1
Midsem Examination Artificial Neuro-Fuzzy Theory AT74.05 March 7, 2008
Time: 10:00-12:00 h.
Open Book
Marks: 100
Attempt all questions.
Q.1 The relation between input, x, and output, y, of a system from an experiment is recorded as
x -1 0 1
y
5 1 3
Apply LMS algorithm to determine the parameters when the relation is modeled by
(a)
y = a
0
(b)
y = a
0
+a
1
x
(c)
y = a
0
+a
1
x+a
2
x
2
Assume that all the data are presented equally. Determine summation of squared errors from each model.
(30)
Solution
Rx
x
h
x
c
x
zz
E
x
tz
E
x
t
E
x
F
T
T
T
T
T
+
−
=
+
−
=
2
)
(
)
(
2
)
(
)
(
2
(1)
[
]
[ ]
35
3
1
3
1
5
3
1
2
2
2
=
+
+
=
c
(2)
In model (a),
[ ] [ ] [ ]
[
]
[ ] [ ]
3
9
3
1
1
3
1
1
1
5
3
1
=
=
+
+
=
h
(3)
[ ][ ] [ ][ ] [ ][ ]
[
]
[ ] [ ]
1
3
3
1
1
1
1
1
1
1
3
1
=
=
+
+
=
R
(4)
[ ] [ ] [ ]
3
3
1
1
1
0
=
=
=
−
−
h
R
a
(5)
3
=
y
(6)
∑
∑
=
−
+
−
+
−
=
−
=
8
)
3
3
(
)
3
1
(
)
3
5
(
)
(
2
2
2
2
2
a
t
e
(7)
In model (b),
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
+
⎥
⎦
⎤
⎢
⎣
⎡
+
⎥
⎦
⎤
⎢
⎣
⎡−
=
9
2
3
1
1
1
3
1
0
1
1
1
5
3
1
h
(8)
2
[
]
[
]
[ ]
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
+
⎥
⎦
⎤
⎢
⎣
⎡
+
−
⎥
⎦
⎤
⎢
⎣
⎡−
=
3
0
0
2
3
1
1
1
1
1
1
0
1
0
1
1
1
1
3
1
R
(9)
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎦
⎤
⎢
⎣
⎡−
⎥
⎦
⎤
⎢
⎣
⎡
=
=
⎥
⎦
⎤
⎢
⎣
⎡
−
−
3
1
9
2
3
0
0
2
1
1
0
1
h
R
a
a
(10)
x
y
−
= 3
(11)
6
)
2
3
(
)
3
1
(
)
4
5
(
)
(
2
2
2
2
2
∑
∑
=
−
+
−
+
−
=
−
=
a
t
e
(12)
In model (c),
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
9
2
8
3
1
1
1
1
3
1
0
0
1
1
1
1
5
3
1
h
(13)
[
]
[
]
[
]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
3
0
2
0
2
0
2
0
2
3
1
1
1
1
1
1
1
1
0
0
1
0
0
1
1
1
1
1
1
3
1
R
(14)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
1
1
3
9
2
8
3
0
2
0
2
0
2
0
2
1
1
0
1
2
h
R
a
a
a
(15)
2
3
1
x
x
y
+
−
=
(16)
0
)
3
3
(
)
1
1
(
)
5
5
(
)
(
2
2
2
2
2
∑
∑
=
−
+
−
+
−
=
−
=
a
t
e
(17)
3
Q.2
A 1-2-1 network is used to model the relation between input and output of the system in Q.1.
Train the network by Momentum Backpropagation (MOBP) algorithm by batching mode for 1 round.
Apply sum squared error as the performance index function. Use the following initial weights and biases
0
,
3
.
0
,
2
.
0
,
1
.
0
,
3
.
0
,
2
.
0
,
1
.
0
2
1
2
12
2
11
1
2
1
1
1
21
1
11
=
=
−
=
=
−
=
=
−
=
b
w
w
b
b
w
w
Assume that log-sigmoid function is applied in the first layer and linear function is applied in the second
layer and
α = 1, γ = 0.5.
