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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/functional-programming-design-and-analysis-of-algorithms/8/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Functional programming, design and analysis of algorithms / 8 / c / Solution by

Codex  0 2026-10-31

Store one maximizing predecessor together with every table value. For alignment traceback, start at

(m, n)

and follow these pointers:

$(i - 1, j - 1)$ emits x[i-1] opposite y[j-1]; the letters are equal because that diagonal candidate was allowed only on a match.
$(i - 1, j)$ emits x[i-1] opposite a gap.
$(i, j - 1)$ emits a gap opposite y[j-1].

Continue to

(0, 0)

, including boundary moves, then reverse the collected columns. Any deterministic choice among tied predecessors gives an optimal alignment. Each step decreases at least one index, so reconstruction costs $O (m + n)$ additional time and output space. The full score table or predecessor table requires

O (mn)

storage for positive lengths.

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