Distributed Ballot Based Protocol

This example protocol implements the task of Quantum E-voting. In this protocol, the election authority prepares and distributes to each voter a blank ballot, and gathers it back after all voters have cast their vote in order to compute the final outcome.

Assumptions

• The tallier is assumed to be trusted to correctly prepare specific states.

Outline

In the beginning, the election authority prepares an N-qudit ballot state where the kth qudit of the state corresponds to $V_{k}$’s blank ballot and sends the corresponding blank ballot to $V_{k}$ together with two option qudits, one for the “yes” and one for the “no” vote. then each voter decides on “yes” or “no” by appending the corresponding option qudit to the blank ballot and performing a 2-qudit measurement, then based on its result she performs a unitary correction and sends the 2-qudits ballot along with the measurement result back to the election authority. At the end of the election, the election authority applies a unitary operation on one of the qudits in the global state and another unitary operation on one of the qudits to find the number of yes votes.

Notations

• $V_{i}:i^{th}$ voter
• c: number of possible candidates
• N: number of voters
• $v_{i}$: vote of $i^{th}$ voter
• T: election authority
• m: number of yes votes

Requirements

• Quantum channel capable of sending qubits -> (qudit) between the election authority and voters
• Qudit Measurement Device for election authority and voters

Properties

This protocol is vulnerable to double voting. Specifically, an adversary can mount a “d-transfer attack”, and transfer d votes for one option of the referendum election to the other.

Protocol Description

• Setup phase:

1. T prepares an N-qudit ballot state $|\Phi \rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}|j\rangle ^{\otimes N}$.

   The states $|j\rangle ,j=0,...,D-1,$ form an orthonormal basis for the D-dimensional Hilbert space, and D > N. The k-th qudit of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Phi } is $V_{k}$'s blank ballot.

2. T sends to $V_{k}$ the corresponding blank ballot and two option qudits,for the "yes" and "no" option:

   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle yes:|\psi (\theta _{y})\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{ij\theta _{y}}|j\rangle }

   ,

   no:Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |\psi (\theta _{n})\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{ij\theta _{n}}|j\rangle } .

   For Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v\in \{y,n\}} we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \theta _{v}=(2\pi l_{v}/D)+\delta } , where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle l_{v}\in \{0,...,D-1\}} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \delta \in [0,2\pi /D)} . Values $l_{y}$ and $\delta$ are chosen uniformly at random from their domain and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle l_{n}} is chosen such that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N(l_{y}-l_{n}{\text{ }}mod{\text{ }}D)} < D.

• Casting phase:

1. Each $V_{k}$ appends the corresponding option qudit to the blank ballot and performs a 2-qudit measurement Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle R=\sum _{r=0}^{D-1}rP_{r}} where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P_{r}=\sum _{j=0}^{D-1}|j+r\rangle \langle j+r|\otimes |j\rangle \langle j|.}

   According to the result Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r_{k},V_{k}} performs a unitary correction Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle U_{r_{k}}=I\otimes \sum _{j=0}^{D-1}|j+r_{k}\rangle \langle j|} and sends the 2-qudits ballot and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r_{k}} back to T

• Tally phase:

1. The global state of the system is: ${\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}\Pi _{k=1}^{N}\alpha _{j,r_{k}}|j\rangle ^{\otimes 2N}$ where , Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha _{j,r_{k}}={\begin{cases}e^{i(D+j-r_{k})\theta _{v}^{k}},{\text{ }}0\leq j\leq r_{k}-1,\\e^{i(j-r_{k})\theta _{v}^{k}}{\text{ }}r_{k}\leq j\leq D-1\end{cases}}}

   For every k, T applies Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle W_{k}=\sum _{j=0}^{r_{k}-1}e^{-iD\delta }|j\rangle |\langle j|+\sum _{j=r_{k}}^{D-1}|j\rangle |\langle j|} on one of the qudits in the global state.

2. By applying the unitary operator Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{j=0}^{D-1}e^{-ijN\theta _{n}}|j\rangle \langle j|} on one of the qudits we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |\phi _{q}\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{2\pi ijq/D}|j\rangle ^{\otimes 2N}} where $q=m(l_{y}-l_{n})$. with the corresponding measurement, T retrieves q and uses values $l_{y},l_{n}$ to compute m.

Further Information

*contributed by Sara Sarfaraz