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 =r r Q x Q x Q x ++⋅+⋅ 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 n n ⨯ grid checkerboard’s initial state vector 0s , the terminated state vector 1s and the initial control matrix A., whether we can attribute the feasible method of Judgment which transforms 0s into 1s to the existence of solutions of equations over finite fields 2F by clicking on the grid: whether there is i x which beyond
2F (l i ,....,3,2,1=) to make the equation 11
02
)(s a x s n i i i =⋅+∑=become true. If there is i x
which beyond 2F , we can transforms 0s into 1s by clicking on the grid. However, If there isn’t i x which beyond 2F , we can ’t transforms 0s into 1s by clicking on the grid .
In order to solve the variable i x of the equation 11
02
)(s a x
s n i i i
=⋅+
∑=, We transform it
into a matrix equation form. Namely, we solve the variable x of the equation 01s s x A += in
which A is equal to ⎪⎪⎪⎪
⎪⎭
⎫ ⎝⎛221n a a a .
In Figure 1, the initial state vector of the window is represented by 0s . Therefore,
s 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 by 1s . 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 0181812211s s a x a x a x a x i i +=⋅+⋅⋅⋅+⋅+⋅⋅⋅+⋅+⋅
in
which
1
s +
s 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 ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭⎫ ⎝
⎛H E
O
O
O
O
O
O
O
E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛=1111111111111111111111111H ⎪⎪⎪⎪
⎪
⎪⎪
⎪⎪
⎪
⎪⎪⎪⎭⎫
⎝⎛=111111111E
By the elementary row transformation, we can obtain the equation
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
⎡987654321I I I I I I I I I H E
O
O
O
O
O
O
O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H
→
→
⎥⎥
⎥
⎥
⎥⎥
⎥⎥⎥
⎥
⎥
⎥⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡12345678923456789I N N N N N N N N H E O O O O O O O P O E O O O O O O P O O E O O O O O P O O O E O O O O P O O O O E O O O P O O O O O E O O P O O O O O O E O P O O O O O O O E P O O O O O O O O
，T I )0,1,1,1,1,1,1,10(1，=T I )1,0,0,0,0,0,0,01(,2，=,T I )1,0,1,0,0,1,0,01(3，=,
T I )1,1,1,1,1,1,1,01(4，=,T I )1,0,1,0,0,0,0,01(5，=T I )1,0,0,0,1,1,1,01(,6，=,
T I )1,0,0,0,0,1,0,01(7，=,T I )1,0,0,0,0,0,0,01(8，=,T I )0,1,1,1,1,1,1,10(9，= .
So the original equation 0181812211s s a x a x a x a x i i +=⋅+⋅⋅⋅+⋅+⋅⋅⋅+⋅+⋅ can
be
written
in
the
form
below
)91(,919293949596979899876543299876543219≤≤⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪
⎪⎭
⎫
⎝⎛=⎪⎪⎪
⎪⎪
⎪
⎪
⎪⎪⎪
⎪
⎪⎪
⎭⎫ ⎝⎛=⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭⎫
⎝
⎛--------i x x x x x x x x x X I I I I I I I I N X X X X X X X X X H E
O
O
O
O
O
O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E O O O O O O O E H E P O O O O O O O
O i i i i i i i i i i ，其中 Through a
direct
derivation,
we
can
obtain
the
equation
.2,
21n k P P H P k k k ≤≤+=--.1,
110
n k I P N i k i i k
≤≤=--=∑
Therefore, 98322119I P I P I P I N ++++= =T
)1,0,1,0,1,0,1,0,1
( Through the mentioned above, we can obtain the equation 999N X P = And also can solve the equation: 9X =T x x x x x x x x ),,,,(8078767481797775+++
9
X `Particular
solution is
T T x x x x x x x x x X )0,0,0,0,0,0,0,0,1(),,,,,,,,(8180797877767574739==
Similarly, we can obtain the equation
T X )0,0,0,0,0,0,0,1,1(8=T X )0,0,1,1,1,1,0,0,1(7=T X )1,0,1,0,0,0,0,1,1(6= T X )0,0,0,0,0,0,0,0,0(5=T X )0,0,0,0,0,0,0,1,0(4=T X )1,1,1,1,1,1,0,1,0(3= T X )1,1,0,1,1,1,0,0,0(2=T X )0,1,1,1,0,1,1,1,1(1=
So
we
can
find
a
particular
solution:
of
the
equations:。

T x )0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,0,1,0,0,1,1,1,1,0,0,1,1,0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,0,1,10,1,0,0,1,1,1,0,0,0,0,1,1,1,1,0,1,1,1,1(= Now we can find all eight linearly independent solutions of the equation
08181=⨯x A .the original equations can be turned into the form:
⎥⎥
⎥
⎥
⎥⎥
⎥⎥⎥
⎥
⎥
⎥⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
⎡12345678998765432123456789I N N N N N N N N X X X X X X X X X H E O O O O O O O P O E O O O O O O P O O E O O O O O P O O O E O O O O P O O O O E O O O P O O O O O E O O P O O O O O O E O P O O O O O O O E P O O O O O O O O
Through solving the equation, we can obtain the fundamental System of
Solutions:
基础解系
1X .
