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Hardness of Reconstructing Multivariate Polynomials. Parikshit Gopalan U. Washington Parikshit Gopalan U. Washington Subhash Khot NYU/Gatech Rishi Saket Gatech/NYU Rishi Saket Gatech/NYU

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Curve Fitting Problem: Given data points, find a low degree polynomial that fits best. Easy if there is a perfect fit. Well studied problem …

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Curve Fitting through the ages

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Statistics: Least Squares

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Polynomial Reconstruction Coding Theory Computational Learning Cryptography PCPs Pseudorandom -ness

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The Reconstruction Problem Input: Degree d. PointsValues x1x1x1x1 f(x 1 ) xixixixi f(x i ) xmxmxmxm f(x m ) Output: A degree d polynomial that best fits the data. In this talk: Finite fields, Hamming distance.

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The Reconstruction Problem Input: Degree d, set S, values f(x) for x 2 S. Output: A degree d polynomial that best fits the data. Parameters that matter: 1. Degree d, Field F. 2. Set S. 3. How good is the best fit? (error-rate )

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Algorithms for Reconstruction Univariate Case [Sudan, Guruswami-Sudan]: Multivariate Case [Goldreich-Levin, Goldreich- Rubinfeld-Sudan, Arora-Sudan, Sudan-Trevisan-Vadhan]: Can tolerate very high error rate. Are these algorithms optimal?

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Hardness Results: Univariate Case Degree d polynomials, n points in F. [Guruswami-Vardy]: NP-hard to tell if some degree d poly. has d +2 agreements. [Guruswami-Sudan]: Can tell if some degree d poly. has (nd) 0.5 agreement.

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Hardness Results: Multivariate Case Linear polynomials over F 2 [Hastad] : NP-hard to tell if Some linear poly. satisfies 1- fraction of points. Some linear poly. satisfies 1- fraction of points. Every linear poly. satisfies less than fraction of points. Every linear poly. satisfies less than fraction of points. Extends to any F and d =1. Implies something for d < F. d ¸ 2 over F 2 : Nothing known.

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Our Results Over F 2 for any d, NP-hard to tell whether Some linear polynomial satisfies 1- fraction of points. Some linear polynomial satisfies 1- fraction of points. Every degree d polynomial satisfies at most d + fraction of points. Every degree d polynomial satisfies at most d + fraction of points. SZ Lemma: For a degree d poly P 0 over F 2, Pr x [ P(x) 0] ¸ 2 -d. d=1 ½ + ½ + d=2 ¾ + ¾ +

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Our Results Over F q for any d, NP-hard to tell whether Some linear polynomial satisfies 1- fraction of points. Some linear polynomial satisfies 1- fraction of points. Every degree d polynomial satisfies at most c(d,q) + fraction of points. Every degree d polynomial satisfies at most c(d,q) + fraction of points. c(d,q): Schwartz-Zippel for polynomials of total degree d over F q.

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Overview of Reduction Reducing from Label-Cover. Reducing from Label-Cover. Dictatorship Testing. Dictatorship Testing. Consistency Testing. Consistency Testing. Putting it all together. Putting it all together.

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Label Cover Graph: G(V,E), |V| =n. Labels: [k] Edges: e ½ [k] £ [k] Goal: Find a labeling satisfying all edges. Thm [PCP + Raz]: It is NP-hard to tell if Some labeling satisfies all edges. Every labeling satisfies · frac. of edges. 1 2 n 3

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X 1 1 X 1 2 … X 1 k The Reduction X 2 1 X 2 2 … X 2 k X 3 1 X 3 2 … X 3 k X n 1 X n 2 … X n k Henceforth d =2, field = F 2. Constraints: Points in {0,1} nk + values. Yes Case: Some L satisfies most constraints. No Case: No Q satisfies many constraints.

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The Reduction X 1 1 X 1 2 … X 1 k X 2 1 X 2 2 … X 2 k X 3 1 X 3 2 … X 3 k X n 1 X n 2 … X n k If l(v) is a good labelling, then L = v X v l(v) will satisfy most points.

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The Reduction X 1 1 X 1 2 … X 1 k X 2 1 X 2 2 … X 2 k X 3 1 X 3 2 … X 3 k X n 1 X n 2 … X n k If l(v) is a good labelling, then L = v X v l(v) will satisfy most points. Any Q that does ¾ + gives a labelling satisfying fraction of edges.

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Dictatorship: Q 1 = Q(X 1 1,…,X 1 k,0,..,0). Q 1 looks like a Dictator X 1 j. Will settle for small list. Consistency: Some pair of labels in the list satisfy. 3, 71, 99 17, 45 Overview of Reduction Constant independent of k.

