Transformation Method for Solving System of Boolean Algebraic Equations

Abstract

In recent years, various methods and directions for solving a system of Boolean algebraic equations have been invented, and now they are being very actively investigated. One of these directions is the method of transforming a system of Boolean algebraic equations, given over a ring of Boolean polynomials, into systems of equations over a field of real numbers, and various optimization methods can be applied to these systems. In this paper, we propose a new transformation method for Solving Systems of Boolean Algebraic Equations (SBAE). The essence of the proposed method is that firstly, SBAE written with logical operations are transformed (approximated) in a system of harmonic-polynomial equations in the unit n-dimensional cube $K_n$ with the usual operations of addition and multiplication of numbers. Secondly, a transformed (approximated) system in $K_n$ is solved by using the optimization method. We substantiated the correctness and the right to exist of the proposed method with reliable evidence. Based on this work, plans for further research to improve the proposed method are outlined.

1. Introduction

The solution of a system of Boolean algebraic equations or, in the general case, algebraic cryptanalysis, plays an important role in cryptanalysis. For example, one of the first striking applications of solving a system of Boolean algebraic equations in cryptography has become the solution of a complex problem—cryptosystems on hidden field mappings (Hidden Fields Equations) in cryptography with a public key. This problem is described by the system of quadratic Boolean polynomials in 80 variables, and, for the first time, its solution was obtained precisely by solving a system of Boolean algebraic equations using the F5 algorithm, later—using the F4 algorithm [1].

Now, one of the most promising methods in modern cryptanalysis is algebraic analysis [1,2,3,4]. For a specific cipher, algebraic cryptanalysis consists of two stages: transform the cipher into a system of polynomial equations (usually over Boolean ring) and solve the resulting system of polynomial equations [1,2,5,6,7,8]. Now, methods and algorithms have been invented for solving systems of Boolean algebraic equations. These methods and algorithms are being researched and improved [1,2,3,4,6,7,8,9,10,11,12,13,14,15,16,17,18]. In recent years, methods of algebraic cryptanalysis have been regarded as the most successful attacks on Linear-feedback shift register (LFSR)-based stream ciphers. These attacks cleverly use over-defined systems of multivariable nonlinear equations to recover the secret key. Algebraic attacks lower the degree of the equations by multiplying a nonzero function; fast algebraic attacks obtain equations of a small degree by a linear combination [15,16].

Most algorithms for the Boolean satisfiability problem (SAT) developed so far solve the problem on the Boolean space. Recently, many Universal SAT problem models that transform the SAT problem into an optimization problem in the real space have been developed. Many optimization techniques, such as the steepest descent methods, Newton’s method, and the coordinate descent, can be used to solve the UniSAT7 problem [18,19,20,21,22,23,24,25].

It has been proven that, when the initial solution is sufficiently close to the optimal solution, the steepest descent method has a linear convergence ratio $\beta <1$, Newton’s method has a convergence ratio of order two, and the convergence ratio of the coordinate descent method is approximately $1-\frac{\beta }{m}$ for the Universal SAT problem with m variables [23,26].

In this paper, we propose a new method for solving systems of Boolean algebraic equations. The essence of the proposed method is that, firstly, systems of Boolean algebraic equations written with logical operations are transformed (approximated) into a system of harmonic-polynomial equations in a unit n-dimensional cube $K_n=\left\{\left(x_1, x_2, x_3, . . . , x_n\right) : 0\le x_1, x_2, x_3, . . . , x_n\le 1\right\}$ with the usual operations of addition and multiplication of numbers. Unlike systems of Boolean algebraic equations, system of harmonic-polynomial equations allows the use of optimization techniques. Secondly, the transformed (approximated) system in $K_n$ is solved by an optimization method. Namely, a minimizable objective function is compiled using a system of equations such that, for a given class, it and its constrictions on the edges and faces of an n-dimensional unit cube $K_n$ will be harmonic functions. Therefore, in $K_n$, such a minimizable objective function does not have a local extremum inside, edges, and faces of an n-dimensional cube $K_n$ and takes its minimum value at the vertices of an $n$-dimensional unit cube $K_n$. This allows for the transformation of the resulting solution of the system of harmonic-polynomial equations back into a solution of the systems of Boolean algebraic equations.

Formulas of transformation (approximation) of logical functions are written as $r\left(x_1, x_2, x_3, . . . , x_n\right)$ $and\left(x_1, x_2, x_3, . . . , x_n\right)$, and their basic properties are proved; it was also proved that, in $K_n$, the formulas of such an approximation are unique. One useful lemma and one theorem are proved: the content of the lemma says that, for any Boolean polynomial with pairwise mutually simple monomials, there is also a single in $K_n$ non-negative harmonic function in each of its variables such that, at the vertices of an n-dimensional unit cube $K_n$, their values are equal, and the content of the theorem says that, if a Boolean system from the studied class has a unique solution, then the corresponding transformed system and the Boolean system are equivalent in $K_n$. One illustrative example shows the method of application of the proposed method and numerical comparative analyses are carried out. However, this new idea should be considered to be at a preliminary stage since it applies to very specific use cases, whereas a thorough complexity analysis is still missing.

2. Approximation Formulas for Logical Functions $\mathit{x}\mathit{o}\mathit{r}\left({\mathit{x}}_1,{\mathit{x}}_2,{\mathit{x}}_3,...,{\mathit{x}}_{\mathit{n}}\right),$$\mathit{a}\mathit{n}\mathit{d}\left({\mathit{x}}_1, {\mathit{x}}_2, {\mathit{x}}_3, . . . , {\mathit{x}}_{\mathit{n}}\right)$ and Some of Their Properties

In this section, we construct elementary approximating harmonic polynomials and list their main properties.

Let $B_n=\left\{\left(b_1,b_2,\dots ,b_n\right):b_1,b_2,\dots ,b_n\in \left\{0,1\right\}\right\}$ be the set of all possible binary words (Boolean vectors) of length $n$, $K_n=\left\{\left(x_1,x_2,\dots ,x_n\right):0\le x_1,x_2,\dots ,x_n\le 1\right\}$ $n$-dimensional cube spanned by Boolean vectors of length $n$.

For clarity, we will define, in what follows, one class of functions.

Definition. 

A function$f\left(x_1, x_2, x_3, . . . , x_n\right)$is called a harmonic function in each of its variables in$K_n$if$\frac{{\partial }^2}{\partial x_k^2}f\left(x_1,x_2,\dots ,x_n\right)=0, \forall k\in \left\{1,\dots ,n\right\}$.

