Simon Algorithm

Posted 2021-07-24 安徽思远

问题：判断一个函数是一对一还是二对一？

• 一对一，函数的输出由唯一的输入得到。
  $$f(1)\to 1,f(2)\to 2,f(3)\to 3,f(4)\to 4$$
• 二对一，函数的输出可由两个不同的输入得到。
  $$f(1)\to 1,f(2)\to 2,f(3)\to 1,f(4)\to 2$$
  $$\begin{gathered} \text{given }{x}_1,x_2:f(x_1)=f(x_2) \\ \text{it is guaranteed }:x_1\oplus x_2=b \end{gathered}$$

给定函数f(x)，如何快速判断其是一对一还是二对一函数？

解法：

Where the query function, ${\text{Q}}_f$acts on two quantum registers as:

$$|x⟩|a⟩\to |x⟩|a\oplus f(x)⟩$$
In the specific case that the second register is in the state$|0⟩=|00\dots 0⟩$ we have:

$$|x⟩|0⟩\to |x⟩|f(x)⟩$$

算法：

1. Two n-qubit input registers are initialized to the zero state:

   $$|{\psi }_1⟩=|0{⟩}^{\otimes n}|0{⟩}^{\otimes n}$$

2. Apply a Hadamard transform to the first register:

   ∣ ψ 2 ⟩ = 1 2 n ∑ x ∈ { 0 , 1 } n ∣ x ⟩ ∣ 0 ⟩ ⊗ n \\lvert \\psi_2 \\rangle = \\frac{1}{\\sqrt{2^n}} \\sum_{x \\in \\{0,1\\}^{n} } \\lvert x \\rangle\\lvert 0 \\rangle^{\\otimes n} ∣ψ2​⟩=2n ​1​x∈{0,1}n∑​∣x⟩∣0⟩⊗n

3. Apply the query function Q f : ∣ ψ 3 ⟩ = 1 2 n ∑ x ∈ { 0 , 1 } n ∣ x ⟩ ∣ f ( x ) ⟩ \\text{Q}_f:\\lvert \\psi_3 \\rangle = \\frac{1}{\\sqrt{2^n}} \\sum_{x \\in \\{0,1\\}^{n} } \\lvert x \\rangle \\lvert f(x) \\rangle Qf​:∣ψ3​⟩=2n ​1​x∈{0,1}n∑​∣x⟩∣f(x)⟩

4. Measure the second register. A certain value of f(x) will be observed. Because of the setting of the problem, the observed value f(x) could correspond to two possible inputs: x and y =$x\oplus b$.
   Therefore the first register becomes: $$|{\psi }_4⟩=\frac{1}{\sqrt{2}}\left(|x⟩+|y⟩\right)$$
   这里的x和y对应得到同一输出的不同输入，通过x和y表示所有状态的叠加态。

5. Apply Hadamard on the first register: ∣ ψ 5 ⟩ = 1 2 n + 1 ∑ z ∈ { 0 , 1 } n [ ( − 1 ) x ⋅ z + ( − 1 ) y ⋅ z ] ∣ z ⟩ \\lvert \\psi_5 \\rangle = \\frac{1}{\\sqrt{2^{n+1}}} \\sum_{z \\in \\{0,1\\}^{n} } \\left[ (-1)^{x \\cdot z} + (-1)^{y \\cdot z} \\right] \\lvert z \\rangle ∣ψ5​⟩=2n+1 ​1​z∈{0,1}n∑​[(−1)x⋅z+(−1)y⋅z]∣z⟩
   Measuring the first register will give an output only if: $$(-1{)}^{x\cdot z}=(-1{)}^{y\cdot z}$$

   which means: $$\begin{gathered} x\cdot z=y\cdot z \\ x\cdot z=\left(x\oplus b\right)\cdot z \\ x\cdot z=x\cdot z\oplus b\cdot z \\ b\cdot z=0\text{ (mod 2)} \end{gathered}$$

6.重复运行算法 $\approx n$ 次, we will be able to obtain n different values of z and the following system of equation can be written: { b ⋅ z

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