Bernstein-Vazirani Algorithm

Posted 2021-07-24 安徽思远

问题：对于函数，
f ( { x 0 , x 1 , x 2 , . . . } ) → 0  or  1  where  x n  is  0  or  1 f(\\{x_0,x_1,x_2,...\\}) \\rightarrow 0 \\textrm{ or } 1 \\textrm{ where } x_n \\textrm{ is }0 \\textrm{ or } 1 f({x0​,x1​,x2​,...})→0 or 1 where xn​ is 0 or 1
该函数的形式为$f(x)=s\cdot x\text{(mod 2)}$，给定x，得到f(x)，求s的值。

该算法步骤：

1. 初始化 |0⟩⊗n state, and output qubit
   to |−⟩ .
2. 对Input寄存器进行Hardmard变化。
3. 对Output进行measure。

对于Hardmard变换，有普遍规律：
∣ a ⟩ → H ⊗ n 1 2 n ∑ x ∈ { 0 , 1 } n ( − 1 ) a ⋅ x ∣ x ⟩ . |a\\rangle \\xrightarrow{H^{\\otimes n}} \\frac{1}{\\sqrt{2^n}} \\sum_{x\\in \\{0,1\\}^n} (-1)^{a\\cdot x}|x\\rangle. ∣a⟩H⊗n ​2n ​1​x∈{0,1}n∑​(−1)a⋅x∣x⟩.

我们对 $|00\dots 0⟩$ 使用n Hadamard gates得到量子叠加态:

∣ 00 … 0 ⟩ → H ⊗ n 1 2 n ∑ x ∈ { 0 , 1 } n ∣ x ⟩ |00\\dots 0\\rangle \\xrightarrow{H^{\\otimes n}} \\frac{1}{\\sqrt{2^n}} \\sum_{x\\in \\{0,1\\}^n} |x\\rangle ∣00…0⟩H⊗n ​2n ​1​x∈{0,1}n∑​∣x⟩

$(-1{)}^{a\cdot x}=0$ , since $a=0,$ and thus $(-1{)}^{a\cdot x}=1$.

之后，我们作用$f_s$函数。这里我们的ouput初始状态a为$|-⟩$.

• 若f(x) = 1，$|a\oplus f(x)⟩=-|-⟩$
• 若f(x) = 0，$|a\oplus f(x)⟩=|-⟩$

$$|x⟩\stackrel{f_s}{\to }(-1{)}^{s\cdot x}|x⟩$$

∣ 00 … 0 ⟩ → H ⊗ n 1 2 n ∑ x ∈ { 0 , 1 } n ∣ x ⟩ → f a 1 2 n ∑ x ∈ { 0 , 1 } n ( − 1 ) a ⋅ x ∣ x ⟩ |00\\dots 0\\rangle \\xrightarrow{H^{\\otimes n}} \\frac{1}{\\sqrt{2^n}} \\sum_{x\\in \\{0,1\\}^n} |x\\rangle \\xrightarrow{f_a} \\frac{1}{\\sqrt{2^n}} \\sum_{x\\in \\{0,1\\}^n} (-1)^{a\\cdot x}|x\\rangle ∣00…0⟩H⊗n ​2n ​1​x∈{0,1}n∑​∣x⟩fa​ ​2n ​1​x∈{0,1}n∑​(−1)a⋅x∣x⟩

由于量子计算是可逆的，我们再使用Hardmard Gate便可以得到a。

1 2 n ∑ x ∈ { 0 , 1 } n ( − 1 ) a ⋅ x ∣ x ⟩ → H ⊗ n ∣ a ⟩ \\frac{1}{\\sqrt{2^n}} \\sum_{x\\in \\{0,1\\}^n} (-1)^{a\\cdot x}|x\\rangle \\xrightarrow{H^{\\otimes n}} |a\\rangle 2n ​1​x∈{0,1}n∑​(−1)a⋅x∣x⟩H⊗n ​∣a⟩

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Coding

# initialization
import matplotlib.pyplot as plt
import numpy as np

# importing Qiskit
from qiskit import IBMQ, Aer
from qiskit.providers.ibmq import least_busy
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister, transpile, assemble

# import basic plot tools
from qiskit.visualization import plot_histogram

n = 3 # number of qubits used to represent s
s = '011'   # the hidden binary string

# We need a circuit with n qubits, plus one auxiliary qubit
# Also need n classical bits to write the output to
bv_circuit = QuantumCircuit(n+1, n)

# put auxiliary in state |->
bv_circuit.h(n)
bv_circuit.z(n)

# Apply Hadamard gates before querying the oracle
for i in range(n):
    bv_circuit.h(i)
    
# Apply barrier 
bv_circuit.barrier()

# Apply the inner-product oracle
s = s[::-1] # reverse s to fit qiskit's qubit ordering
for q in range(n):
    if s[q] == '0':
        bv_circuit.i(q)
    else:
        bv_circuit量子计算求解函数因子：Bernstein-Vazirani Algorithm

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