Boolean Lattice and Symmetric Chain Decompositions Yizhe Zhu - PowerPoint PPT Presentation

Boolean Lattice and Symmetric Chain Decompositions Yizhe Zhu Shanghai Jiao Tong University zyzwstc@sjtu.edu.cn December 30, 2014 Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 1 / 47

Overview Boolean Lattice 1 Definition Parenthesis Matching Conditions for an SCO Necklace Poset 2 Block Code Modified Parenthesis Matching Lyndon Rearrangement Generalization Applications 3 Symmetric Venn Diagrams Symmetric Independent Families Open Problems 4 Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 2 / 47

Symmetric Chain Decomposition (SCD) Let ( P , < ) be a finite poset. A chain in P is a sequence x 1 < x 2 < ... < x n , where each x i ∈ P . For x , y ∈ P , we say y covers x if x < y and there does not exist z ∈ P such that x < z and z < y . A saturated chain in P is a chain where each element is covered by the next. P is ranked if there exists a function r : P → Z ≥ 0 such that x covers y implies r ( y ) = r ( x ) + 1. Suppose min { r ( x ) | x ∈ P } = 0, the rank of P is denoted r ( P ) = max { r ( x ) | x ∈ P } . A saturated chain x 1 < x 2 < ... < x n in a ranked poset P is said to be symmetric if r ( x 1 ) + r ( x n ) = r ( P ). P has a symmetric chain deomposition (SCD) if it can be written as a disjoint union of saturated, symmetric chains. A symmetric chain order (SCO) is a finite ranked poset for which there exists a symmetric chain decomposition. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 3 / 47

Boolean Lattice The Boolean lattice, denoted B n , is the power set of [ n ] = { 1 , 2 , ..., n } ordered by inclusion. r ( A ) = | A | for all A ⊂ [ n ]. An element A ∈ B n can be viewed as an n -bit binary string whose i th bit is 1 if i ∈ A , 0 if i �∈ A . Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 4 / 47

Parenthesis Matching parenthsis matching: start at the left, when a 1 is encounterd, it is matched to the rightmost unmatched zero (if any) with a pair of brackets. Continue in this manner until we reach the end of the string. x = 1011011100010110 the parentesis: 1(01)1(01)110(0(01)(01)1)0. U 0 ( x ) = { 9 , 16 } unmatched zeros U 1 ( x ) = { 1 , 4 , 7 , 8 } unmatched ones M ( x ) = { (2 , 3) , (5 , 6) , (10 , 15) , (11 , 12) , (13 , 14) } matched pairs τ : change the leftmost unmatched 1 to a 0, defined on x ∈ B n , U 0 ( x ) � = ∅ . Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 5 / 47

Parenthesis Matching Theorem (Greene, Kleitman, 1976) For x ∈ B n with | U 0 ( x ) | = k, let C x = { x , τ ( x ) , τ 2 ( x ) , ..., τ k ( x ) } . The following is a symmetric chain decomposition of B n : S = { C x | x ∈ B n , U 1 ( x ) = ∅} Figure: the SCD of B 4 by parenthesis matching Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 6 / 47

Which posets are SCO? Given a ranked poset P , r ( P ) = M . P k = { x ∈ P | r ( x ) = k } . rank-symmetric: | P k | = | P M − k | . rank-unimodal: there exists j such that | P 0 | ≤ | P 1 | ≤ ... ≤ | P j | and | P j | ≥ | P j +1 | ≥ ... ≥ | P M | . antichain: a set of pairwise uncomparable elements of a poset strongly Sperner: for all k = 1 , 2 , ..., M + 1,the union of the k middle levels of P is a unition of k antichains of maximum size. A poset is Peck if it is rank-symmetric, rank-unimodal, and strongly Sperner. Given a group G of automorphisms of a poset P , a quotient poset of P under G , denoted P / G , is orderd in the following way: For orbits of G , A and B , we have A ≤ B iff there are a ∈ A , b ∈ B such that a ≤ b in P . Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 7 / 47

Which posets are SCO? Stanley gave the necessary condition for an SCO. Theorem (Stanley, 1984) (1) An SCO poset P is necessarily Peck. (2) Any quotient of the Boolean lattice is a Peck poset. Griggs showed a sufficient condition for an SCO. Theorem (Griggs, 1977) LYM property, rank-symmetry and rank-unimodality implies that a poset P is SCO. 1 LYM property: for every antichain F , � | P r ( x ) | ≤ 1. x ∈ F Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 8 / 47

Necklace Poset rotation σ on B n : x = ( x 1 , x 2 , ..., x n ), σ ( x ) = ( x n , x 1 , ..., x n − 1 ). For x , y ∈ B n , we say x ∼ y if y = σ k ( x ) for some k . The necklace poset N n is the quotient poset of B n under the equivalence relation ∼ . For X , Y ∈ N n , X ≤ Y iff there exist x ∈ X , y ∈ Y , x ≤ y . For prime p , it is known that N p satisfies the LYM property and has an SCD. For general n , LYM property is unknown. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 9 / 47

