Device-Independent Oblivious Transfer

This example protocol achieves the task of device-independent oblivious transfer in the bounded quantum storage model using a computational assumption.

Assumptions

• The quantum storage of the receiver is bounded during the execution of the protocol
• The device used is computationally bounded - it cannot solve the Learning with Errors (LWE) problem during the execution of the protocol
• The device behaves in an IID manner - it behaves independently and identically during each round of the protocol

Outline

• The protocol consists of multiple rounds, which are randomly chosen for testing or string generation
• The testing rounds are carried out to ensure that the devices used are following the expected behaviour. The self-testing protocol used is a modification of the one used in DIQKD. This modification is necessary as, unlike the DIQKD scenario, the parties involved in OT may not trust each other to cooperate. The self-testing protocol uses the computational assumptions associated with Extended noisy trapdoor claw-free (ENTCF) function families to certify that the device has created the desired quantum states. If the fraction of failed testing rounds exceeds a certain limit, the protocol is aborted.
• At the end of the protocol, the honest sender outputs two randomly generated strings of equal length, and the honest receiver outputs their chosen string out of the two.

Notation

• $S$: The sender
• $R$: The receiver
• $l$: Length of the output strings
• $s_{0},s_{1}$: The strings output by the sender
• $c$: A bit denoting the receiver's choice
• For any bit $r$, [Computational, Hadamard]$_{r}={\begin{cases}{\mbox{Computational, if }}r=0\\{\mbox{Hadamard, if }}r=1\end{cases}}$
• $\sigma _{X}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}$
• $\sigma _{Z}={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$
• For bits $v^{\alpha },v^{\beta }:|\phi ^{(v^{\alpha },v^{\beta })}\rangle =(\sigma _{Z}^{v^{\alpha }}\sigma _{X}^{v^{\beta }}\otimes I){\frac {|00\rangle +|11\rangle }{\sqrt {2}}}$
• An ENTCF family consists of two families of function pairs: $F$ and $G$. A function pair $(f_{k,0},f_{k,1})$is indexed by a public key $k$. If $(f_{k,0},f_{k,1})\in F$, then it is a claw-free pair; and if $(f_{k,0},f_{k,1})\in G$, then it is called an injective pair. ENTCF families satisfy the following properties:
  1. For a fixed $k\in K_{F},f_{k,0}$ and $f_{k,1}$ are bijections with the same image; for every image $y$, there exists a unique pair $(x_{0},x_{1})$, called a claw, such that $f_{k,0}(x_{0})=f_{k,1}(x_{1})=y$
  2. Given a key $k\in K_{F}$, for a claw-free pair, it is quantum-computationally intractable (without access to trapdoor information) to compute both a $x_{i}$ and a single generalized bit of $x_{0}\oplus x_{1}$, where $(x_{0},x_{1})$ forms a valid claw. This is known as the adaptive hardcore bit property.
  3. For a fixed $k\in K_{G},f_{k,0}$ and $f_{k_{1}}$ are injunctive functions with disjoint images.
  4. Given a key $k\in K_{F}\cup K_{G}$, it is quantum-computationally hard (without access to trapdoor information) to determine whether $k$ is a key for a claw-free or an injective pair. This property is known as injective invariance.
  5. For every $k\in K_{F}\cup K_{G}$, there exists a trapdoor $t_{k}$ which can be sampled together with $k$ and with which 2 and 4 are computationally easy.

Protocol Description

Protocol 1: Rand 1-2 OT$^{l}$

1. A device prepares $n$ uniformly random Bell pairs $|\phi ^{(v_{i}^{\alpha },v_{i}^{\beta })}\rangle ,i=1,...,n$, where the first qubit of each pair goes to $S$ along with the string $v^{\alpha }$, and the second qubit of each pair goes to $R$ along with the string $v^{\beta }$.
2. R measures all qubits in the basis $y=[$Computational,Hadamard$]_{c}$ where $c$ is $R$'s choice bit. Let $b\in \{0,1\}^{n}$ be the outcome. $R$ then computes $b\oplus w^{\beta }$, where the $i$-th entry of $w^{\beta }$ is defined by

   $w_{i}^{\beta }:={\begin{cases}0,{\mbox{if }}y={\mbox{ Hadamard}}\\v_{i}^{\beta },{\mbox{if }}y={\mbox{ Computational}}\end{cases}}$

3. $S$ picks uniformly random $x\in \{$ Computational, Hadamard$\}^{n}$, and measures the $i$-th qubit in basis $x_{i}$. Let $a\in \{0,1\}^{n}$ be the outcome. $S$ then computes $a\oplus w^{\alpha }$, where the $i$-th entry of $w^{\alpha }$ is defined by

   $w_{i}^{\alpha }:={\begin{cases}v_{i}^{\alpha },{\mbox{if }}x_{i}={\mbox{ Hadamard}}\\0,{\mbox{if }}x_{i}={\mbox{ Computational}}\end{cases}}$

4. $S$ picks two uniformly random hash functions $f_{0},f_{1}\in F$, announces $x$ and $f_{0},f_{1}$ to $R$ and outputs $s_{0}:=f_{0}(a\oplus w^{\alpha }|_{I_{0}})$ and $s_{1}:=f_{1}(a\oplus w^{\alpha }|_{I_{1}})$ where $I_{r}:=\{i\in I:x_{i}=[$Computational,Hadamard$]_{r}\}$
5. $R$ outputs $s_{c}=f_{c}(b\oplus w^{\beta }|_{I_{c}})$

