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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/functional-programming-design-and-analysis-of-algorithms/6/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Functional programming, design and analysis of algorithms / 6 / d / Solution by

Codex  0 2026-10-31

Each physical item may enter neither knapsack or exactly one of them. Let

F_{i} (a, b)

be the largest combined value using the first

i

items with remaining capacity limits

a, b

. Initialize

F_{0} (a, b) = 0

and compute

F_{i} (a, b) = max ⎩ ⎨ ⎧ F_{i - 1} (a, b), v_{i} + F_{i - 1} (a - w_{i}, b) v_{i} + F_{i - 1} (a, b - w_{i}) if w_{i} \leq a, if w_{i} \leq b .

(1)

An infeasible alternative is omitted. The three alternatives classify the placement of item

i

exhaustively. Every reference uses row

i - 1

, ensuring that no item is placed in both knapsacks. Removing item

i

from an optimum and applying optimal substructure proves the recurrence relation.

The optimum is $F_{n} (W_{1}, W_{2})$ , with time

O (n (W_{1} + 1) (W_{2} + 1))

, conventionally

O (n W_{1} W_{2})

for positive capacities. Two successive layers require

O ((W_{1} + 1) (W_{2} + 1))

space; keep predecessor decisions or recompute them if the actual allocations are wanted. The stipulated uniqueness of values and weights is unnecessary for correctness. The capacity-indexed algorithm is pseudopolynomial.

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