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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/functional-programming-design-and-analysis-of-algorithms-resit/7/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms (resit) / 7 / b / Solution by

Codex  0 2026-10-31

Let

F [x]

be the minimum number of stamps for total

x

. Since denomination 1 is available, every nonnegative total is reachable. The dynamic programming recurrence is

F [0] = 0, F [x] = 1 + min_{d_{i} \leq x} F [x - d_{i}] (x \geq 1) .

(1)

Fill the table in increasing order of

x

F[0] = 0
for x = 1,...,m:
    F[x] = infinity
    for i = 1,...,n:
        if d[i] <= x:
            F[x] = min(F[x], 1 + F[x-d[i]])
return F[m]

Every representation of

x

has a last stamp

d_{i}

, so its remaining count is at least

F [x - d_{i}]

. Conversely, appending

d_{i}

to an optimal residual representation attains each candidate. Thus the recurrence is exact by the principle of optimality. Worst-case time is $O (mn)$ , with

O (m)

storage. This is pseudo-polynomial time in the numerical target, rather than polynomial time in its binary encoding length.

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