A Study Regarding Solution of a Knowledge
Model Based on the Containing-Type
Error Matrix Equation
Xilin Min, Kaizhong Guo, Jize Huang
School of Management, Guangdong University of Technology, Guangzhou, China Email: [email protected]
Received May 19,2012; revised June 22, 2012; accepted July 10, 2012
ABSTRACT
An error matrix equation based on error matrix theory was presented in previous research of the error-eliminating the-ory. The purpose of solving the error matrix equation is to create a decision support on how to switch from bad to good status. A matrix based on error logic is used to express current status u, expectant status u1 and transformation matrix T.
It is u, u1, and T that are used to build error matrix Equation T u
u1. This allows us to find a method whereby bad status “u” changes to good status “u1” by solving the equation. The conversion method that transform from current toexpectant status can be obtained from the transformation matrix T. On this basis, this paper proposes a new kind of error matrix equation named “containing-type error matrix equation”. This equation is more suitable for analyzing the realis-tic question. The method of solving, existence and form of solution for this type of equation have been presented in this paper. This research provides a potential useful new technique for decision analysis.
Keywords: Error Matrix; Set Equation; Containing-Type Error Matrix Equation; Knowledge Model
1. Introduction
The truth is always expected; however, error should be avoided and eliminated firstly in order to seek truth. On the one hand, it should be known why and how the error arises with the aim of avoiding and eliminating errors; on the other hand, it should be known how to eradicate it when error appears. Moreover, within a faultless system, proposition and decision can also be naturally transferred into being defective in normal conditions, because of the passage of time, changes in natural conditions, or devel-opment of science. So, errors are required to be elimi-nated continuously.
Chinese scholar Professor Guo Kaizhong has found the error-eliminating theory since the 1980s, and has taken the study of error to a new level.
The error-eliminating theory is an emerging subject, which is used to study error elimination [1]. The theory of error-eliminating defines the objective system, in which one error factor at least is included to characterize and eliminate errors. Error setting could be termed as
, , ,
C u x u U X f Gu f U R . As men-tioned in the book “Theory of Error Sets” [2], all unary or multiple error sets have been defined in error matrix. In the book “Error Logic” [3], every element
u x, isexpressed as
1 2
, , , , , , ,
, , ,
n
U U
U S t T t L t
x t f u t G t G t
,
p
p
from the standpoint of logic theory, where U is discuss- ing domain; S t
is the objects of u; p
1, 2, ,n
is the space under consideration, and n is used to de-fine the dimensions in the space; is the character-istic of the described objects; is the value of
T t
L t
; f u
,
, U
t
is error function;T t x t t p G
UG t are decision rules. As shown in the above func-tion, object u x
and rule U are set as indepen- dent variables. In addition to this, Guo Kaizhong, Liu Shiyong and Li Min [5-7] have studied Logical relation-ship among various types of error logical words.
t GError matrix based on error logic is a useful method of modeling for scenes. X, A and B could be structured as an error matrix equation for three error matrix, such as XA = B. XA = B is just a concrete form of error matrix equation
1Xilin [8-10] have studied the solution of equation XA = B, in which a containing-type error matrix, such as XA B, was found to be more suited for analyzing the actual question. In this paper, it is attempted to solve the con-taining-type error matrix in order to seek new ways of error-elimination.
2. Error Matrix Equation
2.1. Equational-Type Error Matrix Equation
In the previous research, all of the error matrix equations have been classified into two types (see Table 1). There are five kinds of equations in each type. Meanwhile, there are three constraints on the solution of the equation: The first is an objective condition, named “kg”; the sec- ond is a stipulation that has been confirmed, named “rw”; the third is the restriction of requirement, named “xq”.
In common situations, these three restrictions above can be shown by certain sets. The factors in the matrix and the solving of the matrix equations are also measured in a certain set. Therefore, we can use the intersection method of solving sets which includes the matrix equa-tion and ignores these three restricequa-tions in solving the process.
2.2. Containing-Type Error Matrix Equation
In order to analyze practical issues, we propose a con- taining-type error matrix equation (see Table 2).
3. Solving a Containing-Type Error Matrix
Equation
In the following, let us now consider how to solve the second type of Equation (1): XA B.