(30)
Solution
Determine derivative of transfer function,
1
,
),
)(
1
(
,
)
1
(
2
2
1
1
1
1
1
=
=
−
=
+
=
−
−
f
n
f
a
a
f
e
f
n
&
&
(1)
Present (-1, 5),
[ ]
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
+
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
+
=
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎥
⎦
⎤
⎢
⎣
⎡−
+
−
⎥
⎦
⎤
⎢
⎣
⎡−
=
+
=
−
=
=
−
−
−
−
−
−
4750
.
0
4502
.
0
)
1
(
)
1
(
)
1
(
)
1
(
,
1
.
0
2
.
0
1
.
0
3
.
0
1
2
.
0
1
.
0
,
1
1
1
.
0
1
2
.
0
1
1
1
1
1
1
1
1
2
1
1
e
e
e
e
a
b
p
w
n
p
p
n
n
(2)
[
]
9475
.
4
0525
.
0
5
,
0525
.
0
,
0525
.
0
0
4750
.
0
4502
.
0
3
.
0
2
.
0
1
2
1
2
1
2
1
2
2
1
=
−
=
−
=
=
=
=
=
=
+
⎥
⎦
⎤
⎢
⎣
⎡
−
=
+
=
a
t
e
e
n
a
a
b
a
w
n
(3)
⎥
⎦
⎤
⎢
⎣
⎡
−
=
−
⎥
⎦
⎤
⎢
⎣
⎡−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
=
−
=
−
=
0748
.
0
0495
.
0
]
1
[
3
.
0
2
.
0
4750
.
0
)
4750
.
0
1
(
0
0
4502
.
0
)
4502
.
0
1
(
~
)
(
~
,
1
~
2
2
1
1
2
2
S
w
f
S
f
S
T
&
&
(4)
Σ
purelin
x
1
11
w
2
11
w
1
2
b
2
1
b
y
1
Σ
logsig
1
1
b
1
1
21
w
2
12
w
Σ
logsig
1
4
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
−
−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
Δ
Δ
Δ
Δ
Δ
Δ
Δ
5000
.
0
2375
.
0
2251
.
0
0374
.
0
0248
.
0
0374
.
0
0248
.
0
1
)
4750
.
0
(
1
)
4502
.
0
(
1
0748
.
0
0495
.
0
)
1
(
0748
.
0
)
1
(
0495
.
0
)
1
(
5
.
0
0
0
0
0
0
0
0
5
.
0
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
2
1
1
2
2
1
1
1
2
1
1
2
1
1
1
1
2
1
1
1
2
1
2
12
2
11
1
2
1
1
1
21
1
11
2
1
2
12
2
11
1
2
1
1
1
21
1
11
s
a
s
a
s
s
s
p
s
p
s
b
w
w
b
b
w
w
b
w
w
b
b
w
w
α
γ
γ
(5)
Present (0, 1),
[ ]
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
+
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
+
=
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎦
⎤
⎢
⎣
⎡−
+
⎥
⎦
⎤
⎢
⎣
⎡−
=
+
=
=
=
−
−
−
−
−
−
−
5250
.
0
4256
.
0
)
1
(
)
1
(
)
1
(
)
1
(
,
1
.
0
3
.
0
1
.
0
3
.
0
0
2
.
0
1
.
0
,
0
1
1
.
0
1
3
.
0
1
1
1
1
1
1
2
1
2
1
1
e
e
e
e
a
b
p
w
n
p
p
n
n
(6)
[
]
9276
.
0
0724
.
0
1
,
0724
.
0
,
0724
.
0
0
5250
.
0
4256
.
0
3
.
0
2
.
0
1
2
1
2
1
2
1
2
2
1
=
−
=
−
=
=
=
=
=
=
+
⎥
⎦
⎤
⎢
⎣
⎡
−
=
+
=
a
t
e
e
n
a
a
b
a
w
n
(7)
⎥
⎦
⎤
⎢
⎣
⎡
−
=
−
⎥
⎦
⎤
⎢
⎣
⎡−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
=
−
=
−
=
0748
.
0
0489
.
0
]
1
[
3
.
0
2
.
0
5250
.
0
)
5250
.
0
1
(
0
0
4256
.
0
)
4256
.
0
1
(
~
)
(
~
,
1
~
2
2
1
1
2
2
S
w
f
S
f
S
T
&
&
(8)
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
Δ
Δ
Δ
Δ
Δ
Δ
Δ
5000
.
0
2625
.
0
2128
.
0
0374
.
0
0245
.
0
0
0
1
)
5250
.
