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Rerandomizable Signatures under Standard Assumption Sanjit Chatterjee and R. Kabaleeshwaran Indian Institute of Science, India. INDOCRYPT 2019 Sanjit Chatterjee and R. Kabaleeshwaran (Indian Institute of Science, India. ) Rerandomizable


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Rerandomizable Signatures under Standard Assumption

Sanjit Chatterjee and R. Kabaleeshwaran

Indian Institute of Science, India.

INDOCRYPT 2019

Sanjit Chatterjee and R. Kabaleeshwaran (Indian Institute of Science, India. ) Rerandomizable Signatures under Standard Assumption December 17th 1 / 24

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Outline

1

Motivation

2

Preliminaries

3

Our Constructions

4

Comparison

2 / 24

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Motivation

Camenisch-Lysyanskaya [CL04] - Rerandomizable Signature (RRS)

◮ Signing multiple block messages, ◮ Rerandomizable: Given σ on m under PK, anybody can compute σ′ on

m which is indistinguishable from σ.

◮ Rerandomizability replaces costly zero knowledge proof system in many

privacy preserving protocols.

Used as building block in

◮ Group signature, ◮ Anonymous credentials, ◮ Aggregate signature, ◮ E-Cash, etc.

Goal Fully rerandomizable RRS, unforgeability under standard assumption.

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Literature

[CL04] introduced CL-RRS scheme,

◮ Signature size depends on the message block length ℓ, ◮ UF - LRSW interactive assumption. ◮ [GLOW12] Dual-form of CL-RRS: symmetric composite-order setting,

single message case.

[PS16] introduced PS-RRS scheme,

◮ Constant size signature, ◮ UF - interactive assumption.

[PS18] modified PS-RRS (mPS-RRS) scheme,

◮ Either weak rerandomizability in SM or fully rerandomizable in ROM, ◮ UF - parameterized assumption 4 / 24

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Bilinear pairing

Efficiently computable map e : G × H → GT satisfies Bilinearity: ∀P1, P2 ∈ G, Q1, Q2 ∈ H, e(P1 + P2, Q1 + Q2) = e(P1, Q1)e(P1, Q2)e(P2, Q1)e(P2, Q2), Non-degeneracy: for any P ∈ G, if e(P, Q) = 1, ∀Q ∈ H, then P = 0. Types of pairing: Let N be the order of G, H and GT. If N is prime, it is prime order pairing, otherwise composite order. If G = H, then it is symmetric, otherwise asymmetric pairing,

5 / 24

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Signature Definition

Three PPT algorithms, KeyGen(λ) → (PK, SK), Sign(SK, m) → σ, Ver(PK, m, σ) → 1 or 0.

  • Unforgeability. EUF-CMA model:

C A PK mi ∈ M Sign(SK, mi) → (mi, σi) (m∗, σ∗) i ∈ [1, q] KeyGen(λ) → (SK, PK)

  • 1. m∗ = mi, for i ∈ [1, q]
  • 2. Ver(PK, m∗, σ∗)=1

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Rerandomizable Signature

KeyGen, Sign, Ver + a new PPT algorithm, Rand(PK, m, σ) → (m, σ′). Security: Unforgeability: EUF-CMA model. Randomizability [Gha16]:

C A (SK, PK) (m, σ) If Ver(PK, m, σ) = 1, b ∈R {0, 1} b′ KeyGen(λ) → (SK, PK) b′ = b If b = 0, (m, σ0) ← Sign(SK, m) Else, (m, σ1) ← Rand(PK, m, σ) σb

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Composite-order setting

Let Θ = (N = p1p2, G, H, GT , e) ← GN(λ) with G = g, H = h. Write G ∼ = G1 ⊕ G2, H ∼ = H1 ⊕ H2, pi-order subgroups Gi = gi, Hi = hi, i ∈ [1, 2]. Orthogonal property: e(gi, hj) = 1, for i = j.

◮ Ex: e(g1, h2) = e(g rp2, hsp1) = e(g r, hs)p1p2 = 1.

