Oblivious transfer

In Rabin's oblivious transfer protocol, the sender generates an RSA public modulus N=pq where p and q are large prime numbers, and an exponent e relatively prime to λ(N) = (p − 1)(q − 1). The sender encrypts the message m as m^e mod N.

1. The sender sends N, e, and m^e mod N to the receiver.
2. The receiver picks a random x modulo N and sends x² mod N to the sender. Note that gcd(x,N) = 1 with overwhelming probability, which ensures that there are 4 square roots of x² mod N.
3. The sender finds a square root y of x² mod N and sends y to the receiver.

If the receiver finds y is neither x nor −x modulo N, the receiver will be able to factor N and therefore decrypt m^e to recover m (see Rabin encryption for more details). However, if y is x or −x mod N, the receiver will have no information about m beyond the encryption of it. Since every quadratic residue modulo N has four square roots, the probability that the receiver learns m is 1/2.

In a 1–2 oblivious transfer protocol, Alice the sender has two messages m₀ and m₁, and wants to ensure that the receiver only learns one. Bob, the receiver, has a bit b and wishes to receive m_b without Alice learning b. The protocol of Even, Goldreich, and Lempel (which the authors attribute partially to Silvio Micali) is general, but can be instantiated using RSA encryption as follows.

Alice				Bob
Calculus	Secret	Public		Public	Secret	Calculus
Messages to be sent	${\displaystyle m_{0},m_{1}}$
Generate RSA key pair and send public portion to Bob	${\displaystyle d}$	${\displaystyle N,e}$	${\displaystyle \Rightarrow }$	${\displaystyle N,e}$		Receive public key
Generate two random messages		${\displaystyle x_{0},x_{1}}$	${\displaystyle \Rightarrow }$	${\displaystyle x_{0},x_{1}}$		Receive random messages
					${\displaystyle k,b}$	Choose ${\displaystyle b\in \{0,1\}}$ and generate random ${\displaystyle k.}$
		${\displaystyle v}$	${\displaystyle \Leftarrow }$	${\displaystyle v=(x_{b}+k^{e})\mod N}$		Compute the encryption of ${\displaystyle k}$, blind with ${\displaystyle x_{b}}$ and send to Alice
One of these will equal ${\displaystyle k}$, but Alice does not know which.	${\displaystyle {\begin{aligned}k_{0}&=(v-x_{0})^{d}\mod N\\k_{1}&=(v-x_{1})^{d}\mod N\end{aligned}}}$
Send both messages to Bob		${\displaystyle {\begin{aligned}m'_{0}=m_{0}+k_{0}\\m'_{1}=m_{1}+k_{1}\end{aligned}}}$	${\displaystyle \Rightarrow }$	${\displaystyle m'_{0},m'_{1}}$		Receive both messages
					${\displaystyle m_{b}=m'_{b}-k}$	Bob decrypts the ${\displaystyle m'_{b}}$ since he knows which ${\displaystyle x_{b}}$ he selected earlier.

1. Alice has two messages, ${\displaystyle m_{0},m_{1}}$ and wants to send exactly one of them to Bob. Bob does not want Alice to know which one he receives.
2. Alice generates an RSA key pair, comprising the modulus ${\displaystyle N}$, the public exponent ${\displaystyle e}$ and the private exponent ${\displaystyle d}$.
3. She also generates two random values, ${\displaystyle x_{0},x_{1}}$ and sends them to Bob along with her public modulus and exponent.
4. Bob picks ${\displaystyle b}$ to be either 0 or 1, and selects ${\displaystyle x_{b}}$.
5. Bob generates a random value ${\displaystyle k}$ uses it to blind ${\displaystyle x_{b}}$ by computing ${\displaystyle v=(x_{b}+k^{e})\mod N}$, which he sends to Alice.
6. Alice combines ${\displaystyle v}$ with both of her random values to produce: ${\displaystyle k_{0}=(v-x_{0})^{d}\mod N}$ and ${\displaystyle k_{1}=(v-x_{1})^{d}\mod N}$. Now ${\displaystyle k_{b}}$ will be equal to ${\displaystyle k}$ and the other will be a meaningless random value. However since Alice does not know the value of ${\displaystyle b}$ that Bob chose, she cannot determine which of ${\displaystyle k_{0}}$ and ${\displaystyle k_{1}}$ is equal to ${\displaystyle k}$.
7. She combines the two secret messages with each of the possible keys, ${\displaystyle m'_{0}=m_{0}+k_{0}}$ and ${\displaystyle m'_{1}=m_{1}+k_{1}}$, and sends them both to Bob.
8. Bob knows ${\displaystyle k}$, so he is able to compute ${\displaystyle m_{b}=m'_{b}-k}$. However, since he does not know ${\displaystyle d}$, he cannot compute ${\displaystyle k_{1-b}=(v-x_{1-b})^{d}\mod N}$ and so cannot determine ${\displaystyle m_{1-b}}$.

