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Classical Fully Homomorphic Encryption for Quantum Circuits: Difference between revisions
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Classical Fully Homomorphic Encryption for Quantum Circuits (edit)
Revision as of 23:59, 25 November 2018
, 25 November 2018→Stage 2 Server’s Computation
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####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\in\{0,1\}}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes H^k|\mu_a,r_a\rangle|y\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\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\}^k}(-1)^{e\cdot(\mu_a,r_a) }|e\rangle\bigg)|y\rangle</math>, <math>\because H^k|q\rangle=\sum_{e\in\{0,1\}^k}(-1)^{e\cdot q}|e\rangle</math>, where q has k qubits</br> | ####Server performs Hadamard on second register. The resulting superposition state is:</br><math>\sum_{a,b\in\{0,1\}}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes H^k|\mu_a,r_a\rangle|y\rangle</math></br><math>=\sum_{a,b\in\{0,1\}}\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\}^k}(-1)^{e\cdot(\mu_a,r_a) }|e\rangle\bigg)|y\rangle</math>, <math>\because H^k|q\rangle=\sum_{e\in\{0,1\}^k}(-1)^{e\cdot q}|e\rangle</math>, where q has k qubits</br> | ||
####Server measures the second register to get d. The resulting superposition is:</br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes(-1)^{d\cdot(\mu_a,r_a)}|d\rangle|y\rangle</math></br>The first register could be equivalently written as:</br><math>(-1)^{d\cdot(\mu_0,r_0)}|0,b\rangle+(-1)^{d\cdot(\mu_1,r_1)}|1,b\rangle</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}((-1)^{d\cdot((\mu_0,r_0)\oplus(\mu_0,r_0))}|0,b\rangle+(-1)^{d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))}|1,b\rangle)</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}((-1)^{0\cdot(d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))})|0,b\rangle+(-1)^{1\cdot(d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1)))}|1,b\rangle)</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}((-1)^{a\cdot(d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))})|a,b\rangle</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}(Z^{d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))}|a,b\rangle)</math>, <math>\because Z|q\rangle=(-1)^q|q\rangle</math></br>Thus, the resulting state (upto a global phase) is: </br><math>\approx(Z^{d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))}\otimes X^{\mu_0})CNOT_{12}^s\sum_{a,b\in\{0,1\}}\alpha_{ab}|a,b\rangle</math></br><math>=(Z^{d\cdot ((\mu_0,r_0)\oplus (\mu_1,r_1))}\otimes X^{\mu_0})\mathrm{CNOT}_{1,2}^s|\psi_{12} | ####Server measures the second register to get d. The resulting superposition is:</br><math>=\sum_{a,b\in\{0,1\}}\alpha_{ab}\sqrt{D(\mu_0,r_0)}(I\otimes X^{\mu_0})CNOT_{ab}^s|a,b\rangle\otimes(-1)^{d\cdot(\mu_a,r_a)}|d\rangle|y\rangle</math></br>The first register could be equivalently written as:</br><math>(-1)^{d\cdot(\mu_0,r_0)}|0,b\rangle+(-1)^{d\cdot(\mu_1,r_1)}|1,b\rangle</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}((-1)^{d\cdot((\mu_0,r_0)\oplus(\mu_0,r_0))}|0,b\rangle+(-1)^{d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))}|1,b\rangle)</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}((-1)^{0\cdot(d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))})|0,b\rangle+(-1)^{1\cdot(d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1)))}|1,b\rangle)</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}((-1)^{a\cdot(d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))})|a,b\rangle</math></br><math>=(-1)^{d\cdot (\mu_0,r_0)}(Z^{d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))}|a,b\rangle)</math>, <math>\because Z|q\rangle=(-1)^q|q\rangle</math></br>Thus, the resulting state (upto a global phase) is: </br><math>\approx(Z^{d\cdot((\mu_0,r_0)\oplus(\mu_1,r_1))}\otimes X^{\mu_0})CNOT_{12}^s\sum_{a,b\in\{0,1\}}\alpha_{ab}|a,b\rangle</math></br><math>=(Z^{d\cdot ((\mu_0,r_0)\oplus (\mu_1,r_1))}\otimes X^{\mu_0})\mathrm{CNOT}_{1,2}^s|\psi_{12}\rangle</math> </br>where <math>(\mu_0,r_0)=(\mu_1,r_1)\oplus_H s</math>, as <math>\oplus_H</math> is the homomorphic XOR operation. | ||
####The server uses <math>pk_{i+1}</math> to compute HE.Enc<math>_{pk_{i+1}}(c_{x,z,pk_i})</math> and <math>\mathrm{HE.Enc}_{pk_{i+1}}(\hat{c},y,d)</math>. | ####The server uses <math>pk_{i+1}</math> to compute HE.Enc<math>_{pk_{i+1}}(c_{x,z,pk_i})</math> and <math>\mathrm{HE.Enc}_{pk_{i+1}}(\hat{c},y,d)</math>. | ||
####The server computes the encryption of <math>x,z</math> under <math>pk_{i+1}</math> by homomorphically running the decryption circuit on inputs <math>\mathrm{HE.Enc}_{pk_{i+1}}(sk_i)</math> and <math>\mathrm{HE.Enc}_{pk_{i+1}}(c_{x,z,pk_i})</math>. | ####The server computes the encryption of <math>x,z</math> under <math>pk_{i+1}</math> by homomorphically running the decryption circuit on inputs <math>\mathrm{HE.Enc}_{pk_{i+1}}(sk_i)</math> and <math>\mathrm{HE.Enc}_{pk_{i+1}}(c_{x,z,pk_i})</math>. | ||