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In the field of Gesture Recognition and Speech Recognition (using HMM classifiers), do multiple HMM classifiers (one for each class) share common Codebook (i.e common set of symbols) in Vector Quantization?

If that is not the case, how does the test/classification stage work?

To which Codebook the test input vector belongs to (we need this to determine the symbol, in turn which is fed to the HMM for classification)?

Some background info:

  1. One of the most earliest works done in Action Recognition used different set of symbols for each of the HMM classifier.
  2. This journal in speech recognition by Rabiner suggest using common Codebook (with different HMM classifiers) for identifying 10 digits uttered by large number of users (in page 12),
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do multiple HMM classifiers (one for each class) share common Codebook (i.e common set of symbols) in Vector Quantization?

Yes, for example in large vocabulary speech recognition there is a big codebook of about 4000-10000 gaussians which are shared across 20-30k HMMs

To which Codebook the test input vector belongs to (we need this to determine the symbol, in turn which is fed to the HMM for classification)?

In modern ASR systems codebook defines a set of GMM probability distributions, not the discrete symbols. So "belongs" doesn't really have sense, it's more a probability value. But during the classification a test vector is usually evaluated for the subset of all codebooks which is active for this stage.

Active codebooks are tracked across test vectors based on active HMMs and preceeding history. Whole HMM set is pruned in order to get active ones. Usually there are about 10k active HMMs on each step and about 1k codebooks.

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  • $\begingroup$ when you said "codebooks shared across 20-30k HMMs", only 1 codebook is shared at a time to compute likelihood of set of hmms that take that specific set of symbols (codebook) as input right? $\endgroup$ – rahul Feb 4 '12 at 6:24
  • $\begingroup$ Yes if you prefer that $\endgroup$ – Nikolay Shmyrev Feb 4 '12 at 20:03
  • $\begingroup$ I believe that holds true in Discrete HMM case too. Because, I'm working with Discrete HMMs with respect to gesture recognition. $\endgroup$ – garak Feb 6 '12 at 13:18

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