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Worst case to Average case Reductions for Polynomials

Worst case to Average case Reductions for Polynomials. Shachar Lovett Weizmann Institute / Microsoft Research Joint work with Tali Kaufman. A motivating example. Let p(x 1 ,…,x n )=f(x 1 ,…,x n )g(x 1 ,…,x n ) f,g generic polynomials of degree d over F 2

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Worst case to Average case Reductions for Polynomials

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  1. Worst case to Average case Reductions for Polynomials Shachar Lovett Weizmann Institute / Microsoft Research Joint work with Tali Kaufman

  2. A motivating example Let p(x1,…,xn)=f(x1,…,xn)g(x1,…,xn) f,g generic polynomials of degree d over F2 p(x) is a biased degree 2d polynomial Pr[p(x)=0] ~ 3/4 Reason:p is a biased function of f,g

  3. Another motivating example p(x)=MAJ(f(x),g(x),h(x)) f,g,h generic polynomials of degree d p(x) is unbiased of degree 2d p(x) can be approximated by a lower degree polynomial Pr[p(x)=f(x)] ~ 3/4 Reason:p is a (unbiased) function of f,g,h

  4. Is this generally true? Let p(x1,…,xn) be a degree d polynomial d is constant Assume that: p is biased: Pr[p(x)=0] = ½ + ε or p can be approximated by a lower degree polynomial: Pr[p(x)=f(x)] = ½ + ε Can we deduce a structure theorem for p?

  5. Warm Up – Quadratics Assume p(x) is quadratic Dixon’s Theorem: p(x) = l1(x)l2(x) + l3(x)l4(x) + … + lr-1(x)lr(x) ( + lr+1(x) ) l1,…,lr+1linear and independent All the non-zero Fourier coefficients of p(x) are 2-r/2 Assume that: Pr[p(x)=0] = ½ + ε (p is biased) Or Pr[p(x)=l(x)] = ½ + ε (p is approximated by linear) p(x) has a Fourier coefficient 2ε  r = O(log(1/ε)) p(x) is a function of O(log(1/ε)) linear functions

  6. Main theorem: general degrees p(x1,…,xn) of degree d Assume that: Pr[p(x)=0] = ½ + ε or Pr[p(x)=f(x)] = ½ + ε ( deg(f) ≤ d-1 ) Then: p(x)=C(f1(x),…,fk(x)) f1,…,fk: polynomials of degree at most d-1 C: any combiner function k depends only on d, ε – independent of n !

  7. Functions of polynomials: Computation vs. Approximation pn(x1,...,xn) - family of degree d polynomials The following models are equivalent: pn can be computed by a constant number of lower degree polynomials: • pn can be approximated by a constant number of lower degree polynomials:

  8. Example for applications S4(x1,…,xn) – symmetric polynomial of degree 4 • To refute the Inverse Conjecture for the Gowers Norm, needed to prove: • For any cubicf(x), Pr[S4(x)=f(x)] ≤ ½ + o(1) • Given our theorem, enough to prove: • S4 cannot be computed by a constant number of cubics

  9. Bias implies low rank Let p(x1,…,xn) be a degree d polynomial Bias(p) = E[(-1)p(x)] = Pr[p(x)=0]-Pr[p(x)=1] Bias – a measure for the distance of p(x) from uniformity Pr[p(x)=0] = ½ + ε bias(p) = 2ε Rankd-1(p) = min k s.t. p(x)=C(f(1)(x),…,f(k)(x)) f(1)(x),…,f(k)(x) of degree ≤ d-1 C:F2k  F2any combiner function Theorem: Bias implies low rank |Bias(p)| ≥ ε rankd-1(p) ≤ k(d,ε)

  10. Bias  Approximation “Bias implies low rank” is enough for the general theorem Assume: Pr[p(x)=f(x)] ≥ ½ + ε, deg(p)=d, deg(f) ≤ d-1 Then: Bias(p-f) ≥ 2ε p(x)-f(x) = C(f(1)(x),…,f(k)(x)) deg(f(1)),…,deg(f(k)) ≤ d-1 p(x) = f(x) + C(f(1)(x),…,f(k)(x)) Assume: Pr[p(x)= C(g(1)(x),…,g(k)(x))] ≥ ½ + ε, deg(g(i)) ≤ d-1 Then: There are a1,…,ak F2 s.t. bias(p(x)-(a1g(1)(x)+…+akg(k)(x))) ≥ ε 2-O(k) p(x)-(a1g(1)(x)+…+akg(k)(x)) = C(f(1)(x),…,f(k’)(x))

  11. Bias implies low rank Recall: p(x1,…,xn) is a degree d polynomial Bias(p) = E[(-1)p(x)] Rankd-1(p) = min k : p(x)=C(f(1)(x),…,f(k)(x)), deg(f(i)) ≤ d-1 We want: |Bias(p)| ≥ ε rankd-1(p) ≤ k(d,ε) Green & Tao prove this when d < |F| Used to prove the Inverse Conjecture for the Gower Norm in this case However, The ICGN is false when d >> |F| They conjectured that “bias implies low rank” holds even if d >> |F| We prove “bias implies low rank” for all constant degrees Following Green&Tao proof, with one major change Proof by induction on d. d=1: Trivial – any biased linear function is in fact constant

