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物理代写|量子计算代写Quantum computer代考|Implementation of Boolean Functions

The main task of classical computations is the computation of functions defined in analytical or tabular form. Let us start with Boolean functions that depend on several arguments.

The implementation of Boolean functions with a quantum computer is based on the construction of a quantum circuit that depends on the kind $f(\mathbf{x})$.

We represent the given function using only operations of conjunctions and addition modulo two. This representation is called algebraic normal form (ANF), or ReedMuller expansion, or Zhegalkin polynomial.

The algebraic normal form of the function $f(\mathbf{x})$ is the sum modulo two of several elementary conjunctions of the form
$$G=K_{1} \oplus K_{2} \oplus \ldots \oplus K_{s}$$
where $K_{i}, i=1,2, \ldots, s$ are pairwise different monotone elementary conjunctions over some set of variables $\left{x_{1}, \ldots, x_{n}\right}$, where $n=1,2, \ldots$ One of the $K_{i}$ can be a constant one. The greatest of ranks of elementary conjunctions included in the polynomial $G$ is called a degree of the function. It is known that any Boolean function is uniquely represented as a Reed-Muller decomposition accurate to the order of summands in the sum and the order of cofactors in the conjunctions [1].
Note. More strictly, the decomposition (7.1) is called positive polarity ReedMuller expansion.

There are several methods of constructing an algebraic normal form $G(\mathbf{x})$, expressing the given function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$. Next, let us consider two of these methods in detail.

1. Method of undetermined coefficients.
The Zhegalkin polynomial has the following form:
$$G\left(x_{1}, x_{2}, \ldots, x_{n}\right)=g_{0} \wedge 1 \oplus g_{1} \wedge K_{1} \oplus g_{2} \wedge K_{2} \oplus \ldots \oplus g_{2^{n}-1} \wedge K_{2^{n}-1} \text {, (7.2) }$$
where $K_{i}$ are monotone elementary conjunctions, and coefficients $g_{i} \in \mathbb{B}, i=$ $0,1, \ldots, 2^{n}-1$.

物理代写|量子计算代写Quantum computer代考|To determine the unknown coefficients

To determine the unknown coefficients $g_{0}, g_{1}, \ldots, g_{2^{n}}-1$, we make equations $G\left(\mathbf{x}{j}\right)=f\left(\mathbf{x}{j}\right)$ for all sorts of tuples $\mathbf{x}_{j}$. We get a system of $2^{n}$ equations with $2^{n}$ unknowns; its solution will give coefficients of the polynomial $G(\mathbf{x})$.

1. Method of equivalent transformations.
Let us write down an analytic expression for the function $f(\mathbf{x})$ and reduce it with identity transformations using the equality $A \vee B=\overline{(\bar{A} \wedge \bar{B})}$ to an equivalent expression containing only operations from the set ${-, \wedge}$. Next, we exclude negation operations, for which we will replace everywhere expressions of the form $\bar{A}$ with $A \oplus 1$. Then we expand the brackets using the distribution law $A \wedge(B \oplus C)=$ $(A \wedge B) \oplus(A \wedge C)$ and taking into account the equivalencies
$$\begin{array}{ll} A \wedge A=A, & A \wedge 1=A \ A \oplus A=0, & A \oplus 0=A \end{array}$$
As a result, we get algebraic normal form for the function $f(\mathbf{x})$.
Note that if the function is given by a vector of values, in some cases it is convenient to switch to its disjunctive normal form, and only then to calculate the Reed-Muller expansion using, for example, the method of equivalent transformation.

Example 7.1 Let us build an algebraic normal form for the function $f\left(x_{1}, x_{2}, x_{3}\right)$, defined by the vector of values $\alpha=(10100100)$.
Solution.
We use the method of undetermined coefficients. The algebraic normal form for the function $f\left(x_{1}, x_{2}, x_{3}\right)$ has the following form:
\begin{aligned} G\left(x_{1}, x_{2}, x_{3}\right) &=g_{0} \oplus g_{1} x_{1} \oplus g_{2} x_{2} \oplus g_{3} x_{3} \ & \oplus g_{4} x_{1} x_{2} \oplus g_{5} x_{1} x_{3} \oplus g_{6} x_{2} x_{3} \oplus g_{7} x_{1} x_{2} x_{3} \end{aligned}
where $g_{0}, \ldots, g_{7} \in \mathbb{B}$.

