5knapsack.py

#A naive recursive implementation of 0-1 Knapsack Problem
 
# Returns the maximum value that can be put in a knapsack of
# capacity W
def knapSack(W , wt , val , n):
 
    # Base Case
    if n == 0 or W == 0 :
        return 0
 
    # If weight of the nth item is more than Knapsack of capacity
    # W, then this item cannot be included in the optimal solution
    if (wt[n-1] > W):
        return knapSack(W , wt , val , n-1)
 
    # return the maximum of two cases:
    # (1) nth item included
    # (2) not included
    else:
        return max(val[n-1] + knapSack(W-wt[n-1] , wt , val , n-1),
                   knapSack(W , wt , val , n-1))
 
# end of function knapSack
 
# To test above function
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print knapSack(W , wt , val , n)