Enhancement of the knapsack algorithm with memorization and generalisation #9295

Original file line number	Diff line number	Diff line change
		@@ -1,4 +1,4 @@
	# A naive recursive implementation of 0-1 Knapsack Problem
	# A recursive implementation of 0-N Knapsack Problem

		This overview is taken from:

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65 changes: 43 additions & 22 deletions knapsack/knapsack.py

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Original file line number	Diff line number	Diff line change
		@@ -1,14 +1,23 @@
	"""A naive recursive implementation of 0-1 Knapsack Problem
	"""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) -> int:

	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.
	whereby each weight w has a specific value val
	with option to allow repetitive selection of items

		>>> cap = 50
		>>> val = [60, 100, 120]
Expand All		@@ -17,28 +26,40 @@ def knapsack(capacity: int, weights: list[int], values: list[int], counter: int)
		>>> knapsack(cap, w, val, c)
		220

	The result is 220 cause the values of 100 and 120 got the weight of 50
	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)
Copy link Contributor @jeffreyyancey jeffreyyancey Oct 8, 2024 Choose a reason for hiding this comment The reason will be displayed to describe this comment to others. Learn more. The algorithm has been updated to allow repetition. An example/testcase where repetition is used would be appropriate.
	got the weight of 10*5 which is the limit of the capacity.
		"""

Copy link Contributor @jeffreyyancey jeffreyyancey Oct 8, 2024 Choose a reason for hiding this comment The reason will be displayed to describe this comment to others. Learn more. You might want to consider a few things before getting into the algorithm: Handle any input constraints. Ensure the inputs make sense and raise an exception for the user if they don't non-positive numbers a capacity lower than the weight of any single object etc Before getting into the main algorithm, you can go ahead and remove any items that have a weight above the capacity of the knapsack.
	# 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
	# (2) not included
	if weights[counter - 1] > capacity:
	return knapsack(capacity, weights, values, counter - 1)
	else:
	left_capacity = capacity - weights[counter - 1]
	new_value_included = values[counter - 1] + knapsack(
	left_capacity, weights, values, counter - 1
	)
	without_new_value = knapsack(capacity, weights, values, counter - 1)
	return max(new_value_included, without_new_value)
	@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__":
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12 changes: 11 additions & 1 deletion knapsack/tests/test_knapsack.py

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -30,7 +30,7 @@ def test_base_case(self):

		def test_easy_case(self):
		"""
	test for the base case
	test for the easy case
		"""
		cap = 3
		val = [1, 2, 3]
Expand All		@@ -48,6 +48,16 @@ def test_knapsack(self):
		c = len(val)
		assert k.knapsack(cap, w, val, c) == 220

	def test_knapsack_repetition(self):
	"""
	test for the knapsack repetition
	"""
	cap = 50
	val = [60, 100, 120]
	w = [10, 20, 30]
	c = len(val)
	assert k.knapsack(cap, w, val, c, True) == 300


		if __name__ == "__main__":
		unittest.main()

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Enhancement of the knapsack algorithm with memorization and generalisation #9295

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@jeffreyyancey jeffreyyancey Oct 8, 2024

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