物理代写|量子计算代写Quantum computer代考|On the basis of the algebraic normal

On the basis of the algebraic normal form, construct a quantum circuit implementing an arbitrary Boolean function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)[2,3]$. For this purpose, we use an $n$-qubit register reflecting the input data and a separate qubit $|q\rangle$ for the output of the response. The state of the quantum system is described by the product $\left|x_{0}, x_{1}, \ldots, x_{n}\right\rangle|q\rangle$. Qubit $|q\rangle$ takes the initial value $|0\rangle$.
The method of constructing the quantum circuit is as follows [4].
For each of the Reed-Muller expansion summands (7.1), add a NOT gate controlled by qubits with variables from these summands.

• The constant $f=1$ is represented by a standard NOT gate.
• Variables of the type $x_{i}, 1 \leqslant i \leqslant n$, are represented by the inversion NOT controlled by the qubit $\left|x_{i}\right\rangle$.
• Paired conjunctions of the form $x_{i} x_{j}, 1 \leqslant i, j \leqslant n$, are represented by the inversion controlled by two qubits $\left|x_{i}\right\rangle$ and $\left|x_{j}\right|$.
• Triple conjunctions of the form $x_{i} x_{j} x_{k}, 1 \leqslant i, j, k \leqslant n$, are represented by the inversion controlled by three qubits $\left|x_{i}\right\rangle,\left|x_{j}\right\rangle$, and $\left|x_{k}\right\rangle$.

As it is easy to see, using the NOT, CNOT, CCNOT, …, CC…CNOT operations, it is possible to implement an arbitrary Boolean function. The order of applying elementary operations does not matter, because they commutate with each other. This property is derived from the commutability of the addition modulo two operation in Boolean algebra:
$$\forall x_{1}, x_{2} \in \mathbb{B}\left(x_{1} \oplus x_{2}=x_{2} \oplus x_{1}\right)$$
Example 7.2 The Boolean function of three variables is defined as $f\left(x_{1}, x_{2}, x_{3}\right)=$ $x_{1} \vee x_{1} \bar{x}{2} \vee \bar{x}{1} x_{2} \bar{x}{3}$. Build a quantum circuit that implements this function. Solution. First of all, we calculate the algebraic normal form of the function $f\left(x{1}, x_{2}, x_{3}\right)$ :
$$G_{f}\left(x_{1}, x_{2}, x_{3}\right)=x_{1} \oplus x_{2} \oplus x_{1} x_{2} \oplus x_{2} x_{3} \oplus x_{1} x_{2} x_{3} .$$
Thus, to implement $f\left(x_{1}, x_{2}, x_{3}\right)$, two CNOT elements, one CCNOT element, and one CCCNOT element are required. Their order in the circuit is arbitrary; for certainty, we choose the order defined by the formula ( $7.7)$.

物理代写|量子计算代写Quantum computer代考|Implementation of Boolean Functions

$$G=K_{1} \oplus K_{2} \oplus \ldots \oplus K_{s}$$

$n=1,2, \ldots$ 中的一个 $K_{i}$ 可以是一个常数。多项式中包含的基本连词的最大等级 $G$ 称为函数的度数。众所周知，任何布尔函数都唯一地表 示为 Reed-Muller 分解，精确到和中的被加数的顺序和连词中的辅因子的顺序 [1]。

1. 末定系数的方法。
Zhegalkin 多项式具有以下形式:
$$G\left(x_{1}, x_{2}, \ldots, x_{n}\right)=g_{0} \wedge 1 \oplus g_{1} \wedge K_{1} \oplus g_{2} \wedge K_{2} \oplus \ldots \oplus g_{2^{n_{-1}}} \wedge K_{2^{n}-1},(7.2)$$
在哪里 $K_{i}$ 是单调基本连词和系数 $g_{i} \in \mathbb{B}, i=0,1, \ldots, 2^{n}-1$.

