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We consider training one-dimensional open map of Kohonen with neurons in one-dimensional input space. We assume it is completed the phase of the device and the weights $w_i$, $i = 1,2, \dots, N$, are placed in ascending order. (As we know, this provision will not change in rest of the training.) At each step of the training algorithm, to which one corresponds educational model x, we define the intervals Vi as its regions entrance area with the following property: if x belongs to the interval $V_i$, then the neuron i emerges the winner. We denote by Vi the length of the interval $V_i$.

• What is the length $V_i$, $1 <i <N$, before the update?

During the update the rule $w_i ’= w_i + η (x - w_i)$ applies, where $η <1$. We denote by $c$ the winning neuron. What will be the length $V_c$ after the update, in the following cases of neighborhood definition? The length $V_c$ decreases or increases by the update?

  1. Only the winner $c$, $1<c<N$ is updated.
  2. The winner $c$, $1<c<N$, and the neighbor of $c-1$ are updated.
  3. The winner $c$, $1<c<N$, and the neighbors of $c-1$ and $c + 1$ are updated

The lengths $V$ will be calculated as a function of the values ​​of the weights $w$ and$ n$, $x$, as appropriate.As it turns out, the puzzle refers to nodes that have neighbors. The end nodes $1$ and $N$ are not considered.

KOHONEN MAP

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    $\begingroup$ Can you credit the source of all copied material? $\endgroup$ – Steven Feb 2 at 12:56
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"$If\ x \in Vi,\space then\space\space i = winner\space neuron$"

$V_i = \frac {w_{i+1} - w_i}{2} + \frac {w_i - w_{i-1}}{2} \Rightarrow$

$V_i = \frac{w_{i+1} - w_{i-1}}{2} \space\space\space (1)$


The update equation is:

$W_{i(new)} = W_{i(old)} + η(x-W_{i(old)}) $$\space\space\space(2)$ , where $η$ = learning rate

From now on I will put a ' instead of "new" keyword

$\space\space\space\mathbf1.$ If only the winner's weights are updated then:

The length of $V_c$ remains the same

$\space\space\space\mathbf2.$ The winner $c$, $1<c<N$, and the neighbor of $c-1$ are updated:

$V_c' = \frac{w_{i+1}' - w_{i-1}'}{2} \Rightarrow$ $V_c' = \frac{(w_{i+1} +η(x-w_{i+1}) - w_{i-1})}{2}$, because of (2)

We know that $x<w_i+1$ because $x$ is closer to $w_i$, as $i$ is the winner neuron. So the length $V_c$ is getting smaller

$\space\space\space\mathbf3.$ The winner $c$, $1<c<N$, and the neighbors of $c-1$ and $c + 1$ are updated:

Similar to the previous case, we will use relation (1) combined with (2):

$V_c' = \frac{w_{i+1}' - w_{i-1}'}{2} \Rightarrow$

$V_c' = \frac {W_{i+1} + η(x-W_{i+1}) - W_{i-1} - η(x-W_{i-1})}{2}$

We know:

$w_{i-1} < x < w_{i+1}$ so

$\bullet w_{i-1}' > w_{i-1} $ and $\bullet w_{i+1}' > w_{i+1}$ $ \Rightarrow $

$w_{i+1}' - w_{i-1}' < w_{i+1} - w_{i-1}$

Thus, length $V_c$ is getting smaller

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  • $\begingroup$ Thanks for the answer! Can I ask you please to use MathJax to type all equations? It's a bit hard to read in its current form. $\endgroup$ – Dmitry Mar 1 at 20:59
  • $\begingroup$ You are right, i will try to fix this $\endgroup$ – Ermolai Mar 3 at 11:31

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