0
(
1
)
4256
.
0
(
1
0748
.
0
0489
.
0
)
0
(
0748
.
0
)
0
(
0489
.
0
)
1
(
5
.
0
0
0
0
0
0
0
0
5
.
0
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
2
1
1
2
2
1
1
1
2
1
1
2
1
1
1
1
2
1
1
1
2
1
2
12
2
11
1
2
1
1
1
21
1
11
2
1
2
12
2
11
1
2
1
1
1
21
1
11
s
a
s
a
s
s
s
p
s
p
s
b
w
w
b
b
w
w
b
w
w
b
b
w
w
α
γ
γ
(9)
Present (1, 3),
[ ]
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
+
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
+
=
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎦
⎤
⎢
⎣
⎡−
+
⎥
⎦
⎤
⎢
⎣
⎡−
=
+
=
=
=
−
−
−
−
−
−
−
5744
.
0
4013
.
0
)
1
(
)
1
(
)
1
(
)
1
(
,
3
.
0
4
.
0
1
.
0
3
.
0
1
2
.
0
1
.
0
,
1
1
3
.
0
1
4
.
0
1
1
1
1
1
1
3
1
2
1
1
e
e
e
e
a
b
p
w
n
p
p
n
n
(10)
[
]
9079
.
2
0921
.
0
3
,
0921
.
0
,
0921
.
0
0
5744
.
0
4013
.
0
3
.
0
2
.
0
1
2
1
2
1
2
1
2
2
1
=
−
=
−
=
=
=
=
=
=
+
⎥
⎦
⎤
⎢
⎣
⎡
−
=
+
=
a
t
e
e
n
a
a
b
a
w
n
(11)
⎥
⎦
⎤
⎢
⎣
⎡
−
=
−
⎥
⎦
⎤
⎢
⎣
⎡−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
=
−
=
−
=
0733
.
0
0481
.
0
]
1
[
3
.
0
2
.
0
5744
.
0
)
5744
.
0
1
(
0
0
4013
.
0
)
4013
.
0
1
(
~
)
(
~
,
1
~
2
2
1
1
2
2
S
w
f
S
f
S
T
&
&
(12)
5
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
−
Δ
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
Δ
Δ
Δ
Δ
Δ
Δ
Δ
5000
.
0
2872
.
0
2007
.
0
0367
.
0
0240
.
0
0367
.
0
0240
.
0
1
)
5744
.
0
(
1
)
4013
.
0
(
1
0733
.
0
0481
.
0
)
1
(
0733
.
0
)
1
(
0481
.
0
)
1
(
5
.
0
0
0
0
0
0
0
0
5
.
0
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
2
1
1
2
2
1
1
1
2
1
1
2
1
1
1
1
2
1
1
1
2
1
2
12
2
11
1
2
1
1
1
21
1
11
2
1
2
12
2
11
1
2
1
1
1
21
1
11
s
a
s
a
s
s
s
p
s
p
s
b
w
w
b
b
w
w
b
w
w
b
b
w
w
α
γ
γ
(13)
Determine summation of error square,
7941
.
33
9079
.
2
9276
.
0
9475
.
4
)
(
2
2
2
2
=
+
+
=
=
∑
e
x
F
(14)
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
+
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
+
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
+
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
Δ
Δ
Δ
Δ
Δ
Δ
Δ
+
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
∑
∑
∑
∑
∑
∑
∑
5000
.
1
0872
.
1
4386
.
0
2115
.
0
3773
.
0
1993
.
0
0992
.
0
5000
.
0
2872
.
0
2007
.
0
0367
.
0
0240
.
0
0367
.
0
0240
.
0
5000
.
0
2625
.
0
2128
.
0
0374
.
0
0245
.
0
0
0
5000
.
0
2375
.
0
2251
.
0
0374
.
0
0248
.
0
0374
.
0
0248
.
0
0
3
.
0
2
.
0
1
.
0
3
.
0
2
.
0
1
.