$\left(a\right)$ To construct an approximating polynomial, we proceed from the fact that a logical two-place function $xor\left(x_1,x_2\right)$ (addition by mod 2) can be represented as a polynomial

$$xo{r}_{app}\left(x_1,x_2\right)=x_1+x_2-2x_1{x}_2$$

(1)

with the usual operations of addition and multiplication of numbers.

Proposition 1. 

If$\left(x_1,x_2\right)\in K_2$and$xo{r}_{app}\left(x_1,x_2\right)=x_1+x_2-2x_1{x}_2$then

(i)

$0\le xo{r}_{app}\left(x_1,x_2\right)=x_1+x_2-2x_1{x}_2\le 1, \forall \left(x_1,x_2\right)\in K_2$.

(ii)

$xo{r}_{app}\left(x_1,x_2\right)=x_1+x_2-2x_1{x}_2=0 ⟺ \left(x_1,x_2\right)\in \left\{\left(0,0\right), \left(1,1\right)\right\}$.

(iii)

$xo{r}_{app}\left(x_1,x_2\right)=x_1+x_2-2x_1{x}_2=1 ⟺ \left(x_1,x_2\right)\in \left\{\left(0,1\right), \left(1,0\right)\right\}$.

Proof. 

$(\mathrm{i})$ Indeed, $0=0+0\le x_1\cdot \left(1-x_2\right)+x_2\cdot \left(1-x_1\right)=xo{r}_{app}\left(x_1,x_2\right)\le$ $\sqrt{x_1\cdot \left(1-x_2\right)}+\sqrt{x_2\cdot \left(1-x_1\right)}\le \frac{x_1+\left(1-x_2\right)}{2}+\frac{x_2+\left(1-x_1\right)}{2}=1.$

(ii)

From property $\left(\mathrm{i}\right)$, it follows that $xo{r}_{app}\left(x_1,x_2\right)=0⟺\{\begin{array}{c}{x}_1\cdot \left(1-x_2\right)=0\\ x_2\cdot \left(1-x_1\right)=0\end{array}⟺\left(x_1,x_2\right)\in \left\{\left(0,0\right), \left(1,1\right)\right\}.$

(iii)

From property $\left(\mathrm{i}\right)$, it follows that $xo{r}_{app}\left(x_1,x_2\right)=1 ⟺\{\begin{array}{c}{x}_1\cdot \left(1-x_2\right)=\frac{x_1+\left(1-x_2\right)}{2}\\ x_2\cdot \left(1-x_1\right)=\frac{x_2+\left(1-x_1\right)}{2}\end{array} ⟺ \left(x_1,x_2\right)\in \left\{\left(0,1\right), \left(1,0\right)\right\}.$

In other words, this polynomial $xo{r}_{app}\left(x_1,x_2\right)=x_1+x_2-2x_1{x}_2$ inside the square $K_2$ is a harmonic function and, at the interior points of the square, takes values strictly between $0$ and $1$. In this case, at the vertex of the square, $K_2$ takes the value $0$ if $\left(x_1,x_2\right)\in \left\{\left(0,0\right), \left(1,1\right)\right\}$, and the value $1$ if $\left(x_1,x_2\right)\in \left\{\left(0,1\right), \left(1,0\right)\right\}$. □

Let us construct a multidimensional analog of the polynomial $xo{r}_{app}\left(x_1,x_2\right)$ by the recursive formula

$$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=xo{r}_{app}\left(xo{r}_{app}\left(x_1,\dots ,x_{n-1}\right),x_n\right).$$

(2)

Then the following formula is valid:

$$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=\frac{1}{2}-\frac{1}{2}\left(1-2x_1\right)\left(1-2x_2\right)\cdot \dots \cdot \left(1-2x_n\right).$$

(3)

Indeed, for Formulas $\left(1\right)$ and $\left(2\right)$ with $n=3$, we have:

$$\begin{gathered} xo{r}_{app}\left(x_1,x_2,x_3\right)=xo{r}_{app}\left(x_1,x_2\right)+x_3-2x_{3}xo{r}_{app}\left(x_1,x_2\right) \\ =\frac{1}{2}-\frac{1}{2}\left(1-2xo{r}_{app}\left(x_1,x_2\right)\right)\left(1-2x_3\right) \\ =\frac{1}{2}-\frac{1}{2}\left(1-2\left(\frac{1}{2}-\frac{1}{2}\left(1-2x_1\right)\left(1-2x_2\right)\right)\right)\left(1-2x_3\right) \\ =\frac{1}{2}-\frac{1}{2}\left(1-2x_1\right)\left(1-2x_2\right)\left(1-2x_3\right). \end{gathered}$$

As well, for $n>3$, by induction we derive:

$$\begin{gathered} xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=xo{r}_{app}\left(x_1,\dots ,x_{n-1}\right)+x_n-2x_{n}xo{r}_{app}\left(x_1,\dots ,x_{n-1}\right) \\ =\frac{1}{2}-\frac{1}{2}\left(1-2xo{r}_{app}\left(x_1,\dots ,x_{n-1}\right)\right)\left(1-2x_n\right) \\ =\frac{1}{2}-\frac{1}{2}\left(1-2\left(\frac{1}{2}-\frac{1}{2}\left(1-2x_1\right)\left(1-2x_2\right)\cdot \dots \cdot \left(1-2x_{n-1}\right)\right)\right)\left(1-2x_n\right) \\ =\frac{1}{2}-\frac{1}{2}\left(1-2x_1\right)\left(1-2x_2\right)\left(1-2x_3\right)\cdot \dots \cdot \left(1-2x_n\right). \end{gathered}$$

The polynomial $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$, constructed and defined by Formulas $\left(2\right)$ and $\left(3\right)$, can be interpreted as a multidimensional algebraic analogue of the logical two-place function $xor\left(x_1,x_2\right)$ (addition by mod 2).

Let us formulate and check the main properties of the polynomial $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$.

Proposition 2. 

If$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=\frac{1}{2}-\frac{1}{2}\left(1-2x_1\right)\left(1-2x_2\right)\left(1-2x_3\right)\cdot \dots \cdot \left(1-2x_n\right)$then:

(i)

The polynomial$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$at the vertices of the$n-$dimensional cube$K_n$takes one of the values$0$or$1$;

(ii)

The polynomial$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$and its constrictions on theedges and faces of the$n-$dimensional cube$K_n$are harmonic functions;

(iii)

The polynomial$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$in the$n-$dimensional cube$K_n$takes values$0$and$1$ only at the vertices;

(iv)

The polynomial$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)\in \left\{0,1\right\}$at the vertex of the$n-$dimensional cube$K_n$takes the values$0 \left(1\right)$if and only if the sum of the vertex coordinates is even (odd).

Proof. 

