Recursion

1. Recursion

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 C++

4. Factorial Solution Complexity? int fact( int n ) { if ( n == 0 || n == 1 ) { return 1; } else { return fact(n-1)*n; } // What are some problems here

5. Fibonacci Numbers • Find fib(n) given the definition : • fib(0) = 1 • f ib(1) = 1 • fib(n) = fib(n-1) + fib(n-2)

6. Solution Complexity? public int fib( int n ) { if ( n == 0 || n == 1 ) { return 1; } else { return fib( n - 1 ) + fib( n - 2 ); } } naturally recursive, however better-off being solved iteratively. What are the problems here?

7. An Example Fib(6) Fib(5) fib(4) Fib(4) Fib(3)Fib(3)Fib(2) Fib(3)Fib(2)Fib(2) Fib(1) Fib(2) Fib(1) Fib(1) Fib(0) 1 1 1 1 1 Fib(2) Fib(1) 2 1 2 2

8. Cache the results in the array static int fib(int n) { int[] sols = new int[n + 1]; sols[0] = 1; sols[1] = 1; return fib(n, sols) } static int fib(int n, int[] sols) { if (sols[n] > 0) return sols[n]; int sol = fib(n-1, sols) + fib(n-2, sols); sols[n] = sol; return sol; } Complexity?

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

10. 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.

11. An Approach • To solve such problems ask yourself: "if I had solved the n-1 case could I solve the n case?" • If the answer to this question is positive you proceed under the outrageous assumption that the n-1 case has been solved. Oddly enough this works, so long as there is some base case (often when n is zero or one) which can be treated as a special case.

12. N rings from pole A to pole C? • If you know how to move n-1 rings from one pole to another then simply move n-1 rings to the spare pole - there is only one ring left on the source pole now, simply move it to the destination, then pile the rest of them from the spare pole onto the destination pole.

13. Hanoi… • http://www.mazeworks.com/hanoi/ void moveTower (int numDisks, int start, int end, int temp) { if (numDisks == 1) moveOneDisk (start, end); else { moveTower (numDisks-1, start, temp, end); moveOneDisk (start, end); moveTower (numDisks-1, temp, end, start); } }

14. 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

15. 5 7 2 6 3 N-1 N’th Print reverse - solution • Solution form Write C++ & 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

16. 5 7 2 6 3 N-1 N’th Print forward • Print set of values in normal order • Solution form Write C++ & 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

17. 5 7 2 6 3 first rest S E Print – alternative solution • Print set of values in order • Alternative solution Write C++& 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

18. 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

19. 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)

20. 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)

21. 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

22. 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)

23. 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

24. 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)

25. 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

26. 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?

27. 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!

28. 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)

29. Blob Count • A Blob is the union of adjacent cells • Cells are represented by a two dimensional array: boolean[][] isFull; • The problem is to count the size of a blob starting from coordinates x, & y.

30. Blob Count, the plan • have a recursive method that adds 1 to the count if the given cell is occupied, and recursively adds the numbers of 8 neighbors • we need to remember which cells we already counted, and can use boolean[][] counted for this purpose

31. Blob count, algorithm public int count(boolean[][] cells, int x, int y) { boolean[][] counted = new boolean[cells.length][cells[0].length] return count_(cells, counted, x, y) }

32. Blob Algorithm private int count_(boolean[][] cells, boolean[][] counted, int x, int y) { if ( x < 0 || x >= cells.length) return 0; if (y < 0 || y >= cells[0].length) return 0; if ( counted[x][y]) return 0; if (!isFull[x][y]) return 0; counted[x][y] = true; return 1 + count_(cells, counted, x+1, y) + count_(cells, counted, x, y+1) + count_(cells, counted, x+1, y+1) + … }

33. 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 • palindrome • Count using S only • Maze & blob counting • Fractals? Recursive-descent-parsing? • …on to data structures(stacks, queues, lists, trees, …)