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Consider the following commitment scheme, where $x$ belongs to $\langle g\rangle$ and $u$ is uniformly chosen from $\mathbb{Z}_n$:

$$\mathsf{commit}(u,x) = g^u\cdot x$$

Is it binding and hiding?

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The scheme is clearly hiding: the value $g^u\cdot x$ distributes uniformly at random when $u$ is.

However, it is not at all binding: for a given commitment $c$, the sender, assuming he can compute inverse of an element in the group, can find many pairs, more accurately, as many as the size of the group, $u,x$ for which $c=g^u\cdot x$.

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  • $\begingroup$ It is trivial to find another pair u',x' matching the commitment. It is however difficult to do this with meaningfull x' assuming most x' aren't meaningful. $\endgroup$ – Meir Maor Feb 3 '18 at 6:11
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    $\begingroup$ @MeirMaor In my opinion this is still a security issue, in the same way as when collisions of a hash function are meaningful. $\endgroup$ – Daniel Feb 3 '18 at 9:31
  • $\begingroup$ I'm not saying I recommend using such a scheme. Just trying to provide a more complete answer. $\endgroup$ – Meir Maor Feb 3 '18 at 15:40

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