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Assume furthermore that the probability distribution for facilitatory and
inhibitory contacts are not alike, and let k be a constant for either kind. The
action function
is now:
(
)
=
()
=
Kr r
,
K
a
D
2
12
(
)
-
(
)
2
2
2
2
exp
-
D
2
h
a
exp
-
D
2
h
.
1
1
2
2
The stimulus-response relation is
()
=
Ú
()
()
r
r
K r
D
2
s
d.
A
2
1
1
L
1
For a uniform stimulus distribution
=
()
=
ss
,
o
(
)
r
r
constant
=
2
ps
a h
1
2
-
a h
2
2
,
2
o
1
and if
a
1
h
1
=
a
2
h
2
,
then
r=0.
As one may recall, the interesting feature of giving zero-response to finite
stimuli was obtained earlier for discrete action functions of the binomial
form. A similar result for the Gaussian distribution should therefore not be
surprising, if one realizes that the continuous Gaussian distribution emerges
from a limit operation on the binomial distribution. What are the abstract-
ing properties of this net with a random normal fiber distribution?
Fig. 31 represents the response activity for a stimulus in the form of a
uniformly illuminated square. Clearly, this network operates as a computer
of contours. It may be noted that the structure of this useful network arose
from random perturbation. However, there was an original growth pro-
gram, namely, to grow in parallel bundles. Genetically, this is not too dif-
ficult to achieve; it says: “Repeat”. However, such a net is of little value, as
we have just seen. It is only when noise is introduced into the program that
the net acquires its useful properties. It may be mentioned that this net
works best when the zero-response condition is fulfilled. This, however,
requires adaptation.
(iii) Mapping into a Perendicular Plane
The previous two examples considered action functions with spherical sym-
metry. We shall now explore the properties of an action function of type
K
S
with lateral symmetry. Such an action function may arise in the following
way (see fig. 32):
Assume again two surfaces layers,
L
p
,
L
q
where in
L
p
all neurons are
aligned in parallel with their axis of symmetry perpendicular to the layer's
surface, while in
L
q
they are also in parallel, but with their axis of symme-
try lying in the surface of
L
q
.
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