函数是二对一还是一对一：Simon Algorithm

Posted 2022-11-24 元之田

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:

   $$|{\psi }_2⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in 0,1{}^n}|x⟩|0{⟩}^{\otimes n}$$

3. Apply the query function $${\text{Q}}_f:|{\psi }_3⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in 0,1{}^n}|x⟩|f(x)⟩$$

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: $$|{\psi }_5⟩=\frac{1}{\sqrt{2^{n+1}}}\sum _{z\in 0,1{}^n}\left[(-1{)}^{x\cdot z}+(-1{)}^{y\cdot z}\right]|z⟩$$
   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 1 = 0