1, 1 1 1 1 1 1 1 1 1 1 1
10 10 10 1 2 10 10 10 10 10 10 10
11 11 11 1 2 11 11 11 11 11 11 11
1 1 1 1 2 1
, , , , , , , ,
, , , , ,
, , , , ,
, , ,
U U
X X X n X X X X X U X U X
X X X n X X X X X U X U X
tX tX tX n
X u x U S t T t L t x t f u t G t G t
U S t T t L t x t f u t G t G t
U S t T t L t x t f u t G t G t
U S t T
p p
p p
p p
p tX
t L1tX
t x1tX
t f1tX
u t
, 1tX
,GU tX1
t
GU tX1
t
p
2 2 2 2 2 2 2 2 2 U2 U2
20 20 20 1 2 20 20 20 20 20 20 20
21 21 21 1 2 21 21 21 21 21 21 21
2 2 2 1 2 2 2 2 2 2 2
, , , , , , , ,
, , , , ,
, , , , ,
, , , , ,
n U
n U
t t t n t t t t t U t
A U S t T t L t x t f u t G t G t
U S t T t L t x t f u t G t G t
U S t T t L t x t f u t G t G t
U S t T t L t x t f u t G t G
p p
p p
p p
p p 2
,
U t t
[image:2.595.83.516.295.746.2]U U
Table 1. Classification of equational-type error matrix equation.
Equation Type 1 Type 2 Meaning of operator
1 AX = B XA = B Generic matrix multiplication
2 A●X = B X●A = B Good
3 A▲X = B X▲A = B Bad
4 A∨X = B X∨A = B Or
5 A∧X = B X∧A = B And
Table 2. Classification of containing-type error matrix equation.
Equation Type 1 Type 2 Meaning of operator
1 AX B XA B Generic matrix multiplication
2 A●X B X●A B Good
3 A▲X B X▲A B Bad
4 A∨X B X∨A B Or
[image:2.595.86.510.642.735.2]
1 1 1 1 11 11 12 12
21 21 22 22 2 1 2 1
21 21 22 22 2 1 2 1
201 201 201 1 2 201 201
21 21 21 1 2 21 21
2 2 1 2 2 1 2 2 1 , , , ( , ) ( , ) , , , , , , , , , , m m m m
m m m m m m m m
V V n V V
j V j V j n V j V j
m m V m m V m m
b y
b y b y
b y b y b y
B
b y b y b y
V S t T t L
V S t T t L t
V S t
p p
p
1 2 2 2 1 2 2 1
201 201 201 201 201
21 21 21 21 21
2 2 1 2 2 1 2 2 1 2 2 1 2 2 1
, , ,
, ,
, ,
, ,
n V m m V m m
V V V V V
V j V j V j V j V j
V m m V m m V m m V m m V m m
T t L
y t f v t G t G t
y t f v t G t G t
y t f v t G t G t
p p p t t
Definition 3.1. Suppose
1 1 1 1
11 11 12 12
21 21 22 22 2 1 2 1
21 21 22 22 2 1 2 1
, , , , , , , , , , m m m m
m m m m m m m m
w z
w z w z
w z w z w z
XA
w z w z w z
One element of the matrix
w zij, ij
U1ixU2j S1ix
t S2j
t p1ixp2j T1ix
t T2j
t L1ix
t L2j
t x1ix
t y2j
t GU ix1
t GU j2
t
Therefore
10 20 10 20 10 20 10 20 10 20 10 20 10 20
11 21 11 21 11 21 11 21 11 21 11 21 11 21
1 2 1 2 1 2 1 2 1 2 1 2
X X X X X X U X U
X X X X X X U X U
tX t tX t tX t tX t tX t tX t
U U S t S t t T t T t L t L t x t x t G t G t
U U S t S t t T t T t L t L t x t x t G t G t
U U S t S t t T t T t L t L t x t x t
p p p p
p p
1 2201 201 201 1 2 201 201
21 21 21 1 2 21 21
2 2 1 2 2 1 2 2 1 1 2 2 2 1 2 2 1
201 201 201 201 201
21
, , ,
, , ,
, , ,
, ,
U tX U t
V V n V V
j V j V j n V j V j
m m V m m V m m n V m m V m m
V V V V V
V j
G t G t
V S t T t L t
V S t T t L t
V S t T t L t
y t f v t G t G t
y t f
p p p p
21 21 21 21
2 2 1 2 2 1 2 2 1 2 2 1 2 2 1
, ,
.
