| Kristen Jaskie
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12-12-2007 01:31 AM ET (US)
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The Viterbi algorithm only needs O(m*n) space per word, but for time complexity reasons, we were computing the Viterbi algorithm for all the words together and storing it in a matrix of size O(m*n*numwords). Also, it's necessary to compute g which is of size(numwords*lengthOfLongestWord*4) and to do that, we have to compute four more f matrices (one for 00, 01, 10, 11) which are each (numwords*lengthOfLongestWord*numFeatures) in size. All together there are ton of data structures that have to be stored even to compute U. Unless we are totally confused and doing something wrong...
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