Tower Number Field Sieve Variant of a Recent Polynomial Selection - - PowerPoint PPT Presentation

tower number field sieve variant of a recent polynomial
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Tower Number Field Sieve Variant of a Recent Polynomial Selection - - PowerPoint PPT Presentation

Rump Session 2016 Tower Number Field Sieve Variant of a Recent Polynomial Selection Method Palash Sarkar; Shashank Singh Indian Statistical Institute The Tower Number Field Sieve + SS Polynomial Selection Z [ z ] R = R [ x ] h ( z ) x


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Rump Session 2016

Tower Number Field Sieve Variant

  • f a Recent Polynomial Selection

Method

Palash Sarkar; Shashank Singh

Indian Statistical Institute

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The Tower Number Field Sieve + SS Polynomial Selection

R[x] Q(α) ⊂

R[x] f (x) R[x] g(x) ⊂ Q(β)

R/pR = Fpm α → x x → β α → m

R mod p

m → β

R m

  • d

p

R =

Z[z] h(z)

Barbulescu et al. (Asiacrypt 2015)

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The Tower Number Field Sieve + SS Polynomial Selection

R[x] Q(α) ⊂

R[x] f (x) R[x] g(x) ⊂ Q(β)

R/pR = Fpm α → x x → β α → m

R mod p

m → β

R m

  • d

p

R =

Z[z] h(z)

Barbulescu et al. (Asiacrypt 2015)

Algorithm: A: A new method of polynomial selection for NFS. Input: p, n, d (a factor of n) and r ≥ n/d. Output: f (x), g(x) and ϕ(x). Let k = n/d; repeat Randomly choose a monic irr A1(x) with small coeff.: deg A1 = r + 1; mod p, A1(x) has an irr factor A2(x) of deg k. Choose monic C0(x) and C1(x): deg C0 = d and deg C1 < d. Define f (x) = Resy (A1(y), C0(x) + y C1(x)) ; ϕ(x) = Resy (A2(y), C0(x) + y C1(x)) mod p; ψ(x) = LLL(MA2,r); g(x) = Resy (ψ(y), C0(x) + y C1(x)) . until f (x) and g(x) are irr over Z and ϕ(x) is irr over Fp.; return f (x), g(x) and ϕ(x).

Sarkar-Singh Polynomial Selection Algorithm (Eurocrypt 2016)

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

exTNFS

Taechan Kim and Razvan Barbulescu, Extended Tower Number Field Sieve: A New Complexity for Medium Prime Case - Cryptology ePrint Archive: Report 2015/1027

Setup (FQ):

Q = pn, where n = η × κ and gcd(η, κ) = 1 Complexity of NFS for non-prime field is better for boundary case i.e., p = LQ(2/3, cp). Idea is to leverage the boundary case complexity by increasing p.

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 2 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

Polynomial Selection for TNFS

Palash Sarkar and Shashank Singh, Tower Number Field Sieve Variant of a Recent Polynomial Selection Method. - Cryptology ePrint Archive: Report 2016/401 Polynomial Selection method subsumes GJL method. Polynomial Selection method generalises Conjugation method. It gives the new trade-offs which not covered by GJL and Conjugation method.

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 3 / 9

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Algorithm: B: Polynomial selection for TNFS. Input: p, n = ηκ, d (a factor of κ) and r ≥ κ/d. Output: h(x), f (x), g(x) and ϕ(x). Let k = κ/d; Randomly choose h(z) of deg η with small coeffs and irreducible modulo p. Let R = Z[z]/h(z). repeat Randomly choose a monic irr A1(x) with small coeff.: deg A1 = r + 1; mod p, A1(x) has an irr factor A2(x) of deg k. Choose monic C0(x) and C1(x): deg C0 = d and deg C1 < d. Define f (x) = Resy (A1(y), C0(x) + y C1(x)) ; ϕ(x) = Resy (A2(y), C0(x) + y C1(x)) mod p; ψ(x) = LLL(MA2,r); g(x) = Resy (ψ(y), C0(x) + y C1(x)) . until f (x) and g(x) are irr over R and ϕ(x) is irr over Fpη[z]/h(z).; return h(x), f (x), g(x) and ϕ(x).

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

Example

Let p is a 201-bit prime given below. p = 1606938044258990275541962092341162602522202993782792835301611 and n = 6. Let (η, κ) = (3, 2).

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 5 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

Example

Let p is a 201-bit prime given below. p = 1606938044258990275541962092341162602522202993782792835301611 and n = 6. Let (η, κ) = (3, 2).

Taking d = κ and r = 1, we get the following polynomials. h(x) = x3 + x2 + 15 x + 7 f (x) = x4 − x3 − 2 x2 − 7 x − 3 g(x) =

717175561486984577278242843019 x2 + 2189435313197775056442946543188 x

+2906610874684759633721189386207 Note that g∞ ≈ 2101.

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 5 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

Example

Let p is a 201-bit prime given below. p = 1606938044258990275541962092341162602522202993782792835301611 and n = 6. Let (η, κ) = (3, 2).

Taking d = κ and r = 1, we get the following polynomials. h(x) = x3 + x2 + 15 x + 7 f (x) = x4 − x3 − 2 x2 − 7 x − 3 g(x) =

717175561486984577278242843019 x2 + 2189435313197775056442946543188 x

+2906610874684759633721189386207 Note that g∞ ≈ 2101. If we take d = κ and r = 2, we get the following set of polynomials. h(x) = x3 + x2 + 15 x + 7 f (x) = x6 − 4 x5 − 53 x4 − 147 x3 − 188 x2 − 157 x − 92 g(x) =

15087279002722300985 x4 + 124616743720753879934 x3 + 451785460058994237397 x2

+ 749764394939964245000 x + 567202989572349792620 We have g∞ ≈ 269.

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 5 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

Asymptotic Analysis

Theorem

Let n = ηκ; gcd(η, κ) = 1; κ = kd; r ≥ k; t ≥ 2; p = LQ(a, cp) with 1/3 < a < 2/3 and 0 < cp < 1; and η = cη(ln Q/ ln ln Q)2/3−a. It is possible to ensure that the runtime of the NFS algorithm with polynomials chosen by Algorithm B is LQ(1/3, 2cb) where cb = 2r + 1 3cθkt + 2r + 1 3cθkt 2 + kcθ(t − 1) 3(r + 1) and (2) cθ = cpcη. (3)

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 6 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 7 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

MTNFS and TNFS Combined Plot

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 8 / 9

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Tower Number Field Sieve Variant of a Recent Polynomial Selection Method

MTNFS and TNFS Combined Plot

Sarkar and Singh | Indian Statistical Institute, Kolkata | May, 2016 8 / 9