)0,0,0,0,1,0,1,0,1(;)0,0,0,1,0,1,0,1,0(;)1,0,1,0,0,0,0,0,0(;)0,1,0,1,0,0,0,1,0(;)0,0,1,0,1,0,1,0,1(;)0,1,0,0,0,1,0,1,0(;)1,0,0,0,0,0,1,0,0(;)0,1,0,1,0,1,0,0,0(T
T
T
T
T T T T
2X
.
)1,1,0,0,0,0,0,1,1(;)0,0,1,1,0,1,0,1,1(;)1,0,1,1,0,0,0,0,0(;)1,1,0,1,1,0,1,1,1(;)0,1,1,0,1,0,1,0,1(;)1,1,1,0,1,1,0,1,1(;)1,1,0,0,0,1,1,1,1(;)1,1,0,1,0,1,1,0,0(T
T
T
T
T T T T
3X .)1,0,1,0,0,0,1,0,1(;)0,1,0,1,0,0,0,1,0(;)0,0,1,0,1,0,0,0,0(;)0,1,0,1,0,0,0,0,0(;)0,1,0,1,0,0,0,0,0(;)0,0,0,0,0,1,0,1,0(;)1,0,1,0,1,0,0,0,1(;)0,1,0,0,0,1,0,1,0(T
T
T
T
T T T T
4X
.
)0,1,1,1,0,1,1,1,0(;)1,1,1,0,1,1,1,0,0(;)1,1,0,1,1,1,0,0,0(;)0,0,0,0,0,0,1,1,1(;)1,1,0,1,1,0,1,0,1(;)1,1,1,0,0,0,0,0,0(;)0,1,1,0,1,0,1,0,1(;)0,0,1,1,1,0,1,1,1(T
T
T
T
T T T T
5X
.
)0,0,0,0,0,0,0,0,0(;)0,0,0,1,0,1,0,0,0(;)0,0,1,0,0,0,1,0,0(;)0,1,0,1,0,1,0,1,0(;)1,0,1,0,0,0,1,0,1(;)0,1,0,1,0,1,0,1,0(;)0,0,1,0,0,0,1,0,0(;)0,0,0,1,0,1,0,0,0(T
T
T
T
T T T T
6X .
)0,1,1,1,0,1,1,1,0(;)1,1,0,1,1,0,0,0,0(;)1,0,1,0,1,0,1,1,0(;)1,1,0,1,0,1,1,0,0(;)0,1,1,0,1,1,0,0,0(;)0,0,1,1,0,1,0,1,1(;)0,0,1,1,0,1,0,1,1(;)0,0,0,0,1,1,0,1,1(T
T
T
T
T T T T
7X .)1,0,1,0,0,0,1,0,1(;)0,0,0,1,0,0,0,0,0(;)1,0,0,0,1,0,1,0,1(;)0,1,0,0,0,1,0,0,0(;)0,1,0,0,0,0,0,0,1(;)0,0,0,1,0,0,0,1,0(;)0,0,0,0,1,0,1,0,0(;)0,0,0,0,0,1,0,0,0(T
T
T
T
T T T T
8X .
)1,1,0,0,0,0,0,1,1(;)1,1,1,0,0,0,0,1,1(;)0,1,1,1,0,0,0,0,0(;)0,0,1,1,1,0,0,1,1(;)0,0,0,1,1,1,0,0,0(;)0,0,0,0,1,1,1,0,0(;)0,0,0,0,0,1,1,0,1(;)0,0,0,0,0,0,1,1,1(T
T
T
T
T T T T
9X
.
