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www.bioalgorithms.infoAn Introduction to Bioinformatics Algorithms Block Alignment and the Four-Russians Speedup Presenter: Yung-Hsing Peng Date: 2004.12.03

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Outline Block Alignment Four-Russians Speedup Constructing LCS in Sub-quadratic Time

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Is It Possible to Align Sequences in Subquadratic Time? Dynamic Programming takes O(n 2 ) for global alignment Can we do better? Yes, use Four-Russians Speedup

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Blocks partition n n/tn/t n/tn/t t t n

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Block Alignment: Examples validinvalid

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Block Alignment Problem Goal: Find the longest block path through an edit graph Input: Two sequences, u and v partitioned into blocks of size t. This is equivalent to an n x n edit graph partitioned into t x t subgrids Output: The block alignment of u and v with the maximum score (longest block path through the edit graph

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Constructing Alignments within Blocks n/tn/t Block pair represented by each small square Solve mini-alignmnent problems

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Block Alignment: Dynamic Programming Let s i,j denote the optimal block alignment score between the first i blocks of u and first j blocks of v s i,j = max s i-1,j - block s i,j-1 - block s i-1,j-1 + i,,j block is the penalty for inserting or deleting an entire block i,j is score of block in row i, column j.

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Block Alignment Runtime (cont ’ d) Computing all i,j requires solving (n/t)*(n/t) mini block alignments, each of size (t*t) So computing i,j takes O([n/t]*[n/t]*t*t) = O(n 2 ) time This is the same as dynamic programming How do we speed this up?

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Look-up Table for Four Russians Technique Lookup table “Score” AAAAAA AAAAAC AAAAAG AAAAAT AAAACA … AAAAAA AAAAAC AAAAAG AAAAAT AAAACA … each sequence has t nucleotides size is only n, instead of (n/t)*(n/t)

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info New Recurrency The new lookup table Score is indexed by a pair of t-nucleotide strings, so s i,j = max s i-1,j - block s i,j-1 - block s i-1,j-1 + Score(i th block of v, j th block of u

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Four Russians Technique Let t = log(n), where t is block size, n is sequence size. Instead of having (n/t)*(n/t) mini-alignments, construct 4 t x 4 t mini-alignments for all pairs of strings of t nucleotides (huge size), and put in a lookup table. However, size of lookup table is not rreally that huge if t is small. Let t = (logn)/4. Then 4 t x 4 t = n

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Four Russians Speedup Runtime Each access takes O(logn) time, because we have to transfer the little string into an index. Overall running time: O( [n 2 /t 2 ]*logn ) Since t = logn, substitute in: O( [n 2 /{logn} 2 ]*logn) > O( n 2 /logn )

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info So Far … We know how to speed up when we are solving block alignment In order to speed up the mini-alignment calculations to under n 2, we create a lookup table of size n, which consists of all scores for all t-nucleotide pairs Running time goes from quadratic, O(n 2 ), to subquadratic: O(n 2 /logn)

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Block Alignment vs. LCS Unlike the block partitioned graph, the LCS path does not have to pass through the vertices of the blocks. block alignment longest common subsequence

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Points of Interest block alignment has (n/t)*(n/t) = (n 2 /t 2 ) points of interest LCS alignment has O(n 2 /t) points of interest

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Traversing Blocks for LCS (cont ’ d) If we used this to compute the grid, it would take quadratic, O(n 2 ) time, but we want to do better. we know these scores we can calculate these scores t x t block

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Four Russians Speedup Build a lookup table for all possible values of the four variables: 1.subsequence of sequence u in this block (4 t possibilities) 2.subsequence of sequence v in this block (4 t possibilities) 3.all pairs of possible scores for the first row s *,j 4.all pairs of possible scores for the first column s *,j For each quadruple we store the value of the score for the last row and last column. This will be a huge table, but we can eliminate alignments scores that don’t make sense

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Efficient Encoding of Alignment Scores Instead of recording numbers that correspond to the index in the sequences u and v, we can use binary to encode the differences between the alignment scores 012234 11011 original encoding binary encoding

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info An Example ----CTTCGATGA 0000000000 T0011111111 T0012222222 A0012223333 C0112333333 G0112344444 T0122344555 G0122344566 C0122344566 A012234556 7

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info An Example ----CTTCGATGA T0/01001/0000 T011 A001 C110 G0/10111/10001/0 T001 G001 C000 A0/1101 1011/1

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Reducing Lookup Table Size 2 t possible scores (t = # blocks) 4 t possible strings Lookup table size is (2 t * 2 t )*(4 t * 4 t ) = 2 6t Let t = (logn)/4; Table size is: 2 6((logn)/4) = n (6/4) = n (3/2) Time = O( [n 2 /t 2 ]*logn ) O( [n 2 /{logn} 2 ]*logn) > O( n 2 /logn )

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An Introduction to Bioinformatics Algorithmswww.bioalgorithms.info Summary We take advantage of the fact that for each block of t = log(n), we can pre-compute all possible scores and store them in a lookup table of size n (3/2) We used the Four Russian speedup to go from a quadratic running time for LCS to subquadratic running time: O(n 2 /logn)

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