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Let's say parties A and B have a common secret $k$. Is there a protocol where both the parties jointly release a commitment to $k$ so that later on, neither A or B can deny what the common secret was?

Edit: Specifically, I am interested in the scenario where one of the parties can be malicious and we need to prove to a third party C that a commitment $\Phi$ is actually that of the common secret $k$.

If A and B share two commitments separately, then the malicious party can share a commitment to a completely different $k'$. C would have know way of knowing which one is the commitment to $k$.

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  • $\begingroup$ Obviously, if they both have the secret, they could individually issue commitments to it. Why isn't that sufficient? What else do you require? For example, a proof that they committed to the same thing? $\endgroup$
    – poncho
    Nov 21 at 23:10
  • $\begingroup$ One of the parties (say A) is malicious. If B commits to $k$ truthfully and A commits to a different $k'$ (not known to B), there is no way for a third party C to decide who is being truthful, ie, shared the commitment to $k$ - the common secret. Is there a way to ensure that C can be convinced that a commitment is of the common secret $k$? $\endgroup$
    – Ordinary
    Nov 22 at 0:24
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Well, one approach would be to have both parties generate and publish commitments, and that they jointly publish a zero-knowledge-proof that both commitments are to the same value.

Here's one approach at doing that: both $A$ and $B$ generate and publish Pedersen commitments; for example, $A$ selects a random value $r$ and publishes $C_A = g^k h^r$, while $B$ selects a random value $s$ and publishes $C_B = g^k h^s$ (where this is done in a group where discrete logs are hard, and no one knows the discrete log of $h$ with respect to $g$).

The zero-knowledge-proof that they're committing to the same value is a proof of knowledge of a value $v$ such that $C_A C_B^{-1} = h^v$ (which, for honest commitments, is $v = r-s$); if one side commits to another value, no one will know such a value $v$, and hence no one (not even $A, B$ jointly) will be able to publish such a proof. Note that $C_A C_B^{-1}$ can be computed by anybody with access to the two commitments.

It would appear to be fairly simple for the two sides to work together to generate such a Schnorr proof:

  • $A$ selects a random value $a$ and sends $h^a$ to $B$; $B$ selects a random value $b$ and sends $h^b$ to $A$.

  • They both compute the common value $c = \text{Hash}(h^a h^b)$

  • $A$ computes $x = a + c r$, and publishes $x, h^a$. $B$ computes $y = b - c s$, and publishes $y, h^b$.

The pair $h^ah^b, x+y$ would be a valid Schnorr proof; the verifier would check if $h^{x+y} = (h^ah^b) (C_A C_B^{-1})^{\text{Hash}(h^ah^b)}$

Now:

  • I believe that access to the 'half-proofs' $x, h^a$ and $y, h^b$ does not provide any insight into either commitment.

  • This protocol is protected from a single malicious actor; if (say) $A$ was honest, then if the zero-knowledge proof verifies, the $B$ must have committed to the same value. Actually, even if both sides are malicious, they still cannot individually commit to different values.

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Another, possibly simpler, solution is this:

  • $A$ and $B$ get together and jointly select a large, fixed sized random value $r$, and publish $\text{Hash}( k || r )$

  • They both sign (using their private signature keys) the string $\text{Hash}( k || r )$; both signatures are also published.

Either can open the commitment by publishing $k$ and $r$; anyone can verify that they hash to the commitment. And, anyone with $A$ and $B$'s public keys can verify the signatures.

Obviously, as there's only one commitment which can be opened one way (assuming that $\text{Hash}$ is collision resistant, and $r$ has a well-known length; e.g. it's always 256 bits), there is no opportunity for either side to lie. The only think I can think of for a malicious actor can do is to claim "hey, someone stole my private key; I didn't sign that commitment"

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