$\left(\mathrm{i}\right)$ Indeed, for any Boolean vector $\left(x_1,x_2,\dots ,x_n\right)\in B_n$, we have: $\left(1-2x_1\right)\left(1-2x_2\right)\cdot \dots \cdot \left(1-2x_n\right)\in \left\{-1,1\right\}$, therefore, $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)\in \left\{0,1\right\}$.

(ii) 

This property directly follows from the equality

$\frac{{\partial }^2}{\partial x_k^2}xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=0$, $k=1,\dots ,n$.

From properties $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, it follows that $0\le xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)\le 1$ $\forall \left(x_1,x_2,\dots ,x_n\right)\in K_n$, since it is well known that any function that is harmonic within a bounded region and continuous on the closure of the region takes the largest and smallest values on the border of the region [27,28,29,30,31,32,33,34,35,36].

(iii) 

Indeed, if $\left(x_1,x_2,\dots ,x_n\right)\in K_n$, then the inclusion $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)\in \left\{0,1\right\}$ is equivalent to the inclusion $\left(1-2x_1\right)\left(1-2x_2\right)\cdot \dots \cdot \left(1-2x_n\right)\in \left\{-1,1\right\}$. The last expression is true only if all factors are equal in absolute value to $1$. Hence,

$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)\in \left\{0,1\right\}$, $\left(x_1,x_2,\dots ,x_n\right)\in K_n$ $\iff$ $\left(x_1,x_2,\dots ,x_n\right)\in B_n$.

(iv) 

Checks by analogy with property $\left(\mathrm{iii}\right)$: if $\left(x_1,x_2,\dots ,x_n\right)\in B_n$, then $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=0\iff \left(1-2x_1\right)\left(1-2x_2\right)\cdot \dots \cdot \left(1-2x_n\right)=1\iff$

$\iff {\left(-1\right)}^{x_1}\cdot {\left(-1\right)}^{x_2}\cdot \dots \cdot {\left(-1\right)}^{x_n}=1\iff \left(x_1+x_2+\dots +x_n\right)$. □

$\left(b\right)$ To construct an approximating polynomial, we proceed from the fact that the logical two-place function $and\left(x_1,x_2\right)$ can be represented as a polynomial

$$an{d}_{app}\left(x_1,x_2\right)=x_1{x}_2$$

(4)

with the usual operations of addition and multiplication of numbers. This polynomial inside the square $K_2$ is a harmonic function and, at the inner points of the square, takes values strictly between 0 and 1. At the same time, at the apex of the square, $K_2$ takes the value $0$ if $x_1=0$ or $x_2=0$, and the value 1 if $\left(x_1,x_2\right)=\left(1,1\right)$.

We construct a multidimensional analogue of the polynomial $an{d}_{app}\left(x_1,x_2\right)$ by the recursive formula

$$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=an{d}_{app}\left(an{d}_{app}\left(x_1,\dots ,x_{n-1}\right),x_n\right).$$

(5)

Then, it is obvious that the following formula is valid:

$$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=x_1\cdot x_2\cdot \dots \cdot x_n.$$

(6)

Let us formulate and check the main properties of the polynomial $an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=x_1\cdot x_2\cdot \dots \cdot x_n$.

Proposition 3. 

If$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=x_1\cdot x_2\cdot \dots \cdot x_n$, then:

(i)

The polynomial$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$at the vertices of the$n$-dimensional cube$K_n$takes one of the values$0$or$1$;

(ii)

The polynomial$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$and its constrictions on theedges and faces of the$n$-dimensional cube$K_n$ are harmonic functions;

(iii)

The polynomial$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$in the$n$-dimensional cube$K_n$takes the values 0 if$x_1=0$or$x_2=$0 or … or$x_n=0$;

(i)

The polynomial$an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$at the top of the$n$-dimensional cube$K_n$takes values$1$if and only if$\left(x_1,x_2,\dots ,x_n\right)=\left(1,1,\dots ,1\right)$.

Proof. 

$\left(\mathrm{i}\right)$ Indeed, for any Boolean vector $\left(x_1,x_2,\dots ,x_n\right)\in B_n$, we have: $x_1\cdot x_2\cdot \dots \cdot x_n\in \left\{0,1\right\}$, and therefore $an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)\in \left\{0,1\right\}$.

(ii)

This property directly follows from the equality $\frac{{\partial }^2}{\partial x_k^2}an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=0$, $k=1,\dots ,n$.

(iii) 

Indeed, if $an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=0$, then $x_1=0$ or $x_2=0$ or … or $x_n=0$.

(iv) 

Indeed, if $\left(x_1,x_2,\dots ,x_n\right)\in K_n$, then $an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=1\iff x_1\cdot x_2\cdot \dots \cdot x_n=1\iff \left(x_1,x_2,\dots ,x_n\right)=\left(1,1,\dots ,1\right).$ □

Lemma 1. 

For functions$xor\left(x_1,x_2,\dots ,x_n\right)=x_1\oplus x_2\oplus \dots \oplus x_n, and\left(x_1,x_2,\dots ,x_n\right)=x_1\otimes x_2\otimes \dots \otimes x_n$exist in$K_n$non-negative harmonic functions in each of its variables$f\left(x_1,x_2,\dots ,x_n\right), g\left(x_1,x_2,\dots ,x_n\right)$such that$xor\left(x_1,x_2,\dots ,x_n\right)=f\left(x_1,x_2,\dots ,x_n\right), and\left(x_1,x_2,\dots ,x_n\right)=g\left(x_1,x_2,\dots ,x_n\right)$at$\left(x_1,x_2,\dots ,x_n\right)\in B_n$ and they are unique.

Proof. 

 

Existence: from Propositions 2 and 3, it follows that, as a $f\left(x_1,x_2,\dots ,x_n\right),g\left(x_1,x_2,\dots ,x_n\right)$, you can take $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right),an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$, that is and $g\left(x_1,x_2,\dots ,x_n\right)=an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$.

