Lec04-gaps

Lec04-gaps - Gap Penalties CMSC 423 General Gap Penalties...

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Unformatted text preview: Gap Penalties CMSC 423 General Gap Penalties AAAGAATTCA A-A-A-T-CA vs. AAAGAATTCA AAA----TCA These have the same score, but the second one is often more plausible. A single insertion of “GAAT” into the first string could change it into the second. • Now, the cost of a run of k gaps is GAP × k • A solution to the problem above is to support general gap penalty, so that the score of a run of k gaps is gap(k) < GAP × k. • Then, the optimization will prefer to group gaps together. General Gap Penalties AAAGAATTCA A-A-A-T-CA vs. AAAGAATTCA AAA----TCA Previous DP no longer works with general gap penalties because the score of the last character depends on details of the previous alignment: AAAGAAC AAA---- vs. AAAGAATC AAA----- Instead, we need to “know” how the previous alignment ends in order to give a score to the last subproblem. Three Matrices We now keep 3 different matrices: M[i,j] = score of best alignment of x[1..i] and y[1..j] ending with a charactercharacter match or mismatch. X[i,j] = score of best alignment of x[1..i] and y[1..j] ending with a space in X. Y[i,j] = score of best alignment of x[1..i] and y[1..j] ending with a space in Y. M [i − 1, j − 1] M [i, j ] = match(i, j ) + max X [i − 1, j − 1] Y [i − 1, j − 1] ￿ M [i, j − k ] − gap(k ) X [i, j ] = max Y [i, j − k ] − gap(k ) for 1 ≤ k ≤ j for 1 ≤ k ≤ j ￿ M [i − k, j ] − gap(k ) Y [i, j ] = max X [i − k, j ] − gap(k ) for 1 ≤ k ≤ i for 1 ≤ k ≤ i The M Matrix We now keep 3 different matrices: M[i,j] = score of best alignment of x[1..i] and y[1..j] ending with a charactercharacter match or mismatch. X[i,j] = score of best alignment of x[1..i] and y[1..j] ending with a space in X. Y[i,j] = score of best alignment of x[1..i] and y[1..j] ending with a space in Y. By definition, alignment ends in a match. M [i − 1, j − 1] M [i, j ] = match(i, j ) + max X [i − 1, j − 1] Y [i − 1, j − 1] Any kind of alignment is allowed before the match. A G The X (and Y) matrices k i x G--- ACGT G y j-k j ￿ M [i, j − k ] − gap(k ) X [i, j ] = max Y [i, j − k ] − gap(k ) i x y k G--- -CGT G j-k j k decides how long to make the gap. We have to make the whole gap at once in order to know how to score it. for 1 ≤ k ≤ j for 1 ≤ k ≤ j The X (and Y) matrices k i x G--- ACGT G y j-k j ￿ M [i, j − k ] − gap(k ) X [i, j ] = max Y [i, j − k ] − gap(k ) i x y k G--- -CGT G j-k j k decides how long to make the gap. We have to make the whole gap at once in order to know how to score it. for 1 ≤ k ≤ j for 1 ≤ k ≤ j This case is automatically handled. k i x y ---- GCGT G j-k j Running Time for Gap Penalties M [i − 1, j − 1] M [i, j ] = match(i, j ) + max X [i − 1, j − 1] Y [i − 1, j − 1] ￿ M [i, j − k ] − gap(k ) X [i, j ] = max Y [i, j − k ] − gap(k ) for 1 ≤ k ≤ j for 1 ≤ k ≤ j ￿ M [i − k, j ] − gap(k ) Y [i, j ] = max X [i − k, j ] − gap(k ) for 1 ≤ k ≤ i for 1 ≤ k ≤ i Final score is max {M[n,m], X[n,m],Y[n,m]}. How do you do the traceback? Runtime: • • Assume |X| = |Y| = n for simplicity: 3n2 subproblems 2n2 subproblems take O(n) time to solve (because we have to try all k) O(n3) total time Affine Gap Penalties • O(n3) for general gap penalties is usually too slow... • We can still encourage spaces to group together using a special case of general penalties called affine gap penalties: gap_start = the cost of starting a gap gap_extend = the