A simple recursive algorithm for change making is easily written but inefficient in Java

Printer Code-39 in Java A simple recursive algorithm for change making is easily written but inefficient
A simple recursive algorithm for change making is easily written but inefficient
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For example, let us see how we can make 63 cents in change Clearly, one coin will not suffice We can compute the number of coins required to make 1 cent of change and 62 cents of change independently (these are 1 and 4, respectively) We obtain these results recursively, so they must be taken as optimal (jt happens that the 62 cents is given as two 21-cent pieces and two 10-cent pieces) Thus we have a method that uses five coins If we split the problem into 2 cents and 61 cents, the recursive solutions yield 2 and 4, respectively, for a total of six coins We continue trying all the possibilities, some of which are shown in Figure 822 Eventually, we see a split into 2 1 cents and 42 cents, which is changeable in one and two coins, respectively, thus allowing change to be made in three coins The last split we need to try is 3 1 cents and 32 cents We can change 31 cents in two coins,
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Dynamic Programming
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Some of the subproblems solved recursively in Figure 823
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1 / / Return minimum number of coins to make change 2 / / Simple recursive algorithm that is very inefficient 3 int makechange( const vector<int> & coins, int change ) 4 ( 5 int minCoins = change;
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/ / Look for exact match with any single coin for( int i = 0; i < coinssize( ) ; i++ ) if ( coins[ i ] == change ) return 1; / / No match; solve recursively for( int j = 1; j < = change / 2; j++
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int thiscoins = makechange( coins, j ) + makechange( coins, change - j if( thiscoins < minCoins ) minCoins = thiscoins;
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return mincoins;
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A simple but inefficient recursive procedure for solving the coinchanging problem
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and we can change 32 cents in three coins for a total of five coins But the minimum remains three coins Again, we solve each of these subproblems recursively which yields the natural algorithm shown in Figure 823 If we run the algorithm to make small change, it works perfectly But like the Fibonacci calculations, this
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An alternative recursive algorithm for the coin-changing problem
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Our alternative recursive changemaking algorithm is still inefficient
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algorithm requires too much redundant work, and it will not terminate in a reasonable amount of time for the 63-cent case An alternative algorithm involves reducing the problem recursively by specifying one of the coins For example, for 63 cents, we can give change in the following ways, as shown in Figure 824 One 1-cent piece plus 62 cents recursively distributed One 5-cent piece plus 58 cents recursively distributed One 10-cent piece plus 53 cents recursively distributed One 2 1-cent piece plus 42 cents recursively distributed One 25-cent piece plus 38 cents recursively distributed Instead of solving 62 recursive problems, as in Figure 822, we get by with only 5 recursive calls, one for each different coin Again, a naive recursive implementation is very inefficient because it recomputes answers For example, in the first case we are left with a problem of making 62 cents in change In this subproblem, one of the recursive calls made chooses a 10-cent piece and recursively solves for 52 cents In the third case we are left with 53 cents One of its recursive calls removes the I-cent piece and also recursively solves for 52 cents This redundant work again leads to excessive running time If we are careful, however, we can make the algorithm run reasonably fast The trick is to save answers to the subproblems in an array This dynamic programming technique forms the basis of many algorithms A large answer depends only on smaller answers, so we can compute the optimal way to change 1 cent, then 2 cents, then 3 cents, and so on This strategy is shown in the function in Figure 825
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/ / Dynamic programming algorithm for change-making problem / / As a result, the coinsused array is filled with the minimum
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/ / number of coins needed for change from O->maxChange and / / lastcoin contains one of the coins needed to make the change
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void makechange( const vector<int> & coins, int maxchange, vector<int> & coinsused, vector<int> & lastcoin
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int differentcoins = coinssize( ) ; coinsUsedresize( maxchange + 1 ) ; lastCoinresize( maxchange + 1 ) ; coinsUsed[ 0 ] = 0; lastcoin[ 0 ] = 1; for( int cents = 1; cents <= maxchange; cents++