Protocol 2: Self-testing with a single verifier

1. Alice chooses the state bases $\theta ^{A},\theta ^{B}\in$ {Computational,Hadamard} uniformly at random and generates key-trapdoor pairs $(k^{A},t^{A}),(k^{B},t^{B})$, where the generation procedure for $k^{A}$ and $t^{A}$ depends on $\theta ^{A}$ and a security parameter $\eta$, and likewise for $k^{B}$ and $t^{B}$. Alice supplies Bob with $k^{B}$. Alice and Bob then respectively send $k^{A},k^{B}$ to the device.
2. Alice and Bob receive strings $c^{A}$ and $c^{B}$, respectively, from the device.
3. Alice chooses a challenge type $CT\in \{a,b\}$, uniformly at random and sends it to Bob. Alice and Bob then send $CT$ to each component of their device.
4. If $CT=a$:
   1. Alice and Bob receive strings $z^{A}$ and $z^{B}$, respectively, from the device.
5. If $CT=b$:
   1. Alice and Bob receive strings $d^{A}$ and $d^{B}$, respectively, from the device.
   2. Alice chooses uniformly random measurement bases (questions) $x,y\in$ {Computational,Hadamard} and sends $y$ to Bob. Alice and Bob then, respectively, send $x$ and $y$ to the device.
   3. Alice and Bob receive answer bits $a$ and $b$, respectively, from the device. Alice and Bob also receive bits $h^{A}$ and $h^{B}$, respectively, from the device.

Protocol 3: DI Rand 1-2 OT$^{l}$

Data generation:

1. The sender and receiver execute $n$ rounds of Protocol 2 (Self-testing) with the sender as Alice and receiver as Bob, and with the following modification:

   If $CT_{i}=b$, then with probability $p$, the receiver does not use the measurement basis question supplied by the sender and instead inputs $y_{i}=[$Computational, Hadamard$]_{c}$ where $c$ is the receiver's choice bit. Let $I$ be the set of indices marking the rounds where this has been done.

   For each round $i\in \{1,...,n\}$, the receiver stores:

   • $c_{i}^{B}$
   • $z_{i}^{B}$ if $CT_{i}=a$
   • or $(d_{i}^{B},y_{i},b_{i},h_{i}^{B})$ if $CT_{i}=b$

   The sender stores $\theta _{i}^{A},\theta _{i}^{B},(k_{i}^{A},t_{i}^{A}),(k_{i}^{B},t_{i}^{B}),c_{i}^{A},CT_{i};$ and $z_{i}^{A}$ if $CT_{i}=a$ or $(d_{i}^{A},x_{i},a_{i},h_{i}^{A})$ and $y_{i}$ if $CT_{i}=b$

2. For every $i\in \{1,...,n\},$ the sender stores the variable $RT_{i}$ (round type), defined as follows:
   • if $CT_{i}=b$ and $\theta _{i}^{A}=\theta _{i}^{B}=$Hadamard, then $RT_{i}=$ Bell
   • else, set $RT_{i}=$ Product
3. For every $i\in \{1,...,n\},$ the sender chooses $T_{i}$, indicating a test round or generation round, as follows:
   • if $RT_{i}=$ Bell, choose $T_{i}\in$ {Test, Generate} uniformly at random
   • else, set $T_{i}=$ Test

   The sender sends ($T_{1},...,T_{n}$) to the receiver

   Testing:

4. The receiver sends the set of indices $I$ to the sender. The receiver publishes their output for all $T_{i}=$ Test rounds where $i\notin I$. Using this published data, the sender determines the bits which an honest device would have returned.
5. The sender computes the fraction of test rounds (for which the receiver has published data for) that failed. If this exceeds some $\epsilon$, the protocol aborts

   Preparing data:

6. Let ${\tilde {I}}:=\{i:i\in I$ and $T_{i}=$ Generate} and $n^{\prime }=|{\tilde {I}}|$. The sender checks if there exists a $k>0$ such that $\gamma n^{\prime }\leq n^{\prime }/4-2l-kn^{\prime }$. If such a $k$ exists, the sender publishes ${\tilde {I}}$ and, for each $i\in {\tilde {I}}$, the trapdoor $t_{i}^{B}$ corresponding to the key $k_{i}^{B}$ (given by the sender in the execution of Protocol 2,Step 1); otherwise the protocol aborts.
7. For each $i\in {\tilde {I}},$ the sender calculates $v_{i}^{\alpha }=d_{i}^{A}.(x_{i,0}^{A}\oplus x_{i,1}^{A})$ and defines $w^{\alpha }$ by

   $w_{i}^{\alpha }={\begin{cases}v_{i}^{\alpha },{\mbox{if }}x_{i}={\mbox{Hadamard}}\\0,{\mbox{if }}x_{i}={\mbox{Computational}}\end{cases}}$

   and the receiver calculates $v_{i}^{\beta }==d_{i}^{B}.(x_{i,0}^{B}\oplus x_{i,1}^{B})$ and defines $w^{\beta }$ by

   $w_{i}^{\beta }={\begin{cases}0,{\mbox{if }}y_{i}={\mbox{Hadamard}}\\v_{i}^{\beta },{\mbox{if }}y_{i}={\mbox{Computational}}\end{cases}}$

   Obtaining output:

8. The sender randomly picks two hash functions $f_{0},f_{1}\in F$, announces $f_{0},f_{1}$ and $x_{i}$ for each $i\in {\tilde {I}}$, and outputs $s_{0}=f_{0}(a\oplus w^{\alpha }|_{{\tilde {I}}_{0}})$ and $s_{1}=f_{1}(a\oplus w^{\alpha }|_{{\tilde {I}}_{1}})$, where ${\tilde {I}}_{r}:=\{i\in {\tilde {I}}:x_{i}=[$Computational,Hadamard$]_{r}\}$
9. Receiver outputs $s_{c}=f_{c}(a\oplus w^{\beta }|_{{\tilde {I}}_{c}})$

References

*contributed by Chirag Wadhwa