof). The fact that β d ( d + 1) ≥ d for oddvalues of d follows directly from Lemma 1. (cid:50) We have the following corollary which gives an upper bound to the cardinality of aninduced-bisecting families for arbitrary values of n . Corollary 1.
Let H be any hypergraph on vertex set V ( H ) = { v , . . . , v n } and let d ∈ [ n ] . Let F consist of ( d + 1) -sized subsets of V ( H ) such that for every B ∈ E ( H ) , there exists an A ∈ F with (i) | B ∩ A | ≥ , when d is even; (ii) ≤ | B ∩ A | ≤ d , when d is odd. Then, we can constructan induced-bisecting family of order d of cardinality |F| ( d + 1) for H .Proof. For any subset A ∈ F , using the procedure used in the proof of Theorem 3, we canobtain an induced-bisecting family X A for all the non-trivial subsets of A , where |X A | = d + 1 .When d is even, X A induced-bisects all the d +1 − ( d + 1) − non-empty and non-singletonsubsets of A ; therefore, each B ∈ E ( H ) with | B ∩ A | ≥ is induced-bisected by X A . When d is odd, X A induced-bisects all but the empty set, the singleton sets, and A ; so, each B ∈ E ( H ) with ≤ | B ∩ A | ≤ d is induced-bisected by X A . Repeating the process for each A ∈ F , weget an induced-bisecting family of cardinality |F| ( d + 1) for H . (cid:50) Theorem 3 provides evidence for the following property (which is described in Corollary2) of the odd subsets under any circular permutation of odd number of elements which maybe of independent interest. For any circular permutation σ of [ n ] , a, b ∈ [ n ] , let dist σ ( a, b ) denote the clockwise distance between a and b with respect to σ , which is one more than thenumber of elements residing between a and b in the permutation σ in the clockwise direction. Corollary 2.
Consider any circular permutation σ of [ n ] , where n is odd. For any odd k -sizedsubset A ⊆ [ n ] , let ( a , . . . , a k − ) be the ordering of A with respect to σ . Then, there exists anindex i ∈ { , . . . , k − } such that dist σ ( a i , a i + b k c ) < n and dist σ ( a i + b k c +1 , a i ) < n , wheresummation in the subscript of a is modulo k .Proof. Consider a circular clockwise arrangement of n slots, namely P , . . . , P n in that order.Put vertex σ ( i ) in P i . Now, following the procedure outlined in the proof of Theorem 3, obtaina bicoloring that bisects A . Pick the uncolored vertex residing in slot P d n e with respect to thebicoloring X . Observe that this vertex satisfies the desired property. (cid:50) β d ( n ) and proof of Theorem 1 From Proposition , we know that β d ( n ) ≥ n ( n − d . In this section, we prove an upper boundof (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) for β d ( n ) . Lemma 2. β d ( n ) ≤ (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) . Before proceeding to the proof of the above lemma, we give few definitions that simplifythe proof considerably. Let d be a positive even integer. Let S ( n, d ) = { P , . . . , P d nd e } denote apartition of [ n ] , where each P ∈ S ( n, d ) \ { P d nd e } is of cardinality exactly d , and | P d nd e | ≤ d .Let P d nd e = P d nd e ∪ Q , P d nd e = P d nd e ∪ Q , where Q i denotes a fixed ( d − | P d nd e | ) -sizedsubset of P i . For an even d , we define P ( n, d ) , D ( n, d ) and B ( n, d ) as follows. Definition of P ( n, d ) P ( n, d ) = ( S ( n, d ) , if d divides n S ( n, d ) \ { P d nd e } ∪ { P d nd e , P d nd e } , otherwise. (4) Definition of B ( n, d ) d divides n : For each i, j ∈ (cid:2) nd (cid:3) , i < j , let B i,j : P i ∪ P j → { +1 , − } denote a bicoloring,where B i,j ( x ) = ( +1 , if x ∈ P i − , if x ∈ P j .Let B ( n, d ) = { B i,j | i, j ∈ (cid:2) nd (cid:3) , i < j } denote this set of bicolorings. d does not divide n : For each i, j ∈ (cid:2) d nd e − (cid:3) , i < j , let B i,j : P i ∪ P j → { +1 , − } denotea bicoloring, where B i,j ( x ) = ( +1 , if x ∈ P i − , if x ∈ P j .Let B , d nd e : P ∪ P d nd e → {− , } and B i, d nd e : P i ∪ P d nd e → {− , } , for ≤ i ≤ d nd e− . B , d nd e ( x ) = ( +1 , if x ∈ P − , if x ∈ P d nd e . B i, d nd e ( x ) = ( +1 , if x ∈ P i − , if x ∈ P d nd e , for ≤ i ≤ (cid:24) nd (cid:25) − .Let B ( n, d ) = { B i,j | i, j ∈ (cid:2) d nd e (cid:3) , i < j } denote this set of bicolorings. Definition of D ( n, d ) D ( n, d ) = { D k | D k = P k − ∪ P k , k ∈ (cid:2) d nd e − (cid:3) } ∪ { D d nd e } , where D d nd e = P nd − ∪ P nd , if d divides nP ∪ P d nd e , if d does not divide n and d nd e is odd P d nd e− ∪ P d nd e , if d does not divide n and d nd e is even. (5) Proof. If d = n − , the statement of the lemma follows directly from Theorem 3. So, we assumethat d < n − in the rest of the proof. We prove this lemma considering the exhaustive casesbased on whether d is even or odd, separately. Case 1. d is even Let P = P ( n, d ) , B = B ( n, d ) and D = D ( n, d ) . Observation 3
For any C ⊆ [ n ] , | C | ≥ , if | C ∩ P | ≤ , for all P ∈ P , then C is induced-bisectedby at least one B ∈ B . For any C ⊆ [ n ] , | C | ≥ , it follows from the premise that there exist P i , P j ∈ P , i < j ,such that | C ∩ P i | = | C ∩ P j | = 1 . C is induced-bisected by the bicoloring B i,j , thus completingthe proof of Observation .Let C denote the family of all the subsets of [ n ] that are not induced-bisected by any B ∈ B .Rephrasing Observation , for each C ∈ C , there exists a P ∈ P (and thus, a D ∈ D ) such that | C ∩ P | ≥ (respectively, | C ∩ D | ≥ ). Let D = { D ∪ { j }| j ∈ [ n ] \ D, D ∈ D} . Recall that | D | = d , where d is an even integer less than n − . So, each D ∈ D is a ( d +1) -sized set. UsingCorollary 1, every C ∈ C can be induced-bisected using |D| ( d + 1) bicolorings. Therefore, wehave, β d ( n ) ≤ |B| + |D| ( d + 1) = (cid:0) d nd e (cid:1) + d nd e ( d + 1) , when d is even. Case 2. d is odd Let P = P ( n − , d − , B = B ( n − , d − and D = D ( n − , d − . Since d − is even, P , B and D are well defined. We extend the domain of each B ∈ B to domain ( B ) ∪ { n } , andassign a +1 color to n in each B . Now, each B ∈ B colors exactly d elements of [ n ] . Observation 4
For any C ⊆ [ n ] with | C | ≥ , if n C and | C ∩ P | ≤ for all P ∈ P , then C is induced-bisected by at least one B ∈ B . The proof of this observation is exactly the same as the proof of Observation .Let C denote the family of all the subsets of [ n ] that are not induced-bisected by any B ∈ B . For any D ⊆ [ n ] , let max( D ) denote the maximum integer in the set D . Let D = { D ∪ { n } ∪ { max( D ) + 1 }| D ∈ D} , where the addition is modulo n − . Observation 5
Let D = { D , D , ..., D n − d − e } be the family of subsets constructed as above.Then, | D i ∩ D i +1 | = 2 , if ≤ i ≤ d n − d − e − , and | D n − d − e ∩ D | ≥ . Recall that each D ∈ D is a ( d − -sized subset of [ n − , where d is an odd integerless than n − . So, each D ∈ D is a ( d + 1) -sized set. From Observation , it follows thatfor each C ∈ C , there exists at least one D ∈ D such that | C ∩ D | ≥ . Let C ⊆ C bethe family of subsets of [ n ] such that for each C ∈ C , there exists some D ∈ D such that ≤ | C ∩ D | ≤ d . Using Corollary 1, we can obtain an induced-bisecting family for membersof C of cardinality |D| ( d + 1) . So, it follows that any C ∈ C \ C must contain one or moreelements from { D , D , ..., D n − d − e } as its subsets.For any C ∈ C \ C , if D i ⊆ C , then D i +1 ⊆ C : otherwise, from Observation , ≤| C ∩ D i +1 | ≤ d and from definition of C , C ∈ C . So, it follows that C \ C = { [ n ] } , and [ n ] isa trivial set when d is odd. Therefore, the cardinality of the induced-bisecting family for [ n ] when d is odd is at most |B| + |D| ( d + 1) = (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) . (cid:50) Statement.
Let ≤ d ≤ n , where d and n are integers. Then, n ( n − d ≤ β d ( n ) ≤ (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) . Moreover, β d ( n ) ≥ n − , when d is odd. Proof.
Theorem 1 follows from Proposition , Lemma 1 and Lemma 2. (cid:50) Remark 1.