Uniqueness: let us assume that this is not the case; then let there be others in $K_n$ non-negative harmonic functions in each of its variables $f_1\left(x_1,x_2,\dots ,x_n\right), g_1\left(x_1,x_2,\dots ,x_n\right)$ such that $xor\left(x_1,x_2,\dots ,x_n\right)=f\left(x_1,x_2,\dots ,x_n\right), and\left(x_1,x_2,\dots ,x_n\right)=g\left(x_1,x_2,\dots ,x_n\right)$ by $\left(x_1,x_2,\dots ,x_n\right)\in B_n$. Now consider the functions $d_1\left(x_1,x_2,x_3,\dots ,x_n\right)=f_1\left(x_1,x_2,\dots ,x_n\right)-xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right),$ $d_2\left(x_1,x_2,x_3,\dots ,x_n\right)=g_1\left(x_1,x_2,\dots ,x_n\right)-an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$. First, it is obvious that if $\left(x_1,x_2,x_3,\dots ,x_n\right)\in B_n$, then $d_1\left(x_1,x_2,x_3,\dots ,x_n\right)=f_1\left(x_1,x_2,\dots ,x_n\right)-$

$xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=xor\left(x_1,x_2,\dots ,x_n\right)-xor\left(x_1,x_2,\dots ,x_n\right)=0,$

$d_2\left(x_1,x_2,x_3,\dots ,x_n\right)=g_1\left(x_1,x_2,\dots ,x_n\right)-an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=and\left(x_1,x_2,\dots ,x_n\right)-and\left(x_1,x_2,\dots ,x_n\right)=0$, secondly, the functions $d_1\left(x_1,x_2,x_3,\dots ,x_n\right)$ and $d_2\left(x_1,x_2,x_3,\dots ,x_n\right)$ also harmonic functions in each of its variables, since

$$\begin{gathered} \frac{{\partial }^2}{\partial x_k^2}{d}_1\left(x_1,x_2,x_3,\dots ,x_n\right)=\frac{{\partial }^2}{\partial x_k^2}{f}_1\left(x_1,x_2,\dots ,x_n\right)-\frac{{\partial }^2}{\partial x_k^2}xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)=0-0=0 \forall k\in \left\{1,\dots ,n\right\}, \\ \frac{{\partial }^2}{\partial x_k^2}{d}_2\left(x_1,x_2,x_3,\dots ,x_n\right)=\frac{{\partial }^2}{\partial x_k^2}{g}_1\left(x_1,x_2,\dots ,x_n\right)-\frac{{\partial }^2}{\partial x_k^2}an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)=0-0=0 \forall k\in \left\{1,\dots ,n\right\}. \end{gathered}$$

Now, from the last argument and the maximum principle [12,14,16,17,18,19], it follows that

$$\begin{gathered} 0=\underset{x\in B_n}{\min }{d}_1\left(x_1,x_2,x_3,\dots ,x_n\right)\le d_1\left(x_1,x_2,x_3,\dots ,x_n\right)\le \underset{x\in B_n}{\max }{d}_1\left(x_1,x_2,x_3,\dots ,x_n\right)=0 \\ 0=\underset{x\in B_n}{\min }{d}_2\left(x_1,x_2,x_3,\dots ,x_n\right)\le d_2\left(x_1,x_2,x_3,\dots ,x_n\right)\le \underset{x\in B_n}{\max }{d}_2\left(x_1,x_2,x_3,\dots ,x_n\right)=0⟹ \\ {d}_1\left(x_1,x_2,x_3,\dots ,x_n\right)\equiv 0, d_2\left(x_1,x_2,x_3,\dots ,x_n\right)\equiv 0 ⟹ f_1\left(x_1,x_2,\dots ,x_n\right)\equiv xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right), \end{gathered}$$

$g_1\left(x_1,x_2,\dots ,x_n\right)\equiv an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$, which contradicts our hypothesis. This concludes the proof. □

For clarity, a geometric representation of the polynomials is proposed $xo{r}_{app}\left(x,y\right)=x+y-2xy,an{d}_{app}\left(x,y\right)=x\cdot y,an{d}_{app}\left(x,y\right)=x-x^2+y-y^2+xy$ in the appropriate order as shown in Figure 1.

3. Transformation of a System of Boolean Algebraic Equations into a System of Polynomial Equations

First, for clarity, we will transform one Boolean polynomial and prove the further lemma used.

Lemma 2. 

Let$p\left(x_1,x_2,x_3,\dots ,x_n\right)={\displaystyle \underset{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}{⨁}}g\left(a_1,a_2,\dots ,a_n\right)\cdot x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}$be a Boolean polynomial with pairwise coprime monomials, then there exists$f\left(x_1,x_2,x_3,\dots ,x_n\right)$—in$K_n$non-negative, harmonic function in each of its variables$x_1,x_2,\dots ,x_n$such that$p\left(x_1,x_2,x_3,\dots ,x_n\right)=f\left(x_1,x_2,x_3,\dots ,x_n\right)$by$\left(x_1,x_2,x_3,\dots ,x_n\right)\in B_n$ and is unique.

Proof. 

 

Existence: replacing functions $xor\left(x_1,x_2,\dots ,x_n\right)$ and $and\left(x_1,x_2,\dots ,x_n\right)$ with functions $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$ and $an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$ from the polynomial $p\left(x_1,x_2,x_3,\dots ,x_n\right)={\displaystyle \underset{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}{⨁}}g\left(a_1,a_2,\dots ,a_n\right)\cdot x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}$, we obtain the corresponding function $f\left(x_1,x_2,\dots ,x_n\right)=\frac{1}{2}-\frac{1}{2}\cdot {\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}\left(1-2\cdot g\left(a_1,a_2,\dots ,a_n\right)\cdot x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}\right)$.

(a) 

from the proven properties of formulas $xo{r}_{app}\left(x_1,x_2,\dots ,x_n\right)$ and $an{d}_{app}\left(x_1,x_2,\dots ,x_n\right)$, it follows that, firstly, if $\left(x_1,x_2,\dots ,x_n\right)\in K_n$, then $0\le f\left(x_1,x_2,\dots ,x_n\right)\le 1$, secondly, $f\left(x_1,x_2,\dots ,x_n\right)=p\left(x_1,x_2,x_3,\dots ,x_n\right)$ at $\left(x_1,x_2,x_3,\dots ,x_n\right)\in B_n$

(b) 

$f\left(x_1,x_2,\dots ,x_n\right)$ is a harmonic function in each of its variables, that is $\frac{{\partial }^2}{\partial x_k^2}f\left(x_1,x_2,\dots ,x_n\right)=0\forall k\in \left\{1,\dots ,n\right\}$.

Therefore:

(i) 

if $x_k$ is not included in the polynomial $p\left(x_1,x_2,\dots ,x_n\right)$, in other words, if $p\left(x_1,x_2,\dots ,x_n\right)$ does not depend on $x_k$, then $f\left(x_1,x_2,\dots ,x_n\right)$ also does not depend on $x_k⟹\frac{{\partial }^2}{\partial x_k^2}f\left(x_1,x_2,\dots ,x_n\right)=0$.

