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I am trying to solve this exercise.

Assume an oblivious transfer protocol that uses a trusted third party Trent.

Alice's messages $M_0$ and $M_1$ are binary values of length $k$. Bob's message selection bit is $m$.

  1. Trent -> Alice: $R_0, R_1$ where $R_0, R_1$ are random binary values each of length $k$

  2. Trent -> Bob : $t, R_t $ where $t$ is a random bit

  3. Bob -> Alice : $ b = t \oplus m$

  4. Alice -> Bob : $C_0, C_1 $

How should Alice compute $C_0$ and $C_1$? How should Bob compute $M_m$?

I guess that Alice has to compute $K_0, K_1$ as keys (probably inserted into a hash function) and then encrypt $C_0$, $C_1$ like $C_M = E_{k_m}(M_m)$. Bob then can compute $M_m$ by $M_m = D_{k_m}(C_m)$ where $K_m$ is related to $R_t$ in some way. Does anyone have an idea? Thank you

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  • $\begingroup$ Hint: split it into two cases. In case Bob gets $t=m$, he sends $b=0$ which means "I got from Trent the choice that I wanted". What does this mean in terms of $R_0, R_1, R_t$? In case Bob gets $t \ne m$, he sends $b=1$ which means, "I got from Trent the opposite choice from what I wanted". What does this mean in terms of $R_0, R_1, R_t$? $\endgroup$ – Mikero Dec 8 '18 at 3:04
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  1. Alice computes $C_0$ and $C_1$ as follows:

    If $b=0$ (this means $t=m$), Alice computes $$C_i = M_i \oplus R_i.$$ Otherwise (i.e. when $t=1-m$), Alice computes $$C_i = M_i \oplus R_{1-i},$$ where $i \in \{0,1\}$.

  2. Bob computes $M_m$ as $C_m\oplus R_t$.

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