cost of extending a gap by one more space • Same idea of using 3 matrices, but now we don’t need to search over all gap lengths, we just have to know whether we are starting a new gap or not. Affine Gap Penalties M [i − 1, j − 1] If previous M [i, j ] = match(i, j ) + max X [i − 1, j − 1] alignment ends in match match, this is a Y [i − 1, j − 1] between new gap x and y gap start + gap extend + M [i, j − 1] X [i, j ] = max gap extend + X [i, j − 1] gap in x gap start + gap extend + Y [i, j − 1] gap start + gap extend + M [i − 1, j ] Y [i, j ] = max gap start + gap extend + X [i − 1, j ] gap in y gap extend + Y [i − 1, j ] Affine Base Cases • M[0, i] = “score of best alignment between 0 characters of x and i characters of y that ends in a match” = -∞ because no such alignment can exist. • X[0, i] = “score of best alignment between 0 characters of x and i characters of y that ends in a gap in x” = gap_start + i × gap_extend because this alignment looks like: --------yyyyyyyyy • X[i, 0] = “score of best alignment between i characters of x and 0 characters of y that ends in a gap in X” = -∞ xxxxxxxxx---------- • ← not allowed M[i, 0] = M[0, i] and Y[0, i] and Y[i,0] are computed using the same logic as X[i,0] and X[0,i] Affine Gap Runtime • 3n2 subproblems • Each one takes constant time • Total runtime O(n2), back to the run time of the basic running time. Why do you “need” 3 matrices? • Alternative WRONG algorithm: M[i][j] = max( M[i-1][j-1] + cost(x[i], y[i]), M[i-1][j] + gap + (gap_start if Arrow[i-1][j] != ), M[j][i-1] + gap + (gap_start if Arrow[i][j-1] != ) ) Intuition: we only need to know whether we are starting a gap or extending a gap. The arrows coming out of each subproblem tell us how the best alignment ends, so we can use them to decide if we are starting a new gap. The best alignment up to this cell ends in a gap. The best alignment up to this cell ends in a match. PROBLEM: The best alignment for strings x[1..i] and y[1..j] doesn’t have to be used in the best alignment between x[1..i+1] and y[1..j+1] Why 3 Matrices: Example match = 10, mismatch = -2, gap = -7, gap_start = -15 CART CA-T OPT(4, 3) = optimal score = 30 - 15 - 7 = 8 CARTS CA-T- WRONG(5, 3) = 30 - 15 - 7 - 15 - 7 = -14 CARTS CAT-- OPT(5, 3) = 20 - 2 - 15 - 14 = -11 this is why we need to keep the X and Y matrices around. they tell us the score of ending with a gap in one of the sequences. Side Note: Lower Bounds • Suppose the lengths of x and y are n. • Clearly, need at least Ω(n) time to find their global alignment (have to read the strings!) • The DP algorithms show global alignment can be done in O(n2) time. • A trick called the “Four Russians Speedup” can make a similar dynamic programming algorithm run in O(n2 / log n) time. • • We won’t talk about the Four Russian’s Speedup, but it’s in your book in Sections 7.3 & 7.4. Open questions: Can we do better? Can we prove that we can’t do better? No one knows... Recap • Semiglobal alignment: 0 initial columns or take maximums over last row or column. • Local alignment: extra “0” case. • • General gap penalties require 3 matrices and O(n3) time. Affine gap penalties require 3 matrices, but only O(n2) time. What you should know by now... • Global & local sequence alignment algorithms with basic gap penalties • Alignment with general gap penalties • Alignment with affine gap penalties • Longest common subsequence • Dynamic programming framework ...
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