[IKNP03] Extending Oblivious Transfers Efficiently

论文简介

论文题目： Extending Oblivious Transfers Efficiently
作者： Yuval Ishai, Joe Kilian, Kobbi Nissim, and Erez Petrank
论文出处： CRYPTO [2003]

方案（半诚实）

方案的一些解释

步骤三中 q i q^i qi的解释

q i q^i qi表示矩阵 Q Q Q的第 i i i列。
q i = ( s i ⋅ r ) ⊕ t i q^i=(s_i\cdot r)\oplus t^i qi=(si⋅r)⊕ti
- When S i = 0 S_i=0 Si=0, q i = ( 0 ⋅ r ) ⊕ t i = 0 ⊕ t i = t i q^i=(0\cdot r)\oplus t^i=0\oplus t^i=t^i qi=(0⋅r)⊕ti=0⊕ti=ti.
- When S i = 1 S_i=1 Si=1, q i = ( 1 ⋅ r ) ⊕ t i = r ⊕ t i q^i=(1\cdot r)\oplus t^i=r\oplus t^i qi=(1⋅r)⊕ti=r⊕ti.

步骤三中 q j q_j qj的解释

q j q_j qj表示矩阵 Q Q Q的第 j j j行； q j i q^i_j qji表示矩阵第 i i i列第 j j j行。根据 q i q^i qi将Q展开可得。
Q = [ q 1 1 q 0 2 q 1 3 . . . q 1 k q 2 1 q 1 2 q 2 3 . . . q 2 k q 3 1 q 3 2 q 3 3 . . . q 3 k ⋮ ⋮ ⋮ ⋮ q m 1 q m 2 q m 3 . . . q m k ] = [ ( s 1 ⋅ r 1 ) ⊕ t 1 1 ( s 2 ⋅ r 1 ) ⊕ t 1 2 ( s 3 ⋅ r 1 ) ⊕ t 1 3 . . . ( s k ⋅ r 1 ) ⊕ t 1 k ( s 1 ⋅ r 2 ) ⊕ t 2 1 ( s 2 ⋅ r 2 ) ⊕ t 2 2 ( s 3 ⋅ r 2 ) ⊕ t 2 3 . . . ( s k ⋅ r 2 ) ⊕ t 2 k ( s 1 ⋅ r 3 ) ⊕ t 3 1 ( s 2 ⋅ r 3 ) ⊕ t 3 2 ( s 3 ⋅ r 3 ) ⊕ t 3 3 . . . ( s k ⋅ r 3 ) ⊕ t 3 k ⋮ ⋮ ⋮ ⋮ ( s 1 ⋅ r m ) ⊕ t m 1 ( s 2 ⋅ r m ) ⊕ t m 2 ( s 3 ⋅ r m ) ⊕ t m 3 . . . ( s k ⋅ r m ) ⊕ t m k ] Q=\begin{gathered} \begin{bmatrix} q^1_1 & q^2_0 & q^3_1 & ... & q^{k}_1\\ q^1_2 & q^2_1 & q^3_2 & ... & q^k_2\\ q^1_3 & q^2_3 & q^3_3 & ... & q^k_3\\ \vdots & \vdots & \vdots & & \vdots\\ q^1_m & q^2_m & q^3_m & ... & q^k_m \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix}(s_1\cdot r_1)\oplus t^1_1 & (s_2\cdot r_1)\oplus t^2_1 & (s_3\cdot r_1)\oplus t^3_1 & ... & (s_k\cdot r_1)\oplus t^k_1\\ (s_1\cdot r_2)\oplus t^1_2 & (s_2\cdot r_2)\oplus t^2_2 & (s_3\cdot r_2)\oplus t^3_2 & ... & (s_k\cdot r_2)\oplus t^k_2\\ (s_1\cdot r_3)\oplus t^1_3 & (s_2\cdot r_3)\oplus t^2_3 & (s_3\cdot r_3)\oplus t^3_3 & ... & (s_k\cdot r_3)\oplus t^k_3\\ \vdots & \vdots & \vdots & & \vdots\\ (s_1\cdot r_m)\oplus t^1_m & (s_2\cdot r_m)\oplus t^2_m & (s_3\cdot r_m)\oplus t^3_m & ... & (s_k\cdot r_m)\oplus t^k_m \end{bmatrix} \end{gathered} Q=⎣⎢⎢⎢⎢⎢⎡q11q21q31⋮qm1q02q12q32⋮qm2q13q23q33⋮qm3............q1kq2kq3k⋮qmk⎦⎥⎥⎥⎥⎥⎤=⎣⎢⎢⎢⎢⎢⎡(s1⋅r1)⊕t11(s1⋅r2)⊕t21(s1⋅r3)⊕t31⋮(s1⋅rm)⊕tm1(s2⋅r1)⊕t12(s2⋅r2)⊕t22(s2⋅r3)⊕t32⋮(s2⋅rm)⊕tm2(s3⋅r1)⊕t13(s3⋅r2)⊕t23(s3⋅r3)⊕t33⋮(s3⋅rm)⊕tm3............(sk⋅r1)⊕t1k(sk⋅r2)⊕t2k(sk⋅r3)⊕t3k⋮(sk⋅rm)⊕tmk⎦⎥⎥⎥⎥⎥⎤
很显然，可以总结出， q j = ( r j ⋅ s ) ⊕ t j q_j=(r_j\cdot s)\oplus t_j qj=(rj⋅s)⊕tj

