Information Technology Reference
In-Depth Information
•
μ
Q
1
, obtained by using CRI
with r1, the output
•
μ
Q
2
, obtained by using CRI.
Provided the firing of the rules corresponds to 'fire r1
or
fire r2', then the final
output is given by
with r2, the output
μ
Q
∗
=
max
(μ
Q
1
,μ
Q
2
),
and analogously for more than two rules. For example, in the case of p rules
r
1
,...,
r
p
, the result will be
μ
Q
∗
=
max
(μ
Q
1
,...,μ
Q
p
),
where
μ
Q
i
(
1
i
p
)
is the output obtained with the rule
r
i
and the input
μ
P
∗
.
Example 3.4.6
With
X
=[
0
,
1
]
,
Y
=[
0
,
10
]
, consider the rules
•
If
x
is
big
, then
y
=
2
•
If
x
is
small
, then
y
=
8
•
If
x
is
around
0
.
5, then
y
=
6,
and the input
x
0
=
0
.
4. Which is the final output of this system if the rules are
represented by
J
(
a
,
b
)
=
a
·
b
?
0
.
4
,
if
y
=
2
•
μ
Q
1
(
y
)
=
J
(μ
B
(
0
.
4
), μ
{
2
}
(
y
))
=
0
.
4
.μ
{
2
}
(
y
)
=
0
,
if
y
=
2
0
.
,
=
6
if
y
8
•
μ
Q
2
(
y
)
=
J
(μ
s
(
0
.
4
), μ
{
8
}
(
y
))
=
0
.
6
.μ
{
2
}
(
y
)
=
,
=
0
if
y
8
•
μ
Q
3
(
y
)
=
J
(μ
A
0
.
5
(
0
.
4
), μ
{
6
}
(
y
))
=
μ
A
0
.
5
(
0
.
4
).μ
{
6
}
(
y
)
μ
A
0
.
5
(
0
.
4
),
if
y
=
6
=
0
,
if
y
=
6
,
with
μ
B
(
x
)
=
x
, and
μ
S
(
x
)
=
1
−
x
. Taking as
μ
A
0
.
5
the triangular function
.
3. Hence,
x
−
0
.
35
and since the left side equation is
y
=
,itis
μ
A
0
.
5
(
0
.
4
)
=
0
.
0
.
15
0
3
.
,
if
y
=
6
μ
Q
3
(
y
)
=
0
,
if
y
=
6
,
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