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Published byAyanna Enfield Modified about 1 year ago

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Solusi DP Menggunakan Software Pertemuan 24 : (Off Class) Mata kuliah:K0164-Pemrograman Matematika Tahun:2008

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Learning Outcomes Mahasiswa dapat mendemonstrasikan penyelesaian masalah dinamik programming, dengan menggunakan bantuan program komputer..

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Outline Materi: Penerapan Dinamik Programming (DP) Penyelesaian masalah DP dengan program komputer..

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Alignments -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Three elements: Perfect matches Mismatches Gaps

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Choosing Alignments There are many possible alignments For example, compare: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A to ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Which one is better?

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Scoring Rule Example Score = (# matches) – (# mismatches) – (# gaps) x 2

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Example -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Score: (+1x13) + (-1x2) + (-2x4) = 3 ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Score: (+1x5) + (-1x6) + (-2x11) = -23

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Optimal Alignment Optimal alignment is achieved at best similarity score d, thus is determined by the scoring rule

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Finding the Best Alignment Score The additive form of the score allows to perform dynamic programming to find the best score efficiently Guaranteed to find the best alignment

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The Idea The best alignment that ends at a given pair of bases: the best among best alignments of the sequences up to that point, plus the score for aligning the two additional bases.

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Dynamic Programming Consider the best alignment score of two sequences s, t at base/residue i+1, j+1, respectively:

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Dynamic Programming The best alignment must be in one of three cases: 1. Last position is ( s[i+1],t[j +1] ) 2. Last position is ( -, t[j +1] ) 3. Last position is ( s[i +1], - )

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Dynamic Programming The best alignment must be in one of three cases: 1. Last position is ( s[i+1],t[j +1] ) 2. Last position is ( -, t[j +1] ) 3. Last position is ( s[i +1], - )

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Dynamic Programming The best alignment must be in one of three cases: 1. Last position is ( s[i+1],t[j +1] ) 2. Last position is ( -, t[j +1] ) 3. Last position is ( s[i +1], - )

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Dynamic Programming

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Of course, we first need to handle the base cases in the recursion:

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Dynamic Programming We fill the matrix using the recurrence rule – A G C – – A A A C –

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Dynamic Programming

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Conclusion: d( AAAC, AGC ) = -1

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Reconstructing the Best Alignment AAAC AG-C

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More than one best alignment AAAC A-GC

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Complexity Space: O(mn) Time: O(mn) Filling the matrix O(mn) Backtrace O(m+n)

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Best-aligned Subsequences The best score or start over

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