Efficient multiparty quantum summation protocol in a restricted quantum environment

Abstract

This paper introduces a multiparty semi-quantum summation protocol that leverages graph states within a restricted quantum environment. The protocol enables n classical participants, with the assistance of a semi-honest third party, to compute the summation result modulo 2 while preserving the privacy of their individual inputs. Classical participants have only two quantum capabilities: performing Hadamard operations and Z-basis measurements. The adoption of one-way qubit transmission eliminates the need for additional quantum devices to prevent quantum Trojan horse attacks and reduces transmission costs. Comprehensive security analysis, comparison, and simulation experiments validate the security, efficiency, and feasibility of the proposed protocol.

Appendix 1 Equivalent relation proof

This appendix describes the detailed process of establishing the equivalence between ${\left(\frac{1}{\sqrt{2}}\right)}^{n}\left({\sum }_{j=0}^{{2}^{n}-1}{\left(-1\right)}^{\Lambda }\left|j\rangle \right.\right)$ and Eq. (4). It is evident that the difference between the two quantum system states lies in the exponentiation of − 1. Therefore, if it can be proven that ${\left(-1\right)}^{\Delta }={\left(-1\right)}^{\Lambda }$, then we can demonstrate that the two quantum system states are equivalent. In this appendix, we use mathematical induction to achieve this goal, and the proofs are as follows:

Base Case: Consider a two-qubit complete graph state. There are three situations for $Hw(j)$, i.e., $Hw(j)=0$, 1, or 2. Each situation is analyzed as follows:

$$Hw\left( j \right) = 0\;{\text{because}}\;j = 00,\Lambda = \left\lfloor \frac{0}{2} \right\rfloor \;\bmod \;2 = 0\;{\text{and}}\;\Delta = \sum\nolimits_{0 \le x \le y \le 1} {\left( {j_{x} \times j_{y} } \right)} = 0$$

$$Hw\left( j \right) = 1\;{\text{because}}\;j = 01\;{\text{or10,}}\;\Lambda = \left\lfloor \frac{1}{2} \right\rfloor \;\bmod \;2 = 0\;{\text{and}}\;\Delta = \sum\limits_{0 \le x \le y \le 1} {\left( {j_{x} \times j_{y} } \right)} = 0$$

$$Hw\left( j \right) = 2:\;{\text{because}}\;j = 11{,}\;\Lambda = \left\lfloor \frac{2}{2} \right\rfloor \;\bmod \;2 = 1\;{\text{and}}\;\Delta = \sum\limits_{0 \le x \le y \le 1} {\left( {j_{x} \times j_{y} } \right)} = 1.$$

From the above-mentioned three situations, we can see that ${\left(-1\right)}^{\Delta }={\left(-1\right)}^{\Lambda }$, and the statement is true for the initial value, where the bit length of $j$ is 2.

Inductive Hypothesis: Assume the statement is true for the bit length of $j$ equal to $n$. We denote the statement ${\left(-1\right)}^{{\Delta }_{n}}={\left(-1\right)}^{{\Lambda }_{n}}$.

Inductive Step: Consider the situation where the bit length of $j$ is $n+1$. Here, we discuss the two possible values of the $\left(n+1\right)$-th bit.

The value of this bit is 0: It is evident that ${\Lambda }_{n+1}={\Lambda }_{n}$. Because the $\left(n+1\right)$-th bit, ${j}_{n}$, is 0, ${\Delta }_{n+1}={\sum_{0\le x<y\le n}\left({j}_{x}\times {j}_{y}\right)=\sum_{0\le x<y\le n-1}\left({j}_{x}\times {j}_{y}\right)+\sum_{0\le x\le n-1}\left({j}_{x}\times {j}_{n}\right)=\Delta }_{n}+\sum_{0\le x\le n-1}\left({j}_{x}\times {j}_{n}\right)={\Delta }_{n}+0={\Delta }_{n}$. Because ${\left(-1\right)}^{{\Delta }_{n}}={\left(-1\right)}^{{\Lambda }_{n}}$, we can see that ${\left(-1\right)}^{{\Delta }_{n+1}}={\left(-1\right)}^{{\Lambda }_{n+1}}$. Therefore, this statement holds true in this situation.

The value of this bit is 1: If ${Hw\left(j\right)}_{n}$ is even, ${\Lambda }_{n+1}={\Lambda }_{n}$; otherwise, ${\Lambda }_{n+1}={\Lambda }_{n}+1$, where ${Hw\left(j\right)}_{n}$ is the Hamming weight of $j$ when the bit length of $j$ is n. In these situations, the following equation is obtained:

$$\begin{array}{c}\left\{\begin{array}{c}{\left(-1\right)}^{{\Lambda }_{n+1}}={\left(-1\right)}^{{\Lambda }_{n}}, {Hw\left(j\right)}_{n}\; is\; even. \\ {\left(-1\right)}^{{\Lambda }_{n+1}}={\left(-1\right)}^{{\Lambda }_{n}}\times -1, {Hw\left(j\right)}_{n}\; is\; odd.\end{array}\right.\#\end{array}$$

(A1)

However, we know ${\Delta }_{n+1}={\Delta }_{n}+\sum_{0\le x\le n-1}\left({j}_{x}\times {j}_{n}\right)={\Delta }_{n}+\sum_{0\le x\le n-1}\left({j}_{x}\times 1\right)={\Delta }_{n}+\sum_{0\le x\le n-1}{j}_{x}$. If ${Hw\left(j\right)}_{n}$ is even, $\sum_{0\le x\le n-1}{j}_{x}$ will also be even. Otherwise, $\sum_{0\le x\le n-1}{j}_{x}$ is odd. Therefore, we obtain the following equations for these situations:

$$\begin{array}{c}\left\{\begin{array}{c}\begin{array}{c}{\left(-1\right)}^{{\Delta }_{n+1}}={\left(-1\right)}^{{\Delta }_{n}+\sum_{0\le x\le n-1}{j}_{x}}={\left(-1\right)}^{{\Delta }_{n}}\times 1={\left(-1\right)}^{{\Delta }_{n}}, {Hw\left(j\right)}_{n}\; is\; even. \\ {\left(-1\right)}^{{\Delta }_{n+1}}={\left(-1\right)}^{{\Delta }_{n}+\sum_{0\le x\le n-1}{j}_{x}}={\left(-1\right)}^{{\Delta }_{n}}\times \left(-1\right), {Hw\left(j\right)}_{n}\; is\; odd.\end{array}\#\#\#\#\#\#\#\end{array}\right.\#\end{array}$$

(A2)

From Eqs. (A1) and (A2), we obtain ${\left(-1\right)}^{{\Delta }_{n+1}}={\left(-1\right)}^{{\Lambda }_{n+1}}$, and the statement holds under this situation.

According to these analyses, we can prove that the statement holds for a bit length of $j$ is $n+1$. Therefore, we prove that the equivalent relation exists between ${\left(\frac{1}{\sqrt{2}}\right)}^{n}\left({\sum }_{j=0}^{{2}^{n}-1}{\left(-1\right)}^{\Lambda }\left|j\rangle \right.\right)$ and Eq. (4).