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Bio-medical Ontologies Maintenance and Change Management

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194 I.R. Godínez et al.<br />

3.1.1 New Alpha-Beta Heteroassociative Memory Type Max<br />

LEARNING PHASE<br />

Let A = {0, 1} ,n,p ∈ Z + ,μ ∈{1, 2, ...p} ,i ∈{1, 2, ..., p} <strong>and</strong> j ∈{1, 2, ...n}<br />

<strong>and</strong> let x ∈ An <strong>and</strong> y ∈ Ap be an input <strong>and</strong> output vectors, respectively. The<br />

corresponding fundamental set is denoted by {(xμ , yμ ) | μ =1, 2, ..., p}.<br />

The fundamental set must be built according to the following rules. First,<br />

all the y vectors are built with the one-hot codification. Second, to each yμ vector corresponds one <strong>and</strong> only one xμ vector, this is, there is only one<br />

couple (xμ , yμ ) in the fundamental set.<br />

STEP 1:<br />

For each μ ∈ {1, 2, ...p}, from the couple (xμ , yμ ) build the matrix:<br />

[yμ ⊠ (xμ ) t ] mxn<br />

STEP 2:<br />

Apply the binary ∨ operator to the matrices obtained in step 1 to get<br />

the new alpha-beta heteroassociative memory type max V as follow: V =<br />

p�<br />

[yμ ⊠ (xμ ) t ]withtheij−th component given by:<br />

μ=1<br />

vij =<br />

p�<br />

μ=1<br />

α(y μ<br />

i ,xμ<br />

j )<br />

RECALLING PHASE<br />

STEP 1:<br />

A pattern x ω is presented to V, the Δβ operation is done <strong>and</strong> the resulting<br />

vector is assigned to a vector called z ω : z ω = VΔβx ω<br />

The i−th component of the resulting column vector is:<br />

z ω i =<br />

n�<br />

j=1<br />

β(vij,x ω j )<br />

STEP 2:<br />

It is necessary to build the max sum vector s according to the definition<br />

1, therefore the corresponding yω is given as:<br />

�<br />

1ifsi = �<br />

y ω i =<br />

=1} .<br />

0 otherwise<br />

sk AND z<br />

k∈θ<br />

ω i =1<br />

where θ = {i|zω i<br />

Example 1. Let x1 ,x2 ,x3 ,x4 be the input patterns<br />

x 1 ⎛ ⎞<br />

1<br />

⎜ 0 ⎟<br />

= ⎜ 1 ⎟<br />

⎝ 1 ⎠<br />

0<br />

,x2 ⎛ ⎞<br />

0<br />

⎜ 1 ⎟<br />

= ⎜ 0 ⎟<br />

⎝ 0 ⎠<br />

0<br />

,x3 ⎛ ⎞<br />

1<br />

⎜ 0 ⎟<br />

= ⎜ 0 ⎟<br />

⎝ 1 ⎠<br />

0<br />

,x4 ⎛ ⎞<br />

1<br />

⎜ 1 ⎟<br />

= ⎜ 0 ⎟<br />

⎝ 1 ⎠<br />

1<br />

(1)

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