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The problem is known in the literature as private function evaluation (PFE). A sender has input (a function) $f$; a receiver has input $x$, and only the receiver learns $f(x)$. If you are willing to leak the topology of a circuit that computes $f$ (but not the identity of the gates), then using classical garbled circuits / Yao's protocol will work. These ...


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You can use Oblivious transfer protocol for the answers: https://en.wikipedia.org/wiki/Oblivious_transfer Here is an example with only 2 answers ($m0$ and $m1$) and uses RSA ($e,d,N$) : In your case Alice would have to send $x_0 \ldots x_9$ and Bob would have to pick $b \in \{0,\ldots,9\}$ where $b$ is the number of his question. The operation $m + k$ ...


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Approach 1 The simplest way of doing this is for the receiver, with choice $j \in \{1,\dots,n\}$, to input $1$ in the $j$-th 1-out-of-2 OT and $0$ elsewhere. The sender, with input $(x_1, \dots, x_n)$, inputs $(0,x_i)$ in the $i$-th OT. Approach 2 An alternative protocol (that just came out of a discussion with a colleague, and seems to be actively ...


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There's a new really simple OT protocol based on DH. It's even practical. Watch this video. For the paper and source code, go here.


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That is possible if and only if oblivious transfer is possible. Proof: For the left-to-right implication, Alice just lets the last 8 answers be independent of her inputs. For the right-to-left implication, Alice creates 6 extra answers that are independent of her inputs, splits each answer into 4 shares, and then for i in {0,1,2,3}, obliviously transfers ...


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Kolesnikov & Kumaresan defined a primitive called "string select OT" which basically covers your setting but with a database of 2 items. Sender has $x^1, y^1, x^2, y^2$. Receiver has $x^*$. If $x^* = x^i$ then the receiver learns the corresponding $y^i$. I think a generalization of their protocol would work, at the cost of $n$ 1-out-of-2 string OTs ...



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