Practical Considerations
When t, the supp x~ , then there will be a value of t when
supp x~  folds, it becomes multi-w-to-one-x mapping
Practical Considerations
Then the maximum of all candidate membership value of
w is the membership of x.
Practical Considerations
If supp x~ occupies [-1,1],  x x   1  x  [-1,1] in the state
of complete fuzziness.
~
Fuzzy Numbers
We define a normal, convex fuzzy set on a real line to
be a fuzzy number I
~
Let I~ and J~ be fuzzy numbers
I is a real line in universe Y
~
* Is a set of arithmetic operations ,,,
Z= I * J
~
~
 I *J Z  
~ ~
  x     y 

I
J
~
~


X *Y  Z
Fuzzy Numbers
Example:
1  0.2 / 0  1 / 1  0.2 / 2
~
1 1  0.2 / 0  1 / 1  0.2 / 2   0.2 / 0  1 / 1  0.2 / 2   2
~
~
 min 0.2,0.2  max min 0.2,1, min 1,0.2  



0
1


 max min 0.2,0.2 , min 1,1, min 0.2,0.2  

 
2


 max min 0.2,1, min 1,0.2  min 0.2,0.2  



3
4


 0.2 / 0  0.2 / 1  1 / 2  0.2 / 3  0.2 / 4
~
Fuzzy Numbers


supp I  x |  I x   0
~
~
supp (z) = supp I~ * supp J~
I* J
~
~
=I*J
(crisp intervals)!
I * J   I
~
~ 

* J
They are intervals!
Interval analysis in arithmetic
Fuzzy Numbers
I1 = [a,b] a < b
I2 = [c,d] c < d
I1 * I2 = [a,b] * [c,d]
[a b] + [c d] = [a+c b+d]
[a b] – [c d] = [a-d b-c]
[a b]  [c d] = [min(ac,ad,bc,bd) max(ac,ad,bc,bd)]
[a b] ÷ [c d] = [a b]  [1/d 1/c]
[a b] =
I(J+K)  I  J + I  K
Note: A A  0
~
~
0  [c,d]
[a b]
>0
[b a]
>0
Fuzzy Numbers
-3[1,2] = [-6,-3]
[0,1] – [0,1] = [-1,1]
[1,3]  [2,4] = [min(1.2,1.4,3.2,3.4)
max(1.2,1.4,3.2,3.4)]
=[2,1.2]
[1 2] ÷ [1 2] = [1 2]  [1/2 1] = [1/2 2]
If I = [1,2]
J = [2,3]
K = [1,4]
I  (J-K) = [1,2]  [-2,2] = [-4,4]
IJ – IK = [1,2][2,3] – [1,2][1,4] = [2,6] – [1,8] = [-6,5]
[-4,4]  [-6,5]
Approximate Methods of Extension
When discretization of continuous-valued function, it may
have irregular and error membership values, which will be
propagated from input to output by extension principle.
To overcome the above problem, several methods are
studied.
Approximate Methods of Extension
Vertex Method
Combining the -cut and standard interval analysis.

For B~  f A~
We can decompose A into a series of -cut and standard
interval I. If f(x) is continuous and monotonic on I = [a,b]
the interval representing B at a particular .
~
B = f(I) = [min(f(a),f(b)) max(f(a),f(b))]
Approximate Methods of Extension
If y = f(x1,x2,…,xn)
Each input variable can be described by an interval Ii
Ii = [ai bi]
i = 1,2,…,n
Approximate Methods of Extension
As seen in the fig. above, the endpoint pairs of each interval
intersect in the 3D space and form the vertices (corners) of
the Cartesian space. The coordinates of these vertices are
the values used in the vertex method when determining the
output interval for each -cut. The number of vertices, N, is a
quantity equal to N = 2n, where n is the number of fuzzy input
variables. When the mapping y = f(x1,x2,…,xn) is continuous
in the n-dimensional Cartesian region
Approximate Methods of Extension
B  f I1 , I 2  ,..., I n 
 min  f c j , max  f c j 
 j

j  1,2,..., N
If there are extreme points
B  min  f c j , f Ek , max  f c j , f Ek 
 j ,k