)1,0,0,0,0,0,0,0,1(;)0,1,0,0,0,0,0,0,0(;)0,0,1,0,0,0,0,0,1(;)0,0,0,1,0,0,0,0,0(;)0,0,0,0,1,0,0,0,1(;)0,0,0,0,0,1,0,0,0(;)0,0,0,0,0,0,1,0,1(;)0,0,0,0,0,0,0,1,0(T
T
T
T
T T T T
3、the classification of the board of the ninth order
Similar to the board of the fifth order, the relation of j f (1≤j ≤81)and i V (1≤i ≤56) is the
chart below:
k f 所属集合
k 所属集合
k f 所属集合
k 所属集合
k f 所属集合
k 所属集合
1V 1U ={41} 20V 20U ={63} 39V 39U ={12,28} 2V 2U ={74} 21V 21U ={72} 40V 40U ={40,42} 3V 3U ={75} 22V 22U ={44,66} 41V 41U ={33,65} 4V 4U ={56,76} 23V 23U ={8} 42V 42U ={15} 5V 5U ={5,37,45,77} 24V 24U ={4,20} 43V 43U ={25,73} 6V 6U ={62,78} 25V 25U ={6,26} 44V 44U ={16,36} 7V 7U ={79} 26V 26U ={29} 45V 45U ={52} 8V 8U ={80} 27V 27U ={67} 46V 46U ={53} 9V 9U ={21,81} 28V 28U ={39,43} 47V 47U ={47} 10V 10U ={22} 29V 29U ={27} 48V 48U ={34} 11V 11U ={58} 30V 30U ={9,57} 49V 49U ={32,50} 12V 12U ={3} 31V 31U ={14,68} 50V 50U ={64} 13V 13U ={23,59} 32V 32U ={2} 51V 51U ={11,51} 14V 14U ={19} 33V 33U ={69} 52V 52U ={10} 15V 15U ={38,44} 34V 34U ={55} 53V 53U ={17,49} 16V 16U ={60} 35V 35U ={54,70} 54V 54U ={18} 17V 17U ={7} 36V 36U ={48} 55V 55U ={30} 18V
18U ={1,61}
37V
37U ={31,71}
56V
56U ={35}
19V
19U ={24}
38V
38U ={13}
This gives a partition of set V: V =
i i V ⋃
∈}
4,3,2,1{ and =≠∀j i V V j i ,∅. So we can derive the
general rules of every specific element i V )561(≤≤k . The chart3 give a case
表3
2m 3m 4m 5m 6m 7m 8m 9m 10m 11m 12m
奇数 奇数 偶数 偶数 偶数 奇数 奇数 偶数 奇数 奇数 奇数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数
13m 14m 15m 16m 17m 18m 19m 20m 21m 22m 23m
偶数 奇数 偶数 奇数 奇数 偶数 奇数 奇数 奇数 偶数 奇数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数
24m 25m 26m 27m 28m 29m 30m 31m 32m 33m 34m
偶数 偶数 奇数 奇数 偶数 奇数 偶数 偶数 奇数 奇数 奇数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数
35m 36m 37m 38m 39m 40m 41m 42m 43m 44m 45m
偶数 奇数 偶数 奇数 偶数 偶数 偶数 奇数 偶数 偶数 奇数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数
46m 47m 48m 49m 50m 51m 52m 53m 54m 55m 56m
奇数 奇数 奇数 偶数 奇数 偶数 奇数 偶数 奇数 奇数 奇数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数 偶数
Here the value of 1m have no impact on t. we can give the general rules of every
specific element i S )561(≤≤k .
17 31 11 23 04 24 16 22 29 51 50 38 37 30 41 43 52 53 13 23 08 09 12 18 42 24 28 38 25 54 36 48 40 47 55 43 04 14 27 39
39 27 14 04
21 46 35 52 48 50 44 45 34 33 03 29
10 12 15 17 05 19
We can derive the general rules of every specific element i S )561(≤≤k in which the
general rules of i S is in the chart4 below:
表4 1S 中元素的规律
N(01) N(02) N(03) N(04) N(05) N(06) N(07) N(08) N(09) N(10) N(11) 奇数 奇数 偶数 偶数 偶数 奇数 奇数 偶数 奇数 奇数 奇数 偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
N(12) N(13) N(14) N(15) N(16) N(17) N(18) N(19) N(20) N(21) N(22) 偶数 奇数 偶数 奇数 奇数 偶数 奇数 奇数 奇数 偶数 奇数 偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
N(23) N(24) N(25) N(26) N(27) N(28) N(29) N(30) N(31) N(32) N(33) 偶数 偶数 奇数 奇数 偶数 奇数 偶数 偶数 奇数 奇数 奇数 偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
N(34) N(35) N(36) N(37) N(38) N(39) N(40) N(41) N(42) N(43) N(44) 偶数 奇数 偶数 奇数 偶数 偶数 偶数 奇数 偶数 偶数 奇数 偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
N(45) N(46) N(47) N(48) N(49) N(50) N(51) N(52) N(53) N(54) N(55) 奇数 奇数 奇数 偶数 奇数 偶数 奇数 偶数 奇数 奇数 奇数 偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
偶数
Through the chart4, we must be able to click with the mouse to transform the state of chart1 into completely white. Based on the chart4, we can pick a form of the
most simple state as a representative from each of the specific (i = 1, 2, 3, 4).we can called them type I, type II,….,type 56 in which type I is completely white. 4、conclusion
The model transforms the light issue into the solving problem of equations over
finite fields and we can give a promotion. However, when the control matrix A class is larger, the solution is complicated. We can only use the computer program to solve it.
49 40 21 26 30 32 34 36 20 42 01 02
03 04 05 06 07 08。