2017-01-03 11:20:13 +00:00
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"""
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2019-08-19 05:39:39 +00:00
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Given weights and values of n items, put these items in a knapsack of
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capacity W to get the maximum total value in the knapsack.
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Note that only the integer weights 0-1 knapsack problem is solvable
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using dynamic programming.
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2017-01-03 11:20:13 +00:00
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"""
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2019-08-19 05:39:39 +00:00
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def MF_knapsack(i, wt, val, j):
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2019-10-05 05:14:13 +00:00
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"""
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2018-07-22 19:06:53 +00:00
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This code involves the concept of memory functions. Here we solve the subproblems which are needed
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unlike the below example
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F is a 2D array with -1s filled up
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2019-10-05 05:14:13 +00:00
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"""
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2018-07-22 19:06:53 +00:00
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global F # a global dp table for knapsack
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if F[i][j] < 0:
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2019-10-05 05:14:13 +00:00
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if j < wt[i - 1]:
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val = MF_knapsack(i - 1, wt, val, j)
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2018-07-22 19:06:53 +00:00
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else:
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2019-10-05 05:14:13 +00:00
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val = max(
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MF_knapsack(i - 1, wt, val, j),
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MF_knapsack(i - 1, wt, val, j - wt[i - 1]) + val[i - 1],
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)
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2018-07-22 19:06:53 +00:00
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F[i][j] = val
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return F[i][j]
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2019-08-19 05:39:39 +00:00
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2017-01-03 11:20:13 +00:00
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def knapsack(W, wt, val, n):
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2019-10-05 05:14:13 +00:00
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dp = [[0 for i in range(W + 1)] for j in range(n + 1)]
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2017-01-03 11:20:13 +00:00
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2019-10-05 05:14:13 +00:00
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for i in range(1, n + 1):
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for w in range(1, W + 1):
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if wt[i - 1] <= w:
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dp[i][w] = max(val[i - 1] + dp[i - 1][w - wt[i - 1]], dp[i - 1][w])
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2017-01-03 11:20:13 +00:00
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else:
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2019-10-05 05:14:13 +00:00
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dp[i][w] = dp[i - 1][w]
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2017-01-03 11:20:13 +00:00
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2019-08-19 05:39:39 +00:00
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return dp[n][W], dp
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2019-10-05 05:14:13 +00:00
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def knapsack_with_example_solution(W: int, wt: list, val: list):
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2019-08-19 05:39:39 +00:00
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"""
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Solves the integer weights knapsack problem returns one of
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the several possible optimal subsets.
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Parameters
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---------
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W: int, the total maximum weight for the given knapsack problem.
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wt: list, the vector of weights for all items where wt[i] is the weight
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of the ith item.
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val: list, the vector of values for all items where val[i] is the value
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of te ith item
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Returns
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-------
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optimal_val: float, the optimal value for the given knapsack problem
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example_optional_set: set, the indices of one of the optimal subsets
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which gave rise to the optimal value.
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Examples
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-------
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>>> knapsack_with_example_solution(10, [1, 3, 5, 2], [10, 20, 100, 22])
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(142, {2, 3, 4})
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>>> knapsack_with_example_solution(6, [4, 3, 2, 3], [3, 2, 4, 4])
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(8, {3, 4})
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>>> knapsack_with_example_solution(6, [4, 3, 2, 3], [3, 2, 4])
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Traceback (most recent call last):
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...
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ValueError: The number of weights must be the same as the number of values.
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But got 4 weights and 3 values
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"""
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if not (isinstance(wt, (list, tuple)) and isinstance(val, (list, tuple))):
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2019-10-05 05:14:13 +00:00
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raise ValueError(
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"Both the weights and values vectors must be either lists or tuples"
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)
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2019-08-19 05:39:39 +00:00
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num_items = len(wt)
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if num_items != len(val):
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2019-10-05 05:14:13 +00:00
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raise ValueError(
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"The number of weights must be the "
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"same as the number of values.\nBut "
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"got {} weights and {} values".format(num_items, len(val))
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)
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2019-08-19 05:39:39 +00:00
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for i in range(num_items):
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if not isinstance(wt[i], int):
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2019-10-05 05:14:13 +00:00
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raise TypeError(
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"All weights must be integers but "
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"got weight of type {} at index {}".format(type(wt[i]), i)
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)
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2019-08-19 05:39:39 +00:00
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optimal_val, dp_table = knapsack(W, wt, val, num_items)
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example_optional_set = set()
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_construct_solution(dp_table, wt, num_items, W, example_optional_set)
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return optimal_val, example_optional_set
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2019-10-05 05:14:13 +00:00
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def _construct_solution(dp: list, wt: list, i: int, j: int, optimal_set: set):
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2019-08-19 05:39:39 +00:00
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"""
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Recursively reconstructs one of the optimal subsets given
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a filled DP table and the vector of weights
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Parameters
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---------
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dp: list of list, the table of a solved integer weight dynamic programming problem
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wt: list or tuple, the vector of weights of the items
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i: int, the index of the item under consideration
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j: int, the current possible maximum weight
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optimal_set: set, the optimal subset so far. This gets modified by the function.
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Returns
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-------
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None
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"""
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# for the current item i at a maximum weight j to be part of an optimal subset,
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# the optimal value at (i, j) must be greater than the optimal value at (i-1, j).
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# where i - 1 means considering only the previous items at the given maximum weight
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if i > 0 and j > 0:
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if dp[i - 1][j] == dp[i][j]:
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_construct_solution(dp, wt, i - 1, j, optimal_set)
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else:
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optimal_set.add(i)
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2019-10-05 05:14:13 +00:00
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_construct_solution(dp, wt, i - 1, j - wt[i - 1], optimal_set)
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2019-08-19 05:39:39 +00:00
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2018-07-22 19:06:53 +00:00
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2019-10-05 05:14:13 +00:00
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if __name__ == "__main__":
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"""
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2018-07-22 19:06:53 +00:00
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Adding test case for knapsack
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2019-10-05 05:14:13 +00:00
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"""
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2019-08-19 05:39:39 +00:00
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val = [3, 2, 4, 4]
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wt = [4, 3, 2, 3]
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2018-07-22 19:06:53 +00:00
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n = 4
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w = 6
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2019-08-19 05:39:39 +00:00
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F = [[0] * (w + 1)] + [[0] + [-1 for i in range(w + 1)] for j in range(n + 1)]
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2019-10-05 05:14:13 +00:00
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optimal_solution, _ = knapsack(w, wt, val, n)
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2019-08-19 05:39:39 +00:00
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print(optimal_solution)
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2019-10-05 05:14:13 +00:00
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print(MF_knapsack(n, wt, val, w)) # switched the n and w
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2019-08-19 05:39:39 +00:00
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# testing the dynamic programming problem with example
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# the optimal subset for the above example are items 3 and 4
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optimal_solution, optimal_subset = knapsack_with_example_solution(w, wt, val)
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assert optimal_solution == 8
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assert optimal_subset == {3, 4}
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print("optimal_value = ", optimal_solution)
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print("An optimal subset corresponding to the optimal value", optimal_subset)
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