Parameter-Hiding property: Chinese Remainder Theorem (CRT) ensures that, for a ∈R ZN, a mod p1 does not reveal a mod p2, i.e., ga

1 ga 2 ≈ ga 1 ga1 2 , for a1 ∈R ZN.

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Subgroup Hiding (SGH) assumption

Used in EUF-CMA security. SGHH

p1→p1p2: Given g1, h1, h2, ˆ

T, hard to decide ˆ T ∈ H1 or ˆ T ∈ H, SGHG

p1→p1p2: Given g1, g2, h1, T, hard to decide T ∈ G1 or T ∈ G,

SGHH

p2→p1p2: Given g2, h1, h2, ˆ

T, hard to decide ˆ T ∈ H2 or ˆ T ∈ H.

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Our RRS Scheme in Composite-order setting

Construction is inspired from PS-RRS [PS16] (Single message case). KeyGen(λ): Let (N = p1p2, G, H, GT , e, µ = {G1, H1})

$

← GN(λ). Choose g1 ∈R G1, h1 ∈R H1 and x, y ∈R ZN. Return SK = {g1, x, y}, PK = {h1, X = hx

1, Y = hy 1}.

Sign(SK, m): Choose r ∈R ZN, compute A := gr

1, B := gr(x+my) 1

. Return (m, σ = (A, B)). Verify(PK, m, σ): Accept only if e(A, h1) = 1 and e(B, h1) = e(A, XY m). Rand(PK, m, σ): If Ver(PK, m, σ)=1, then choose s ∈R ZN and compute A′ := As, B′ := Bs. Return (m, σ′ = (A′, B′)). Correctness: e(A, h1) = 1 ensures A contain a non-zero exponent of g1. e(B, h1) = e(gr(x+my)

1

, h1) = e(gr

1, h(x+my) 1

) = e(A, XY m).

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Randomizability

Fully randomizable.

C A (SK, PK) (m, σ) If Ver(PK, m, σ) = 1, b ∈R {0, 1} b′ g1 ∈R G1, h1 ∈R H1, x, y ∈R ZN b′ = b If b = 0, σ0 = (A0 = gr

1, B0 = gr(x+my) 1

) Else, σ1 = (A1 = As, B1 = Bs) σb SK = {g1, x, y} PK = {h1, hx

1, hy 1}

Both σ0 and σ1 are distributed identically!

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Unforgeability

Use Dual-form signature technique [GLOW12]. SignA=Sign, SignB(SK∪{g2}, m): Choose r, δ1, δ2 ∈R ZN and return m and σ = (A := gr

1gδ1 2 , B := gr(x+my) 1

gδ2

2 ).

Forgery Class: V = {(m∗, σ∗) ∈ ZN × G 2 : Ver(PK, m∗, σ∗) = 1},

◮ Type-I VI = {(m∗, σ∗) ∈ V : (A∗)p1 = 1, (B∗)p1 = 1}, ◮ Type-II VII = {(m∗, σ∗) ∈ V : (A∗)p1 = 1 or (B∗)p1 = 1}. 12 / 24

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Theorem

SGH assumptions ⇒ RRS scheme is EUF-CMA secure.

Proof.

We use a hybrid argument. GameR: Real EUF-CMA game, here A → V, Game0: ∼ GameR, except A → VI, Gamek: ∼ Game0, except 1st k queries answered using SignB, E event that A → VII in Game0. Then we prove

1

|Adv GameR

A

− Adv Game0

A

| ≤ Pr[E] ≤ Adv

SGHH

p1→p1p2

B

+ 1/N,

2

|Adv Gamek−1

A

− Adv Gamek

A

| ≤ Adv

SGHG

p1→p1p2

B

,

3

Adv Gameq

A

≤ Adv

SGHH

p2→p1p2

B

. Hence, Adv GameR

A

≤ Adv

SGHH

p1→p1p2

B

+ Adv

SGHG

p1→p1p2

B

+ Adv

SGHH

p2→p1p2

B

+ 1/N.

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Lemma

Pr[E] ≤ Adv

SGHH

p1→p1p2

B

+ 1/N.

Proof.