  12. First step: bias amplification Assume: bias(p(x)) ≥ ε We will generate degree d-1 polynomials f(1)(x),…,f(k)(x) s.t. Prx[p(x)=C(f(1)(x),…,f(k)(x))] ≥ 1 -  k=k(, ε) Will use with =2-O(d) Derivatives: pa(x) = p(x+a) – p(x) a  F2n pa(x) of degree d-1 Proof: Fix x, and consider

  13. First step: bias amplification Sampling: • There exists a1,…,ak

  14. First step: bias amplification Originally – Lemma of Bogdanov & Viola Used to build PRG for low degree polynomials We will prove: If f(1),…,f(k) are “random enough”, then in fact p(x) = MAJ(f(1)(x),…,f(k)(x)) for all x  F2n Otherwise we “make them random enough” Derivatives f(1),…,f(k) of degree ≤ d-1  use “bias implies low rank” inductively You can think of this as a generalization of: q(x) of degree d-1, Pr[p(x)=q(x)] > 1-2-d p=q

  15. f(1),…,f(k) partition the space F2n into 2k “equal”regions Partitioning the space

  16. Partitioning the space • f(1),…,f(k) partition the space F2n into 2k “equal”regions

  17. Partitioning the space f(1),…,f(k) partition the space F2n into 2k “equal” regions

  18. F assigns a value to each region p is equal to F almost everywhere  On most regions, p is almost constant (and equal to F) Partitioning the space F(x) Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  19. Partitioning the space Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  20. Good regions: bad areas are very small ( 2-O(d) ) Almost all regions are good ( 1 – 2-O(d) ) Good & bad regions Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  21. Proof strategy The proof has two steps: Good regions are excellent: p=F on all points in region (i.e. p(x) is constant on good regions) Assuming almost all regions are excellent, we will prove all regions are excellent (i.e. p(x) is constant on all regions) Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  22. Proof: Step 1 • We will prove:p(x) is constant on good regions • Let R be a good region • p(x) = F(R) (=const) for almost all x  R • Let x0  R be arbitrary • We will prove: p(x0) = F(R) • We will use: • p is a low degree polynomial • Regions defined by lower degree polynomials • Induction on “bias implies low rank”

  23. The derivatives identity • p(x) of degree d • Derivatives reduce degree: • py(x)= p(x+y)-p(x) of degree d-1 • py_1,…,y_{d+1}(x) 0 • Thus we have the identity:

  24. Using the derivatives identity • Let R be a good region • Take arbitrary x0 R • Assume there are y1,…,yd+1 s.t. x0 + iS yi are in the “good part” of R, for all non-empty S • Then p(x0 + iS yi) = F(R) for all non-empty S • Then also: p(x0) = F(R) !

  25. Using the derivatives identity x0 x0+y1 x0+y2 x0+y1+y2 Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  26. Getting all the points in R • We need to show: y1,…,yd+1: x0 +iS yi R • In fact, we need them to be in the “good part” of R • We will handle this later • Solution: • choose y1,…,yd+1 randomly and uniformly • show the required condition occurs with positive probability • Each separate variable x0 +iS yi is uniform in F2n • Problem: handling the dependencies

  27. Structure of regions • We need: x0+iS yi R (for all non-empty S) • Recall: regions defined by polynomials f(1),f(2),…,f(k) • R = {x  F2n: f(1)(x)=c1, f(2)(x)=c2,…} (c1,c2,…  F2) • So, we actually need: { f(j) (x0+iS yi)=cj }j=1..k, non-empty S  [d+1] • We need to find “randomness” conditions on f(j) s.t. all the events are “almost independent” • Thus all occur simultaneously with positive probability

  28. Randomness conditions (1st attempt) • We need: for any x0, if y1,…,yd+1 are uniform, then the set of random variables {f(j) (x0+iS yi) : j=1..k, S  [d+1], S non-empty} are almost independent • Actually, this can never be true  • Reason:f(j) are polynomial of degree ≤ d-1 • d derivations zeros f(j) • Random variables are linearly dependent • These dependencies can be handled

  29. Randomness conditions (2nd attempt) • We need: for any x0, if y1,…,yd+1 are uniform, then the set of random variables {f(j) (x0+iS yi) : j=1..k, S  [d+1], S non-empty, |S| ≤ deg(f(j))} are almost independent • It turns out its enough to prove this for random x • Proof: Cauchy-Schartz • Even for random x, this can still be false • Reason: non-linear dependencies