, ,
V j V j V j V j
V m m V m m V m m V m m V m m
v t G t G t
y t f v t G t G t
p p
According to the definition of matrix equivalent:
w zij, ij
b yij, ij
. So
1 2 1 2 1 1 2 2 1 2
1 2 1 1 1 1 2 1 2
, , ,
, , ,
iX j iX j iX n j iX j
iX j iX iX iX U iX j U iX U j ij ij
U U S t S t T t T t
L t L t x t f u t p G t x t G t G t b y
That is
1 2 1 2 1 1 2 2 1 2
1 2 1 1 1 1 2 1 2
2 2 2 2 2 2 2
, , ,
, ,
.
iX j iX j iX n j iX j
iX j iX iX iX U iX j U iX U j
ij V ij V ij V ij V ij V ij V ij
U U S t S t T t T t
L t L t x t f u t G t x t G t G t
V S t T t L t y t G t
p p
p p
i.e., according to the corresponding factors in these two matrixes, the equations below are obtained:
tion of minimization.
In fact, the equation sets above are not irrelevant. After calculation, the factors of matrix can compose a complete proposition. This can be defined as
10X 20 V20,
U U V
10X 20 V20 ,S t S t S t
10X 1, 2, ,n 20 V20 1, 2, , ,
p p p
1 2 1 1 2
2 1 1 2 2 3 1 2
4 1 2 5 1 2
6 1 2
( , , )
.
iX j iX j
iX n j iX j
iX j iX j
U iX U j
U U h S t S t
h h T
h L t L t h x t x t
h G t G t
p p
n t T t
10X 20 V20 ,T t T t T t
10X 20 V20 ,L t L t L t
10X 20 V20 ,x t x t y t “h
i, i = 1, 2, ···, 6” means a complete matrix element composition can be composed after calculating. The methods of composing are decided according to the spe-cific situation. One way is the multiplication of m × 7- order error matrix, from which is composed a new error set or error logical proposition by using each parameter from calculation to be the corresponding parameter.
10 20 20 ;
U X U V
G t G t G t
1iX 2j V j2 ,
U U V
1iX 2j V j2 ,S t S t S t
1iX 1, 2, ,n 2j V j2 1, 2, ,n
p p p
, solving error matrix equationTheorem 1. A necessary and sufficient condition of XA B is that equation
1iX 2j V j2 ,
T t T t T t X Ai Bi, i = (1,2, ···, m2) has a solution.
Proof: if XA´=B has a solution, according to the equivalent equationsXA B and X Ai Bi, i = (1,2, ···, m2), the equationX Ai Bi, i = (1,2, ···, m2) should have a solution first; Vice versa, if X Ai Bi, i = (1,2, ···, m2) has a solution, we also can know XAB has a solution by the same principle.
1iX 2j V j2 ,L t L t L t
1iX 2j V j2 ,x t x t y t
1 2 2 .
U iX U j V j
G t G t G t
So, we can discuss the solution of equation XA B, the same as when solving
If both sides of the equation are all sets, operator ∧
“ ”means the operation of intersection. If both sides of ∧
the equation are values, operator “ ”means the opera-
i i
X A B, i = (1,2, ···, m2). With respect to X Ai Bi,
1 1 1 1 2 1 1 1 1 1 1 1
1 20 1 20 1 1 2 20 1 20
1 20 1 20 1 20
1 2 1 2 1 1 2 2 1
, , , , ,
[ , ,
( , , )
iX iX iX n iX iX iX iX iX U iX U iX
iX iX iX n iX
iX iX U iX U
iX j iX j iX n j iX
U S t T t L t x t f u t G t G t A
U U S t S t T t T t
L t L t x t x t G t G t
U U S t S t T t
p p
p p
p p
2
1 2 1 2 1 2
1 2 1 1 2 1 1 1 2 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1 1 2 2 1 1
( , , )
, , ,
j
iX j iX j U iX U j
iX m iX m iX n m iX m
iX m iX m U iX U m i i i i im im
T t
L t L t x t x t G t G t
U U S t S t T t T t
L t L t x t x t G t G t b y b y b y
.