B A PK = {h1, X = hx

1, Y = hy 1}

mi SignA(SK, mi) → (mi, σi) (m∗, σ∗ = (A∗, B∗)) C g1, h1, h2, ˆ T x, y ∈R ZN SK = {g1, x, y} S := B∗(A∗)−(x+m∗y) = gγ2−γ1(x+m∗y)

2

= 1 with non-neg prob. If e(S, ˆ T) = 1, then ˆ T ∈ H1, else ˆ T ∈ H i ∈ [1, q] CRT x, y mod p2 is random hidden to A

Here A∗ = gr

1gγ1 2 , B∗ = gr(x+m∗y) 1

gγ2

2 .

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Lemma

|Adv Gamek−1

A

− Adv Gamek

A

| ≤ Adv

SGHG

p1→p1p2

B

.

Proof.

B A PK = {h1, X = hx

1, Y = hy 1}

mi SignB(SK, mi) → (mi, σi) (m∗, σ∗ = (A∗, B∗)) C g1, g2, h1, T = gt1

1 gt2 2

x, y ∈R ZN SK = {g1, x, y}∪{g2} i ∈ [1, k − 1] mk (mk, σk = (Ak, Bk)) mt SignA(SK, mt) → (mt, σt) t ∈ [k + 1, q] 1 or 0 Ak = gt1

1 gt2 2 , Bk = (gt1 1 gt2 2 )(x+mky)

CRT x, y mod p2 is random

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Lemma

Adv Gameq

A

≤ Adv

SGHH

p2→p1p2

B

.

Proof.

B A PK = {h1, X = hx

1, Y = hy 1}

mi SignB(SK, mi) → (mi, σi = (Ai, Bi)) (m∗, σ∗ = (A∗, B∗)) C g2, h1, h2, ˆ T gτ1

1 gτ2 2 , ∈R G, x, y ∈R ZN

SK = {x, y}∪{g2} If e(A∗, ˆ T) = 1, then ˆ T ∈ H2, else ˆ T ∈ H Ai = (gτ1

1 gτ2 2 )rgδ′

1

2 ,

Bi = (gτ1

1 gτ2 2 )r(x+miy)gδ′

2

2

i ∈ [1, q]

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Composite To Prime-order setting

Use Dual Pairing Vector Space (DPVS) [OT10] in the prime-order setting. Orthogonal property via Dual basis (B, B∗) ← Dual(λ, F4

p), where

B = { bi}4

i=1, B∗ = {

b∗

i }4 i=1.

  • bi ·

b∗

j =

ψ if i = j, if i = j. Parameter Hiding (PH) property [Lew12]: (B, B∗)

A∈GL(2)

− → (D, D∗) such that (D, D∗) is independent of A.

◮ Ex:

d1 = b1, d2 = b2, d∗

1 =

b∗

1,

d∗

2 =

b∗

2,

( d3, d4)⊤ = A−⊤( b3, b4)⊤, ( d∗

3 ,

d∗

4 )⊤ = A(

b∗

3,

b∗

4)⊤.

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Composite To Prime-order setting

Decisional Subspace (DS) Assumptions: DDHH: Given g, h, ha, hb, hc, hard to decide c = ab mod p or not. DSH: g

b1, g b2, h b∗

1 , h

b∗

2 , h

b∗

3 , h

b∗

4 , U1 = gµ1

b1+µ2 b3, U2 = gµ1 b2+µ2 b4

T1 = hτ1

b∗

1 +τ2

b∗

3 , T2 = hτ1

b∗

2 +τ2

b∗

4 , τ2 = 0 or not. ◮ Similar to SGHH

p1→p1p2,

DSG: g

b1, g b2, g b3, g b4, h b∗

1 , h

b∗

2 , U1 = hµ1

b∗

1 +µ2

b∗

3 , U2 = hµ1

b∗

2 +µ2

b∗

4 ,

T1 = gτ1

b1+τ2 b3, T2 = gτ1 b2+τ2 b4, τ2 = 0 or not.

◮ Similar to SGHH

p1→p1p2,

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RRS scheme in prime-order setting

KeyGen(λ): Let P(λ) → (p, G, H, GT , e). Choose (D, D∗) ∈R Dual(λ, F4

p), x, y ∈R Zp. Return SK = {g d1, g d2, x, y},

PK = {h

d∗

1 , h

d∗

2 , X = hx

d∗

1 , Y = hy

d∗

1 }.

