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The Complexity of Linear Dependence Problems in Vector Spaces David Woodruff IBM Almaden Joint work with Arnab Bhattacharyya, Piotr Indyk, and Ning Xie from MIT

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The 3-SUM Problem Given a set S containing r real numbers, are there: a, b, c 2 S with a+b+c = 0? Solve in O(r 2 ) time –Interview question Conjectured to require (r 2 ) time Useful for hardness results in P. Many problems are 3-SUM Hard

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Generalizations We study generalizations of this problem: –Replace 3 summands with k summands R –Replace real field R with a finite field –Replace sum of field elements with sum of vectors –Replace sum with a fixed linear combination –Replace sum with any linear combination –Require vectors be minimally linearly dependent –Replace target 0 with an arbitrary vector –and so on…

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Applications Maximum Likelihood Decoding - Given x 1, …, x r in F q n and z in F q n, do there exist x i 1, …, x i k that contain z in their span? - x i are the columns of a parity-check matrix - z is the syndrome - there is a codeword corrupted in at most k positions with syndrome z iff the k-span contains z Weight Distribution Problem –Let A be an n x r matrix over F 2 –Define the code C = {x | Ax = 0} –C has a codeword of weight k iff k columns of A sum to 0

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Formal Definitions In this talk, we focus on two problems: (k,r)-LinDependence: given r elements x 1, …, x r in F 2 n and z in F 2 n, do there exist x i 1, …, x i k that span z? (k,r)-ZeroSum: given r elements x 1, …, x r in F 2 n, do there exist x i 1, …, x i k with x i 1 + x i 2 + … + x i k = 0? We allow k and r to be functions of n First problem at least as hard as second

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Results Assume 3-SAT cannot be solved in time less than 2 cn for a constant c > 0 Then (k,r)-ZeroSum requires min(r k, 2 n ) time, up to polynomial factors –So, (k,r)-LinDependence requires min(r k, 2 n ) time –Other variants also require this time Have matching upper bound: –r k is trivial. Can get roughly r k/2 –Can get 2 n with the FFT

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Implications (k,r)-LinDependence reduces to Maximum Likelihood Decoding, so min(r k, 2 n ) lower bound (k,r)-ZeroSum reduces to the Weight Distribution Problem, so min(r k, 2 n ) lower bound Results improve previous best r k 1/4 lower bounds for these coding theory problems [Downey, Fellows, Vardy, Whittle] Hold for r and k functions of n

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R Our starting point: [PW] showed an r bound for k-SUM over R assuming 3-SAT on n variables requires 2 cn time: 3-SAT formula F with n variables and m clauses Á1Á1 Á2Á2 … ÁsÁs - s = 2 εn. Each Á i has n variables and O(n) clauses [CIP] Ã1Ã1 Ã2Ã2 … ÃsÃs - Á i replaced with 1- in-3-SAT formula Ã i - Ã i converted to k-SUM instance Each k-SUM instance on a set of r = 2 Θ(n/k) real numbers. If can solve k-SUM in time r o(k), can solve 3-SAT in time r o(k) ¢ 2 εn This ensures bit complexity of resulting numbers is small

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Reducing a 1-in-3-SAT formula Ã i on n variables and O(n) clauses to k-SUM on r = 2 Θ(n/k) real numbers G1G1 …GiGi …GkGk v i,1 v i,2 v i,3 …v i, 2 n/k In each group G i, create a real number v i,j for each possible assignment to its n/k variables v i,j : k group indicator digitsO(n) clause digits Base-k representation i-th indicator digit is 1 iff v 2 G i j-th clause digit is 1 iff A(v) sets exactly 1 literal of j-th clause to 1 All other digits are 0 Partition variables into k groups G 1, …, G k of n/k variables Ã i is true iff there are k real numbers that sum to 1 k + O(n)

