Next, we continue the process for the second iteration as follows:
Y
Y
Y
Y
f
1
2
3
4
040404
404040
=
−−−
−−−
0004040
404040
1
1
1
1
1
1
−−
−−−
×
−
−
=
−
f
4
112
4
12
1
1
1
1−
=
−
−
X
X
X
X
X
X
f
1
2
3
4
5
6
1111
=−−
×
−−−
−−−
−−
−−−
040404
404040
004040
404040
×
−
−
1
1
1
1
1
1
=
−
−
−
f
8
8
8
8
8
8
=
−
−
−
1
1
1
1
1
1
At this point, we see that the network converges since there are no state
changes. Since the result is the same as the first training vector (1, –1, 1, –1,
1, –1), the desired output becomes (1, –1, 1, –1), which can be either defined
or interpreted as ...
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