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As fgrieu hints at in the comments, it is not in general possible to find $\operatorname{Enc}$. Otherwise you would be able to break an arbitrary block cipher, because the key of any block cipher can be cast as part of the algorithm instead. Even if you ignore the computational cost, there is no way to find $\operatorname{Enc}$ given values for only one ...

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I played with an idea for an asymmetric proof of work scheme that I put a small amount of thought into making parallel resistant. I will explain the basic idea of the proof of work scheme first, followed by how I attempted to make it parallel resistant. The challenge creator hashes a small number of random bytes with a random iv. A desired number of ...

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The following paragraph applies even if the honest prover does not use iteration. Strictly speaking, only a constant-size part of the input can be read from in constant time, so one could always find a suitable result in constant time. For a deterministic verifier that makes at most k probes to an alleged result consisting of M w-bit words, one can find ...

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