p p
1 20 1 20 1 1 2 20 1 20 1 20 1 20
1 20 20 20 20 1 2 20 20 20 20
1 2 1 2 1 1 2 2 1 2 1 2 1 2
, ,
, , ;
, ,
iX iX iX n iX iX iX
U iX U V V n V V V V
iX j iX j iX n j iX j iX j iX j
U
U U S t S t T t T t L t L t x t x t
G t G t V S t T t L t y t G t
U U S t S t T t T t L t L t x t x t
G
p p
p
p p
1 2 2 2 2 1 2 2 2 2 2
1 2 1 1 2 1 1 1 2 2 1 1 2 1 1 2 1
1 2 1 1 2 1 2 2 2 1 2 2 2 2 2
, , ;
( , , )
, , .
iX U j j V j V j n V j V j V j V j
tX m tX m tX n m tX m tX m
tX m U tX U m t V t V t n V t V t V t V t
t G t V S t T t L t y t G t
U U S t S t T t T t L t L t
x t x t G t G t V S t T t L t y t G t
p
p p
p
There are equations:
U1iX U20
VV20,
S1iX t S20 t
SV20
t ,
p1iX 1, 2, ,n p20
pV20
1, 2, ,n
,
T1iX t T20 t
TV20
t ,
L1iX t L20 t
LV20
t ,
x1iX t x20 t
yV20
t ,
GU iX1 t GU20 t
GV20
t ;
U1iX U2j
VV j2 ,
S1iX t S2j t
SV j2
t ,
p1iX 1, 2, ,n p2j
pV j2
1, 2, ,n
,
T1iX t T2j t
TV j2
t ,
L1iX t L2j t
LV j2
t ,
x1iX t x2j t
yV j2
t ,
GU iX1 t GU j2 t
GV j2
t ;
U1tX U2 1m
VV t2 ,,
S1iX t S2 1m t
SV t2
t ,
p1iX 1, 2, ,n p2 1m
pV t2
1, 2, ,n
T1iX t T2 1m t
TV t2
t ,
L1iX t L2 1m t
LV t2
t ,
x1iX t x2 1m t
yV t2
t ,
GU iX1 t GU m2 1 t
GV t2
t .Theorem 2. Necessary and sufficient condition of solv- ing error matrix equation X Ai Bi is:
20 V20,
U V
20 V20 ,S t S t
20 V20 1, 2, ,n ,
p p
20 V20 ,T t T t
20 V20 ,L t L t
20 V20 ,x t y t
20 20 ;
U V
G t G t
2j V j2 ,
U V
2j V j2 ,S t S t
2j V j2 1, 2, ,n , p p
2j V j2 ,T t T t
2j V j2 ,L t L t
2j V j2 ,x t y t
2 2 ;
U j V j
G t G t
2 1m V2,
U V t
2 1m V t2 ,S t S t
2 1m V t2 1, 2, ,n , p p
2 1m V t2 ,T t T t
2 1m V t2 ,L t L t
2 1m V t2 ,
2 1 2 .
U m V t
G t G t
Proof of necessity: If one of above conditions is not satisfied, for example, condition S2j
t SV j2
t , is not satisfied, then in equation
S1iX t S2j t
SV j2
t , no matter what the value of S1iX
t is,
S1iX
t S2j
t
SV j2
t cannot becalculated.
Proof of sufficiency: The condition ABi is satis-fied in error matrix equation X Ai Bi, so the element of Xi, which corresponds to element A, take union of set. i.e.,
1iX 20 21 2j 2t,
U U U U U
1iX 20 21 2j 2t,
S S S S S
1iX 1, 2,n 20 21 2j 2t
p p p p p ,
1iX 20 21 2j 2t,
T T T T T
1iX 20 21 2j 2t,
L L L L L
1iX 20 21 2j 2t ,
x t x x x x x x x t
1
20 21 2 2 .
