Digital Signal Processing Reference
In-Depth Information
cluster at the presentation of the
n
th training pattern during the
k
th training
session. At the beginning,
m
j
,
S
j
, and
n
j
are appropriately initialized. For
each pattern presentation
x
(
n
)
,
n
=
1
,
2
,
...
,M
, during the
k
th session, the
winner neuron is found:
x
=
x
.
N
(
k
)
(
)
min
j
n
w
(
k
)
c
w
(
k
)
j
(
n
)
−
(
n
)
(
n
)
−
(
n
)
(11.94)
=
1
In Equation 11.94,
N
(
k
)
(
is the number of output neurons, when the
n
th train-
ing vector is presented to the input of SOM during the
k
th training session.
For the sake of simplicity, let us first describe the test for detecting outliers in
the case of the first training session (
k
n
)
1), i.e., in case (i) of Algorithm 2.
70
,
71
=
More specifically, we decide that
x
is not an outlier and, therefore, can be
merged with the patterns already represented by the winner neuron if
z
(
k
)
c
(
n
)
d
T
S
(
k
)
c
)
−
1
d
(
n
−
1
)
−
p
(
n
−
1
≤ F
p,z
(
k
)
c
05
,
(11.95)
(
n
−
1
)
−
p
;0
.
p
F
where
05
denotes the upper 5% level of significance for the
F
-distribution
p,z
−
p
;0
.
with
p
and
z
−
p
degrees of freedom, and
d
is given by
1
x
)
.
1
m
(
k
)
c
d
=−
(
n
)
−
(
n
−
1
(11.96)
z
(
k
)
(
n
−
1
)
+
c
It should be noted that the test (Equation 11.95) cannot be applied when
z
(
k
)
c
(
is unconditionally merged with
the cluster of the training vectors represented by
w
(
k
)
c
n
−
1
)<
p
. Therefore, the input pattern
x
(
n
)
. The same analysis is
also applied to case ii outlined in Algorithm 2 ( i.e., when
c
(
k
)
(
(
n
)
)
=
c
(
k
−
1
)
(
)
n
n
,
with
c
(
k
)
(
)
denoting the index of the winner neuron at the presentation of
the
n
th pattern during
k
th session,
k
n
2).
If Equation 11.95 is satisfied, the winner vector is updated as SOM suggests:
6
≥
)
x
)
.
w
(
k
)
c
w
(
k
)
c
w
(
k
)
c
(
n
+
1
)
=
(
n
)
+
α(
k
(
n
)
−
(
n
(11.97)
Furthermore, the number of patterns, the sample mean, and the sample dis-
persion matrix of the cluster associated with either the winner
c
(
k
)
(
n
)
in case i
or the previous winner
c
(
k
−
1
)
(
in case ii are updated. Next, we consider
case iii of Algorithm 2. It refers to the remaining patterns of a cluster, which has
been modified due to a removal of another pattern. Because
c
(
k
)
(
n
)
c
(
k
−
1
)
(
n
)
=
n
)
,
for a moment we exclude the pattern
x
from the cluster of patterns that is
represented by the winner, and we verify whether its inclusion to this cluster
is still valid by applying a test similar to Equation 11.95. For the remaining
neurons, all the corresponding parameters are left intact. In the following, we
consider what happens when
x
(
n
)
is found to be an outlier. It is reasonable
then to examine whether the cluster represented by the winner neuron can
be split into two subclusters.
(
n
)
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