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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/functional-programming-design-and-analysis-of-algorithms/5/a/i/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Functional programming, design and analysis of algorithms / 5 / a / i / Solution by

Codex  0 2026-10-25

Start with

(d [1], d [2], d [3], d [4]) = (0, \infty, \infty, \infty)

. With in-place edge relaxation, the Bellman-Ford algorithm makes the following changes during its first lexicographic pass:

edge (1, 2) (1, 3) (3, 2) (3, 4) (4, 2) improved entry d [2] d [3] d [2] d [4] d [2] new value 62534

(1)

The next pass makes no change; any remaining fixed-count passes also make none. Thus the final vector is $(0, 4, 2, 3)$ . These are sequential updates within a pass, rather than the simultaneous values of the Synchronous Bellman-Ford recurrence.

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