U iX
U U U j U t
G t
G x G x G x G t
Q.ed.
Then discussing the whole solution of equations
i i
X A B and XA B.
After solving X x x
1, ,2 xn
of X Ai Bi, by tak-ing the intersection of X, kg, rw and xq, we can acquire
1, ,2 n
X x x x X .
4. Application Example of Error Matrix
Equation
For example: suppose 11 12 ,
21 22
a a
A
a a
11 201 201 201 1, 2, , n 201 201 201 201 , 201 , U201 201 ,
a U S t p T t L t x t f u t p G t GU t
12 202 202 202 1, 2, , n 202 202 202 202 , 202 , U202 202 ,
a U S t p T t L t x t f u t p G t GU t
21 211 211 211 1, 2, , n 211 211 211 211 , 211 , U211 211
a U S t p T t L t x t f u t p G t GU t ,
22 212 212 212 1, 2, , n 212 212 212 212 , 212 , U212 212
a U S t p T t L t x t f u t p G t GU t .
,
,
p
Each element included in a11, a12, a21, a22 is A set type.
For instance, an element of discussing domain U201 has
several elements under it. So U201 can be expressed as
. Thus, we could express all the elements as follows: 201
201 202 20, , , n
U u u u
201 201, 202, , 20n
U u u u
201 201, 202, , 20n
S t s s s
201 1, 2, ,n 201, 202, , 20n ,
p p p
201 201, 202, , 20n
T t t t t ,
,
,
,
,
,
,p
,
201 201,202, , 20n
L t l l l
201 201, 202, , 20n
x t x x x
201 201, 202, , 20
U n
G t g g g
n = 11;
202 201, 202, , 20n
U u u u
202 201, 202, , 20n
S t s s s
202 1, 2, ,n 201, 202, , 20n
p p p
202 201,202, , 20n
T t t t t
202 201,202, , 20n
L t l l l ,
202 201, 202, , 20n
x t x x x ,
202 201, 202, , 20n
G t g g g ,
n = 9;
211 211, 212, , 21n
U u u u ,
211 211, 212, , 21n
S t s s s ,
211 1, 2, ,n 211, 212, , 21n ,
p p p p
211 1, 2, ,n 211, 212, , 21n ,
p p p p
211 211,212, , 21n
T t t t t ,
211 211,212, , 21n
L t l l l ,
211 211, 212, , 21n
x t x x x ,
211 211, 212, , 21n
G t g g g ,
n = 10;
212 211, 212, , 21n
U u u u ,
212 211, 212, , 21n
212 1, 2, ,n 211, 212, , 21n ,
p p p p
,
,
212 211,212, , 21n
T t t t t
212 211, 212, , 21n
L t l l l
212 211, 212, , 21n
x t x x x ,
212 211, 212, , 21n
G t g g g ,
n = 15;
Suppose X
x x1 2
,
1 10X 10X 10X 1, 2, , n 10X 10X 10X 10X , 201 , U10X U10X ,
x U S t p T t L t x t f u t p G t G t
2 11X 11X 11X 1, 2, , n 11X 11X 11X 11X , 201 , U11X U11X ,
x U S t p T t L t x t f u t p G t G t
and 11 12 ,
21 22
b b
B
b b
11 201 V201 V201 1, 2, , n V201 V201 V201 V201 , V201 , V201 V201
b V S t p T t L t y t f u t p G t G t ,
12 202 V202 V202 1, 2, , n V202 V202 V202 V202 , V202 , V202 V202
b V S t p T t L t y t f u t p G t G t ,
21 211 V211 V211 1, 2, , n V211 V211 V211 V211 , V211 , V211 V211 ,
b V S t p T t L t y t f u t p G t G t
22 212 V212 V212 1, 2, , n V212 V212 V212 V212 , V212 , V212 V212
b V S t p T t L t y t f u t p G t G t .
