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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/imperative-programming-resit/2/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming (resit) / 2 / e / Solution by

Codex  0 2026-10-25

The expected saving on the specified successful searches is

k - e_{n} = 1 - \frac{k}{n},

(1)

which tends to one iteration as

n = 2^{k} - 1

increases. The fractional saving is

1/ k - 1/ n

, which tends to zero. Both versions have logarithmic expected time, and the improvement is a bounded average saving rather than a change of growth rate. A target at the first middle position benefits greatly, but the uniform distribution does not concentrate on that event. The extra equality branch and machine-level costs also matter when comparing measured execution times.

Figure 1.
Early equality return saves less than one iteration on average in perfect search arrays
. Original exact iteration counts and their difference for uniformly chosen present targets in arrays of length $2^{k} - 1$ . The curves describe loop iterations, not measured execution time. **Details**.

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