1
and Y
1
are one exemplar pair, the third component of Y
1
is what we need as the third component of
the output in the current cycle of operation; therefore, we fed X
1
and received Y
1
. If we feed Y
1
at the other
end (B field) , the activations in the A field will be (2, –2, –2, 2), and the output vector will be (1, 0, 0, 1),
which is X
1
.
With A field input X
2
you get B field activations (4, −4, 0), giving the output vector as (1, 0, 1), which is Y
2
.
Thus X
2
and Y
2
are heteroassociated with each other.
Let us modify our X
1
to be (1, 0, 1, 1). Then the weight matrix W becomes
1 [−1 1 1] −1 [1 −1 1] −2 2 0
W = −1 + 1 = 2 −2 0
1 1 0 0 2
1 −1 −2 2 0
and
−2 2 0 −2
W
T
= 2 −2 0 2
0 0 2 0
Now this is a different set of two exemplar vector pairs. The pairs are X
Do'stlaringiz bilan baham: |