# 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.

## AssumptionsEdit

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

## OutlineEdit

In the beginning, the election authority prepares an N-qudit ballot state where the kth qudit of the state corresponds to ${\displaystyle V_{k}}$ ’s blank ballot and sends the corresponding blank ballot to ${\displaystyle V_{k}}$  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.

## NotationsEdit

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

## RequirementsEdit

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

## PropertiesEdit

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 DescriptionEdit

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

The states ${\displaystyle |j\rangle ,j=0,...,D-1,}$  form an orthonormal basis for the D-dimensional Hilbert space, and D > N. The k-th qudit of ${\displaystyle \Phi }$  is ${\displaystyle V_{k}}$ 's blank ballot.

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

${\displaystyle yes:|\psi (\theta _{y})\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{ij\theta _{y}}|j\rangle }$

,

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

For ${\displaystyle v\in \{y,n\}}$  we have ${\displaystyle \theta _{v}=(2\pi l_{v}/D)+\delta }$ , where ${\displaystyle l_{v}\in \{0,...,D-1\}}$  and ${\displaystyle \delta \in [0,2\pi /D)}$ . Values ${\displaystyle l_{y}}$  and ${\displaystyle \delta }$  are chosen uniformly at random from their domain and ${\displaystyle l_{n}}$  is chosen such that ${\displaystyle N(l_{y}-l_{n}{\text{ }}mod{\text{ }}D)}$  < D.
• Casting phase:
1. Each ${\displaystyle V_{k}}$  appends the corresponding option qudit to the blank ballot and performs a 2-qudit measurement ${\displaystyle R=\sum _{r=0}^{D-1}rP_{r}}$  where ${\displaystyle P_{r}=\sum _{j=0}^{D-1}|j+r\rangle \langle j+r|\otimes |j\rangle \langle j|.}$

According to the result ${\displaystyle r_{k},V_{k}}$  performs a unitary correction ${\displaystyle U_{r_{k}}=I\otimes \sum _{j=0}^{D-1}|j+r_{k}\rangle \langle j|}$  and sends the 2-qudits ballot and ${\displaystyle r_{k}}$  back to T

• Tally phase:
1. The global state of the system is: ${\displaystyle {\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}\Pi _{k=1}^{N}\alpha _{j,r_{k}}|j\rangle ^{\otimes 2N}}$  where ,
${\displaystyle \alpha _{j,r_{k}}={\begin{cases}e^{i(D+j-r_{k})\theta _{v}^{k}},{\text{ }}0\leq j\leq r_{k}-1,\\e^{i(j-r_{k})\theta _{v}^{k}}{\text{ }}r_{k}\leq j\leq D-1\end{cases}}}$

For every k, T applies ${\displaystyle W_{k}=\sum _{j=0}^{r_{k}-1}e^{-iD\delta }|j\rangle |\langle j|+\sum _{j=r_{k}}^{D-1}|j\rangle |\langle j|}$  on one of the qudits in the global state.

2. By applying the unitary operator ${\displaystyle \sum _{j=0}^{D-1}e^{-ijN\theta _{n}}|j\rangle \langle j|}$ on one of the qudits we have ${\displaystyle |\phi _{q}\rangle ={\dfrac {1}{\sqrt {D}}}\sum _{j=0}^{D-1}e^{2\pi ijq/D}|j\rangle ^{\otimes 2N}}$  where ${\displaystyle q=m(l_{y}-l_{n})}$ . with the corresponding measurement, T retrieves q and uses values ${\displaystyle l_{y},l_{n}}$  to compute m.

## Further InformationEdit

*contributed by Sara Sarfaraz