专业英语翻译好的材料

1、Introduction

Lights, which are popular in a very interesting game on the Internet in

recent years, are defined as follows: In an n × n grid board, every square have

two states: white (open) and black (closed). When you click on any one of

these squares with the mouse, the box and all its adjacent boxes change

states. Namely, the black boxes become the white boxes. At the edge of the

board, the box can`t have the four adjacent boxes. Therefore, we only consider

those existed boxes.

using the method of algebra and mathematical modeling to give the

mathematical modeling, based on the linear equations of finite field, Zhou Hao

gave all the solutions for the case of n=5.Furthermore, he used the

"Classification" method of algebra to give the intuitive description of four

equivalence classes of the game so that the players can immediately

determine which type the “mess” is.

Of course, before Zhou Hao, scholars added the control vector for a

problem of PRG game on a similar light and analyzed and proved the problem

by a mathematical model. After that, researchers made general promotion for

a control problem in the RPG game and give a more comprehensive solution.

Furthermore, some foreign scholars used the dynamic programming

methods to prove a variety of matrix reconstruction problems and also proved

some complex results on reconstructing neighborhood binary matrix.

The mathematical knowledge that Zhou Hao use to build the model on the control

state issues of lights is relatively complicated. After that, scholars made general

promotion for the Zhou Hao’s mathematical model in order to be accepted.

However, if the control variable is larger, solving the effective matrix “Q” and the

combination “0s+1s=rrQxQxQx2211” is more complicated. It’s wise to

use the program or mathematical Software such as Mathematic.

2、The light issue when n=9

For the above defined rules of the game, we study the following two issues:

When n=9, the board's initial state is a mess: part of the boxes is white and part of

the boxes is black. If you continue to click on it, whether there is a way to make the mess

eventually become completely white or completely black.

Such light issues are control issues. Because light only have opened or closed states,

we can make the definition such as binary vector to study the issue, and ultimately

translate it into the existence of linear equations on a limited domain.

Under the conditions of the Known nn grid checkerboard’s initial state vector0s,

the terminated state vector 1sand the initial control matrix A., whether we can attribute the

feasible method of Judgment which transforms 0sinto1s to the existence of solutions of

equations over finite fields 2F by clicking on the grid: whether there isixwhich beyond 2F(li,....,3,2,1) to make the equation 1102)(saxsniiibecome true. If there is ix

which beyond 2F, we can transforms 0sinto1s by clicking on the grid. However, If there

isn’t ix which beyond2F, we can’t transforms 0sinto1s by clicking on the grid.

In order to solve the variableixof the equation1102)(saxsniii, We transform it

into a matrix equation form. Namely, we solve the variable x of the equation01ssxA in

which A is equal to

221naaa.

In Figure 1, the initial state vector of the window is represented by0s. Therefore,

0s is equal to

,)1,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,0,0,1,1,1,1,0,1,1,0,0,1,1,1,0,0,0,1,0,0,1,0,1,1,1,1,1,0,0,0,0,0,0,0,0,1,0,0,1,0,1,1,0,1,1,0,0,1,1,1,1,1,1,1,0,1,0,0,0,0,0,0,0,1(T

The terminated state vector of the window is represented by1s. Therefore, 1s is

equal to ,)1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1(T

We can list the equation 0181812211ssaxaxaxaxii

in which1s+0s is equal to

,)0,1,1,1,1,1,1,1,0,1,0,0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,1,1,0,0,0,1,1,1,0,1,1,0,1,0,0,0,0,0,1,1,1,1,1,1,1,1,0,1,1,0,1,0,0,1,0,0,1,01,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,0(T The coefficient matrix A is HEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEH

1111111111111111111111111H111111111E

By the elementary row transformation, we can obtain the equation

987654321IIIIIIIIIHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEHEOOOOOOOEH