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Simple Affine Extractors using Dimension Expansion .

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:

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Simple Affine Extractors using Dimension Expansion .

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  1. Simple Affine Extractors using Dimension Expansion. Matt DeVos and Ariel Gabizon

  2. Pseudorandomness 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 cliques or large independent sets. The field of pseudorandomness aims to explicitly constructpseudorandom objects.

  3. Efficient Det. Alg. Explicitly constructing pseudorandom objects Universe of exp(n) objects good object bad objects

  4. 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

  5. NP machine PNP by explicitly constructing pseudorandom objects functions on n bits function in NP without poly-size circuits functions with poly-size circuits

  6. The nice property can usually be phrased as avoiding a not too large set of bad events. Example: A functionof high circuit complexity avoids the event `being computed by circuitC’ for all small circuitsC. Circuits are hard to understand – let’s first work with bad events that are easier to understand. The bad event in this paper– a functionthat is biased on an affine subspace.

  7. Affine Extractors

  8. Finite fieldF, with |F|=q (q=plfor prime p) Vector SpaceFn An affine extractor is a coloring of Fn such that any large enough affine subspace is colored in a balanced way Fn For simplicity assume only 2 colors

  9. Just to make sure.. An affine subspaceXµFn of dim. k Defined by vectors a(1),…,a(k),b2Fnwhere a(1),…,a(k)are independent X={(j=1 to k) tj¢a(j) + b|t1,…,tk2F}

  10. Now, more formally.. An affine extractor for dim k, field size q and error² is a function D:Fn{0,1} such that for anyaffine subspaceXµFn of dimk |PrxX(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

  11. Feeling the parameters.. k-dimension ofsubspace q- 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 functionD:Fn{0,1}isw.h.panaffine extractorwhenq=2andk = 5¢logn

  12. 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)

  13. Warm Up Suppose q>n. How can we get a function f:FnF that is non-constant on lines? i.e, for everya0, b2Fnwantg(t) , f(a¢t + b) = f(a1¢t + b1,…,an¢t + bn)to be a non-constant function

  14. 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: ai0 for some i. Suppose that an0. • g(t) is a non-constant polynomial of degree n. as q>n, this is a non-constant function on F. (from G-Raz)

  15. Weil’s Theorem 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..

  16. Subspace X of dimk defined by a(1),…,a(k),b For f:FnF, define f|X (t1,..,tk) = f((j=1 to k) tj¢a(j) + b ) Using Weil: Polyf(X1,..,Xn) of degreed such that: f|X constant for all X of dimk • Affine Extractor for dimk and q»d2

  17. Vector Space\Field Duality `trick’: Using this view can multiply vectors x,y2(Fq)n- not just add them!

  18. Fix 1-1 Φ:(Fq)n -->Fqn s.t. ∀a,b∈Fqns,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

  19. Suppose we allow f|X to have coeff. in the `big field’ Fqn can takef(x) = x. For any subspaceX 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|Xspan Fqn. over Fq– we can `project down to Fq’ without becoming zero\constant

  20. `dimension expansion of products of subspaces’ A,Blinear 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)

  21. 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 subspaceX of dimk, f|Xis a non-constant poly of degreed with coeffin Fq. Proofidea: When Char(F) is large enough, coefficients of f|Xare`independent products’ofbasis elements.

  22. Open question: Similar results over F2 Relates to following: n is prime.V a linear subspace of dimk in (F2)n , k>min{100logn,n/100}.t=┌2n/k┐.Vt ={x1+2+4+..+2^{t} | x2V}. Show that Vtspans(F2)n over F2.

  23. Cauchy – Davenport A,B½Zp A+B , {a+b| a2A, b2B} C-D: |A+B| ¸ min{|A|+|B|-1,p}

  24. 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

  25. 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=Zpwe 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)

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