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As a part of a school project, I had to read a paper by Steven Lawrence about using SOMs and CCNs to detect faces. For those of you who are curious, heres the paper: http://clgiles.ist.psu.edu/papers/UMD-CS-TR-3608.face.hybrid.neural.nets.pdf

On page 12 of the paper, Lawrence describes how he uses the SOMs to reduce the dimensionality of the face data. However, I do not understand how this works. In this example, Lawrence uses a 5x5x5 SOM, with input vectors that are 25D. If my understanding is correct, when the training process is done, you will be left with a 25D vector attached to each of the neurons in the net. So, how did this reduce the dimensions of the data? Where exactly is the reduced-dimension data on a self organizing map? Ive researched in alot of places, but for some reason, I could not find the answer to this question. As this question has been bugging me for awhile now, it would be be greatly appreciated if someone could answer it for me.

user1231745
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1 Answers1

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Each training set consist in ten images, sampled in a 5x5 grid (this is the 25D vector), so each training set (face) can be viewed as a 250D vector.

Pag 2

There are 10 different images of 40 distinct subjects.

Lawrence used a 3D som with 5 nodes per dimension, wich conforms a (5x5x5) 125D vector that preserves the topological information of the original data.

Pag 12

The SOM quantizes the 25-dimensional input vectors into 125 topologically ordered values. The three dimensions of the SOM can be thought of as three features

You may appreciate that data dimensionality have been reduced by two and space dimensionality have been reduced by five.

Hope this helps

b3ltsazar
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  • Thanks for your answer! But im still confused about where exactly this data would be located in terms of the SOM. Where can I retrieve this compressed data, I think, is my main question. Please excuse me if its a really basic question. – user1231745 Jul 21 '13 at 21:14
  • "Compressed data", or output vector is conformed by the weights of the trained SOM. – b3ltsazar Jul 21 '13 at 23:16
  • But the dimensions of the weights of the trained SOM would be the same dimension as the input. Correct me if I'm wrong, but how has this reduced the dimensions of the data? – user1231745 Jul 21 '13 at 23:35
  • SOM size is independent from input size, in SOM training you not present the whole input data vector to the net (like other algorithms like backpropagation, hopfield...), instead of this, you present single components in each iteration and SOM topology adapts itself to represent the input space. – b3ltsazar Jul 22 '13 at 08:08
  • Thank you so much for your help! My last question is, on page 12 in the paper, in step four, Lawrence says he creates a 23x28 image from the output of the trained SOM. Could you please clarify this for me? – user1231745 Jul 22 '13 at 12:18
  • My response was not good enough, and maybe is confusing you, i was in a mistake with dimensionality reduction, let me study better the article. Ill write the response tonight ;) – b3ltsazar Jul 22 '13 at 13:52
  • By now as I understand point step 3 is feature extraction dimension reduction takes place in point 2. Images are 92x112 then are sampled with a 5x5 grid stepped 4 pixels in each sample, making 644 vectors of 25D that will serve to train a 5x5x5 SOM, dimension reduction take place here reducing 644 vectors to 125. – b3ltsazar Jul 22 '13 at 13:54
  • I promise this is my last question.... :) How will the 125 vectors be used to form the 23x28 "map" mentioned in point 4? – user1231745 Jul 22 '13 at 14:48