4961b3aa89
* enhance knapsack problem * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * wei/refactor code * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update test_knapsack.py * Update knapsack.py * Update test_knapsack.py * Update knapsack.py * Update knapsack.py * Update knapsack.py * Update knapsack.py * Update knapsack.py * Update test_knapsack.py --------- Co-authored-by: weijiang <weijiang@weijiangs-MacBook-Pro.local> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Maxim Smolskiy <mithridatus@mail.ru>
69 lines
2.1 KiB
Python
69 lines
2.1 KiB
Python
"""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()
|