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Dictatorship: Q 1 = Q(X 1 1,…,X 1 k,0,..,0). Q 1 looks like a Dictator X 1 j. Will settle for small list. Can enforce this for frac. of vertices. Consistency: Some pair of labels in the list satisfy. Can enforce this for all edges. 3, 71, 99 17, 45 Overview of Reduction

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3, 71, 99 17, 45 Overview of Reduction If Q does ¾ + Small list for frac. of vertices. Consistency for all edges. Assign random labels from list. Satisfies constant fraction of edges.

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Overview of Reduction Dictatorship Testing. Dictatorship Testing. Consistency Testing. Consistency Testing. Putting it all together. Putting it all together.

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Overview of Reduction Dictatorship Testing. Dictatorship Testing. Consistency Testing. Consistency Testing. Putting it all together. Putting it all together.

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Dictatorship Testing for low- degree Polynomials. Input: Q(X 1,…,X k ) of degree 2. Goal: Design a test s.t Every dictatorship X i passes w.p close to 1. Every dictatorship X i passes w.p close to 1. If Q does better than ¾, it is close to a dictatorship. If Q does better than ¾, it is close to a dictatorship. Test: Pick a random point x 2 {0,1} k. Check if Q(x) = y. Mini reconstruction problem! Small List

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Dictatorship Testing for low- degree Polynomials. Dictatorships Quadratic polys. All polys.

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Dictatorship Testing [Hastad, Bourgain, MOO] Dictatorships All polys. Hard to do with just 2 queries.

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Dictatorship Testing for low- degree Polynomials. Dictatorships Quadratic polys. Poly. is of low degree. Allowed one query (!)

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Dictatorship Test (0,…,0) (1,…,1) Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. Uniform dist: Quadratic polys. are 3:1 balanced. -biased: Dictatorships are highly skewed. Is there a converse? Each i =1 independently. w.p

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Dictatorship Test Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. X i passes w.p 1-. X i X j passes w.p X 1 (X 1 + … + X ) + X 2 (X + …) passes w.p 1 - 2

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Dictatorship Test Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. Define G(Q) to be the graph of Q. Q = X 1 X 2 + X 2 X 3, G(Q) = Thm: If Q passes w.p ¾ +, then G(Q) has no large matchings

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Dictatorship Test Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. Thm: If Q passes w.p ¾ +, then G(Q) has no large matchings. 1. Large matching: Independent monomials. 2. Only small matchings: Small vertex cover. X 1 L 1 + X 2 L 2

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Dictatorship Test Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. Thm: If Q does better than ¾, then G(Q) has no large matchings. QQ X i = 0 w.p 1- 2 X i 2 R {0,1} c = ? 0 If G(Q) has a large matching, then Q 0 w.h.p. If Q 0, then c =1 w.p ¸ ¼ (SZ lemma). If Q does well, G(Q) has no large matchings.

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Dictatorship Test Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. Thm: If Q does better than ¾, then G(Q) has no large matchings. If G(Q) has a large matching, then Q 0 w.h.p. Each edge survives w.p 4 2. Events for each matching edge are independent.

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Dictatorship Test Dictatorship Test: Pick 2 {0,1} k from the -biased distribution. Check if Q( ) = 0. Define G(Q) to be the graph of Q. Q = X 1 X 2 + X 2 X 3, G(Q) = Thm: If Q passes w.p ¾ +, then G(Q) has no large matchings Small List: Vertex set of a maximal matching.

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Dictatorship: Assign a small list to a vertex. Consistency: Some pair of labels in the list satisfy. 3, 71, 99 17, 45 Overview of Reduction

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Overview of Reduction Dictatorship Testing. Dictatorship Testing. Consistency Testing. Consistency Testing. Putting it all together. Putting it all together.

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Consistency Testing l(x) = l(y)

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Consistency Testing X 1 X 2 … X k Y 1 Y 2 … Y k l(x) = l(y) Given Q(X 1,…,X k,Y 1,…,Y k ) s.t Q(X i ) and Q(Y j ) both pass the dict. Test. Want Q(X 1,..,X k,0,…,0) = Q(0,…,0,Y 1,…,Y k ). Test: Q(r,0) = Q(0,r) for r 2 R {0,1} k. Two queries!

