Calculating the value of a function for given inputs, especially, in the context of secure multi-party computation and/or homomorphic encryption, without disclosing the inputs to some or all parties carrying out the calculation.

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Fully Homomorphic Encryption over the Integers - perform an operation on an encrypted data

In Fully Homomorphic Encryption scheme represented here Fully Homomorphic Encryption over the Integers In the Evaluate process (see section “3.1 The Construction” of the paper): $$Evaluate(pk, C, c1, ...
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Outsourcing arbitrary computations securely

Consider the following scheme. Alice wants Bob to make some computations for her, but she doesn't want to reveal the data on which he's going to do it. So, she encrypts the data, sends them to Bob, he ...
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How hard is to invert the function that computes the middle-bits of (x^2)?

I'm designing a function f that should be moderately hard to invert and very fast to evaluate in a modern CPU. The function will be used in a proof-of-work function. I've read that the middle-bits of ...
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What criteria make the theta step of Keccak's round function reversible?

From what I've been reading, Keccak's round function is reversible. That's pretty obvious for the $\rho$, $\pi$ and $\iota$ transforms. For $\chi$ to be reversible, $x$'s range has to be odd — but ...
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What's efficient MPC protocol for determining if sum's bigger than y?

My secure multi-party computation (MPC) in need is simply to determine if a sum of two private variable is bigger than a given value $y$, as $f(x_0, x_1) = [(x_0 + x_1) > y]$ in which the value ...
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Alternatives to FHE for secure function evaluation

As a followup to a previous question I asked which was more related to Fully Homomorphic Encryption (FHE), what other cryptographic methods are available for computing a private function on public ...
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Background for modular arithmetic function

I'm investigating this function: $a := ((b\cdot c) \bmod k) - (b \cdot c)/k$ where $/$ indicates integer division. Two things I've noticed: It's equivalent to multiplying a·b, and then ...
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Why use a 1-2 Oblivious Transfer instead of a 1 out of n Oblivious Transfer?

When initiating an oblivious transfer, why would someone use a 1-2 oblivious transfer rather than going for an 1 out of n oblivious transfer? Perhaps a slight time overhead for the extra message ...