Solving Bernstein and Vazirani’s Problem with the 2-bit Permutation Function

Abstract

In this paper, we propose a novel quantum algorithm, based on the Bernstein–Vazirani algorithm, for finding $a$ if the function $f\left( x \right) = a \cdot {\uppi }\left( x \right)$, where $a,x \in \left\{ {0,1} \right\}^{2}$ and ${\uppi }\left( x \right)$ is a 2-bit permutation function. Note that the Bernstein–Vazirani algorithm cannot find $a$ when we select the function $f\left( x \right) = a \cdot {\uppi }\left( x \right)$, where ${\uppi }\left( x \right) = \left( {\begin{array}{*{20}c} {\begin{array}{*{20}c} 0 & 1 \\ \end{array} } & {\begin{array}{*{20}c} 2 & 3 \\ \end{array} } \\ {\begin{array}{*{20}c} 2 & 0 \\ \end{array} } & {\begin{array}{*{20}c} 1 & 3 \\ \end{array} } \\ \end{array} } \right)$. Our algorithm can be further applied to Nagata et al.’s problem. Our algorithm will output $g\left( {A_{1} } \right)$ and $g\left( {A_{0} } \right)$ if the function $f\left( x \right) = \left( {g\left( {A_{1} } \right)g\left( {A_{0} } \right)} \right)_{2} \cdot {\uppi }\left( x \right)$, where $A_{1}$, $A_{0} \in \left\{ {0,1} \right\}^{m}$, $x \in \left\{ {0,1} \right\}^{2}$ and $g: \left\{ {0, 1} \right\}^{m} \to \left\{ {0, 1} \right\}.$

Appendix 1 The simulation of our example in Julia language and the simulation.

The simulation of our example have been implemented in Julia language. The process of executing Algorithm 1 for finding $a$ if the function $f\left(x\right)=a\cdot {\uppi }_{41}\left(x\right)$, where $a=3$. The steps show that Algorithm 1 can output $a=3$.