By removing some duplicate bicolorings, we can actually improve the upper boundfor β d ( n ) from (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) to (cid:0) d n − d − e (cid:1) + d n − d − e d . Theorem 1 asserts an upper bound of O ( n ) on β d ( n ) when d ∈ Ω ( √ n ) . Let k ( G ) denotethe minimum cardinality of any hyperedge of the hypergraph G , i.e., k ( G ) = min e ∈ E ( G ) | e | .For any hypergraph G , the upper bound for β d ( G ) can be improved to O ( n ) even if d ∈ o ( √ n ) provided ( d − k ( G ) > n − in the following way. Since ( d − k ( G ) > n − , every hyperedgeis large enough so that the family D constructed in all the cases of proof of Lemma 2 satisfiesthe conditions of the family requirements of Corollary 2. Therefore, the set of bicoloringsgiven by B = B ( n, d ) (or B ( n − , d − ) can be completely avoided. Thus, we have thefollowing theorem. Theorem 4.
For any hypergraph G , let k ( G ) = min e ∈ E ( G ) | e | . If ( d − k ( G ) > n − , then β d ( G ) ≤d n − d − e ( d + 1) .Remark 2. The proof of Theorem 1 is algorithmic: it yields an induced bisecting family ofcardinality at most (cid:0) d n − d − e (cid:1) + d n − d − e ( d + 1) with a running time of O ( n d + n ) . Observe thatthe running time of our algorithm is asymptotically equivalent to the cardinality of the familyof bicolorings it outputs. Therefore, the asymptotic running time of our algorithm is optimalwhenever it outputs an asymptotically optimal solution. Recall that Theorem 1 asserts tightbounds for β d ( n ) except for the even values of d ∈ Ω ( n . (cid:15) ) , for any (cid:15) , < (cid:15) ≤ . .We note that if d = O (1) , then Theorem 1 asserts that β d ( n ) = Θ ( n ) . However, thecorresponding coefficients are not the same: the constant factor in the Ω ( n ) lower boundis d whereas the constant factor in the upper bound is d − . It would be interesting todetermine the exact coefficient in this case. Moreover, when d is even and d ∈ Ω ( n . (cid:15) ) , forany (cid:15) , < (cid:15) ≤ . , we have an upper bound of O ( n ) on β d ( n ) ; the lower bound for this caseis o ( n ) . We believe that β d ( n ) is more close to the upper bound and tightening of the boundfor β d ( n ) in this case remains open. References
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Appendix A Proof of Theorem 3 when d is odd Statement. d ≤ β d ( d + 1) ≤ d + 1 , d is an odd integer. Proof.
As in the proof of Theorem 3, the slots P to P d +12 − are colored with +1, slots P d +12 +1 to P d +1 are colored with -1, and only slot P d +12 remains uncolored. Note that along withthe empty set and the singleton sets, the set [ d + 1] becomes trivial under this restriction.Each slot can contain exactly one vertex and each vertex takes the color of the slot it re-sides in. As the initial configuration, let v i ∈ P i , for ≤ i ≤ d + 1 . This configuration givesthe coloring X , where (i) X (+1) = { v , . . . , v d +12 − } , (ii) X ( −
1) = { v d +12 +1 , . . . , v d +1 } ,and, (iii) the vertex v d +12 +1 remains uncolored. We obtain the second coloring X from X by one clockwise rotation of the vertices in the circular arrangement. Therefore, we have, X (+1) = { v d +1 , v , . . . , v d +12 − } , X ( −
1) = { v d +12 , . . . , v d } ; the vertex v d +12 − remains un-colored. Similarly, repeating the rotation d times, we obtain the set X = { X , . . . , X d +1 } ofbicolorings.The proof for X being an induced-bisecting family for any odd hyperedge A o (cid:40) [ d + 1] is exactly similar to that given in the proof of Theorem 3. So, we consider only the evenhyperedges. Let c i = | A ∩ X i +1 (+1) |−| A ∩ X i +1 ( − | , for ≤ i ≤ d . All additions/subtractionsin the subscript of c are modulo d + 1 . For the sake of contradiction, assume that A is an evenhyperedge not induced-bisected by X . If we can show that some c j , ≤ j ≤ d , is zero, thenwe get the desired contradiction. Observation 6 | c i − c i +1 | ≤ , for ≤ i ≤ d . The proof of Observation follows from the construction. Consider the sequence ( c i , c i +1 , . . . ,c i + d +1 ) , where c i ≤ c j , j ∈ { i + 1 , . . . , i + d + 1 } , and the addition is modulo d + 1 . Sincethere is a full rotation of the vertex set with respect to the slots, it follows that (i) c i ≤ , and(ii) there exists another index j such that c j is positive. From Observation , it follows that ifnone of the c j , j ∈ { , . . . d } , is zero, there exists an index p such that c p = − and c p +1 = 1 .Note that c p = − asserts that A ∩ P d +12 is non-empty. However, under this configuration, c p +1 can never become 1. This yields the desired contradiction.can never become 1. This yields the desired contradiction.