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G1G1 …GiGi …GkGk v i,1 v i,2 v i,3 …v i, 2 n/k In each group G i, create a real number v i,j for each possible assignment to its n/k variables v i,j : k group indicator digitsO(n) clause digits Base-k representation i-th indicator digit is 1 iff v 2 G i j-th clause digit is 1 iff A(v) sets exactly 1 literal of j-th clause to 1 All other digits are 0 Partition variables into k groups G 1, …, G k of n/k variables Can we do the same for F 2 ? In each group G i, create a vector v i,j for each possible assignment to its n/k variables k group coordinatesO(n) clause coordinates k + O(n) coordinates i-th indicator coordinate is 1 iff v 2 G i j-th clause coordinate is 1 iff A(v) sets exactly 1 literal of j-th clause to 1 All other coordinates are 0 - A sum of k vectors over F 2 can equal 1 k+O(n), but just means an odd number of literals in each clause are true - Odd-SAT is easy - A sum of k vectors over F 2 can equal 1 k+O(n), but just means an odd number of literals in each clause are true - Odd-SAT is easy

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Our Modifications 3-SAT formula F with n variables and m clauses Á1Á1 Á2Á2 … ÁsÁs - s = 2 εn. Each Á i has n variables and O(n) clauses [CIP] Ã1Ã1 Ã2Ã2 … ÃsÃs - Á i replaced with NAE-SAT Formula Ã i - Ã i converted to (k,r)-ZeroSum Each (k,r)-ZeroSum instance on a set of r = 2 Θ(n/k) vectors. If can solve (k,r)-ZeroSum in time r o(k), solve 3-SAT in time r o(k) ¢ 2 εn - A NAE-SAT formula Ã i is 1 if for each clause, at least one but not all literals are true - - A NAE-SAT formula Ã i is 1 if for each clause, at least one but not all literals are true - R - With 1-in-3-SAT over R, variables in different groups independently update the clause digit We need interaction between groups - Before this was used for bit complexity. - Now it determines the number of dimensions - Before this was used for bit complexity. - Now it determines the number of dimensions

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Interacting Variables We can replace duplicates of a variable with distinct variables and introduce equality constraints –preserve NAE-SAT and · 3 literals per clause –each variable occurs in a constant number of clauses For each clause (a Ç b Ç c), we introduce pairvairs –1 variable is [a, b], 1 variable is [b, c], and 1 variable is [c, a] Partition original n variables into k groups G i of n/k variables For a pairvar [a,b], –if original variables a and b occur in the same group G i, place [a,b] in G i –else, if a 2 G i and b 2 G j, place [a,b] in G min(i, j) G i still has O(n/k) variables

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New Reduction In each group G i, create a vector v i,j for each assignment to its n/k variables as well as variables in G i s pairvars v i,j : k group coordinates O(n) clause coordinates k+O(n) coordinates O(n) consistency coordinates i-th group coordinate is 1, the others are 0 clause coordinates more complicated depend on variables and pairvars assigned to the group consistency coordinates allow for assignments to the same variable from different groups to be patched together 1 pair of consistency coordinates for each pairvar (a,b)

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Clause Coordinates Clause coordinates are set so that for a consistent assignment (i.e., group and consistency coordinates are ok), then for clause with literals a, b, c –v(a) + v(b) + v(c) – v(a) ¢ v(b) – v(b) ¢ v(c) – v(a) ¢ v(c) –v(.) denotes the value assigned Case analysis –Clause only equals 1 if exactly 1 or 2 literals are true

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Upper Bounds Consider functions f: Z 2 n ! {0,1} RFourier transform F: Z 2 n ! R is F(x) = 2 -n ¢ y f(y) ¢ (-1) Fast Fourier Transform computes F from f in O(n ¢ 2 n ) time Let f be indicator function of input set of r vectors. Then sum v 1 + v 2 + … v k = 0 f(v 1 ) ¢ f(v 2 ) f(v k ) is what we want This is just 2 n times the 0 n -Fourier coefficient of f k So we can get O(n ¢ 2 n ) time instead of the trivial 2 nk

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Conclusion Assuming 3-SAT cannot be solved in time less than 2 cn for a constant c > 0, –(k,r)-LinDependence and (k,r)-ZeroSum require min(r k, 2 n ) time (up to polynomial factors) –Same bound holds for many similar problems –Almost matching upper bounds –New way to prove hardness in coding theory –Optimal hardness of basic problems in coding theory

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