,
,
, k p,
,
,
,
,
,
k p,
,
,
,
The representation of all the elements of b11, b12, b21,
b22 is the same as a11, a12, a21, a22. So they can be
ex-pressed as follows:
201 201, 202, , 20k
V u u u
201 201, 202, , 20
V k
S t s s s
201 1, 2, , 201, 202, , 20
V n
p p p
201 201,202, , 20
V k
T t t t t
201 201, 202, , 20
V k
L t l l l
201 201, 202, , 20
V k
y t y y y
201 201, 202, , 20
V k
G t g g g
k = 7;
202 201, 202, , 20k
V u u u
202 201, 202, , 20
V k
S t s s s
202 1, 2, , 201, 202, , 20 ,
V n
p p p
202 201,202, , 20
V k
T t t t t
202 201,202, , 20
V k
L t l l l
202 201, 202, , 20
V k
y t y y y
202 201, 202, , 20
V k
G t g g g
k = 8;
211 211, 212, , 21k
V u u u ,
211 211, 212, , 21
V k
S t s s s ,
211 1, 2, , 211, 212, , 21 ,
V n
p p p p k
211 211, 212, , 21
V k
T t t t t ,
211 211,212, , 21
V k
L t l l l ,
211 211, 212, , 21
V k
y t y y y ,
211 211, 212, , 21
V k
G t g g g ,
k = 2;
212 211, 212, , 21k
V u u u ,
212 211, 212, , 21
V k
S t s s s ,
212 1, 2, , 211, 212, , 21
V n
p p p p k ,
212 211, 212, , 21
V k
T t t t t ,
212 211, 212, , 21
V k
L t l l l ,
212 211, 212, , 21
V k
y t y y y ,
212 211, 212, , 21
V k
G t g g g ,
k = 3.
below.
1 2
: , , , , ,
, , ,
kg kg kg n
kg kg kg U
kg U S t
T t L t X t G t
p ,
, ,
, p,
,
, ,
201, 202, , 20 , 201, 201, , 20
kg n j
U u u u v v v
201, 202, , 20
kg n
S t s s s
1, 2, ,
201, 202, , 20kg n n
p p p
201, 202, , 20
kg n
T t t t t
201,202, , 20
kg n
L t l l l
201, 202, , 20 , 211, 212, , 21
kg n j
y t x x x y y y
201, 202, , 20
U n
G t g g g
n = 4, j = 5.
1 2
: , , , , ,
, , ,
rw rw rw n
rw rw rw U
rw U S t
T t L t X t G t
.
p ,
, ,
,p
, ,
, ,,
,
,
,p
,
,
,
201, 202, , 20
rw n
U u u u
201, 202, , 20
rw n
S t s s s
1, 2, ,
201, 202, , 20rw n n
p p p
201,202, , 20
rw n
T t t t t
201, 202, , 20
rw n
L t l l l
201, 202, , 20 , 211, 212, , 21rw n j
y t x x x y y y
201, 202, , 20
U n
G t g g g
n = 10.
1 2
: , , , , ,
, , ,
xq xq xq n
xq xq xq U
xq U S t
T t L t X t G t
p
201, 202, , 20
xq n
U u u u
201, 202, , 20
xq n
S t s s s
1, 2, ,
201, 202, , 20xq n n
p p p
201, 202, , 20
xq n
T t t t t
201,202, , 20
xq n
L t l l l
201, 202, , 20 , 211, 212, , 21
xq n j
y t x x x y y y
201, 202, , 20
U n
G t g g g ,
n = 8.
By Theorem 2, the solution of XA B is:
10X 201, 202, , 20n
U u u u ,
10X 201, 202, , 20n
S t s s s ,
10X 1, 2, ,n 201, 202, , 20
p p p p n ,
10X 201, 202, , 20n
T t t t t ,
10X 201,202, , 20n
L t l l l ,
10X 201, 202, , 20n
x t x x x ,
10 201, 202, , 20
U X n
G t g g g ,
n = 11.
Combining the three constrains of kg, rw, xq to U10X mentioned above, the solution satisfied X ∩ kg ∩ rw ∩ xq is:
10X 201, 202, , 20n
U u u u ,
10X 201, 202, , 20n
S t s s s ,
10X 1, 2, ,n 201, 202, , 20
p p p p n ,
10X 201, 202, , 20n
T t t t t ,
10X 201,202, , 20n
L t l l l ,
10X 201, 202, , 20n
x t x x x ,
10 201, 202, , 20
U X n
G t g g g ,
n = 4.
