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 $\Phi$ corresponds to $V_{k}$'s blank ballot.

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

   $yes:|\psi (\theta _{y})\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{ij\theta _{y}}|j\rangle$, no:$|\psi (\theta _{n})\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{ij\theta _{n}}|j\rangle$.

   For $v\in \{y,n\}$ we have $\theta _{v}=(2\pi l_{v}/D)+\delta$, where $l_{v}\in \{0,...,D-1\}$ and $\delta \in [0,2\pi /D)$. Values $l_{y}$ and $\delta$ are chosen uniformly at random from their domain and $l_{n}$ is chosen such that $N(l_{y}-l_{n}{\text{ }}mod{\text{ }}D)$ < D.

• Casting phase:

???

• 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 ,
   $$\alpha _{j,r_{k}}={\begin{cases}e^{i(D+j-r_{k})\theta _{v}^{k}},{\text{ }}0\leq j\leq r_{k}-1,\\e^{i(j-r_{k})\theta _{v}^{k}}{\text{ }}r_{k}\leq j\leq D-1\end{cases}}$$

Further Information

*contributed by Sara Sarfaraz