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

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

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 construct pseudorandom objects. Pseudorandomness

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

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

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

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.

Affine Extractors

Finite field F, with |F|=q (q=p l for prime p) Vector Space F n An affine extractor is a coloring of F n such that any large enough affine subspace is colored in a balanced way For simplicity assume only 2 colors FnFn

Just to make sure.. An affine subspace X µ F n of dim. k Defined by vectors a (1),…,a (k),b 2 F n where a (1),…,a (k) are independent X={  (j=1 to k) t j ¢ a (j) + b|t 1,…,t k 2 F}

Now, more formally.. An affine extractor for dim k, field size q and error ² is a function D:F n  {0,1} such that for any affine subspace X µ F n of dim k |Pr x  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

Feeling the parameters.. k-dimension of subspace 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 function D:F n  {0,1} is w.h.p an affine extractor when q=2 and k = 5 ¢ logn

Previous results and ours: (explicit) G-Raz: Affine Extractor for all k ¸ 1, when q>n 2. 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)

Warm Up Suppose q>n. How can we get a function f:F n  F that is non-constant on lines? i.e, for every a  0, b 2 F n want g(t), f(a ¢ t + b) = f(a 1 ¢ t + b 1,…,a n ¢ t + b n ) to be a non-constant function

Answer: Take f(x 1,..,x n ) =   i=1 to n) x i i. g(t), f(a ¢ t + b) =   i=1 to n) (a i ¢ t + b i ) i Note: a i  0 for some i. Suppose that a n  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)

Quadratic Residue Function: QR:F  {0,1}, QR(a) = 1 \$9 b 2 F such that b 2 =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 t 2 F randomly.. QR(g(t)) has bias at most d/q 1/2 works for multivariate g too.. Weil’s Theorem

Subspace X of dim k defined by a (1),…,a (k),b For f:F n  F, define f| X (t 1,..,t k ) = f(  (j=1 to k) t j ¢ a (j) + b ) Using Weil: Poly f(X 1,..,X n ) of degree d such that: f| X  constant for all X of dim k  Affine Extractor for dim k and q » d 2

`trick’: Using this view can multiply vectors x,y 2 (F q ) n - not just add them! Vector Space\Field Duality

Fix 1-1 Φ:(F q ) n -->F q n s.t. ∀ a,b ∈ F q n s,t ∈ F q : Φ(at+ bs) = Φ(a)∙t + Φ(b)∙s We identify the source output with an element of F q n : ∑a j ∙t j +b --> Φ[∑ a j ∙t j +b] =∑Φ(a j )∙t j +Φ(b) (as t j ∈ F q )  our source coincides with a multivariate polynomial with coeff in F q n (from now omit Φ and think of a j ∈ F q n ) Viewing the source over the `big’ field

Suppose we allow f| X to have coeff. in the `big field’ F q n  can take f(x) = x. For any subspace X f| X (t 1,..,t k ) =  (j=1 to k) a j ¢ t j + b is non- constant. but to use Weil need f| X with coeff. in F q Idea- if coeff. of f| X span F q n. over F q – we can `project down to F q ’ without becoming zero\constant

A,B linear subspaces in F q n Dfn: A ¢ B, span{a ¢ b|a 2 A, b 2 B} (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 Z p ) `dimension expansion of products of subspaces’

Thm: Suppose n is prime. Let T: F q n  F q be any non-trivial F q -linear map. Let d=n/(k-1). Suppose Char(F)>d. Let f(x)=T(x d ). Then for any affine subspace X of dim k, f| X is a non-constant poly of degree d with coeff in F q. Proof idea: When Char(F) is large enough, coefficients of f| X are `independent products’ of basis elements.

Open question: Similar results over F 2 Relates to following: n is prime.V a linear subspace of dim k in (F 2 ) n, k>min{100logn,n/100}. t= ┌ 2n/k ┐. V t ={x 1+2+4+..+2^{t} | x 2 V}. Show that V t spans (F 2 ) n over F 2.

Cauchy – Davenport A,B ½ Z p A+B, {a+b| a 2 A, b 2 B} 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 ; ( A Å B ( A Using Inclusion-Exclusion + Ind. Hyp |A Å B + A [ B| ¸ min{|A Å B| + |A [ B| -1,p} = min{|A| +|B| -1,p} Done as A Å B + A [ B ½ A+B

justify assumption ; ( A Å B ( A: w.l.g: 0 2 A,B (can replace A by –a +A, for some a 2 A. This does not change |A+B|) |A|>1, so can fix 0≠a 2 A. If B=Z p we are done. Otherwise, fix first c s.t. c ∙ a ∉ B. Replace B by –(c-1) ∙ a + B. We have 0 2 B but a ∉ B. (which justifies above assumption)

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