Knapsack_with_memoization

knapsack/knapsack_memoization.py

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+"""
+A shopkeeper has bags of wheat that each have different weights and different profits.
+eg.
+no_of_items : 5
+profit [15, 14,10,45,30]
+weight [2,5,1,3,4]
+max_weight that can be carried : 7
+
+Constraints:
+    max_weight > 0
+    profit[i] >= 0
+    weight[i] >= 0
+
+Calculate:
+     The maximum profit that the shopkeeper can make given maxmum weight that can
+be carried.
+
+This problem is implemented here with MEMOIZATION method using the concept of
+Dynamic Programming
+"""
+"""
+for more information visit https://en.wikipedia.org/wiki/Memoization
+"""
+
+
+def knapsack(
+    values: list, weights: list, num_of_items: int, max_weight: int, dp: list
+) -> int:
+    """
+    Function description is as follows-
+    :param weights: Take a list of weights
+    :param values: Take a list of profits corresponding to the weights
+    :param number_of_items: number of items available to pick from
+    :param max_weight: Maximum weight that could be carried
+    :param dp: it is a list of list, i.e, a table whose (i,j)
+                cell represents the maximum profit earned
+                for i items and j as the maximum weight allowed, it
+                is an essential part for implementing this problem
+                using memoization dynamic programming
+    :return: Maximum expected gain
+
+    Testcase 1:
+    >>> values = [1, 2, 4, 5]
+    >>> wt = [5, 4, 8, 6]
+    >>> n = len(values)
+    >>> w = 5
+    >>> dp = [[-1 for x in range(w+1)] for y in range(n+1)]
+    >>> knapsack(values,wt,n,w,dp)
+    2
+
+    Testcase 2:
+    >>> values = [3 ,4 , 5]
+    >>> wt = [10, 9 , 8]
+    >>> n = len(values)
+    >>> w = 25
+    >>> dp = [[-1 for x in range(w+1)] for y in range(n+1)]
+    >>> knapsack(values,wt,n,w,dp)
+    9
+
+    Testcase 3:
+    >>> values = [15, 14,10,45,30]
+    >>> wt = [2,5,1,3,4]
+    >>> n = len(values)
+    >>> w = 7
+    >>> dp = [[-1 for x in range(w+1)] for y in range(n+1)]
+    >>> knapsack(values,wt,n,w,dp)
+    75
+    """
+    # no profit gain if any of these two become zero
+    if max_weight == 0 or num_of_items == 0:
+        dp[num_of_items][max_weight] = 0
+        return 0
+    # if this case is previously encountered => maximum gain for this case is already
+    elif dp[num_of_items][max_weight] != -1:
+        # in dp table
+        return dp[num_of_items][max_weight]
+
+    # if the item can be included in the bag
+    elif weights[num_of_items - 1] <= max_weight:
+        # ans1 stores the maximum profit if the item at
+        #  index num_of_items -1 is included in the bag
+        incl = knapsack(
+            values,
+            weights,
+            num_of_items - 1,
+            max_weight - weights[num_of_items - 1],
+            dp,
+        )
+        ans1 = values[num_of_items - 1] + incl
+        # ans2 stores the maximum profit if the item at
+        # index num_of_items -1 is not included in the bag
+        ans2 = knapsack(values, weights, num_of_items - 1, max_weight, dp)
+        # the final answer is the maximum profit gained from any of ans1 or ans2
+        dp[num_of_items][max_weight] = max(ans1, ans2)
+        return dp[num_of_items][max_weight]
+
+    # if the item's weight exceeds the max_weight of the bag
+    #  => it cannot be included in the bag
+    else:
+        dp[num_of_items][max_weight] = knapsack(
+            values, weights, num_of_items - 1, max_weight, dp
+        )
+        return dp[num_of_items][max_weight]
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod(name="knapsack", verbose=True)