46
edits
Quantum voting based on conjugate coding: Difference between revisions
→Protocol Description
| Line 35: | Line 35: | ||
==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> that will be kept secret from T until the end of the ballot casting phase. | #EA picks a vector <math> \bar{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_{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> | #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> | ||
#EA sends one blank ballot to each <math> V_k </math> over an authenticated channel. | #EA sends one blank ballot to each <math> V_k </math> over an authenticated channel. | ||