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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/8/a/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 / 8 / a / Solution by

Codex  0 2026-10-25

The three answers, in source order, are:

$T (n) = Θ (n^{2})$ . In the Master theorem, $a = 6$ , $b = 3$ , and $n^{2}$ grows polynomially faster than $n^{l o g_{3} 6}$ . The regularity condition holds because $6 (n /3)^{2} = (2/3) n^{2}$ . For powers of three, expansion even gives $T (n) = 3 n^{2} - 2 n^{l o g_{3} 6}$ .
$T (n) = n^{2}$ , and hence $Θ (n^{2})$ . The telescoping sum is $T (n) = 1 + \sum_{j = 2}^{n} (2 j - 1) = \sum_{j = 1}^{n} (2 j - 1) = n^{2}$ .
$T (n) = Θ (n^{l o g_{2} 3})$ . Set $m = n - 2$ and $U (m) = T (m + 2)$ . Then $U (m) = 3 U (m /2) + m + 2$ , with constant base cost for $m \leq 2$ . The Master theorem with $a = 3$ , $b = 2$ applies because $m + 2 = Θ (m)$ is polynomially smaller than $m^{l o g_{2} 3}$ . Finally $m = n - 2 = Θ (n)$ .

The change of variable in the third case handles the additive shift in the recursive argument exactly; it is not an unjustified deletion of the +1. These Big theta notation conclusions use the usual admissible input sizes, or bounded integer rounding of subproblem sizes.

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