物理代写|量子计算代写Quantum computer代考|To determine the unknown coefficients

1. 等效变换方法。
让我们写下函数的解析表达式 $f(\mathbf{x})$ 并使用等式通过身份转换来减少它 $A \vee B=(\bar{A} \wedge \bar{B})$ 到仅包含集合中的操作的等效表达式 $-, \wedge$. 接下来，我们排除否定操作，我们将替换所有形式的表达式 $\bar{A}$ 和 $A \oplus 1$. 然后我们使用分布定律扩展括号 $A \wedge(B \oplus C)=$ $(A \wedge B) \oplus(A \wedge C)$ 并考虑到等价物
$$A \wedge A=A, \quad A \wedge 1=A A \oplus A=0, \quad A \oplus 0=A$$
结果，我们得到函数的代数范式 $f(\mathbf{x})$.
请注意，如果函数由值向量给出，在某些情况下，切换到其析取范式很方便，然后才使用例如等效变换的方法计算 Reed-Muller 展 开。
例 $7.1$ 让我们为函数建立一个代数范式 $f\left(x_{1}, x_{2}, x_{3}\right)$ ，由值向量定义 $\alpha=(10100100)$.
解决方案。
我们使用待定系数的方法。函数的代数范式 $f\left(x_{1}, x_{2}, x_{3}\right)$ 具有以下形式:
$$G\left(x_{1}, x_{2}, x_{3}\right)=g_{0} \oplus g_{1} x_{1} \oplus g_{2} x_{2} \oplus g_{3} x_{3} \quad \oplus g_{4} x_{1} x_{2} \oplus g_{5} x_{1} x_{3} \oplus g_{6} x_{2} x_{3} \oplus g_{7} x_{1} x_{2} x_{3}$$
在哪里 $g_{0}, \ldots, g_{7} \in \mathbb{B}$.

物理代写|量子计算代写Quantum computer代考|On the basis of the algebraic normal

• 常数 $f=1$ 由标准非门表示。
• 类型的变量 $x_{i}, 1 \leqslant i \leqslant n$, 由不受量子位控制的反转表示 $\left|x_{i}\right\rangle$.
• 形式的成对连词 $x_{i} x_{j}, 1 \leqslant i, j \leqslant n_{t}$ 由两个量子位控制的反转表示 $\left|x_{i}\right\rangle$ 和 $\left|x_{j}\right|$.
• 形式的三连词 $x_{i} x_{j} x_{k}, 1 \leqslant i, j, k \leqslant n$ ，由三个量子位控制的反转表示 $\left|x_{i}\right\rangle,\left|x_{j}\right\rangle$ ，和 $\left|x_{k}\right\rangle$.
很容易看出，使用 NOT、CNOT、CCNOT、…、CC…CNOT 操作，可以实现任意布尔函数。应用基本操作的顺序无关紧要，因为它们相互 交换。该属性源自布尔代数中加法模二运算的可交换性：
$$\forall x_{1}, x_{2} \in \mathbb{B}\left(x_{1} \oplus x_{2}=x_{2} \oplus x_{1}\right)$$
例 7.2 三个变量的布尔函数定义为 $f\left(x_{1}, x_{2}, x_{3}\right)=x_{1} \vee x_{1} \bar{x} 2 \vee \bar{x} 1 x_{2} \bar{x} 3$. 构建实现此功能的量子电路。解决方案。首先，我们计算函数的 代数范式 $f\left(x 1, x_{2}, x_{3}\right)$ :
$$G_{f}\left(x_{1}, x_{2}, x_{3}\right)=x_{1} \oplus x_{2} \oplus x_{1} x_{2} \oplus x_{2} x_{3} \oplus x_{1} x_{2} x_{3} .$$
因此，要实现 $f\left(x_{1}, x_{2}, x_{3}\right)$ ，需要两个 CNOT 元素、一个 CCNOT 元素和一个 CCCNOT 元素。它们在电路中的顺序是任意的；可以肯定 的是，我们选择公式定义的顺序 (7.7).

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