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* Create recursive_approach_knapsack.py Added a new naïve recursive approach to solve the knapsack problem. * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update recursive_approach_knapsack.py Updated the code * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update recursive_approach_knapsack.py Updated * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
53 lines
1.4 KiB
Python
53 lines
1.4 KiB
Python
# To get an insight into naive recursive way to solve the Knapsack problem
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"""
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A shopkeeper has bags of wheat that each have different weights and different profits.
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eg.
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no_of_items 4
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profit 5 4 8 6
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weight 1 2 4 5
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max_weight 5
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Constraints:
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max_weight > 0
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profit[i] >= 0
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weight[i] >= 0
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Calculate the maximum profit that the shopkeeper can make given maxmum weight that can
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be carried.
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"""
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def knapsack(
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weights: list, values: list, number_of_items: int, max_weight: int, index: int
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) -> int:
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"""
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Function description is as follows-
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:param weights: Take a list of weights
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:param values: Take a list of profits corresponding to the weights
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:param number_of_items: number of items available to pick from
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:param max_weight: Maximum weight that could be carried
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:param index: the element we are looking at
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:return: Maximum expected gain
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>>> knapsack([1, 2, 4, 5], [5, 4, 8, 6], 4, 5, 0)
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13
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>>> knapsack([3 ,4 , 5], [10, 9 , 8], 3, 25, 0)
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27
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"""
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if index == number_of_items:
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return 0
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ans1 = 0
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ans2 = 0
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ans1 = knapsack(weights, values, number_of_items, max_weight, index + 1)
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if weights[index] <= max_weight:
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ans2 = values[index] + knapsack(
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weights, values, number_of_items, max_weight - weights[index], index + 1
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)
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return max(ans1, ans2)
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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