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Suppose I have 2 bit commitment schemes, C1 and C2 which are computationally hiding.Suppose I commit the same bit using both and send them to an adversary who knows that the same bit was committed. Is the scheme still computationally hiding, or are there counterexamples?

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It is certainly safe for any commitment scheme we would want to use. However, I believe that we can make a stronger statement; we can show that if there is such a method, then either C1 or C2 is not computationally hiding.

Suppose we have an oracle $\theta( C_1^x, C_2^y )$ that would take a commitment in C1 to the bit $x$ and a commitment in C2 to the bit $y$, and if those two commitments were to the same bit ($x=y$), it would output that common value; if they were to different bits, then the behavior is unspecified.

Then, suppose we were given a commitment $C_1^b$ in scheme C1 to a bit $b$. When we could do is generate two commitments in C2, namely $C_2^0$ and $C_2^1$, and evaluate both $\theta(C_1^b, C_2^0)$ and $\theta(C_1^b, C_2^1)$. Now, if $b=0$, we know that $\theta(C_1^b, C_2^0)$ must evaluate to zero (by the definition of the oracle); and so if the behavior of $\theta(C_1^b, C_2^1)$ was anything other than evaluating to one with high probability, we would be able to deduce that $b=0$. Similarly, if $b=1$, then we see that $\theta(C_1^b, C_2^0)$ must evaluate to zero with high probability.

Hence, for C1 to be hiding, we must have $\theta(C_1^b, C_2^x) = x$ almost all the time.

This gives us a computationally efficient method to evaluate C2 commitments, that is, C2 is not hiding.

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