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Example text

In this way, the space complexity becomes O(m). This method is called the cost-only Needleman-Wunsch algorithm. Although the space complexity is reduced from O(nm) to O(m), we only compute the optimal alignment score. Can we also reconstruct the optimal alignment? The answer is YES! m] in O(n + m) space. m]) for some j. 1 illustrates this idea. Note that AS(A, B) denotes the optimal alignment score between sequences A and B. 1) 1≤j≤m The integer j, which maximizes the sum, is called the mid-point.

The global alignment problem and the string edit distance problem are in fact a dual problem. The lemma below shows that the solutions reported by them are in fact the same. 1 Let σ be the cost matrix of the edit distance problem and δ be the score matrix of the global alignment problem. If δ(x, y) = −σ(x, y) for all x, y ∈ Σ { }, the solution to the edit distance problem is equivalent to the solution to the string alignment problem. 32 Algorithms in Bioinformatics — A Practical Introduction PROOF Let (x, y), x, y ∈ Σ ∪ { }, be an operation that transforms x to y.

J2 ] using linear space. If we divide the problem into two halves based on the mid-point and recursively deduce the alignments for the two halves, we can compute the optimal alignment while using O(n + m) word space. 7. Below, we analyze the time and the space complexities. For algorithm FindMid, both steps 1 and 2 take O(nm/2) time. Step 3 finds the maximum of m sums, which requires O(m) time. In total, FindMid takes O(nm) time. m]) be T ime(n, m). T ime(n, m) = time for finding the mid-point + time for solving the two subproblems = O(nm) + T ime(n/2, j) + T ime(n/2, m − j).

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