In the same way, we can acquire:
11X 201, 202, , 20n
U u u u ,
11X 201, 202, , 20n
S t s s s ,
11X 1, 2, , n 201, 202, , 20 ,
p p p p n
11X 201,202, , 20n
T t t t t ,
11X 201,202, , 20n
L t l l l ,
11X 201, 202, , 20n
x t x x x ,
11 201, 202, , 20
U X n
G t g g g ,
n = 8.
Combining the three constrains of kg, rw, xq to U11X mentioned above, the solution satisfying X ∩ kg ∩ rw ∩ xq is:
11X 201, 202, , 20n
,
np
,
,
, ,
REFERENCES
11X 201, 202, , 20n
S t s s s
[1] K. Z. Guo and S. Y. Liu, “Introduction for the Theory of Error-Eliminating,” Advances in Modelling & Analysis A: Mathematical, General Mathematical Modelling, Vol. 39, No. 4, 2002, pp. 39-66.
11X 1, 2, , n 201, 202, , 20 ,
p p p
11X 201, 202, , 20n
T t t t t
11X 201, 202, , 20n
L t l l l [2] K. Z. Guo and S. Q. Zhang, “Theory of Error Sets,” Cen-
tral South University Press, Changsha, 2001.
11X 201, 202, , 20n
x t x x x [3] K. Z. Guo, “Error Logic,” Science Press, Beijing, 2008.
[4] K. Z. Guo and S. Y. Liu, “Fuzzy Error System with Change of Time and Space—The Effect of Change of Time and Space on Decision Making,” Advances in Modeling and Analysis B, Vol. 45, No. 3, 2002, pp. 49-60.
11 201, 202, , 20
U X n
G t g g g
n = 4.
5. Conclusions
[5] K. Z. Guo and S. Y. Liu, “Logical Relationship betweenFuzzy Error Logical Decomposition Words and a Pre-De- composed No Words,” Advances in Modelling & Analy- sis A: Mathematical, General Mathematical Modelling, Vol. 39, No. 4, 2002, pp. 1-15.
In order for a decision to be correct, we need to know the rule of how errors generate or convert. Each object in real world can be indicated as (u,x). Every element (u,x) is expressed as
1 2
, , , , , , ,
, , ,
n
U U
U S t T t L t
x t f u t G t G t
,
p p
[6] K. Z. Guo and S. Y. Liu, “Exploration and Application of Redundancy System in decision Making Relation be- tween Fuzzy Error Logic Increase Transformation Word and Connotative Model Implication Word,” Advances in Modelling & Analysis A: Mathematical, General Mathe- matical Modelling, Vol. 39, No.4, 2002, pp. 17-27.
from the standpoint of logic theory. An error matrix can be constructed when the seven parameters mentioned above in (u,x) are used as matrix columns. And three error matrixes X, A, B can create an error matrix equation
XAB. Matrix A is express current status and matrix B is expectant status, X is what we want to achieve. X al-lows us to find a method, in which bad status “A” changes into good status “B”.
[7] M. Li and K. Z. Guo, “Research on Decomposition of Fuzzy Error Set,” Advances in Modelling & Analysis A: Mathematical, General Mathematical Modelling, Vol. 43, No.2, 2006, pp. 15-26.
[8] X. L. Min, K. Z. Guo, “Knowledge Model Based on Error Logic for Intelligent System,” World Congress on Com- puter Science and Information Engineering, Los Angeles, 31 March-2 April 2009, pp. 216-220.
[9] X. L. Min and K. Z. Guo, “A Knowledge Discovery Meth- od Based on Error Matrix Equation,” The Sixth Interna- tional Conference on Fuzzy Systems and Knowledge Dis- covery, Tianjin, 14-16 August 2009 pp. 151-155.
doi:10.1109/FSKD.2009.104
Based on the error matrix equation, which is a kind of mathematical tool to be used to describe error itself and the transformation rules of errors, this paper proposes a new kind of error matrix equation named “containing- type error matrix equation”. The method of solving, ex-istence and form of solution for this type of equation have been presented in this paper. Our research provides a new method of decision making.