(ii) 

If $x_k$ is included in the polynomial $p_i\left(x_1,x_2,\dots ,x_n\right)=0$, in other words, if $p\left(x_1,x_2,\dots ,x_n\right)$ depends on $x_k$, then $x_k$ is included in only one monomial, since all monomials $p\left(x_1,x_2,x_3,\dots ,x_n\right)$ are pairwise coprime. Let it be a monomial $x_1^{a_1^{*}}{x}_2^{a_2^{*}}\dots x_n^{a_n^{*}}$ (of course, that $g_1\left(a_1^{*},a_2^{*},\dots ,a_n^{*}\right)=1$ and $a_k^{*}=1$), then

$$\begin{gathered} \frac{{\partial }^2}{\partial x_k^2}f\left(x_1,x_2,\dots ,x_n\right)=\frac{{\partial }^2}{\partial x_k^2}\left(\frac{1}{2}-\frac{1}{2}\cdot {\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}\left(1-2\cdot g\left(a_1,a_2,\dots ,a_n\right)x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}\right)\right) \\ =-\frac{1}{2}\cdot \frac{{\partial }^2}{\partial x_k^2}\left({\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}\left(1-2\cdot g\left(a_1,a_2,\dots ,a_n\right)x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}\right)\right) \\ =\left({\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n\backslash \left\{\left(a_1^{*},a_2^{*},\dots ,a_n^{*}\right)\right\}}\left(1-2\cdot g\left(a_1,a_2,\dots ,a_n\right)x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}\right)\right)\cdot \\ \left(-\frac{1}{2}\right)\cdot \frac{{\partial }^2}{\partial x_k^2}\left(\left(1-2\cdot g\left(a_1^{*},a_2^{*},\dots ,a_n^{*}\right)x_1^{a_1^{*}}{x}_2^{a_2^{*}}\dots x_n^{a_n^{*}}\right)\right) \\ =\left({\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n\backslash \left\{\left(a_1^{*},a_2^{*},\dots ,a_n^{*}\right)\right\}}\left(1-2\cdot g\left(a_1,a_2,\dots ,a_n\right)x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}\right)\right)\cdot 0=0. \end{gathered}$$

Thus, the existence is proven.

Uniqueness: Let us assume that this is not the case; then let there be others in $K_n$ non-negative harmonic function $h\left(x_1,x_2,\dots ,x_n\right) \left(h\left(x_1,x_2,\dots ,x_n\right)\ne f\left(x_1,x_2,\dots ,x_n\right)\right)$ in each of its variables such that $h\left(x_1,x_2,\dots ,x_n\right)=p\left(x_1,x_2,x_3,\dots ,x_n\right)$ by $\left(x_1,x_2,x_3,\dots ,x_n\right)\in B_n$. Now consider the function $d\left(x_1,x_2,x_3,\dots ,x_n\right)=f\left(x_1,x_2,\dots ,x_n\right)-h\left(x_1,x_2,\dots ,x_n\right)$. First, it is obvious that if $\left(x_1,x_2,x_3,\dots ,x_n\right)\in B_n$, then $d\left(x_1,x_2,x_3,\dots ,x_n\right)=f\left(x_1,x_2,\dots ,x_n\right)-$$h\left(x_1,x_2,\dots ,x_n\right)=p\left(x_1,x_2,x_3,\dots ,x_n\right)-p\left(x_1,x_2,x_3,\dots ,x_n\right)=0$, secondly, the function $d\left(x_1,x_2,x_3,\dots ,x_n\right)$ harmonic function in each of its variables, because $\frac{{\partial }^2}{\partial x_k^2}d\left(x_1,x_2,\dots ,x_n\right)=\frac{{\partial }^2}{\partial x_k^2}f\left(x_1,x_2,\dots ,x_n\right)-\frac{{\partial }^2}{\partial x_k^2}h\left(x_1,x_2,\dots ,x_n\right)=0-0=0$ $\forall k\in \left\{1,\dots ,n\right\}$.

Now, it follows from the last argument and maximum principle [27,28,29,30,31,32,33,34,35,36] that $0=\underset{x\in B_n}{\min } d\left(x_1,x_2,x_3,\dots ,x_n\right)\le d\left(x_1,x_2,x_3,\dots ,x_n\right)\le \underset{x\in B_n}{\max }d\left(x_1,x_2,x_3,\dots ,x_n\right)=0⟹d\left(x_1,x_2,x_3,\dots ,x_n\right)\equiv 0⟹h\left(x_1,x_2,\dots ,x_n\right)\equiv f\left(x_1,x_2,\dots ,x_n\right)$ which contradicts our hypothesis. This concludes the proof. □

Now we can transform a system of Boolean algebraic equations into a system of harmonic-polynomial equations into a unit n-dimensional cube $K_n=\left\{\left(x_1,x_2,x_3,...,x_n\right):0\le x_1,x_2,x_3,...,x_n\le 1\right\}$ with the usual operations of addition and multiplication of numbers.

Let there be a system of Boolean algebraic equations

$$\{\begin{array}{l}{p}_1(x_1,x_2,\dots ,x_n)=\underset{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}{\oplus }{g}_1(a_1,a_2,\dots ,a_n)·x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}=0\\ p_2(x_1,x_2,\dots ,x_n)=\underset{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}{\oplus }{g}_2(a_1,a_2,\dots ,a_n)·x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}=0\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\dots \\ p_m(x_1,x_2,\dots ,x_n)=\underset{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}{\oplus }{g}_m(a_1,a_2,\dots ,a_n)·x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n}=0\end{array}$$

(7)

where $p_i\left(x_1,x_2,\dots ,x_n\right)$ Boolean polynomial, $i\in \left\{1,2,\dots ,m\right\}$.

Replacing functions $xor\left(x_1,x_2,\dots ,x_n\right)$ and $and\left(x_1,x_2,\dots ,x_n\right)$ from the system $\left(7\right)$, we obtain the following new polynomial in $K_n$, a transformed (approximated) system with the usual operations of addition and multiplication of numbers:

$$\{\begin{array}{l}{f}_1(x_1,x_2,\dots ,x_n)=\frac{1}{2}-\frac{1}{2}·{\displaystyle {\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}}(1-2·g_1(a_1,a_2,\dots ,a_n)·x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n})=0\\ f_2(x_1,x_2,\dots ,x_n)=\frac{1}{2}-\frac{1}{2}·{\displaystyle {\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}}(1-2·g_2(a_1,a_2,\dots ,a_n)·x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n})=0\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\dots \\ f_m(x_1,x_2,\dots ,x_n)=\frac{1}{2}-\frac{1}{2}·{\displaystyle {\displaystyle \prod }_{a_1,a_2,\dots ,a_n\in {\mathbb{Z}}_2^n}}(1-2·g_m(a_1,a_2,\dots ,a_n)·x_1^{a_1}{x}_2^{a_2}\dots x_n^{a_n})=0\end{array}$$

(8)

Let us define the objective function in $K_n$ by system $\left(8\right)$:

$$tf\left(x_1,x_2,\dots ,x_n\right)={\displaystyle \sum }_{i=1}^m{f}_i\left(x_1,x_2,\dots ,x_n\right)$$

Theorem 1. 

if the monomials of each Boolean polynomial$p_i\left(x_1,x_2,\dots ,x_n\right)$systems (7) are pairwise coprime$\forall i=1,2,\dots ,m$and system (7) has a unique solution$x_0=\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)$, then, in$K_n$, system (8) also has only the unique solution$x_0=\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)$.

