on the size of dissociated bases
DESCRIPTION
On the size of dissociated bases. Raphael Yuster University of Haifa Joint work with V sevolod Lev University of Haifa. Dissociated bases. Recall, that subset sums of a subset of an abelian group are group elements of the form: where B - PowerPoint PPT PresentationTRANSCRIPT
On the size of dissociated bases
Raphael YusterUniversity of Haifa
Joint work withVsevolod Lev
University of Haifa
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• Recall, that subset sums of a subset of an abelian group are group elements of the form:
where B
• Note: there are at most 2|| distinct subset sums.
• Famous conjecture of Erdös (80 years ago, $500):
If all subset sums of an integer set [1,n] are pairwisedistinct, then || ≤ log n+O(1).
• Similarly, one can investigate the largest possible size of subsets of other natural sets in abelian groups, possessing the distinct subset sums property.
Dissociated bases
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• Example:What is the largest possible size of a set {0,1}n n with all subset sums pairwise distinct?
• Definition: a subset of an abelian group, all of whose subset sums are pairwise distinct, is called dissociated.
• Dissociated sets are useful due to the fact that if is a maximal dissociated subset of a given set A, then every element of A is representable as a linear combination of the elements of with the coefficients in {-1,0,1}.
• Hence, maximal dissociated subsets of a given set can be considered as its ``linear bases” over the set {-1,0,1}.
• This interpretation naturally makes one wonder whether, and to what extent, the size of a maximal dissociated subset of a given set is determined by this set?
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Theorem 1
For a positive integer n, the set {0,1}n possesses adissociated subset of size:
Is it true that all maximal dissociated subsets of a given finite set in an abelian group are of about the same size?
The following two theorems give a satisfactory answer:
Theorem 2
If and M are maximal dissociated subsets of a subset Aof an abelian group, then
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Since the standard basis is a maximal dissociated subset of the set {0,1}n , comparing Theorems 1 and 2 we conclude that:
• Theorem 2 is sharp in the sense that the logarithmic factors cannot be dropped or replaced with a slower growing function.
• Theorem 1 is sharp in the sense that n log n is the true order of magnitude of the size of the largest dissociated subset of {0,1}n.
Why is this a satisfactory answer:
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• Recall: we want to prove that {0,1}n possesses a dissociated subset of size:
• This is the same as showing that if n > (log 9+o(1))m/log mthen {0,1}n possesses an m-element dissociated subset.
• The trick is to switch to the dual setting:
• We prove that there exists a set D {0,1}m with |D|=n such that for every non-zero vector s S:={-1,0,1}m there is an element of D, not orthogonal to s.
• Once this is done, we consider the n m matrix whose rows are the elements of D; the columns of this matrix form an m-element dissociated subset of {0,1}n, as required.
Outline of the proof of Theorem 1
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• Explanation: Suppose the sum of the red vectors is equal to the sum of the blue vectors. Then each row is orthogonal to the vector:
• We construct D by choosing uniformly at random, and independently of each other, n vectors from the set {0,1}m .
• We will show that for every s S:={-1,0,1}m, the probability that all vectors from D are orthogonal to s is very small.
n
m
1 -1 0 1 -1 1 -1 0 -1
The rows are the elements of D {0,1}m
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• We say that a vector from S is of type (m+,m-) if it has m+
coordinates equal to +1, and m- coordinates equal to -1.
• If sS is of type (m+,m-) then a vector d {0,1}m is orthogonal to s if and only if there exists j≥0 such that d has:
- exactly j non-zero coordinates in the (+1)-locations of s, - exactly j non-zero coordinates in the (-1)-locations of s.
• Example: s=(1,-1,0,1,1,-1) is of type (3,2) and d=(0,1 ,1,0,1,0 ) is orthogonal to s, here with j=1.
• The probability for a randomly chosen d {0,1}m to be orthogonal to s is
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• It follows that the probability for all n elements of D to be simultaneously orthogonal to s is smaller than
• Since the number of elements of a given type (m+,m-) is
to conclude the proof it suffices to prove that
• To this end we rewrite this sum as
and split it into two parts, according to whether t<T or t > T, where T := m/(log m)2. Denote the parts by ∑1 and ∑2 .
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• We prove that ∑1 < ½ and ∑2 < ½.
• For ∑1 we have
• As T := m/(log m)2 and n > (log 9+o(1))m/log m we have
and therefore ∑1< ½ .
• Proving that ∑2< ½ is only slightly more involved.
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• Recall: we want to prove that if and M are maximal dissociated subsets of a subset A of an abelian group, then
• Here we will only prove the lower bound:
• The upper bound is only slightly more complicated.
• By maximality of , every element of A, and thereby every element M, is a linear combination of the elements of with coefficients in {-1,0,1}.
• Hence, every subset sum of M is a linear combination of the elements of with coefficients in {-|M|,-|M|+1,…,|M|}.
Outline of the proof of Theorem 2
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• There are 2|M| subset sums of M, all distinct from each other.
• There are (2|M|+1)|| linear combinations of the elements of with the coefficients in {-|M|,-|M|+1,…,|M|}.
• we have: 2|M| · (2|M|+1)|| and follows.
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• For a positive integer n, let Ln denote the largest size of a dissociated subset of the set {0,1}n n . What are the limits
• Notice, that by Theorems 1 and 2 we have
Open problem
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Thanks!