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+# 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)