Proof. 

By Lemma 2 $p_i\left(x_1,x_2,x_3,\dots ,x_n\right)=f_i\left(x_1,x_2,x_3,\dots ,x_n\right)$ by $\left(x_1,x_2,x_3,\dots ,x_n\right)\in B_n$

$\forall i\in \left\{1,2,\dots ,m\right\}$. In particular, it follows that $p_i\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)=$$f_i\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)=0 \forall i\in \left\{1,2,\dots ,m\right\}$—this means that $x_0=\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)$—system solution $\left(8\right)$. Now, we prove that in $K_n$ is the only solution to the system $\left(8\right)$. Let us assume that this is not the case; then, let $c=\left(c_1,c_2,\dots ,c_n\right)\in K_n$ be the other solution system $\left(8\right)$. By Lemma 2 $c=\left(c_1,c_2,\dots ,c_n\right)\in K_n$—solution system $\left(8\right) ⟹tf\left(c_1,c_2,\dots ,c_n\right)=0$.

Since the system $\left(7\right)$ has a unique solution and a target function $tf\left(x\right)=tf\left(x_1,x_2,\dots ,x_n\right)$ harmonic function in each of its variables $x_1,x_2,\dots ,x_n$, and it follows that $\left(c_1,c_2,\dots ,c_n\right)\in B_n$, this means that $f_i\left(c_1,c_2,\dots ,c_n\right)=p_i\left(c_1,c_2,\dots ,c_n\right)=0$

$\forall i\in \left\{1,2,\dots ,m\right\}⟹c=\left(c_1,c_2,\dots ,c_n\right)$—solution system $\left(7\right)$, since, by the condition of theorem, system (7) has a unique solution $x_0=\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)⟹x_0=\left(x_1^{*},x_2^{*},\dots ,x_n^{*}\right)=$

$\left(c_1,c_2,\dots ,c_n\right)=c$ which contradicts our hypothesis. This concludes the proof. □

Consequence: If the monomials of each Boolean polynomial $p_i\left(x_1,x_2,\dots ,x_n\right)$ systems (7) are pairwise coprime $\forall i=1,2,\dots ,m$, then:

$\left(a\right)$ target function $tf\left(x_1,x_2,\dots ,x_n\right)$ does not have a local extremum inside, edges, and faces of an n-dimensional cube $K_n$;

$\left(b\right)$ it takes its extremes at the vertices of an n-dimensional cube $K_n$, that is, on $B_n$.

Suffice it to note that $\frac{{\partial }^2}{\partial x_k^2}tf\left(x_1,x_2,\dots ,x_n\right)=0\forall k\in \left\{1,2,\dots ,n\right\}.$

$$\frac{{\partial }^2}{\partial x_k^2}tf\left(x_1,x_2,\dots ,x_n\right)=\frac{{\partial }^2}{\partial x_k^2}\left({\displaystyle \sum }_{i=1}^m{f}_i\left(x_1,x_2,\dots ,x_n\right)\right)={\displaystyle \sum }_{i=1}^m\frac{{\partial }^2}{\partial x_k^2}{f}_i\left(x_1,x_2,\dots ,x_n\right)={\displaystyle \sum }_{i=1}^{m}0=0$$

4. The Application of the Proposed Method

In this section, using one illustrative example, we will show the methodology of applying the proposed method and conducting numerical comparative analyses.

Consider a system of Boolean algebraic equations with a unique solution:

$$\{\begin{array}{c}{p}_1(x_1,x_2,\dots ,x_{14})=1\oplus x_1\oplus x_3=0\\ p_2(x_1,x_2,\dots ,x_{14})=x_6\oplus x_1\otimes x_2=0\\ p_3(x_1,x_2,\dots ,x_{14})=x_1\oplus x_2\oplus x_7=0\\ p_4(x_1,x_2,\dots ,x_{14})=1\oplus x_2\oplus x_5=0\\ p_5(x_1,x_2,\dots ,x_{14})=x_3\oplus x_2\oplus x_4=0\\ p_6(x_1,x_2,\dots ,x_{14})=1\oplus x_5\oplus x_9=0\\ p_7(x_1,x_2,\dots ,x_{14})=x_4\oplus x_5\oplus x_{11}=0\\ p_8(x_1,x_2,\dots ,x_{14})=1\oplus x_6\oplus x_{10}=0\\ p_9(x_1,x_2,\dots ,x_{14})=x_6\oplus x_7\oplus x_8=0\\ p_{10}(x_1,x_2,\dots ,x_{14})=x_8\oplus x_6\oplus x_{12}=0\\ p_{11}(x_1,x_2,\dots ,x_{14})=x_{14}\oplus x_9\otimes x_{10}=0\\ p_{12}(x_1,x_2,\dots ,x_{14})=x_9\oplus x_{10}\oplus 1=0\\ p_{13}(x_1,x_2,\dots ,x_{14})=x_{11}\oplus x_{12}\oplus x_{13}=0\\ p_{14}(x_1,x_2,\dots ,x_{14})=x_{13}\oplus x_{14}\oplus 1=0\end{array}$$

(9)

where $\oplus$ is logical operation $xor$, $\otimes$ is logical operation $and,$ $x_i\in \left\{0,1\right\}.$

