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

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

2 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

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

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 with poly-size circuits functions on n bits function in NP without poly-size circuits

6 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.

7 Affine Extractors

8 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

9 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}

10 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

11 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

12 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)

13 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

14 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)

15 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

16 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

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

18 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

19 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

20 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’

21 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.



24 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.

25 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}

26 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

27 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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