simple affine extractors using dimension expansion
DESCRIPTION
Simple Affine Extractors using Dimension Expansion . Matt DeVos and Ariel Gabizon. Pseudorandomness. Vague Definition: A pseudorandom object(e.g. graph, function) has some nice property a random object would have with high probability . For example: - PowerPoint PPT PresentationTRANSCRIPT
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Simple Affine Extractors using Dimension Expansion.
Matt DeVos and Ariel Gabizon
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Vague Definition: A pseudorandom object(e.g. graph, function) has some nice property a random object would have with high probability.
For example: A graph that has no large cliquesor large independent sets. The field of pseudorandomness aims to
explicitly construct pseudorandom objects.
Pseudorandomness
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Efficient
Det. Alg.
Explicitly constructing pseudorandom objects
bad objects
Universe of exp(n) objects
good object
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Why do we want to explicitly construct pseudorandom objects?
-Insight into the computational power(lessnes) of randomness
-Useful tools in derandomizing algorithms (good example-expanders!)
Still, is constructing pseudorandom objects more meaningful than making money, or trying to become famous?
Thm: Pseudorandomness is meaningless Theoretical Computer Science is meaningless
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NP machine
PNP by explicitly constructing pseudorandom objects
functions with poly-size circuits
functions on n bits
function in NP without poly-size
circuits
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The nice property can usually be phrased as avoiding a not too large set of bad events.
Example: A function of high circuit complexity avoids the event `being computed by circuit C’ for all small circuits C.
Circuits are hard to understand – let’s first work with bad events that are easier to understand.
The bad event in this paper – a function that is biased on an affine subspace.
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Affine Extractors
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Finite field F, with |F|=q (q=pl for prime p)Vector Space Fn
An affine extractor is a coloring of Fn such that any large enough affine subspace is colored in a balanced way
For simplicit
y assume only 2 colors
Fn
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Just to make sure..
An affine subspace XµFn of dim. k Defined by vectors a(1),…,a(k),b2Fn where a(1),
…,a(k) are independent
X={ (j=1 to k) tj¢a(j) + b|t1,…,tk2F}
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Now, more formally.. An affine extractor for dim k, field size q
and error ² is a function D:Fn{0,1} such that for any affine
subspace XµFn of dim k |PrxX(D(x) =1 ) - ½|·²(We will omit ² from now on, think of it as 1/100)
Intuition: D `extracts’ a random bit for the uniform distribution on X.
1/100
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Feeling the parameters..k-dimension of subspaceq- field size
k larger problem easier (need to be unbiased only on larger subspaces)
q smaller problem harder(subspaces have less structure - are closed under scalar multiplication from smaller field)
Random function D:Fn{0,1} is w.h.p an affine extractor when q=2 and k = 5¢logn
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Previous results and ours: (explicit)G-Raz: Affine Extractor for all k¸1, when
q>n2.Bourgain: Affine Extractor for k=®¢n, for
any constant ®>0, and q=2. (exponentially small error)
Our result: Affine Extractor for all k¸1 , when q=((n/k)2)
Simple Construction and Proof! However: need char(F)=(n/k) (have weaker
result for arbitrary characteristic)
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Warm UpSuppose q>n. How can we get a function
f:FnF that is non-constant on lines?
i.e, for every a0, b2Fn want g(t) , f(a¢t + b) = f(a1¢t + b1,…,an¢t + bn) to be a non-constant function
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Answer: Take f(x1,..,xn) = i=1 to n) xii.
g(t) , f(a¢t + b) = i=1 to n) (ai¢t + bi)i
Note: ai0 for some i. Suppose that an0. g(t) is a non-constant polynomial of degree n.as q>n, this is a non-constant function on F.
(from G-Raz)
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Quadratic Residue Function:QR:F{0,1} , QR(a) = 1 $9b2F such that b2=a
Thm[Weil]: Let F be a field of odd size q.Let g(t) be a non-constant polynomial over F of
odd degree d. Choose t2F randomly.. QR(g(t)) has bias at most d/q1/2
works for multivariate g too..
Weil’s Theorem
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Subspace X of dim k defined by a(1),…,a(k),b
For f:FnF, define f|X (t1,..,tk) = f((j=1 to k) tj¢a(j) + b )
Using Weil: Poly f(X1,..,Xn) of degree d such that: f|X
constant for all X of dim kAffine Extractor for dim k and q»d2
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`trick’: Using this view can multiply vectors
x,y2(Fq)n - not just add them!
Vector Space\Field Dualitynq
nq FF
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Fix 1-1 Φ:(Fq)n -->Fqn s.t. ∀a,b∈Fqn s,t∈Fq: Φ(at+ bs) = Φ(a)∙t + Φ(b)∙s We identify the source output with an element
of Fqn:∑aj∙tj+b --> Φ[∑ aj∙tj+b] =∑Φ(aj)∙tj+Φ(b)(as tj ∈ Fq ) our source coincides with a multivariate
polynomial with coeff in Fqn
(from now omit Φ and think of aj∈Fqn )
Viewing the source over the `big’ field
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Suppose we allow f|X to have coeff. in the `big field’ Fqn
can take f(x) = x.For any subspace X f|X (t1,..,tk) = (j=1 to k) aj¢tj + b is non-
constant.but to use Weil need f|X with coeff. in Fq
Idea- if coeff. of f|X span Fqn. over Fq – we can `project down to Fq’ without becoming zero\constant
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A,B linear subspaces in Fqn
Dfn: A¢B,span{a¢b|a2A, b2B} (enough to take products of basis elements)
[Heur-Lieng-Xiang]Suppose n is prime. Then dim(A¢B)¸ min{dim(A)+dim(B)-1,n}
(analogous to the classic Cauchy-Davenport on Zp)
` dimension expansion of products of subspaces’
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Thm: Suppose n is prime. Let T: Fqn Fq be any non-trivial Fq-linear map. Let d=n/(k-1). Suppose Char(F)>d. Let f(x)=T(xd).
Then for any affine subspace X of dim k,f|X is a non-constant poly of degree d with
coeff in Fq.Proof idea: When Char(F) is large enough,
coefficients of f|X are `independent products’ of basis elements.
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Open question: Similar results over F2
Relates to following: n is prime.V a linear subspace of dim k
in (F2)n , k>min{100logn,n/100}. t=┌2n/k┐. Vt ={x1+2+4+..+2^{t} | x2V}. Show that Vt spans (F2)n over F2.
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Cauchy – DavenportA,B½Zp
A+B , {a+b| a2A, b2B}
C-D: |A+B| ¸ min{|A|+|B|-1,p}
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C-D: |A+B| ¸ min{|A|+|B|-1,p}Proof: Induction on |A|.
|A|=1 : |A+B| = |B| (=|A|+|B|-1)
Induction step: Assume first that ; ( AÅB ( A
Using Inclusion-Exclusion + Ind. Hyp |AÅB + A[B| ¸ min{|AÅB| + |A[B| -1,p}
= min{|A| +|B| -1,p}Done as AÅB + A[B ½ A+B
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justify assumption ; ( AÅB (A:w.l.g: 02A,B (can replace A by –a +A, for
some a2A. This does not change |A+B|)|A|>1 , so can fix 0≠a2A.If B=Zp we are done.Otherwise, fix first c s.t. c∙a ∉B.Replace B by –(c-1)∙a + B.We have 02B but a∉B. (which justifies
above assumption)