Quantum voting based on conjugate coding: Difference between revisions

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==Protocol Description==
==Protocol Description==
*'''Setup phase''':
*'''Setup phase''':
#EA picks a vector <math> b = (b_1, . . . , b_{n+1}) \in \{0, 1\}^{n+1}</math> and it will be kept secret from T until the end of the ballot casting phase.
#EA picks a vector <math> b = (b_1, . . . , b_{n+1}) \in \{0, 1\}^{n+1}</math> that will be kept secret from T until the end of the ballot casting phase.
#For each <math>V_k</math>, EA prepares w = poly(n) blank ballot fragments each of the form <math>||\phi_{\bar{a_j},\bar{b}}\rangle</math>
#For each <math>V_k</math>, EA prepares w = poly(n) blank ballot fragments each of the form <math>||\phi_{a_j^1,b_1}\rangle=|\psi_{a_j^1,b_1}\rangle \otimes ...\otimes |\psi_{a_j^{n+1},b_{n+1}}\rangle,j \in \{1,...,w\},\text{ where }\bar{a}_j=(a_j^1,...,a_j^{n+1}) \text { such that }(a_j^1,...,a_j^{n})\in \{0,1\}^n,a_j^{n+1}=a_j^1 \oplus ...\oplus a_j^{n+}\text{ and } |\psi_{0,0}\rangle=|0\rangle,|\psi_{1,0}\rangle=|1\rangle,|\psi_{0,1}\rangle=\dfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle),|\psi_{1,1}\rangle=\dfrac{1}{\sqrt{2}}(|0\rangle-|1\rangle)  </math>


==Further Information==
==Further Information==


<div style='text-align: right;'>''*contributed by Sara Sarfaraz''</div>
<div style='text-align: right;'>''*contributed by Sara Sarfaraz''</div>