optical architecture for (restricted) exponential time hard problems nova fandina ben-gurion...

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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

Ben Gurion University of the Negev 19/5/2011

Ben Gurion University of the Negev 2

Outline

Searching for Hard Problem

Succinct Permanent (mod p) is NEXP Time Hard

Succinct Permanent (mod p) Has “Many ” Hard Instances

Holographic Based Optical Architecture

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



Succinct representation of Graphs[GW83]

Small Circuit Representation output

0 1 2 3 n-1 0 1 2

n-1 log n log n

1

0010

0011

0010 0011

1

1


Computational problems with succinctly represented inputs

Succ_ problem:𝚷input: succinct representation of the graph Goutput: (G)𝚷 [PY86] If 3SAT 𝚷 then Succ_ 𝚷 is NEXP time hard


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

11 1

1

n

n n

n nn

a a

A

a a

n

iσ(i)σ i 1

Perm(A) a

|


Succinct Permanent modulo (small) prime p

input: small boolean circuit representing an integer matrix A with bounded (positive and negative)

entries, prime p (in binary representation)

k- constant c -constant

output: perm A (mod p)

NEXP hard & Big Set of Hard Instances

Outline

8Ben Gurion University of the Negev






Zero Succinct Permanent :input: small boolean circuit C representing integer matrix A with bounded entriesoutput: permanent(A)==0

Zero Succinct Permanent (mod p):input: small boolean circuit C representing integer matrix A with bounded entries, small prime poutput: permanent(A)==0 (mod p)


Roadmap

• Zero Succinct Permanent NEXP time hard

• Zero Succinct Permanent

Zero Succinct Permanent (mod p)

• Zero Succinct Permanent (mod p ) ≤ Succinct Permanent (mod p)


[PY86]• Succinct 3SAT is NEXP hard

[Val77]• #3SAT Permanen

Zero Succinct Permanent is NEXP hard


Roadmap

• Zero Succinct Permanent NEXP time hard

• Zero Succinct Permanent

Zero Succinct Permanent (mod p)

• Zero Succinct Permanent (mod p ) ≤ Succinct Permanent (mod p)


• C represent an integer matrix A with integer values constant |Permanent (A)|

• Chinese Reminder Theorem: Permanent(A) can be computed by computing Permanent(A) modulo each prime p {p


Define X be a set of first 2|U| primes

The number of prime number in [1,x] is:

The length of representation of each prime in X is polylogarithmical


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


Pick a prime p’ uniformly at random from the set X

• pick p’ uniformly at random from [1, ]

• while(! primality test(p’) ) p’ = pick [1, ]

Primality test: AKS[04]Expected number of attempts: O(logn)

Zero Succinct Permanent (mod p) is NEXP hard (via randomized reduction)

Outline




Succinct Permanent (mod p) Has Many Hard Instances



Computing Permanent over

XRPeraRPeraRPeraAPer nn

iiii )()()()( loglog

22

1

********

********

********

********

****

********

********

********

4321 tttt

********

********

********

********

****

********

********

********

42

32

222 aaaa

********

********

********

********

****

********

********

********

43

33

233 aaaa

********

********

********

********

****

********

********

********

44

34

244 aaaa

********

********

********

********

****

********

********

********

41

31

211 aaaa

********

********

********

********

****

********

********

********

45

35

255 aaaa


Given an answers for (log n +1) matrices chosen at random from the set, compute an answer for matrix A in polynomial time :

solve a system of equations to find

Unique solution exists Computations mod p

naaa log1

2111

naaa log2

2221

nnnn aaa log1log

21log1log1

)( 1APerm

)( 2APerm

)( 1log nAPerm

Outline







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


Preprocessing Procedure

A A A A

O O O O

1 1 1 0 1 0 0 0

1 1 0 1 0 1 0 0

1 0 1 1 0 0 1 0

0 1 1 1 0 0 0 1


Preprocessing Procedure

1 1 1 0 1 0 0 0

1 1 0 1 0 1 0 0

1 0 1 1 0 0 1 0

0 1 1 1 0 0 0 1


Holographic implementation

Recording phase:


Reconstruction phase:


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


Thank you!

optical architecture for (restricted) exponential time hard problems nova fandina ben-gurion...

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