A large version of the small parsimony problem
Given a multiple alignment over k sequences, an evolutionary tree relating the sequences, and a subadditive gap penalty function (e.g. an affine function), we reconstruct the internal nodes of the tree optimally we find the optimal explanation in terms of indels of the observed gaps and find the mos...
| Main Authors: | , , |
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| Format: | Conference item |
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2003
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| Summary: | Given a multiple alignment over k sequences, an evolutionary tree relating the sequences, and a subadditive gap penalty function (e.g. an affine function), we reconstruct the internal nodes of the tree optimally we find the optimal explanation in terms of indels of the observed gaps and find the most parsimonious assignment of nucleotides. The gaps of the alignment are represented in a so-called gap graph, and through theoretically sound preprocessing the graph is reduced to pave the way for a running time which in all but the most pathological examples is far better than the exponential worst case time. E.g. for a tree with nine leaves and a random alignment of length 10.000 with 60% gaps, the running time is on average around 45 seconds. For a real alignment of length 9868 of nine HIV-1 sequences, the running time is less than one second. © Springer-Verlag Berlin Heidelberg 2003. |
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