can we upgrade OT(oblivious transfer) from 1

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can we upgrade OT(oblivious transfer) from 1

#can we upgrade OT(oblivious transfer) from 1| 来源: 网络整理| 查看: 265

reading paper: ‘even s. A randomized protocol for signing contracts [J]. ACM SIGACT news, 1983’, there has been a question why this scheme cannot be directly extended from 1-out-of-2 OT to 1-out-of-n OT. The OT extension in the later paper adopts very complex methods to expand.

Brief description of the original scheme: Alice is the sender and Bob is the receiver. Alice has messages M0, M1; Bob has choice bit: $b \in \{0, 1\}$;

-Step 1: Alice generates random numbers x0, x1 and sends them to Bob;

-Step 2: Bob generates a random number k and calculates $c_b = x_b + ENC_{PKA}(k)$, and send $c_b$ to Alice;

-Step 3: Alice calculates $k0 = DEC_{ska}(c_b - x0),k1 = DEC_{ska}(c_b - x1)$, and then calculates $e0 = M0 \bigoplus k0$, $e1 = M1 \bigoplus k1$, and sends e0, e1 to Bob;

-Step 4: Bob calculate the result: $Mb = eb \bigoplus k$.

I understand that this protocol should be directly extended to 1-out-of-N OT protocol by: Alice is the sender and Bob is the receiver. Alice has messages M1,M2,..., Mn; Bob has choice number $b \in [1, n]$;

-Step 1: Alice generates random number x1, ..., xn, and send all of them to Bob;

-Step 2: Bob generates a random number k and calculates $c_b = x_b + ENC_{PKA}(k)$,and send $c_b$ to Alice;

-Step 3: Alice calculates: $k1 = DEC_{ska}(c_b - x1),..., kn = DEC_{ska}(c_b - xn)$, and then calculates $e1 = M1 \bigoplus k1, ..., en = Mn \bigoplus kn$, and sends e1,...en to Bob;

-Step 4: Bob calculate the result: $Mb = eb \bigoplus k$.

My question is, why do the subsequent papers not do this, but usually use very complex methods to achieve 1-out-of-N OT? Is it for safety or efficiency?



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