j ,k
where j = 1,2,…,N and k = 1,2,…,m for m extreme points
in the region.
Approximate Methods of Extension
Example:
We wish to determine the fuzziness in the output of a
simple nonlinear mapping given by the expression, y =
f(x) = x(2 – x), seen in the fig 6.7a, where the fuzzy input
variable, x, has the membership function shown in fig
6.7b.
Approximate Methods of Extension
We shall solve this problem using the fuzzy vertex method
at three -cut levels, for  = 0+,0.5,1. As seen in fig 6.7b,
the intervals corresponding to these -cuts are I0 = [0.5,2],
I0.5 = [0.75,1.5], I1 = [1,1] (a single point).
I 0  0.5,2
c1  0.5, c 2  2, E1  1
f c1  0.52  0.5  0.75
f c 2  22  2  0
f E1  12  1  1
B0  min 0.75,0.1, max 0.75,0,1  0,1
Approximate Methods of Extension
I 0.5  0.75,1.5
c1  0.75, c 2  1.5, E1  1
f c1  0.752  0.75  0.9375
f c 2  1.52  1.5  0.75
B0.5  min 0.9375,0.75,1, max 0.9375,0.75,1
 0.75,1
I1  1,1
c1  1  c 2  E1
f c1  f c 2  f E   1  (2  1)  1
DSW Algorithm
1. Select a , 0 <  < 1
2. Find the interval(s) in the input membership function(s)
corresponding to .
3. Using standard binary interval operations, compute the
interval for the output membership function for the
selected -cut.
4. Repeat 1 – 3 for different values of 
Example:
DSW Algorithm
y  2x  x2
I 0  0.5,2


B0  20.5,2  0.52 ,2 2  1.25,8
I 0.5  0.75,1.5


B0.5  20.75,1.5  0.752 ,1.52  1.5,3  0.5625,2.25
 2.0625,5.25
I1  1,1


B1  21,1  12 ,12  2,2  1,1  3,3  3
DSW Algorithm
Example 2:
Suppose the domain of the input variable x is changed to
include negative numbers… the computations for each cut will be as follows:
I 0   0.5,1
 
B0  2 0.5,1  0,12   1,2  0,1   1,3
Note: The zero marked with the arrow is taken as the
minimum, since (-0.5)2 > 0; because zero is contained in
the interval [-0.5,1] the minimum of squares of any
number in the interval will be zero.
DSW Algorithm
I 0.5   0.25,0.5


B0.5  2 0.25,0.5  0,0.52   0.5,1  0,0.25   0.5,1.25
I1  0,0
B1  20,0  0,0  0,0
Restricted DSW Algorithm
I = [a,b]
J = [c,d]
a,b,c,d > 0
No Subtraction
Then, I  J = [a,b]  [c,d] = [a  c,b  d]
I/J = [a,b] ÷ [c,d] = [a/d,b/c]
Comparisons
Comparisons
Comparisons
Comparisons
Comparisons
Comparisons
Comparisons
Comparisons
Fuzzy Vectors
Formally, a vector, a  a1 , a2 ,..., an , is called a fuzzy vector
~
if for any element we have 0 < aI < 1 for I = 1,2,…,n.
T
Similarly, the transpose of a fuzzy vector a ,denoted by, a is
~
~
a column vector if is a row vector, i.e.,
a1 
a 
T
a   2
~
 
 
 an 
Fuzzy Vectors
Let us define a and b as fuzzy vectors of length n, and
~
~
define
n
a b   ai  bi 
T
~
~
i 1
as the fuzzy inner product of the two fuzzy vectors and
n
a b   ai  bi 
T
~
~
i 1
as the fuzzy outer product of the two vectors
Fuzzy Vectors
Example:
Find the inner and outer product of the given
fuzzy vectors of length 4
a  0.3,0.7,1,0.4
~
b  0.5,0.9,0.3,0.1
~
 0.5 
 
 0.9 
a b  0.3,0.7,1,0.4 
~ ~
0.3
 
 0.1 
 
 .3  .5  .7  .9  1  .3  .4  .1  .3  .7  .3  .1  0.7
a b  .3  .5  .7  .9  1  .3  .4  .1  .5  .9  1  .4  0.4
~
~
Note: outer product is different from the one in inner algebra.