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Prepare-and-Send Quantum Fully Homomorphic Encryption (edit)
Revision as of 08:49, 11 July 2019
, 11 July 2019→Stage 1 Client’s Preparation
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<math>\sum_{a,b\epsilon\{0,1\}}\frac{1}{4}\rho(HE.Enc_{pk_0^{[i]}}(a^{[i]}),HE.Enc_{pk_0^{[i]}}(b^{[i]}))\otimes X^aZ^b\sigma Z^bX^a</math> | <math>\sum_{a,b\epsilon\{0,1\}}\frac{1}{4}\rho(HE.Enc_{pk_0^{[i]}}(a^{[i]}),HE.Enc_{pk_0^{[i]}}(b^{[i]}))\otimes X^aZ^b\sigma Z^bX^a</math> | ||
#Client sends encryptions (<math>\tilde{a}^{[i]},\tilde{b}^{[i]}</math>) and the quantum one time padded (QOTP) state<math> X^{a^{[1]}}Z^{b^{[1]}}\otimes.....\otimes X^{a^{[n]}}Z^{b^{[n]}}\rho Z^{b^{[1]}}X^{a^{[1]}}\otimes.....\otimes X^{a^{[n]}}Z^{b^{[n]}} \forall i</math>, to the Server with the evaluation keys and public keys. | #Client sends encryptions (<math>\tilde{a}^{[i]},\tilde{b}^{[i]}</math>) and the quantum one time padded (QOTP) state<math> X^{a^{[1]}}Z^{b^{[1]}}\otimes.....\otimes X^{a^{[n]}}Z^{b^{[n]}}\rho Z^{b^{[1]}}X^{a^{[1]}}\otimes.....\otimes X^{a^{[n]}}Z^{b^{[n]}} \forall i</math>, to the Server with the evaluation keys and public keys. | ||
'''Gadget Construction (<math>\text{QFHE.GenGadget}_{pk_{i+1}}(sk_i)</math>)''' | |||
# Generate 4m EPR pairs ( | # Generate <math>4m</math> EPR pairs (<math>|\phi\rangle=\frac{1}{\sqrt{2}}(00+11))</math>, <math>\{(a_1,b_1),...,(a_{4m},b_{4m})\}</math> | ||
# Choose 2m pairs | # Choose <math>2m</math> pairs <math>\epsilon \{a_1, a_2,....,a_{4m}\}</math> using sk | ||
## If (sk = 0) then {( | ## If <math>(sk=0)</math> then <math>\{(a_1,a_2),(a_2,a_3),...,(a_{4m-1},a_{4m})\}</math> | ||
# For j=1 to 2m, | ## If <math>(sk=1)</math> then <math>\{(a_1,a_3),(a_2,a_4),...,(a_{4m-2},a_{4m})\}</math> | ||
## Choose p[j] | # For j=1 to 2m, | ||
## Perform Bell Measurement on | ## Choose p[j] <math>\epsilon_R \{0,1\}</math> | ||
## Thus, new EPR pairs are | ## Perform Bell Measurement on <math>j^{th}</math> pair with an extra <math>(P^\dagger)^p</math> operation, get outcomes (x[j],z[j]) | ||
## Thus, new EPR pairs are | |||
### If <math>(sk=0)</math> then <math>\{(b_1,b_2),(b_2,b_3),...,(b_{4m-1},b_{4m})\}</math> | |||
# Encrypt (x[j],z[j]), p[j] for all j and sk using | ### If <math>(sk=1)</math> then <math>\{(b_1,b_3),(b_2,b_4),...,(b_{4m-2},b_{4m})\}</math> | ||
## Denote the <math>2m</math> entangled pairs be denoted by <math>\{(s_1,t_1),(s_2,t_2),...,(s_{2m},t_{2m})\}</math>, such that | |||
## The classical information of gadget be g(sk)<math>=(\{(s_1,t_1),(s_2,t_2),...,(s_{2m},t_{2m}),p,sk\}</math>. | |||
## The quantum state of gadget can be written as <math>\gamma_{x,z}(g(sk))=\pi_{j=1}^mX^{x[i]}Z^{z[i]}(P^\dagger){p[i]}|\phi\rangle\langle\phi|_{s_jt_j}(P^\dagger){p[i]}Z^{z[i]}X^{x[i]}</math> | |||
# Encrypt (x[j],z[j]), p[j] for all j and sk using <math>pk_{i+1}</math>. Resulting Gadget is the classical-quantum (CQ) state, | |||
<math>\Gamma_{pk_{i+1}}(sk_i)=\rho(HE.Enc_{pk_{i+1}}(g(sk))\otimes \frac{1}{2^{2m}}\sum_{x,z\epsilon\{0,1\}^m}\rho(HE.Enc_{pk_{i+1}}(x,z)\otimes \gamma_{x,z}(g(sk))</math> | |||
=== Stage 2 Server’s Computation=== | === Stage 2 Server’s Computation=== | ||