Transforming the system $\left(9\right)$ in $K_{14}$, we obtain the following system:

$$\{\begin{array}{c}{f}_1(x_1,x_2,\dots ,x_{14})=\frac{1}{2}+\frac{1}{2}·(1-2·x_1)(1-2·x_3)=0\\ f_2(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_6)(1-2·x_1·x_2)=0\\ f_3(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_1)(1-2·x_2)(1-2·x_7)=0\\ f_4(x_1,x_2,\dots ,x_{14})=\frac{1}{2}+\frac{1}{2}·(1-2·x_2)(1-2·x_5)=0\\ f_5(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_3)(1-2·x_2)(1-2·x_4)=0\\ f_6(x_1,x_2,\dots ,x_{14})=\frac{1}{2}+\frac{1}{2}·(1-2·x_5)(1-2·x_9)=0\\ f_7(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_4)(1-2·x_5)(1-2·x_{11})=0\\ f_8(x_1,x_2,\dots ,x_{14})=\frac{1}{2}+\frac{1}{2}·(1-2·x_6)(1-2·x_{10})=0\\ f_9(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_6)(1-2·x_7)(1-2·x_8)=0\\ f_{10}(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_8)(1-2·x_6)(1-2·x_{12})=0\\ f_{11}(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_{14})(1-2·x_9·x_{10})=0\\ f_{12}(x_1,x_2,\dots ,x_{14})=\frac{1}{2}+\frac{1}{2}·(1-2·x_9)(1-2·x_{10})=0\\ f_{13}(x_1,x_2,\dots ,x_{14})=\frac{1}{2}-\frac{1}{2}·(1-2·x_{11})(1-2·x_{12})(1-2·x_{13})=0\\ f_{14}(x_1,x_2,\dots ,x_{14})=\frac{1}{2}+\frac{1}{2}·(1-2·x_{13})(1-2·x_{14})=0\end{array}$$

(10)

$$tf\left(x_1,x_2,\dots ,x_{14}\right)={\displaystyle \sum }_{i=1}^{14}{f}_i\left(x_1,x_2,\dots ,x_{14}\right)$$

Let us list and check the main properties of the target function $tf\left(x_1,x_2,\dots ,x_{14}\right)$.

$\left(a\right)$ Function $tf\left(x_1,x_2,\dots ,x_{14}\right)$ is harmonic in a 14 -dimensional cube $K_{14}$;

$\left(b\right)$ Function $tf\left(x_1,x_2,\dots ,x_{14}\right)$ and its constrictions on the edges and faces of a 14-dimensional cube $K_{14}$ are harmonic functions.

It is enough to show that $\frac{{\partial }^2}{\partial x_k^2}tf\left(x_1,x_2,\dots ,x_{14}\right)=0\forall k\in \left\{1,\dots ,14\right\}$.

$$\frac{{\partial }^2}{\partial x_k^2}tf\left(x_1,x_2,\dots ,x_{14}\right)=\frac{{\partial }^2}{\partial x_k^2}\left({\displaystyle \sum }_{i=1}^{14}{f}_i\left(x_1,x_2,\dots ,x_{14}\right)\right)={\displaystyle \sum }_{i=1}^{14}\frac{{\partial }^2}{\partial x_k^2}{f}_i\left(x_1,x_2,\dots ,x_{14}\right)={\displaystyle \sum }_{i=1}^{14}0=0.$$

From properties $\left(a\right)–\left(b\right)$, it follows that

• target function $tf\left(x_1,x_2,\dots ,x_{14}\right)$ does not have a local extremum inside, edges, and faces of a 14-dimensional cube $K_{14}$;
• It takes its extremums at the vertices of a 14-dimensional cube $K_{14}$.

To optimize the target function $tf\left(x_1,x_2,\dots ,x_{14}\right)$ in a 14-dimensional cube $K_{14}$, we use the coordinate descent algorithm. The coordinate descent algorithm is described as follows:

1.  

The initial approximation ${\overrightarrow{x_0}}^{\left[0\right]}=\left(x_1{\left(0\right)}^{\left[0\right]},x_2{\left(0\right)}^{\left[0\right]},\dots ,x_{14}{\left(0\right)}^{\left[0\right]}\right)\in K_n$

2. 

Compute, $\{\begin{array}{c}{\overrightarrow{x_1}}^{[k]}={\overrightarrow{x_0}}^{[k]}-{\lambda }_1^{[k]}·t{f}_{x_1}^{\prime }\left({\overrightarrow{x_0}}^{[k]}\right)·\overrightarrow{e_1}\\ {\overrightarrow{x_2}}^{[k]}={\overrightarrow{x_1}}^{[k]}-{\lambda }_2^{[k]}·t{f}_{x_2}^{\prime }\left({\overrightarrow{x_1}}^{[k]}\right)·\overrightarrow{e_2}\\ {\overrightarrow{x_3}}^{[k]}={\overrightarrow{x_2}}^{[k]}-{\lambda }_3^{[k]}·t{f}_{x_3}^{\prime }\left({\overrightarrow{x_2}}^{[k]}\right)·\overrightarrow{e_3}\\ \hspace{1em}\hspace{1em}\hspace{1em} \dots \dots \dots \dots \\ {\overrightarrow{x_{13}}}^{[k]}={\overrightarrow{x_{12}}}^{[k]}-{\lambda }_{13}^{[k]}·t{f}_{x_{13}}^{\prime }\left({\overrightarrow{x_{12}}}^{[k]}\right)·\overrightarrow{e_{13}}\\ {\overrightarrow{x_{14}}}^{[k]}={\overrightarrow{x_{13}}}^{[k]}-{\lambda }_{14}^{[k]}·t{f}_{x_{14}}^{\prime }\left({\overrightarrow{x_{13}}}^{[k]}\right)·\overrightarrow{e_{14}}\end{array},$

• where ${\lambda }_i^{\left[k\right]}=\underset{{\lambda }^{\left[k\right]}\ge 0}{\mathrm{argmin}} tf\left({\overrightarrow{x_{i-1}}}^{\left[k\right]}-{\lambda }^{\left[k\right]}\cdot t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{i-1}}}^{\left[k\right]}\right)\cdot \overrightarrow{e_i}\right)$, $\overrightarrow{e_i}$ this is a unit vector with coordinates on $i-th$ location $1$, with other coordinates being $0$.

3. 

Check the stop condition:

(i) 

If $tf\left({\overrightarrow{x_{14}}}^{\left[k\right]}\right)=0,$ that $x=\left(x_1,x_2,\dots ,x_{14}\right)={\overrightarrow{x_{14}}}^{\left[k\right]}$ and stop;

(ii) 

If $tf\left({\overrightarrow{x_{14}}}^{\left[k\right]}\right)\ne 0$, that $k≔k+1$, in the quality ${\overrightarrow{x_0}}^{\left[k+1\right]}$ take either a point ${\overrightarrow{x_{14}}}^{\left[k\right]}$ or another point, if the point ${\overrightarrow{x_{14}}}^{\left[k\right]}$—locally-vertex and transition to step 2.

