Recursion and Dynamic Programming CSC 172 SPRING 2004 LECTURE 9

Grad TA help sessions?  What is a good time?  Office hours are kind of early  Would people like 5ish?  Induction  Big-Oh  Recurrence  Code Style review

Project 1  General Observations  Your “style” should travel with you  Go to the writing place  Some people paraphrased Bently, but got it wrong  While big-Oh analysis was not required, some notion of absraction and generalization is appropriate  Some people lost large amounts because of no disclaimer

Recursive definition WWhat is a list? AA list can be “empty” AA list can be an item, followed by a list.

Recursive definition  What is a Tree?  A tree can be “empty”  A tree can be a node with “children” that are trees.

Recursive definition  What is a number? (integer)  A single digit is a number {0,1,2,3,4,5,6,7,8,9}  A number is a digit followed by a number

Recursive Definition  What is a simple (4 func) arithmetic expression ?  (7+5)* (3-2)  A number is an arithmetic expression  If X is an ae, then (X) is an ae  If X and Y are both aes,  then X+Y, X-Y, X*Y, X/Y are all ae

Recursion  We often think of recursion as a “programming technique”  Recursion is also a design pattern for data structures  Recursion is a way to think about problem solving

A Theological Aside Disclaimer: you don’t have to believe  Can computer science inform theology?  Christian notion of “Trinity”?  "A three-fold personality existing in one divine being or substance; the union in one God of Father, Son, and Holy Spirit as three infinite, co-equal, co- eternal persons; one God in three persons."  from about 400 AD to about 1800 AD, countless people were put to death for refusing to believe in the idea of "one God in three persons."

Serpinski Triangle This design is called Sierpinski's Triangle, after the Polish mathematician Waclaw Sierpinski who described it in Among these is its fractal or self-similar character. The large blue triangle consists of three smaller blue triangles, each of which itself consists of three smaller blue triangles, each of which..., a process of subdivision which could, with adequate screen resolution, be seen to continue indefinitely.

Serpinski Triangle What is a Serpinski Triangle? A Serpinski Triangle is a triangle made up of 3 Serpinski triangles.

Weiss’s fundamental rules of recursion 1. Base case: always have at least one case that can be solved without recursion 2. Make progress: Any recursive call must progress toward the base case 3. “You gotta believe”: Always assume that the recursive call works. (In proofs “BTIH”). 4. Compound Interest rule: Never duplicate work by solving the same instance of a problem in separate recursive calls.

Example : Fibonacci Number // Bad algorithm public static long fib (int n) { if (n <= 1) return n; else return fib (n-1) + fib(n-2); }

Why is it bad?

Dynamic Programming  If you can mathematically express a problem recursively, then you can express it as a recursive algorithm.  However, sometimes, this can be inefficiently expressed by a compiler  Fibonacci numbers  To avoid this recursive “explosion” we can use dynamic programming

Example Problem: Making Change  For a currency with coins C 1,C 2,…C n (cents) what is the minimum number of coins needed to make K cents of change?  US currency has 1,5,10, and 25 cent denominations  Anyone got a 50-cent piece?  We can make 63 cents by using two quarters & 3 pennies  What if we had a 21 cent piece?

63 cents  25,25,10,1,1,1  Suppose a 21 cent coin?  21,21,21 is optimal

Recursive Solution 1. If we can make change using exactly one coin, then that is a minimum 2. Otherwise for each possible value j compute the minimum number of coins needed to make j cents in change and K – j cents in change independently. Choose the j that minimizes the sum of the two computations.

public static int makeChange (int[] coins, int change){ int minCoins = change; for (int k = 0;k<coins.length;k++) if (coins[k] == change) return 1; for (int j = 1;j<= change/2;j++) { int thisCoins = makeChange(coins,j) +makeChange(coins,change-j); if (thisCoins < minCoins) minCoins = thisCoins; } return minCoins; }// How long will this take?

How many calls? 63¢ 62¢2¢2¢ 1¢1¢ 61¢31¢32¢...

How many calls? 63¢ 3¢3¢ 1¢1¢ 4¢4¢61¢62¢... 2¢2¢

How many calls? 63¢ 1¢1¢ 3¢3¢ 1¢1¢ 4¢4¢61¢62¢... 2¢2¢ 1¢

How many calls? 63¢ 1¢1¢ 3¢3¢ 1¢1¢ 4¢4¢61¢62¢... 2¢2¢ 1¢ 3¢3¢1¢1¢4¢4¢ 61¢... 2¢2¢

How many times do you call for 2¢? 63¢ 1¢1¢ 3¢3¢ 1¢1¢ 4¢4¢61¢62¢... 2¢2¢ 1¢ 3¢3¢1¢1¢4¢4¢ 61¢... 2¢2¢

Some Solutions 1(1) & 62(21,21,10,10) 2(1,1) & 61(25,25,10,1).. 21(21) & 42(21,21) …. 31(21,10) & 32(21,10,1)

Improvements?  Limit the inner loops to the coins 1 & 21,21,10,10 5 & 25,21,10,1,1 10 & 21,21,10,1 21 & 21,21 25 & 25,10,1,1,1 Still, a recursive branching factor of 5 How many times do we solve for 52 cents?

public static int makeChange (int[] coins, int change){ int minCoins = change; for (int k = 0;k<coins.length;k++) if (coins[k] == change) return 1; for (int j = 1;j<= coins.length;j++) { if (change < coins[j]) continue; int thisCoins = 1+makeChange(coins,change-coins[j]); if (thisCoins < minCoins) minCoins = thisCoins; } return minCoins; }// How long will this take?

How many calls? 63¢ 58¢53¢62¢42¢38¢ 48¢43¢52¢32¢13¢ 57¢52¢61¢41¢37¢

Tabulation aka Dynamic Programming  Build a table of partial results.  The trick is to save answers to the sub-problems in an array.  Use the stored sub-solutions to solve the larger problems

DP for change making  Find optimum solution for 1 cent  Find optimum solution for 2 cents using previous  Find optimum solution for 3 cents using previous  …etc.  At any amount a, for each denomination d, check the minimum coins for the (previously calculated) amount a-d  We can always get from a-d to a with one more coin

public static int makeChange (int[] coins, int differentcoins, int maxChange, int[] coinsUsed, int [] lastCoin){ coinsUsed[0] = 0; lastCoin[0]=1; for (int cents = 1; cents <= maxChange; cents++) { int minCoins = cents; int newCoin = 1; for (int j = 0;j<differentCoins;j++) { if (coins[j] > cents) continue; if (coinsUsed[cents – coins[j]]+1 < minCoins){ minCoins=coinsUsed[cents – coins[j]]+1; newCoin = coins[j]; } coinsUsed[cents] = minCoins; lastCoin[cents] = newCoin; }

Dynamic Programming solution  O(NK)  N denominations  K amount of change  By backtracking through the lastCoin[] array, we can generate the sequence needed for the amount in question.