shubham sahai srivastava - cse factorization_ shubham sahai srivastava (iitk) factoring integers...

Download Shubham Sahai Srivastava - CSE Factorization_ Shubham Sahai Srivastava (IITK) Factoring Integers January

If you can't read please download the document

Post on 17-Jun-2020

0 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • Factoring Integers via Diophantine Approximation

    Shubham Sahai Srivastava

    Indian Institute of Technology, Kanpur

    ssahai@cse.iitk.ac.in

    January 16, 2014

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 1 / 14

  • Introduction and Surview

    The task of factoring large composite integer N has a long history and is still a challenging problem.

    Here, this task is reduced to the following diophantine approximation :

    Definition (Problem)

    Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

    where, c > 1 and p1, ...pt are first t prime numbers.

    Whats next ??

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 2 / 14

  • Introduction and Surview

    The task of factoring large composite integer N has a long history and is still a challenging problem.

    Here, this task is reduced to the following diophantine approximation :

    Definition (Problem)

    Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

    where, c > 1 and p1, ...pt are first t prime numbers.

    Whats next ??

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 2 / 14

  • Introduction and Surview

    The task of factoring large composite integer N has a long history and is still a challenging problem.

    Here, this task is reduced to the following diophantine approximation :

    Definition (Problem)

    Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

    where, c > 1 and p1, ...pt are first t prime numbers.

    Whats next ??

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 2 / 14

  • Introduction and Surview

    Given these t + 2 diophantine approximations of log N, we can factorize N as follows:

    The integer u := ∏

    ej>0 p ej j must be close approximation to vN, where

    v := ∏

    ej 1, β, γ ≥ 0 be fixed and let pt < N. If (e1, ..., et) ∈ Zt satisfies the inequalities

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp β+o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2δ log pt

    then we have for u := ∏

    ej>0 p ej j , v :=

    ∏ ej

  • Introduction and Surview

    Given these t + 2 diophantine approximations of log N, we can factorize N as follows:

    The integer u := ∏

    ej>0 p ej j must be close approximation to vN, where

    v := ∏

    ej 1, β, γ ≥ 0 be fixed and let pt < N. If (e1, ..., et) ∈ Zt satisfies the inequalities

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp β+o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2δ log pt

    then we have for u := ∏

    ej>0 p ej j , v :=

    ∏ ej

  • Introduction and Surview

    Given these t + 2 diophantine approximations of log N, we can factorize N as follows:

    The integer u := ∏

    ej>0 p ej j must be close approximation to vN, where

    v := ∏

    ej 1, β, γ ≥ 0 be fixed and let pt < N. If (e1, ..., et) ∈ Zt satisfies the inequalities

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp β+o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2δ log pt

    then we have for u := ∏

    ej>0 p ej j , v :=

    ∏ ej

  • Introduction and Surview

    So, we have |u − vN| ≤ p1+o(1)t

    Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

    And we obtain a non-trivial congruence∏ ej>0

    p ej j = ±

    ∏t j=1 p

    ej j (mod N).

    Given t + 2 of these congruences we compute x , y satisfying x2 = y2

    (mod N)

    So, we can compute a factor of N as gcd(x+y, N).

    This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

  • Introduction and Surview

    So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

    And we obtain a non-trivial congruence∏ ej>0

    p ej j = ±

    ∏t j=1 p

    ej j (mod N).

    Given t + 2 of these congruences we compute x , y satisfying x2 = y2

    (mod N)

    So, we can compute a factor of N as gcd(x+y, N).

    This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

  • Introduction and Surview

    So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

    And we obtain a non-trivial congruence∏ ej>0

    p ej j = ±

    ∏t j=1 p

    ej j (mod N).

    Given t + 2 of these congruences we compute x , y satisfying x2 = y2

    (mod N)

    So, we can compute a factor of N as gcd(x+y, N).

    This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

  • Introduction and Surview

    So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

    And we obtain a non-trivial congruence∏ ej>0

    p ej j = ±

    ∏t j=1 p

    ej j (mod N).

    Given t + 2 of these congruences we compute x , y satisfying x2 = y2

    (mod N)

    So, we can compute a factor of N as gcd(x+y, N).

    This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

  • Introduction and Surview

    So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

    And we obtain a non-trivial congruence∏ ej>0

    p ej j = ±

    ∏t j=1 p

    ej j (mod N).

    Given t + 2 of these congruences we compute x , y satisfying x2 = y2

    (mod N)

    So, we can compute a factor of N as gcd(x+y, N).

    This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

  • Reduction to Lattice problem

    So, we are good to go, if we are able to solve the following problem:

    Definition (Diophantine Approximation Problem)

    Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

    where, c > 1 and p1, ...pt are first t prime numbers.

    The above problem can be formulated as a nearly closest vector problem in the 1-norm.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 5 / 14

  • Reduction to Lattice problem

    So, we are good to go, if we are able to solve the following problem:

    Definition (Diophantine Approximation Problem)

    Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

    where, c > 1 and p1, ...pt are first t prime numbers.

    The above problem can be formulated as a nearly closest vector problem in the 1-norm.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 5 / 14

  • Reduction to Lattice problem

    So, we are good to go, if we are able to solve the following problem:

    Definition (Diophantine Approximation Problem)

    Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

    1. | ∑t

    i=1 ei log pi − log N| ≤ N−cp o(1) t

    2. | ∑t

    i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

    where, c > 1 and p1, ...pt are first t prime numbers.

    The above problem can be formulated as a nearly closest vector problem in the 1-norm.

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 5 / 14

  • Reduction to Lattice problem

    We associate with N a point N ∈ Rt+1

    and with the primes p1, ..., pt a lattice L ⊂ Rt+1 of rank t and basis B.

    B =

     log p1 0 · · · 0

    ... . . .

    ...

    0 . . . 0

    0 0 · · · log pt Nc log p1 N

    c log p2 · · · Nc log pt

     , N = 

    0 0 ... 0

    Nc ln N ′

     , c ≥ 1

    Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 6 / 14

  • Reduction to Lattice problem

    We associate with N a point N ∈ Rt+1 and with the primes p