Block Codes The explicit construction of an SCD for N p was given by the idea of block code together with parenthesis matching. A surprising application of such SCD of N p is to construct symmetric Venn diagrams. Theorem (Griggs, Killian, Savage, 2004) For prime n, there is a way to select a set R n consisting of one representative from each necklace in N n such that the induced necklace-representative subposet ( R n , ≤ ) of B n has an SCD. It is a stronger than the claim that N p has an SCD. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 10 / 47

Block Codes block code β ( x ) of a binary string x : If x starts with 0 or ends with 1, β ( x ) = ( ∞ ). If x = 1 a 1 0 b 1 1 a 2 0 b 2 ... 1 a t 0 b t , where t > 0 , a i > 0 , b i > 0 , 1 ≤ i ≤ t , then β ( x ) = ( a 1 + b 1 , a 2 + b 2 , ..., a t + b t ). Consider the rotation of x = 0011011, we have β (0011011) = ( ∞ ) , β (0110110) = ( ∞ ) , β (1101100) = (3 , 4), β (1011001) = ( ∞ ) , β (0110011) = ( ∞ ) , β (1100110) = (4 , 3) , β (1001101) = ( ∞ ) . Lemma (Griggs, Killian, Savage, 2004) If n is prime, no two strings of { 0 , 1 } n in the same necklace have the same finite block code. The representative ρ ( x ) of necklace containing x is the rotation y of x for which β ( y ) is minimum. By the lemma, ρ ( x ) is the unique representative. For example, ρ (0011011) = 1101100. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 11 / 47

Block Codes R n = { ρ ( x ) | x ∈ { 0 , 1 } n } = { x ∈ { 0 , 1 } n | ρ ( x ) = x } is a subposet of B n , called necklace-representative poset. Consider R ∗ n = R n − { 0 n , 1 n } , we construct an SCD for R n by showing R ∗ n has an SCD. Lemma (Griggs, Killian, Savage, 2004) If x ∈ R ∗ n and | U 0 ( x ) | ≥ 2 , then τ ( x ) ∈ R ∗ n . Similarly, if | U 1 ( x ) | ≥ 2 , then τ − 1 ( x ) ∈ R ∗ n . Define the set S ∗ = { z ∈ R ∗ n | U 1 ( z ) = { 1 }} . For z ∈ S ∗ , define the chain of z by J z = z , τ ( z ) , τ 2 ( z ) , ...τ k − 1 ( z ). Every x ∈ R ∗ n is in the chain J z for some z ∈ S ∗ . The set of chains { J z | z ∈ S ∗ } is an SCD of R ∗ n . Extend { J z | z ∈ S ∗ } to an SCD in R n by extending the chain 10 n − 1 < 110 n − 2 < ... < 1 n − 2 00 < 1 n − 1 0 to the chain 0 n < 10 n − 1 < 110 n − 2 < ... < 1 n − 2 00 < 1 n − 1 0 , 1 n . Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 12 / 47

Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 13 / 47

Is N n an SCO for all n ≥ 1? Theorem (Jiang, Savage, 2009) N n is an SCO for all prime n and for composite n ≤ 18 . They considered the periodic block code but failed to generalize the construction for lager n . Theorem (Jordan, 2010) The necklace poset is a symmetric order for all n ≥ 1 . The proof is constructive and the most important step is to choose the representative of each necklace. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 14 / 47

Modified Parenthesis Matching Consider a set M n consisting of x ∈ B n such that x achieves the maximum number of unmatched ones over all rotations. M n = { y ∈ B n : | U 1 ( y ) | = max {| U 1 ( σ k ( y )) | : k = 1 , 2 , ..., n }} Figure: the SCD for B 6 with memebers of M 6 in bold Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 15 / 47

Lemma (Jordan, 2010) Let x ∈ M n . (1)If | x | < n 2 , τ i ( x ) ∈ M n , 1 ≤ i ≤ n − 2 | x | . (2) If | x | > n 2 , τ − i ( x ) ∈ M n , 1 ≤ i ≤ 2 | x | − n. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 16 / 47

Modified Parenthesis Matching If x ∈ M n and C x is the chain containing x in the SCD of B n , the smallest symmetric subchain of C x containing x is also in M n . Note that the resulting chains still contain at least one representative of each necklace. Figure: the SCD for M 6 with duplicate representatives of N 6 Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 17 / 47

Lemma (Jordan, 2010) Let x , y ∈ M n with x ∼ y. (1) If | x | ≥ n 2 , then τ ( x ) ∼ τ ( y ) or { τ ( x ) , τ ( y ) } ∩ M n = ∅ . (2) If | x | ≤ n 2 , then τ − 1 ( x ) ∼ τ − 1 ( y ) or { τ − 1 ( x ) , τ − 1 ( y ) } ∩ M n = ∅ . Lemma (Jordan, 2010) Let x , y ∈ B n with | x | = | y | ≤ n 2 , then x ∼ y ⇔ τ n − 2 k ( x ) ∼ τ n − 2 k ( y ) The three lemmas above are proven by the idea of circular matching. Yizhe Zhu (SJTU) Symmetric Chain Decomposition December 30, 2014 18 / 47