Sign(SK, m): Choose r ∈R ZN, compute σ = gr

d1−r(x+my)

  • d2. Return

(m, σ). Verify(PK, m, σ): Accept if e(σ, h

d∗

1 ) = 1 and e(σ, XY mh

d∗

2 ) = 1.

Rand(PK, m, σ): If Ver(PK, m, σ)=1, then choose s ∈R ZN and return (m, σ′ := σs). Correctness: e(σ, XY mh

  • d∗

2 ) = e(gr

d1−r(x+my) d2, hx d∗

1 (hy

d∗

1 )mh

  • d∗

2 )

= e(g, h)r(x+my)

d1· d∗

1 −r(x+my)

d2· d∗

2 = 1. 19 / 24

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Security

Fully rerandomizable: σ0 = gt

d1−t(x+my) d2 and σ1 = gsr d1−sr(x+my) d2

are identically distributed. Unforgeability under SXDH assumption.

◮ Use dual-form signature technique, ◮ replace SGHH

p1→p1p2 with DSH assumption,

⋆ DDHH ⇒ DSH, ◮ replace SGHG

p1→p1p2 with DSG assumption,

⋆ DDHG ⇒ DSG, ◮ replace SGHH

p2→p1p2 with DDHH assumption.

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Summary

Our RRS scheme is secure under SXDH assumption:

◮ This variant inspired from [PS16], ◮ Main signature exponent (x + ℓ

j=1 mjyj),

◮ Constant-size signature, while signing multiple block messages, ◮ Fully rerandomizable.

Another RRS variant inspired from [Gha16].

◮ Main signature exponent: (x + m1 + ℓ

j=2 mjyj)/y1.

◮ Fully rerandomizable, ◮ Unforgeability - same set of assumptions. 21 / 24

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Comparison

Table : Comparing rerandomizable signatures for multiple block messages.

|PK| |σ| Signing Cost Verification Cost Rand. Assum. CL-RRS (ℓ + 2)|G| (2ℓ + 1)|G| (2ℓ + 1)EG 4ℓP + ℓEG + ℓMG Full LRSW mCL-RRS (ℓ + 3)|G| (2ℓ + 3)|G| (2ℓ + 3)EG 4(ℓ + 1)P + (ℓ + 1)EG Weak q-MSDH-1 +1|Zp| +(ℓ + 1)MG PS-RRS (ℓ + 2)|H| 2|G| 2EG 2P + ℓEH + ℓMH Full PS mPS-RRS (ℓ + 3)|H| 2|G| + 1|Zp| 2EG 2P + (ℓ + 1)EH Weak q-MSDH-2 +(ℓ + 1)MH PS-RRS (4ℓ + 13)|H| + 1|G| 4|G| 8EG + 4MG 8P + 6MGT + 4ℓEH Full SXDH RRS (4ℓ + 9)|H| + 1|G| +4(ℓ + 1)MH 22 / 24

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References

[CL04] Jan Camenisch and Anna Lysyanskaya, “Signature schemes and anonymous credentials from bilinear maps”, CRYPTO 2004, [Gha16] Essam Ghadafi, “Short structure-preserving signatures”, CT-RSA 2016, [GLOW12] Michael Gerbush, Allison B. Lewko, Adam O’Neill and Brent Waters, “Dual form signatures: An approach for proving security from static assumptions”, ASIACRYPT 2012, [Lew12] Allison B. Lewko, “Tools for simulating features of composite order bilinear groups in the prime order setting”, EUROCRYPT 2012, [OT10] Tatsuaki Okamoto and Katsuyuki Takashima. “Fully secure functional encryption with general relations from the decisional linear assumption”, CRYPTO 2010, [PS16] David Pointcheval and Olivier Sanders, “Short randomizable signatures”, CT-RSA 2016, [PS18] David Pointcheval and Olivier Sanders, “Reassessing security of randomizable signatures”, CT-RSA 2018,

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THANK YOU

See https://ia.cr/2019/1144.

Question?

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