On the size of dissociated bases - PowerPoint Presentation

Slide1

On the size of dissociated bases

Raphael

Yuster

University

of Haifa

Joint work with

V

sevolod Lev

University of HaifaSlide2

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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 pairwise distinct, 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 basesSlide3

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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? Slide4

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Theorem 1

For a positive integer

n, the set {0,1}n possesses adissociated subset of size: Theorem 1Is 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 2If  and M are maximal dissociated subsets of a subset Aof an abelian group, then

Theorem 2Slide5

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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:Slide6

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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 1Slide7

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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

Slide8

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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 sS 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 Slide9

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It follows that the probability for

all

n elements of D to be simultaneously orthogonal to s is smaller thanSince the number of elements of a given type (m+,m-) isto conclude the proof it suffices to prove thatTo this end we rewrite this sum asand 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 .Slide10

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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 haveand therefore ∑1< ½ .Proving that ∑2< ½ is only slightly more involved.Slide11

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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, thenHere 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 2Slide12

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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. Slide13

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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 limitsNotice, that by Theorems 1 and 2 we have Open problemSlide14

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Thanks!