On the size of dissociated bases

Raphael YusterUniversity of Haifa

Joint work with

Vsevolod 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!