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Knapsack solved with memoization

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Srinjan Dutta 2023-10-09 19:03:45 +05:30
parent 954f5d1705
commit 0e789ce3ec
1 changed files with 25 additions and 13 deletions
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knapsack/knapsack_memoization.py
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@ -15,21 +15,25 @@ 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
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:
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 problems
: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
@ -60,26 +64,34 @@ def knapsack(values:list, weights:list, num_of_items:int, max_weight:int, dp:lis
>>> knapsack(values,wt,n,w,dp)
75
"""
if max_weight == 0 or num_of_items == 0: #no profit gain if any of these two become zero
#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
elif dp[num_of_items][max_weight] != -1: #if this case is previously encountered => maximum gain for this case is already
#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]
elif weights[num_of_items-1] <= max_weight: #if the item can be included in the bag
# ans1 stores the maximum profit if the item at index num_of_items -1 is included in the bag
ans1 = values[num_of_items - 1] + knapsack(values, weights, num_of_items-1, max_weight-weights[num_of_items-1], dp)
# ans2 stores the maximum profit if the item at index num_of_items -1 is not included in the bag
#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
# 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)
dp[num_of_items][max_weight] = knapsack(values, weights,
num_of_items-1, max_weight, dp)
return dp[num_of_items][max_weight]