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[37] | 1 | %subseq_longest Find the longest continuous subsequences in columns of MM. |
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| 2 | % |
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| 3 | % Put the initial image of the longest continuous subsequence of known |
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| 4 | % points in column ``p'' to ``b(p)''. |
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| 5 | % |
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| 6 | % [b, len] = subseq_longest(I) |
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| 7 | % |
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| 8 | % Parameters: |
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| 9 | % len ... len(p) is length of the longest continuous |
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| 10 | % subsequence of known points in column ``p'' |
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| 11 | |
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| 12 | function [b, len] = subseq_longest(I) |
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| 13 | |
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| 14 | [m n] = size(I); |
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| 15 | |
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| 16 | b(n) = 0; % memory allocation |
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| 17 | len(n) = 0; % " " |
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| 18 | |
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| 19 | for p = 1:n |
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| 20 | seq(1:m) = 0; |
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| 21 | l = 1; |
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| 22 | for i = 1:m |
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| 23 | if I(i,p) |
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| 24 | seq(l) = seq(l) +1; |
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| 25 | else |
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| 26 | l = i+1; |
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| 27 | end |
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| 28 | end |
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| 29 | len(p) = max(seq); |
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| 30 | best = find(seq == len(p)); |
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| 31 | b(p) = best(1); |
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| 32 | end |
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| 33 | |
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| 34 | %b(:) = 1; % debug |
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