CS102 – Recursion

1. CS102 – Recursion David Davenport Bilkent University Ankara – Turkey email: david@bilkent.edu.tr Spring 2003

2. What is Recursion? • Method of solving problems • Alternative to iteration • All recursive solutions can be implemented iteratively • Characteristic... • recursive methods call themselvesrecursive functions defined in terms of (simpler versions) themselves • General recursive case & stopping cases

3. Factorial N • Iterative • N! = N * N-1 * N-2 … 2 * 1 • Recursive • 1! = 1 - stopping case • N! = N * (N-1)! - general case implement & trace in Java

4. So, why use recursion? • Advantages... • Interesting conceptual framework • Intuitive solutions to difficult problems • But, disadvantages... • requires more memory & time • requires different way of thinking!

5. Towers of Hanoi • The Problem… • Rules • only move one disk at a time • can’t put a disk on a smaller one • Monks believed world would end once solved for 64 disks! Takes 580 billion years! at one move per sec.

6. 5 7 2 6 3 Print reverse • Print first N values in reverse • Simple iterative solution… • Recursive? • Hint - what is the simplest case? • Write out sequence of cases& look for previous case within each casegeneralise & note stopping case! Given... print... 3, 6, 2, 7, 5

7. 5 7 2 6 3 N-1 N’th Print reverse - solution • Solution form Write Java & trace To print N values in reverse order if N > 0 Print the N’th value & then Print the preceding N-1 values in reverse order

8. 5 7 2 6 3 N-1 N’th Print forward • Print set of values in normal order • Solution form Write Java & trace To print N values in normal order if N > 0 Print the first N-1 values in normal order & then Print the N’th value

9. 5 7 2 6 3 first rest S E Print – alternative solution • Print set of values in order • Alternative solution Write Java & trace To print values btw S & E in order if S <= E Print the value at S & then Print the values btw S+1 & E in order

10. Think about solution in terms of these two sub-problems 5 7 2 6 3 N-1 N’th Find Max • Find max of first N values To find max of N values if N = 1 return N’th value else return greater of N’th & max of preceding N-1 values

11. Think about simplest cases of problem & more general solution in terms of these two sub-problems 5 7 2 6 3 N-1 N’th Sequential search Search( first N values of X for target) if N = 0 return not found else if N’th value is target return found at N’th else return search( first N-1 values of X for target) O(N)

12. 5 7 2 6 3 N-1 N’th Summation • Find the sum of the first N values To sum first N values of X if N =0 return 0 else return N´th value + sum of preceding N-1 values

13. 5 7 2 -1 3 rest first Summation 2 • Find sum of all values before sentinel (-1)

14. 7 9 1 2 3 5 6 first half second half S E middle Binary Search O(log2N) • Think about using telephone directory • Values must be in order! • Have knowledge of data distribution • If unknown? (number guessing game)

15. Binary Search (continued…) Search( X btw S & E for target) if S > E return not found else if ( middle value is target) return found at middle else if ( target < middle value) return search( first half of X for target) else return search( second half of X for target) First half isS, middle-1 Second half is middle+1, E

16. 5 7 2 6 3 N-1 N’th locOfMax Selection sort O(N2) selectionSort( first N values of X) if N > 1 locOfMax = findmax( first N values of X) exchange N’th value with value at locOfMax selectionSort( first N-1 values of X)

17. 9 1 6 6 pivot partition 5 2 7 1 3 3 9 5 7 2 Sort first part Sort second part pivot QuickSort O(Nlog2N) QuickSort( X btw S & E) if S < E pivot loc. = partition( X btw S & E) QuickSort( first part of X) QuickSort( second part of X) First part is S to pivot loc.-1 Second part is pivot loc.+1 to E

18. Sort second half 7 3 9 6 Sort first half 1 1 5 2 3 7 5 9 2 6 merge MergeSort O(Nlog2N) Merge is the basis of sequential processing MergeSort( X btw S & E) if S < E MergeSort( first half of X) MergeSort( second half of X) Merge( first & second halves of X)

19. 5 7 2 0 3 first rest S Counting • Count number of values before zero Zero is sentinel value& so is guaranteed to exist. This is a common form of representation for strings! count no. values before zero starting at S if value at S is zero return 0 else return 1 + count no. values starting at S+1

20. Common mistakes! (1) sum = 0; public void count( int[] X, int S) { if ( X[S] != 0) { sum++; count( X, S+1); } } • Problems: • Need another method to getCount() • What happens if called again?

21. Common mistakes! (2) public int count( int[] X, int S) { int sum; if ( X[S] == 0) return sum; else { sum++; return count( X, S+1); } } • Problem: • New instance of local variable for each instantiation!

22. R A D A R middle first last Palindrome isPalindrome( word) if first >= last return true else if first char of word != last char of word return false else return isPalindrome( middle of word)

23. The plan… • Print set of N values forward & backward • Repeat using S & E • find max value • search for target - sequential • binary search • sort - selection, quick, merge • Count using S only • palindrome • Maze & blob counting • Fractals? Recursive-descent-parsing? • …on to data structures(stacks, queues, lists, trees, …)