"""A recursive implementation of 0-N Knapsack Problem https://en.wikipedia.org/wiki/Knapsack_problem """ from __future__ import annotations from functools import lru_cache def knapsack( capacity: int, weights: list[int], values: list[int], counter: int, allow_repetition=False, ) -> int: """ Returns the maximum value that can be put in a knapsack of a capacity cap, whereby each weight w has a specific value val with option to allow repetitive selection of items >>> cap = 50 >>> val = [60, 100, 120] >>> w = [10, 20, 30] >>> c = len(val) >>> knapsack(cap, w, val, c) 220 Given the repetition is NOT allowed, the result is 220 cause the values of 100 and 120 got the weight of 50 which is the limit of the capacity. >>> knapsack(cap, w, val, c, True) 300 Given the repetition is allowed, the result is 300 cause the values of 60*5 (pick 5 times) got the weight of 10*5 which is the limit of the capacity. """ @lru_cache def knapsack_recur(capacity: int, counter: int) -> int: # Base Case if counter == 0 or capacity == 0: return 0 # If weight of the nth item is more than Knapsack of capacity, # then this item cannot be included in the optimal solution, # else return the maximum of two cases: # (1) nth item included only once (0-1), if allow_repetition is False # nth item included one or more times (0-N), if allow_repetition is True # (2) not included if weights[counter - 1] > capacity: return knapsack_recur(capacity, counter - 1) else: left_capacity = capacity - weights[counter - 1] new_value_included = values[counter - 1] + knapsack_recur( left_capacity, counter - 1 if not allow_repetition else counter ) without_new_value = knapsack_recur(capacity, counter - 1) return max(new_value_included, without_new_value) return knapsack_recur(capacity, counter) if __name__ == "__main__": import doctest doctest.testmod()