I’m trying to devise a protocol, complimentary to a private-set-intersection, involving three parties, namely Alice, Bob and Charlie.

Alice has a public and a private key. And receives website logs from Charlie.

Charlie runs a website, and he sends his website logs to Alice, encrypting the user-ids using Alice’s public key. Let’s say they are using RSA with padding for each log entry, this way the same user-id always looks random when encrypted.

Bob is managing a small part of Charlie’s website. Charlie has agreed to forward a related portion of the logs to Bob. But it’s directly sampled from the same logs Alice got, so they carry the user-ids encrypted with Alice’s public key.

Naturally, Bob doesn’t have Alice’s private key, and wants to learn the user-ids that visited his section. He can ask Alice to decrypt the user-ids he has, however he doesn’t want to give Alice the information about the users he’s interested in.

Is it possible to devise a protocol where Alice decrypts the messages encrypted by her public key, without really knowing what the message was. Hence the name Oblivious Decryption...

So, I have thought of the commutative property of RSA. If Bob had a private and public key both secret to himself, before sending the message to Alice he might have encrypted and then decrypt what Alice has decrypted, getting access to the user-id himself only. However for Bob to be able to create such a key-pair he needs to know the prime factors of the modulus.

Is there a way, this can be achieved. And If not can we prove that this is not possible?

up vote 2 down vote accepted

This can be done by exploiting the homomorphic property of RSA. Let's say Alice's key is $(e,N,d)$ where $e$ is the public exponent, $N$ the modulus, and $d$ the private exponent.

To decrypt $x$, Bob samples $r$ randomly from $\{1,\cdots,N\}$ and computes $xr^e\mod N$ and sends it to Alice. Alice computes $(xr^e)^d=x^dr\mod N$ and sends it back to Bob. Bob multiplies by $r^{-1}$ to extract $x^d$.

Note that Alice does not know $r$ and thus cannot determine $x^d$ from $x^dr$. This does lead to at least one vulnerability: Bob can send Alice whatever he likes. In particular, he could ask for the decryptions of other private messages to Alice that he somehow acquired.

  • I've just implemented a test for this, seems to work. However can you tell anything about the security of this? Can one factor $x^dr$ to get the $x^d$ value? – zetaprime Oct 3 at 12:50
  • see crypto.stackexchange.com/questions/62846/… for the question I have asked related to the security of this. – zetaprime Oct 3 at 13:34
  • Since we are working over a field, $x^dr$ does not have a unique factorization. In fact, for every sequence $a_1,a_2,\cdots, a_{k-1}$ that are relatively prime to $N$, there exists an $a_k$ such that $a_1\cdot a_2, \cdots a_{k}=x^dr$. Thus 'factorization' yields no useful information. – rikhavshah Oct 3 at 15:31
  • can we conclude that, the extraction of $r$ and $x^d$ from the multiplication $x^dr$ is essentially equivalent to the discrete log problem? – zetaprime Oct 3 at 16:26

For each user-id, Charlie can perform the first half of a Diffie Hellman exchange with Alice using an ephemeral keypair. Charlie will then use a symmetric cypher to encrypt the user-id, using as a key the cryptographic hash of the shared secret that is established via this DH exchange. The logs will include the encrypted user-ids as well as the ephemeral public key for each encrypted user-id.

Alice will use her private key and the ephemeral public key for each record to determine the shared secret to decrypt each user-id. Alice will send all shared secrets to Bob.

Bob can then use this list of shared secrets to decrypt the user-ids that are in his subset of the logs.

In order to avoid leaking the total number of user-ids in her log files, Alice could transmit to Bob a randomly ordered list of: (Hash(shared secret), Hash(encrypted user-id)). Alice can generate a large volume of dummy data to include in this list, so that Bob could tell the upper-bound of the number of user-ids in Alice's logs, but not exactly how many.

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