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Subset Sum of Coins for Exact Change

#======================================================================= # Author: Isai Damier # Title: Subset Sum of Coins for Exact Change # Project: geekviewpoint # Package: algorithms # # Statement: # Given a set of coins, find the shortest sequence of coins that # adds up to a given total t. # # Time Complexity: O(n*t) # where n is the number of coins and t is the total to make change for. # # Sample Input: coin=[1,5,10,25]; t=167 # Sample Output: [1, 1, 5, 10, 25, 25, 25, 25, 25, 25] # # Dynamic Programming Strategy: # # Our approach is to compute the number of coins in each subtotal # amount of money from 0 to t, inclusive. The technique is to # add coins to a previously found subtotal so to get the next # subtotal, until we solve for the minimum number of coins in t. # After that, we simply walk the list of subtotals to get the # exact change (you'll see). # # 0] Create an array N of size t+1: # The indices/keys of the array represent different subtotals. # And the element/value at each index represents the number of # coins that adds up to that given subtotal: # N[subtotal] = number of coins # 1] Initialize N to infinity: # Before any computation is performed, assume/pretend that each # subtotal takes an infinite number of coins to compute. We know # that's not true, but that's what we have to prove. # 2] Set N at index 0 to 0: # Recall that for the N array, each index represents a subtotal # and the element at that index represents the number of coins # that add up to that subtotal. Therefore, we can set N[0] to 0 # because we know for certain that it takes 0 coins to get 0 # amount of money (this is the anchor/base case). # # 3] For subtotals ranging from 1 to t, inclusive, do: # For each coin: # If coin <= subtotal and N[subtotal-coin]+1 < subtotal, then: # N[subtotal] = N[subtotal-coin]+1 # # Step 3 is known as the relaxation step. We started with an extreme # and unrealistic assumption (i.e. that it takes infinite number of # coins to make change for the money t), then we keep relaxing # our assumptions until we get to the truth. # # EXAMPLE: You should really skip the following manual example and go # straight to the code. Then, if you need help seeing how the code # is computing the number of coins for each subtotal, come back here. # # For t=10 and coins = [1,3,5], the final version of N is: # # N[0] = 0; //base case : 0 coins for 0 cent # N[1] = N[0]+coins[0] = 0+1 = 1 : 1 coin for 1 cent [1] # N[2] = N[1]+coins[0] = 1+1 = 2 : 2 coins for 2 cents [1,1] # N[3] = coins[1] = 1 : 1 coin for 3 cents [3] # N[4] = N[1]+coins[1] = 1+1 = 2 : 2 coins for 4 cents [3,1] # N[5] = coins[2] = 1 : 1 coin for 5 cents [5] # N[6] = N[5]+coins[0] = 1+1 = 2 : 2 coins for 6 cents [5,1] # N[7] = N[6]+coins[0] = 2+1 = 3 : 3 coins for 7 cents [5,1,1] # N[8] = N[5]+coins[2] = 1+1 = 2 : 2 coins for 8 cents [5,3] # N[9] = N[8]+coins[0] = 2+1 = 3 : 3 coins for 9 cents [5,1,3] # N[10] = N[5]+coins[2] = 1+1 = 2 : 2 coins for 10 cents [5,5] #======================================================================= # return makeExactChange(numberOfCoinsPerSubtotal(t, coins), t); def subsetSumOfCoins( coins, t ) : C = numberOfCoinsPerSubtotal( t, coins ) return makeExactChange( C, t ) #====================================================================== # Compute the number of coins in each subtotal amount of money # from 0 to t. #====================================================================== from sys import maxint def numberOfCoinsPerSubtotal( t, coins ): C = [maxint] * ( t + 1 ) C[0] = 0 # base case for coin in coins: for subtotal in range( 1, t + 1 ): if coin <= subtotal and C[subtotal] > C[subtotal - coin] + 1: C[subtotal] = C[subtotal - coin] + 1 return C #====================================================================== # Now that we know how many coins it takes to get each subtotal # amount of money up to t, we can track the denomination of each coin: # keep subtracting the next subtotal that takes fewer coins than t. # For instance in the manual example above where t=10 and # coins = [1,3,5]; since t N[t=10] takes 2 coins, the next subtotal # that takes fewer coins than N[10] is N[5], and the next subtotal # that takes fewer coins than N[5] is N[0]. Hence the denominations # for N[10] are 10-5 and 5-0: [5,5]. #====================================================================== def makeExactChange( C, t ): result = [] while 0 < C[t]: big = t while C[big] <= C[t]: t -= 1 result.append( big - t ) return result;

import unittest from dynamic_programming import SubsetSum as algorithm class Test( unittest.TestCase ): def testSubsetSumOfCoins( self ): coin = [1, 5, 10, 25] t = 167 expResult = [1, 1, 5, 10, 25, 25, 25, 25, 25, 25] result = algorithm.subsetSumOfCoins( coin, t ) self.assertEquals( expResult, result )