A 1-out-of-n oblivious transfer protocol can be defined as a natural generalization of a 1-out-of-2 oblivious transfer protocol. Specifically, a sender has n messages, and the receiver has an index i, and the receiver wishes to receive the i-th among the sender's messages, without the sender learning i, while the sender wants to ensure that the receiver receive only one of the n messages.

1-out-of-n oblivious transfer is incomparable to private information retrieval (PIR). On the one hand, 1-out-of-n oblivious transfer imposes an additional privacy requirement for the database: namely, that the receiver learn at most one of the database entries. On the other hand, PIR requires communication sublinear in n, whereas 1-out-of-n oblivious transfer has no such requirement. However, assuming single server PIR is a sufficient assumption in order to construct 1-out-of-2 Oblivious Transfer.[5]

1-out-of-n oblivious transfer protocol with sublinear communication was first constructed (as a generalization of single-server PIR) by Eyal Kushilevitz and Rafail Ostrovsky.[6] More efficient constructions were proposed by Moni Naor and Benny Pinkas,[7] William Aiello, Yuval Ishai and Omer Reingold,[8] Sven Laur and Helger Lipmaa.[9] In 2017, Kolesnikov et al.,[10] proposed an efficient 1-n oblivious transfer protocol which requires roughly 4x the cost of 1-2 oblivious transfer in amortized setting.

Brassard, Crépeau and Robert further generalized this notion to k-n oblivious transfer,[11] wherein the receiver obtains a set of k messages from the n message collection. The set of k messages may be received simultaneously ("non-adaptively"), or they may be requested consecutively, with each request based on previous messages received.[12]

k-n Oblivious transfer is a special case of generalized oblivious transfer, which was presented by Ishai and Kushilevitz.[13] In that setting, the sender has a set U of n messages, and the transfer constraints are specified by a collection A of permissible subsets of U. The receiver may obtain any subset of the messages in U that appears in the collection A. The sender should remain oblivious of the selection made by the receiver, while the receiver cannot learn the value of the messages outside the subset of messages that he chose to obtain. The collection A is monotone decreasing, in the sense that it is closed under containment (i.e., if a given subset B is in the collection A, so are all of the subsets of B). The solution proposed by Ishai and Kushilevitz uses the parallel invocations of 1-2 oblivious transfer while making use of a special model of private protocols. Later on, other solutions that are based on secret sharing were published – one by Bhavani Shankar, Kannan Srinathan, and C. Pandu Rangan,[14] and another by Tamir Tassa.[15]

In the early seventies Stephen Wiesner introduced a primitive called multiplexing in his seminal paper "Conjugate Coding", which was the starting point of quantum cryptography.[16] Unfortunately it took more than ten years to be published. Even though this primitive was equivalent to what was later called 1–2 oblivious transfer, Wiesner did not see its application to cryptography.

Protocols for oblivious transfer can be implemented with quantum systems. In contrast to other tasks in quantum cryptography, like quantum key distribution, it has been shown that quantum oblivious transfer cannot be implemented with unconditional security, i.e. the security of quantum oblivious transfer protocols cannot be guaranteed only from the laws of quantum physics.[17]

• k-anonymity
• Secure multi-party computation
• Zero-knowledge proof
• Private information retrieval

• Helger Lipmaa's collection of Web links on the topic