  30. Non-linear dependencies Example 1: • f(1)decomposes: f(1)(x) = g(x) h(x) • f(1) is biased  f(1)(x) is not uniform Example 2: • A derivative of f(1)decomposes • f(1)y_1,y_2 (x) = f(1)(x) - f(1)(x+y1) - f(1)(x+y2) + f(1)(x+y1+y2) = g(x,y1,y2)h(x,y1,y2) • f(1)(x) - f(1)(x+y1) - f(1)(x+y2) + f(1)(x+y1+y2) is biased • {f(1)(x), f(1)(x+y1), f(1)(x+y2), f(1)(x+y1+y2)} is not uniform

  31. Solving non-linear dependencies Example 1:f(1)(x)is ε-far from uniform • Bias(f(1)(x))  ε • Degree of f(1)≤ d-1 • We can use induction on “bias implies low rank” • Decompose f(1)into a constant number of polynomials

  32. Solving non-linear dependencies • f(1)(x)= G(g(1)(x),…,g(t)(x)) • deg(g(i)) ≤ deg(f(1)) – 1 • f(1),…,f(k) were used to approximate p(x) • Pr[MAJ(f(1)(x),…,f(k)(x)) = p(x)]  1 – 2-O(d) • Replace f(1) by g(1),…,g(t) • Pr[ MAJ( G(g(1)(x),…,g(t)(x)),f(2)(x)…,f(k)(x)) = p(x)]  1 – 2-O(d) • Replace by a single combiner function C1: Pr[ C1( g(1)(x),…,g(t)(x) ,f(2)(x)…,f(k)(x)) = p(x)]  1 – 2-O(d) • Got a set of “smaller degree” polynomials approximating p

  33. Solving non-linear dependencies Example 2: f(1)(x) - f(1)(x+y1) - f(1)(x+y2) + f(1)(x+y1+y2) is ε-far from uniform • Bias(f(1)(x)+f(1)(x+y1)+f(1)(x+y2)+f(1)(x+y1+y2))  ε • f(1)(x)-f(1)(x+y1)-f(1)(x+y2)+f(1)(x+y1+y2) is a polynomial of degree ≤ d-1 (in variables x,y1,y2) • Again, we can use induction • Here we deviate from the original Green & Tao proof • In large fields, it is enough to consider just Example 1

  34. Solving non-linear dependencies • General solution: If {f(j) (x+iS yi)} is non-uniform • Find a biased linear combination • Decompose it • In each step, we replace a polynomial with a constant number of smaller degree polynomials • We choose what is “non-uniform” adaptively: • If we have T polynomials, we need bias ≤ 2-O(T) of all linear combinations to be close to uniform  Required bias is a function of T • Still, the process stops after finitely many steps • We end with a constant number of polynomials

  35. Getting back to the big picture The proof has two steps: Good regions are excellent: p=F on all points in region (i.e. p(x) is constant on good regions) Assuming almost all regions are excellent, we will prove all regions are excellent (i.e. p(x) is constant on all regions) Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  36. Proof of step 1 • Let x0 be in a good region R • Using the “randomness” of f(1),…,f(k) • {x0+iS yi  R} are almost independent • Joint event occurs with positive probability • In fact, x0+iS yi are almost pairwise-independent, even given that they all lie in R • Since R is good, with positive probability, they all lie in the “good part” of R • Proof: union bound

  37. Proof of step 2 • Assume almost all regions are excellent • i.e. p is constant on these regions • Let x0,x1 be in a (bad) region R • We need to show: p(x0)=p(x1) • Consider x0+iS yi, x1+iS yi • Assume that for any non-empty S: • Region(x0+iS yi) = Region(x1+iS yi) • Assume also this is an excellent region • Then p(x0+iS yi) = p(x1+iS yi) for all non-empty S  p(x0)=p(x1) • For random y1,…,yd+1, this happens with positive probability

  38. Proof of step 2 x0+y1+y2 x1+y1+y2 x0+y2 x0 x0+y1 x1+y2 x1 x1+y1 Good areas: p(x)=F(x) Bad areas: p(x)≠F(x)

  39. Summary of results Bias implies low rank • If: • p(x1,…,xn) degree d polynomial over F • The distribution of p(x) is ε-far from uniform • Then: p(x)=C(f(1)(x),…,f(k)(x)) deg(f(i)) ≤ d-1, k=k(F,d,ε) • In fact: f(j) are derivatives of p f(j)(x) = pa_j(x) = p(x+aj)-p(x)

  40. Summary of results Approximation and computation are equivalent • If: Pr[p(x) = C(g(1)(x),…,g(k)(x))]  1/F + ε deg(g(i)) ≤ d-1 • Then: p(x) = C’(f(1)(x),…,f(k’)(x)) deg(f(i)) ≤ d-1, k’=k’(F,d,ε,k) • In fact: f(j)(x)=pa_j(x) or f(j)=g(j)(x+aj)

  41. Open problems • Give a good bound on the constants • Even in the case of d<<|F| • Given p, can we compute its rank? • Assume p(x)=f1(x) f2(x)+ f3(x) f4(x) • Can we find f1,…,f4 efficiently? • Or is it NP-hard? • Generalization to other “constant-depth” models • E.g. constant-depth circuits

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