0
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
2
1
2
12
2
11
1
2
1
1
1
21
1
11
2
1
2
12
2
11
1
2
1
1
1
21
1
11
2
1
2
12
2
11
1
2
1
1
1
21
1
11
b
w
w
b
b
w
w
b
w
w
b
b
w
w
b
w
w
b
b
w
w
(15)
6
Q.3
Linear associator network with Pseudoinverse rule is used to perform as a fuzzy controller for a
mobile robot. 3 sensors, used to detect the existence of obstacles, are equipped at the left, middle and
right of the robot. The reading from the sensor varies between -1 to 1; the reading indicates 1 if there
exists definitely the obstacle, -1 if there is no obstacle. The output from the network is the commands
sending to 2 motors at the left and right wheels. The motor command of 1 means move forward at full
speed, the command of 0 means no move, while the command of -1 means move backward at full speed.
Assume that the following rules are desired.
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
motor
right
motor
left
,
sensor
right
sensor
middle
sensor
left
t
p
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
=
0
0
,
1
1
1
,
1
0
,
1
1
1
,
0
1
,
1
1
1
3
3
2
2
1
1
t
p
t
p
t
p
Find the weight from Pseudoinverse rule. Determine the output when the sensor reading indicates
(a)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−1
1
1
, (b)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡−
1
1
1
, (c)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
1
1
1
, (d)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
1
1
1
, and (e)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
1
1
1
.
(20)
Solution
By Pseudoinverse rule,
W = TP
+
(1)
Where
P
+
= (P
T
P)
-1
P
T
(2)
and
⎥
⎦
⎤
⎢
⎣
⎡
=
0
1
0
0
0
1
T
,
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
=
1
1
1
1
1
1
1
1
1
P
(3)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
=
3
1
1
1
3
1
1
1
3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
P
P
T
(4)
7
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
=
−
−
5
.
0
25
.
0
25
.
0
25
.
0
5
.
0
25
.
0
25
.
0
25
.
0
5
.
0
3
1
1
1
3
1
1
1
3
)
(
1
1
P
P
T
(5)
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
−
5
.
0
0
5
.
0
0
5
.
0
5
.
0
5
.
0
5
.
0
0
1
1
1
1
1
1
1
1
1
5
.
0
25
.
0
25
.
0
25
.
0
5
.
0
25
.
0
25
.
0
25
.
0
5
.
0
)
(
1
T
T
P
P
P
(6)
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
−
⎥
⎦
⎤
⎢
⎣
⎡
=
=
+
0
5
.
0
5
.
0
5
.
0
5
.
0
0
5
.
0
0
5
.
0
0
5
.
0
5
.
0
5
.
0
5
.
0
0
0
1
0
0
0
1
TP
W
(7)
(a)
when the input is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡−
1
1
1
,
⎥
⎦
⎤
⎢
⎣
⎡
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
=
1
0
1
1
1
0
5
.
0
5
.
0
5
.
0
5
.
0
0
a
(8)
(b)
when the input is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−1
1
1
,
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
=
0
1
1
1
1
0
5
.
0
5
.
0
5
.
0
5
.
0
0
a
(9)
(c)
when the input is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
1
1
1
,
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
=
0
0
1
1
1
0
5
.
0
5
.
0
5
.
0
5
.
0
0
a
(10)
(d)
when the input is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
1
1
1
,
8
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
=
1
1
1
1
1
0
5
.
0
5
.
0
5
.
0
5
.
0
0
a
(11)
(e)
when the input is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
1
1
1
,
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
=
1
1
1
1
1
0
5
.
0
5
.
0
5
.
0
5
.
0
0
a
(12)
9
Q.4
Design an LVQ network which can recognize classes A and B based on the vector spaces shown
below.
(20)
Solution
LVQ network can be used.
In the first layer of LVQ,
⎥
⎦
⎤
⎢
⎣
⎡
=
10
5
1
w
(1)
⎥
⎦
⎤
⎢
⎣
⎡
=
5
10
2
w
(2)
⎥
⎦
⎤
⎢
⎣
⎡
=
10
15
3
w
(3)
⎥
⎦
⎤
⎢
⎣
⎡
=
5
20
4
w
(4)
The weight matrix of the first layer is thus
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
=
5
20
10
15
5
10
10
5
1
W
(5)
0 5 10 15 20 25 x
y
15
10
5
0
A
A
B
B
10
The weight matrix of the second layer combines and recognizes subclasses 1 and 4 as class A (1),
subclasses 2 and 3 as class B (2) respectively.
⎥
⎦
⎤
⎢
⎣
⎡
=
0
1
1
0
1
0
0
1
2
W
(6)
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