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#A naive recursive implementation of 0-1 Knapsack Problem | ||
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# Returns the maximum value that can be put in a knapsack of | ||
# capacity W | ||
def knapSack(W , wt , val , n): | ||
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# Base Case | ||
if n == 0 or W == 0 : | ||
return 0 | ||
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# 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) | ||
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# 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)) | ||
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# end of function knapSack | ||
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# To test above function | ||
val = [60, 100, 120] | ||
wt = [10, 20, 30] | ||
W = 50 | ||
n = len(val) | ||
print knapSack(W , wt , val , n) |