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Optical Architecture for (Restricted) Exponential Time Hard Problems Nova Fandina Ben-Gurion University of the Negev, Israel Joint work with: Prof. Shlomi Dolev & Prof. Joseph Rosen Ben-Gurion University of the Negev, Israel 1Ben Gurion University of the Negev19/5/2011

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Outline 2Ben Gurion University of the Negev Searching for Hard Problem Succinct Permanent (mod p) is NEXP Time Hard Succinct Permanent (mod p) Has “Many ” Hard Instances Holographic Based Optical Architecture

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Modern Cryptographic Schemes Based on unproven complexity assumptions… what happens if P = NP ? NEXP hard: don’t have a polynomial time solution Hard on the average: randomly chosen instance is hard with high probability Ben Gurion University of the Negev3

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Succinct representation of Graphs [GW83] 1 4Ben Gurion University of the Negev

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Computational problems with succinctly represented inputs 5Ben Gurion University of the Negev

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Succinct Permanent Permanent problem where the summation is over all permutations σ of {1,…n} #P - Complete [Val77] Hard on Average as on the Worst Case [Lip91] Succinct Permanent the output can be too big 6Ben Gurion University of the Negev

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Outline 8Ben Gurion University of the Negev Searching for Hard Problem Succinct Permanent (mod p) is NEXP Time Hard Succinct Permanent (mod p) Has “Many ” Hard Instances Holographic Based Optical Architecture

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Zero Succinct Permanent : input: small boolean circuit C representing integer matrix A with bounded entries output: permanent(A)==0 Zero Succinct Permanent (mod p): input: small boolean circuit C representing integer matrix A with bounded entries, small prime p output: permanent(A)==0 (mod p) 9Ben Gurion University of the Negev

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Randomized algorithm: pick a prime p’ uniformly at random from the set X compute Perm(A) mod p’ if (Perm(A) mod p’ == 0) return Perm(A)==0 else return Perm(A)!=0 If Per(A)==0 the answer is correct with probability 1 If Per(A)!=0 the answer is correct with probability > ½ 15Ben Gurion University of the Negev

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Outline 17Ben Gurion University of the Negev Searching for Hard Problem Succinct Permanent (mod p) is NEXP Time Hard Succinct Permanent (mod p) Has Many Hard Instances Holographic Based Optical Architecture

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Outline 20Ben Gurion University of the Negev Searching for Hard Problem Succinct Permanent (mod p) is NEXP Time Hard Succinct Permanent (mod p) Has “Many ” Hard Instances Holographic Based Optical Architecture

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Optical Device for restricted Succinct Permanent( mod p ) Solves the instances of the balanced form Preprocessing unit: generates and records all matrices that can be represented by balanced small boolean circuits (holographic based implementation) Optical Solver: given an instance outputs an encoded matrix. Forward matrix as an instance to the existing Permanent Solver. Applies mod p operation to the result 21Ben Gurion University of the Negev

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Preprocessing Procedure AAAA O OO O 11101000 11010100 10110010 01110001 22Ben Gurion University of the Negev

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Preprocessing Procedure 11101000 11010100 10110010 01110001 23Ben Gurion University of the Negev

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11…1 11…1 11..1 11…1 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 00…0 G1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 R(G1) 11…1 G1 11…1 G1 11…1 G1 11…1 G1 11…1 G1 11…1 G1 11…1 G1 11…1 G1 11…1 G1 11…1 24Ben Gurion University of the Negev

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Holographic implementation Recording phase: 25Ben Gurion University of the Negev

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Reconstruction phase: 26Ben Gurion University of the Negev

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Conclusions Establishing a computational complexity of Succinct Permanent Problem mod p – NEXP time hard via randomized reduction – Average case complexity detect a hard instance and compose many hard instances Optical Solver device – restricted inputs – existence of the Permanent Solver 27Ben Gurion University of the Negev

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Thank you! 28Ben Gurion University of the Negev

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1 1. Which of these sequences correspond to Hamilton cycles in the graph? (a) 0 1 3 5 4 2 (b) 0 1 2 4 3 5 (c) 0 1 4 3 1 2 (d) 0 2 3 5 4 1 (e) 0 1 3 5 6.

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