# Is there a way to do fair exchange between two parties who don't trust each other?

Let's suppose we have an Alice who knows a secret key A, and Bob who knows key B. Using their own keys, they each encrypt a message (Alice encrypts $m_A$, Bob encrypts $m_B$) with their own key, and send the encrypted messages to one another.

Now is there some cryptographic scheme thinkable that lets Alice only decrypt her message (using key B) when Bob is able to decrypt his (via key A)? To guarantee that Alice will tell Bob her key after Bob told his?

The simplest solution would be a third-party which releases both keys at exactly the same time, but maybe there is also a cryptographic way to do this?

• Are you familiar with en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange? If yes, and this wouldn't meet your specific requirements, please explain why. Commented May 13, 2013 at 10:09
• @Joshua: $\:$ How should Bob verify the key he gets? $\;\;\;$
– user991
Commented May 14, 2013 at 1:07

I think what you want to look at is "fair exchange". There's a giant catalog of protocols for fair exchange. They've been designed for doing exactly what (I think) you want.

See, e.g.,

Some fair exchange protocols require a third party who Alice and Bob trust, and involve the third party in the protocol. Others (called optimistic fair exchange) require that there be a trusted third party, but normally the third party does not have to be involved in any way -- the third party only gets involved to settle disputes, if something goes wrong. Still others try to avoid any third party (e.g., gradual fair exchange); they're not perfect, but they try to ensure that if Alice can learn Bob's message with some amount of effort, then Bob can learn Alice's messages with not too much more effort.

To learn more about fair exchange, do a literature search on fair exchange. There are lots of research papers on this topic.

• I've read the references you gave, but it seems there has been done very little research in this area. I cannot find any well-known protocol or implementation. Commented May 14, 2013 at 13:30
• "there has been done very little research in this area" - Not true. There have been tons of papers written on fair exchange protocols. (Google Scholar turns up over 100.) Have you done a literature search? You should.
– D.W.
Commented May 15, 2013 at 1:26

Note that it is probably technically impossible to meet your requirement that both secrets have to be revealed at exactly the same time, if interpreted literally. However, if we interpret this requirement as meaning that each party needs guarantees that the other party selected its own message, before decrypting the message of the other party, then this might be achieved using a scheme by which each party commits to a message without revealing it, before actually sending it.

Note: In order to guarantee that Alice will get $M_B$ when revealing $M_A$, you need a trusted third party. However, if your requirements are consistent with simply giving Alice the ability to abort further interactions with Bob if Bob decides to cheat, the following scheme should be sufficient.

For simplicity, suppose each party has a RSA key pair that can be used for both signing and encryption, and that Alice already has the public key of Bob, and vice versa.

1. Alice selects a message $M_A$. Bob selects a message $M_B$.
2. Alice selects a symmetric AES key $K_A$. Bob selects a symmetric AES key $K_B$.
3. Alice generates $S_A = Sign_{Priv_A}(K_A)$. Bob generates $S_B = Sign_{Priv_B}(K_B)$.
4. Alice generates $C_A = AESEnc_{K_A}(M_A)$. Bob generates $C_B = AESEnc_{K_B}(M_B)$.
5. Alice sends $S_A,C_A$ to Bob. Bob sends $S_B,C_B$ to Alice.
6. Alice generates $T_A = Encrypt_{Publ_B}(K_A)$. Bob generates $T_B = Encrypt_{Publ_A}(K_B)$
7. Alice sends $T_A$ to Bob. Bob sends $T_B$ to Alice.
8. Alice generates $K'_B = Decrypt_{Priv_A}(T_B)$ and aborts if $Verify_{Publ_B}(S_B,K'_B)$ fails. Bob generates $K'_A = Decrypt_{Priv_B}(T_A)$ and aborts if $Verify_{Publ_A}(S_A,K'_A)$ fails.
9. Alice generates $M'_B = AESDec_{K'_B}(C_B)$. Bob generates $M'_A = AESDec_{K'_A}(C_A)$.

The above scheme should only be regarded as a general outline.

• I think you know this, but just to be very explicit: The protocol in this answer doesn't meet the requirement of the original question. If Alice is malicious, she can arrange to learn Bob's message without revealing her own message.
– D.W.
Commented May 13, 2013 at 17:46
• Yes, that perhaps should be made explicit. The best you can do without a trusted third party, is to get the ability to abort if the other party tries to change its message after you reveal yours. For some scenarios that is enough. Commented May 13, 2013 at 17:57
• I think that "designated aborter" (as you described) and "an aborter can't learn the honest party's $\;\;$ message with much less effort than the honest party can learn the aborter's message" are the two optimal properties for the setting without a third party (and that a protocol in that setting can't have both of those). $\:$
– user991
Commented May 13, 2013 at 20:59
• @RickyDemer: Without additional assumptions and constraints, protocols with the latter property are not possible without a third party. cs.utexas.edu/~shmat/courses/cs395t_fall04/pagnia.pdf Commented May 13, 2013 at 21:24
• I suppose the assumption/constraint needed for the latter property is that $\hspace{1.33 in}$ computation is not much easier for the adversary. $\:$
– user991
Commented May 13, 2013 at 21:37