In our case, using the specifics of the function $tf\left(x_1,x_2,\dots ,x_{14}\right)$, slightly modify (simplify) the algorithm of coordinate descent. First, there is no need to perform one-dimensional numerical optimization ${\lambda }_i^{\left[k\right]}=\underset{{\lambda }^{\left[k\right]}\ge 0}{\mathrm{argmin}}tf\left({\overrightarrow{x_{i-1}}}^{\left[k\right]}-{\lambda }^{\left[k\right]}\cdot t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{i-1}}}^{\left[k\right]}\right)\cdot \overrightarrow{e_i}\right)$, and it can be found explicitly, since the function $tf\left(x_1,x_2,\dots ,x_{14}\right)$ harmonic function in each of its variables $⟹i$—th vector coordinate ${\overrightarrow{x_i}}^{\left[k\right]}={\overrightarrow{x_{i-1}}}^{\left[k\right]}-{\lambda }^{\left[k\right]}\cdot t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{i-1}}}^{\left[k\right]}\right)\cdot \overrightarrow{e_i}$ there will be the following $x_i(i{)}^{[k]}=\{\begin{array}{c}1,\mathit{if} t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{l-1}}}^{[k]}\right)<0\\ x_i(i-1{)}^{[k]},\mathit{if} t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{l-1}}}^{[k]}\right)=0=\\ 0,\mathit{if} t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{l-1}}}^{[k]}\right)>0\end{array}$ $\frac{1-\mathit{sign}\left(t{f}_{x_i}^{\prime }\left({\overrightarrow{x_{l-1}}}^{[k]}\right)\right)}{2}\left|\mathit{sign}\left(t{f}_{x_i}^{\prime }(\underset{x_{l-1}}{\to }[k])\right)\right|+x_i(i-1{)}^{[k]}·(1-\left|\mathit{sign}\left(t{f}_{x_i}^{\prime }(\overrightarrow{x_{l-1}}[k])\right)\right|)$, secondly, as a ${\overrightarrow{x_0}}^{\left[k+1\right]}$ let us take a new point.

We applied a modified (simplified) coordinate descent algorithm to the objective function $tf\left(x_1,x_2,\dots ,x_{14}\right)$ in $K_{14}$ and received the following results in the appropriate order:

• Modified algorithm of coordinate descent from 533 points (out of $2^{14}$—all possible options) after the first iteration will find the solution $x=\left(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10},x_{11},x_{12},x_{13},x_{14}\right)=\left(1,1,0,1,0,1,0,1,1,0,1,0,1,0\right)$, you can directly check;
• The modified algorithm of coordinate descent was run 1000 times from a random vertex of a 14-dimensional cube $K_{14},$ and every time it will find the solution $x=\left(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10},x_{11},x_{12},x_{13},x_{14}\right)=\left(1,1,0,1,0,1,0,1,1,0,1,0,1,0\right)$, and in an average of 32 iterations.

Verifying directly the solution is easily found to be $x=\left(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10},x_{11},x_{12},x_{13},x_{14}\right)=\left(1,1,0,1,0,1,0,1,1,0,1,0,1,0\right)$, which is the solution to system $\left(9\right)$;

• Now, we solve this system by the other most well-known methods and make a

  Table 1. Results of the comparative analysis.

  Methods	Brute Force	Linearization	XSL	Gröbner’s Basis	The Proposed Method
  Number of iterations	13674 by lexicographic order	71	1362	57 according to Buchberger’s algorithm	32 on average
  	9804 in order of Gray code			21 according to Faugère’s algorithm (F4)

  .

The method of brute force: in lexicographic order, it finds a solution in 13674 iterations, in the order of Gray’s code in 9804 iterations.

Linearization method: let $x_{15}=x_1\otimes x_2, x_{16}=x_9\otimes x_{10}$, then the system (9) is linear concerning $x_1, x_2, x_3,\dots ,x_{14}, x_{15}, x_{16}$ and we apply the Gauss method and, in 15 iterations, we obtain that all variables $x_1, x_2, x_3,\dots ,x_{14}, x_{15}, x_{16}$ will depend on $x_{12}$ and $x_{16}$. Now, substituting the values $x_{15}=x_1\otimes x_2, x_{16}=x_9\otimes x_{10}$, the resulting system can be solved analytically or the resulting system is solved by iterating over $x_{12}$ and $x_{16}$ no more than for $15+2^2\cdot 14=71$ iterations.

$XSL$: all linear equations of system (9) multiplied by monomials $x_1, x_2, x_3,\dots ,x_{14}$, we add in the system the linearization method to the system and apply it in 1362 iterations: on the one hand, all variables in the extended system will depend only on $x_{104}=x_{12}\otimes x_{14}$, on the other hand, the value $x_{12}\left(x_{104}\right)=0⨁0⨂x_{104}=0⟹x_{104}=0$. This means that, at this level, the XSL method solves the system in full.

The Grobner basis: the improved Buchberger algorithm calculates the Grobner basis in 57 iterations (that is, it will take 57 times to calculate the S-polynomial), and the F4 algorithm in 21 iterations.

Remark. 

We have an understanding (another option) that, in the unit square$K_2$, the logical function$and\left(x_1, x_2\right)$polynomial$x_1-x_1^2+x_2-x_2^2+x_1{x}_2$describes well what$an{d}_{app}\left(x_1, x_2\right)=x_1{x}_2$. Since$\left(x_1,x_2\right)\in K_2$, then

$$(a) x_1-x_1^2+x_2-x_2^2+x_1{x}_2=1 ⟺\left(x_1,x_2\right)=\left(1,1\right)$$

$$(b) x_1-x_1^2+x_2-x_2^2+x_1{x}_2=0 ⟺\left(x_1,x_2\right)\in \left\{\left(0,0\right), \left(0,1\right),\left(1,0\right)\right\}.$$

The function $an{d}_{app}\left(x_1, x_2\right)=x_1{x}_2$ does not have the property $(b)$, but, firstly, it is harmonic, and, secondly, in these classes of systems the obtained objective functions $tf \left(x_1, x_2, x_3,\dots , x_n\right)$ and their restrictions on the edges and faces of the $n$-dimensional cube $K_n$ were harmonic functions and, therefore, we used $an{d}_{app}\left(x_1, x_2\right)=x_1{x}_2$.

5. Conclusions

In this paper, we have proposed a new transformation method for solving a system of Boolean algebraic equations. We substantiated the correctness and the right to exist of the proposed method with reliable evidence. One illustrative example showed the method of applying the proposed method and carried out a comparative analysis. Comparative analysis also shows that the results of the proposed method are good. However, this new idea should be considered to be at a preliminary stage since it applies to very specific use cases, whereas a thorough complexity analysis is still missing. Therefore, in order to improve the proposed method based on this work, in the near future we plan to research and prepare for publication the following works: transformation of an arbitrary system of Boolean algebraic equations into a system of harmonic-polynomial equations in $K_n$, investigate in which cases the system of Boolean algebraic equations and the system harmonic-polynomial equations in $K_n$ will be equivalent, estimate and find the asymptotic complexity of the proposed algorithm, and then apply the improved method to solve specific applied problems.