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Restrictions on sums over. Yiannis Koutis Computer Science Department Carnegie Mellon University. Vectors over. p: prime d: dimension of the vectors point-wise multiplication mod p p = 3, d=3. Work on sum of vectors.
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Restrictions on sums over Yiannis Koutis Computer Science Department Carnegie Mellon University
Vectors over • p: prime • d: dimension of the vectors • point-wise multiplication mod p • p = 3, d=3
Work on sum of vectors • f(n,d) : the minimum number such that every subset of has a sub-subset that sums up to • f(n,1) = 2n-1 [ Erdos, Ginzburg, Ziv] • f(n,d) · cd n [Alon, Dubiner] • f(n,d) ¸ (9/8)d/3 (n-1)2d+1 [Elsholtz] • f(n,2) = ? [conjectured equal to 4n-3]
dimensionality restrictions • is an arbitrary subset of size p(d+2) • numbers of subsets of A summing up to • what is the probability that the number of zero-sum subsets of A is odd, when p=2? • how smaller can the number of zero-sum subsets be, than then number of v-sum subsets?
dimensionality restrictions • is an arbitrary subset • numbers of subsets of A summing up to • what is the probability that the number of 0-sum subsets of A is odd, when p=2? [ prob = 1 ] • how smaller can the number of zero-sum subsets be, than then number of v-sum subsets? [zero is attractive : worst case one less ]
motivation • The Set Packing problem: Given a collection C of sets on a universe U of n elements, is there a sub-collection of k mutually disjoint sets ? • Algebraization: • Assign variables xi to the elements of U • for each set S, let
motivation • Let • If there are k disjoint terms, there is a multilinear term. If not, fk is in the ideal <x12,x2,2,..>
example • Define the sets • Then :
motivation • Let • If there are k disjoint terms, there is a multilinear term. If not, fk is in the ideal <x12,x2,2,..> • Basic idea: evaluate fkover a ‘small’ commutative ring with a polynomial number of operations and exploit the squares
example • Assign distinct to element i and substitute v0+xviin xi • Then for every i • If there is no set packing of size k, then fk is a multiple of (1+x)2 • How large must d be so that the multilinear term is not a multiple of (1+x)2 ? [must be linear, unfortunately]
representation theory for • each element is represented by a matrix • addition is isomorphic to matrix multiplication • 1-1: elements with entries in the first row
representation theory for • The coefficient of xi in H(1,j) is the number of vj-sum sets of cardinality i in A. • For x=1, H(1,1) = #zero-sum+1 H(1,j) = #vj-sum
representation theory for • All matrices are simultaneously diagonalizable • V is a Hadamard matrix, every entry is 1 or -1 • () is diagonal, containing the eigenvalues which are all 1 and -1
parity of zero sum subsets For x=1, H(1,1) = #zero-sum+1, H(1,j) = #vj-sum Each matrix (I+() ) has eigenvalues 0 and 2 For d +1 terms in the product, the eigenvalues are either 0 or 2d+1. All entries of H are even. #zero-sum+1 = even , #vj-sum=even
number of zero-sum subsets 2dH(1,1) = trace(H) =sum of eigenvalues 2dH(1,j) = weight eigenvalues by 1 and -1 H(1,1)¸ H(1,j) [zero is attractive : #zero sum +1 ¸ #vj-sum]
restrictions on sums • Let N(v,k) be the number of v-sum subsets of cardinality k Theorem: Given N(v,2t) mod 2,for 1· t · 2log n, the numbers N(v,2t) mod 2, for t>2log n can be determined completely .
restrictions on sums - outline • Form • Also • v’ has only a 1 in the extra dimension • The coefficient aj of xj in H’, is zero mod 2 when j¸ 4d • aj is a linear combination of the coefficients of xj for j\leq 4d in H
restrictions on sums • Let N(v,k) be the number of v-sum subsets of cardinality k Theorem: Given N(v,2t) mod 2,for 1· t · 2log n, the numbers N(v,2t) mod 2, for t>2log n can be completely determined. Question:What are the ‘admissible’ values for the 2log n free numbers, over selections ?
conclusions – back to motivation • Assign distinct to element i and substitute v0+xviin xi • Then for every i • If there is no set packing of size k, then fk is a multiple of (1+x)2 • How large must d be so that the multilinear term is not a multiple of (1+x)2 ? [must be linear, unfortunately]
conclusions – back to motivation • If there is no set packing of size k, then fk is a multiple of (1+x)2 • But now, we know that fk must also satisfy many linear restrictions • Question: Can we exploit this algorithmically?