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SOM - Self Organized Map, every input dimension maps to all output nodes, nodes compete with each other for scoring - vector quantization. PCA and other clustering methods can be seen as simplified special cases of this process.

There is only ever a single winning node in a SOM. However, what happens when an input strongly resembles two established 'clusters'? Could it so happen that the first neuron wins over a second neuron by a small margin and yet the two are very far apart? If so, would it not also be extremely useful information?

If so, then it means the entire activation pattern with all its various outputs would be useful in classifying an input.

The reason I'm asking is because I'm considering plugging SOMs into other neural networks and then maybe back again into SOMs. And when plugging in, I wish to know if it would be safe to just carry over the entire lattice with all its outputs instead of just the winning node.

I have tried checking the math of the SOM, when training it only considers the winning neuron, but nothing seems to indicate that if a new input is used, only the winning node is of importance to the operator.

John Stroni
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The goal of the algorithm at the end of training is to have the first and second winning nodes of each input pattern in adjacent positions in the lattice. This is referred as Topology Preservation of the input data space. The inverse case is considered as bad training and is calculated by the topological error. One simple measure of this error is the ratio of input vectors for which the first and second winning nodes are not adjacent.

Search for SOM and topology preservation. Here is a quick link .

Keep in mind that small maps generally produce a smaller topological error but increased quantization error where larger maps tend to inverse this situation. So there is a trade of between topology preservation and quantization accuracy. There isn't a golden rule for this. It always depends on the domain, the application and the expected results.

pater
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  • Thanks for that paper. Yes I see how the BMUs must be adjacent. But the overall question wasn't answered I think, does the overall pattern of activation strength hold no meaning? Are the BMUs distributed equally with the first BMU as the center? – John Stroni Mar 26 '15 at 16:49