步骤四解释（正确性证明）

当 r j = 0 r_j=0 rj=0时， q j = ( r j ⋅ s ) ⊕ t j = t j q_j=(r_j\cdot s)\oplus t_j=t_j qj=(rj⋅s)⊕tj=tj，
z j = y j 0 ⊕ H ( j , t j ) = x j 0 ⊕ H ( j , q j ) ⊕ H ( j , t j ) = x j 0 ⊕ H ( j , t j ) ⊕ H ( j , t j ) = x j 0 \begin{aligned} z_j&= y_{j0}\oplus H(j,t_j)\\ &= x_{j0}\oplus H(j,q_j)\oplus H(j,t_j) \\ &= x_{j0}\oplus H(j,t_j)\oplus H(j,t_j) \\ &= x_{j0}\\ \end{aligned} zj=yj0⊕H(j,tj)=xj0⊕H(j,qj)⊕H(j,tj)=xj0⊕H(j,tj)⊕H(j,tj)=xj0
当 r j = 1 r_j=1 rj=1时， q j = s ⊕ t j q_j=s\oplus t_j qj=s⊕tj，
z j = y j 1 ⊕ H ( j , t j ) = x j 1 ⊕ H ( j , q j ⊕ s ) ⊕ H ( j , t j ) = x j 1 ⊕ H ( j , s ⊕ t j ⊕ s ) ⊕ H ( j , t j ) = x j 0 ⊕ H ( j , t j ) ⊕ H ( j , t j ) = x j 1 \begin{aligned} z_j&= y_{j1}\oplus H(j,t_j)\\ &= x_{j1}\oplus H(j,q_j\oplus s)\oplus H(j,t_j) \\ &= x_{j1}\oplus H(j,s\oplus t_j\oplus s)\oplus H(j,t_j) \\ &= x_{j0}\oplus H(j,t_j)\oplus H(j,t_j) \\ &= x_{j1}\\ \end{aligned} zj=yj1⊕H(j,tj)=xj1⊕H(j,qj⊕s)⊕H(j,tj)=xj1⊕H(j,s⊕tj⊕s)⊕H(j,tj)=xj0⊕H(j,tj)⊕H(j,tj)=xj1

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[IKNP03] Extending Oblivious Transfers Efficiently

论文简介

方案（半诚实）

方案的一些解释

步骤三中 q i q^i qi的解释

步骤三中 q j q_j qj的解释

步骤四解释（正确性证明）

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[IKNP03] Extending Oblivious Transfers Efficiently

论文简介

方案（半诚实）

方案的一些解释

步骤三中 q i q^i qi的解释

步骤三中 q j q_j qj​的解释

步骤四解释（正确性证明）

[IKNP03] Extending Oblivious Transfers Efficiently相关推荐

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步骤三中 q j q_j qj的解释