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Consistency via Folding X 1 X 2 … X k Y 1 Y 2 … Y k l(x) = l(y) Yes case: Q = X i + Y i for some i. All of them vanish over H = (r,r). Constant on each coset of H. Enforce this on Q even in the No case. H

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Consistency via Folding Def: Q is folded over subspace H µ {0,1} k if Q is constant on every coset of H. Examples: Linear polys., juntas. H Thm: Q is folded over H iff for some nice basis ( 1,…, t, 1,..., k-t ), Q = R( 1,…, t ) is a t-junta for t = k – dim(H) In the nice basis ( 1,…, t, 1,..., k-t ) i s: coset of H, j s: position in coset.

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Template for Folding Want Q folded over a subspace H. Compute nice basis ( i, j ). Ask for R( 1,…, t ). To test if Q(x) = y o Let x = ( ); test R( ) = y. For analysis: Rewrite R( ) as Q(x). Now Q is folded. H {0,1} n /H

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Consistency via Folding l(x) = l(y) Fold over H = (r,r) for r 2 {0,1} k. Polys. folded over H can be written as: Q(X 1,…,X k,Y 1,…,Y k ) = R(X 1 +Y 1, …, X k + Y k ) Gives Q(X 1,…,X k ) = Q(Y 1,…,Y k ).

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Dictatorship: Assign a small list to a vertex. Consistency: Some pair of labels in the list satisfy. 3, 71, 99 17, 45 Overview of Reduction

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Consistency via Folding l(x) = l(y) Fold over H = (r,r) for r 2 {0,1} k. Polys. folded over H can be written as: Q(X 1,…,X k,Y 1,…,Y k ) = R(X 1 +Y 1, …, X k + Y k ) Gives Q(X 1,…,X k ) = Q(Y 1,…,Y k ). List of X i s: Vertex set of maximal matching. Every two maximal matchings intersect.

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Consistency Test 1.Graphs of restrictions are the same. 2.Graph has no large matchings. 3. List = Vertex set of maximal matching. G(Q(X 1,…,X k )) G(Q(Y 1,…,Y k ))

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Consistency Test 1.Graphs of restrictions are the same. 2.Graph has no large matchings. 3. List = Vertex set of maximal matching. G(Q(X 1,…,X k )) G(Q(Y 1,…,Y k ))

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Summary of Reduction Each constraint gives H ½ {0,1} nk. Fold over the span of all H. Run Dict. test on every vertex. No explicit consistency tests. If Q passes w.p ¾ +, fraction of vertices do well on Dict. test. Consistency for all edges by folding.

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Overview of Reduction Dictatorship Testing. Dictatorship Testing. Consistency Testing. Consistency Testing. Putting it all together. Putting it all together.

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Projections … X 1 1 X 1 2 … X 1 k X 2 1 X 2 2 … X 2 k X 3 1 X 3 2 … X 3 k X n 1 X n 2 … X n k Can handle equality, permutations. Need perfect completeness: no UGC. Have to deal with projections.

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Projections … 1, 2, k (l u ) = (l v ) 1, 2, k 1,,t [k] ! [t]

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Projections … 1, 2, k (l u ) = (l v ) 1, 2, k 1,,t [k] ! [t]

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Projections … Decoding is a vertex cover for G(Q i ). Need to show that every two vertex covers intersect.

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Projections … Do every two vertex covers of G intersect? No:

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Projections … Do every two vertex covers of G intersect? … but in any three VCs, some pair intersects. No:

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Hypergraph Label Cover 1, 2, k 1,,t [k] ! [t] 1, 2, k [k] ! [t] Strongly satisfied: all 3 projections agree. Weakly satisfied: some 2 agree. Thm: NP-hard to tell if all edges are strongly satsified or at most are weakly satified.

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Main Theorem Over F 2 for any d, NP-hard to tell whether Some linear polynomial satisfies 1- fraction of points. Some linear polynomial satisfies 1- fraction of points. Every degree d polynomial satisfies at most d + fraction of points. Every degree d polynomial satisfies at most d + fraction of points.

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Better Hardness? Problem: Can we improve soundness to ? Bottleneck: Dictatorship test. Present analysis is optimal in general: Q = (X X k )(X k+1 + … +X 2k ) passes w.p ¾. Can assume that Q is balanced.

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Thank You!

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Curve Fitting in Deep Space

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G : graph on k vertices. Alice and Bob have G. Want to pick a common vertex. G Consistency Testing

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G : graph on k vertices. Pick a maximal matching; output a random vertex. G Consistency Testing

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Overall Reduction X 1 1 X 1 2 … X 1 k X 2 1 X 2 2 … X 2 k X 3 1 X 3 2 … X 3 k Fold over consistency constraints. Test dictatorship on every vertex. If fraction do better than ¾, many edges are satisfied.

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