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Classical Fully Homomorphic Encryption for Quantum Circuits: Difference between revisions
Classical Fully Homomorphic Encryption for Quantum Circuits (edit)
Revision as of 21:41, 25 November 2018
, 25 November 2018→Stage 2 Server’s Computation
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####Server measures the last register to get <math>y =\mathrm{AltHE.Enc}(\mu_0,r_0)=\mathrm{AltHE.Enc}_{pk}(\mu_1,r_1)\oplus_H AltHE.Enc_{pk}(s)</math>.</br> The resulting superposition state is:<math>\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}|a,b\rangle|\mu_a,r_a\rangle|\mathrm{AltHE.Enc}(\mu_0,r_0)\rangle=\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}|a,b\rangle|\mu_a,r_a\rangle|y\rangle</math> | ####Server measures the last register to get <math>y =\mathrm{AltHE.Enc}(\mu_0,r_0)=\mathrm{AltHE.Enc}_{pk}(\mu_1,r_1)\oplus_H AltHE.Enc_{pk}(s)</math>.</br> The resulting superposition state is:<math>\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}|a,b\rangle|\mu_a,r_a\rangle|\mathrm{AltHE.Enc}(\mu_0,r_0)\rangle=\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}|a,b\rangle|\mu_a,r_a\rangle|y\rangle</math> | ||
***'''Encrypted CNOT operation:''' | ***'''Encrypted CNOT operation:''' | ||
<math>\sum_{a,b\in\{0,1\}}\alpha_{ab}CNOT_{a,b}^s|a,b\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}|a,b\oplus a\cdot s\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}|a,b\oplus a\cdot(\mu_0\oplus\mu_1)\rangle</math></br><math>=\sum_{b\in\{0,1\}}\alpha_{0b}|0,b\oplus \mu_0\oplus\mu_0\rangle+\alpha_{1b}|1,b\oplus \mu_0\oplus\mu_1\rangle</math>, <math>\because q\oplus q=0</math></br><math>=\sum_{b\in\{0,1\}}\alpha_{0b}|0\rangle\otimes X^{\mu_0}|b\oplus \mu_0\rangle | <math>\sum_{a,b\in\{0,1\}}\alpha_{ab}CNOT_{a,b}^s|a,b\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}|a,b\oplus a\cdot s\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}|a,b\oplus a\cdot(\mu_0\oplus\mu_1)\rangle</math></br><math>=\sum_{b\in\{0,1\}}\alpha_{0b}|0,b\oplus \mu_0\oplus\mu_0\rangle+\alpha_{1b}|1,b\oplus \mu_0\oplus\mu_1\rangle</math>, <math>\because q\oplus q=0</math></br><math>=\sum_{b\in\{0,1\}}\alpha_{0b}|0\rangle\otimes X^{\mu_0}|b\oplus \mu_0\rangle+\alpha_{1b}|1\rangle \otimes X^{\mu_0}|b\oplus \mu_1\rangle</math>, <math>\because |q\oplus y\rangle=X^y|q\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}|a\rangle\otimes X^{\mu_0}|b\oplus \mu_a\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}(I\otimes X^{\mu_0})|a,b\oplus \mu_a\rangle</math></br> | ||
####Server XORs the second qubit of first register with <math>\mu_a</math> to get:</br><math>\sum_{a,b\in\{0,1\}}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{a,b}^s|a,b\rangle\otimes|\mu_a,r_a\rangle|y\rangle</math> | ####Server XORs the second qubit of first register with <math>\mu_a</math> to get:</br><math>\sum_{a,b\in\{0,1\}}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{a,b}^s|a,b\rangle\otimes|\mu_a,r_a\rangle|y\rangle</math> | ||
####Server performs Hadamard on second register. The resulting superposition state is:</br><math>\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes H|\mu_a,r_a\rangle|y\rangle</math></br><math>=\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes\bigg(\sum_{e\in\{0,1\}}(-1)^{e\cdot(\mu_a,r_a) }|e\rangle\bigg)|y\rangle</math>, <math>\because H|q\rangle=\sum_{e\in\{0,1\}}(-1)^{e\cdot q}|e\rangle</math></br> | ####Server performs Hadamard on second register. The resulting superposition state is:</br><math>\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes H|\mu_a,r_a\rangle|y\rangle</math></br><math>=\sum_{a,b,\mu\in\{0,1\},r}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes\bigg(\sum_{e\in\{0,1\}}(-1)^{e\cdot(\mu_a,r_a) }|e\rangle\bigg)|y\rangle</math>, <math>\because H|q\rangle=\sum_{e\in\{0,1\}}(-1)^{e\cdot q}|e\rangle</math></br> | ||