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Prepare-and-Send Quantum Fully Homomorphic Encryption (edit)
Revision as of 08:56, 11 July 2019
, 11 July 2019→Stage 3 Client’s Correction
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===Stage 3 Client’s Correction=== | ===Stage 3 Client’s Correction=== | ||
'''Decryption (QFHE.Dec()) | |||
*Input: QOTP Circuit Output State ( | *Input: QOTP Circuit Output State (<math> X^{a'^{[1]}}Z^{b'^{[1]}}\otimes.....\otimes X^{a'^{[k]}}Z^{b'^{[k]}}\rho' Z^{b'^{[1]}}X^{a'^{[1]}}\otimes.....\otimes X^{a'^{[k]}}Z^{b'^{[k]}}</math>), Corresponding Encrypted Pad key (<math>\tilde{a'},\tilde{b'}</math>) | ||
*Output: Final outcome of the computation | *Output: Final outcome of the computation C<math>\rho</math>C<math>^\dagger=\rho'</math> | ||
# Client uses <math>sk_L</math> to restore the pad key from sent encryption: (<math>{a'}^{[i]},{b'}^{[i]}</math>)=(HE.Dec<math>_{sk_L^{[i]}}(a'^{[i]}</math>),HE.Dec<math>_{pk_L^{[i]}}(b'^{[i]}))</math> | |||
# Client uses pad key and operates | # Client uses pad key and operates <math>X^{a'^{[i]}}Z^{b'^{[i]}}</math> on single qubits i separately just like encryption. | ||
# Client repeats this for all single qubits and hence gets the quantum state | ##Let single qubit representation of the output state be <math>X^{a'^{[i]}}Z^{b'^{[i]}}\sigma' Z^{b'^{[i]}}X^{a'^{[i]}}</math>, then operation of Pauli X,Z gates as above yields <math>\sigma'</math> | ||
# Client repeats this for all single qubits and hence gets the quantum state <math>\rho'</math>, final outcome of the computation. | |||
==Further Information== | ==Further Information== | ||
<div style='text-align: right;'>''*contributed by Shraddha Singh''</div> | <div style='text-align: right;'>''*contributed by Shraddha Singh''</div> | ||