Why use recursion? Some problems are more easily expressed in a recursive definition

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1 CS1706 Intro to Object Oriented Dev II - Spring 04 Announcements Week 6 Program 1 due Friday Grades I will have exam 1 graded Wednesday Material Recursion

2 Recursion Definitions Recursion: see Recursion a process in which the result of each repetition is dependent upon the result of the next repetition. Simplifies program structure at a cost of function calls Recursive function is a function that calls itself in the process of calculating its value Why use recursion? Some problems are more easily expressed in a recursive definition

3 More Definitions Hofstadter's Law It always takes longer than you expect, even when you take into account Hofstadter's Law. Sesquipedalian a person who uses words like sesquipedalian. Yogi Berra: Its déjà vu all over again.

4 Example Factorial (mathematical) definition 1! = 1 N! = N * (N-1)! for N > 1 So... 10! = 10 * 9! = 10 * 9 * 8! =... And the code for it is: public int factorial(int n) { assert(n >= 1); if (n == 1) return 1; else return n * factorial(n-1);

5 Anatomy of Recursive f()s Conditional processing one or more steps in the solution Base case Recursive call Divide and conquer Consume what you produce Leap of faith! public int fact(int n) { assert(n >= 1); if (n == 1) return 1; else return n * fact(n-1);

6 Anatomy of Recursive f()s Conditional processing Base case at least one of the steps does not involve a recursive call Recursive call Divide and conquer Consume what you produce Leap of faith! public int fact(int n) { assert(n >= 1); if (n == 1) return 1; else return n * fact(n-1);

7 Anatomy of Recursive f()s Conditional processing Base case Recursive call at least one of the steps does involve a recursive call Divide and conquer Consume what you produce Leap of faith! public int fact(int n) { assert(n >= 1); if (n == 1) return 1; else return n * fact(n-1);

8 Anatomy of Recursive f()s Conditional processing Base case Recursive call Divide and conquer The recursive call should make the problem smaller It should get it closer to the base case. Consume what you produce Leap of faith! public int fact(int n) { assert(n >= 1); if (n == 1) return 1; else return n * fact(n-1);

9 Anatomy of Recursive f()s Conditional processing Base case Recursive call Divide and conquer Consume what you produce Leap of faith! You must believe that your function will work Your definition is RECURSIVE public int fact(int n) { assert(n >= 1); if (n == 1) return 1; else return n * fact(n-1); Example of of Tail Recursion

10 Recursive Execution Trace Factorial (5) return * Factorial (4) return 24 4 * Factorial (3) 3 * Factorial (2) return 6 return 2... and then the return sequence First the recursive descent... 2 * Factorial (1) 1 return 1

11 Summation Example public int sum(int num) { if (num == 1) return 1; else return num + sum(num - 1); Example of of Tail Recursion N k = 1 k = N ( N 2 + 1)

12 Recursion Attributes Every recursive algorithm can be implemented nonrecursively. recursion <==> iteration Eventually, the routine must not call itself, allowing the code to "back out". Recursive routines that call themselves continuously are termed: infinite recursion <==> infinite loop Recursion is inefficient at runtime.

13 Recursive Array Summation Here is a recursive function that takes an array of integers and computes the sum of the elements: // X[] array of integers to be summed // Start start summing at this index... // Stop... and stop summing at this index // int SumArray(int X[], int Start, int Stop) { // error check if (Start > Stop Start < 0 Stop < 0) return 0; else if (Start == Stop) // base case return X[Stop]; else // recursion return (X[Start] + SumArray(X, Start + 1, Stop));

14 Recursive Array Summation Trace The call: final int Size = 5; int X[Size] = {37, 14, 22, 42, 19; SumArray(X,0,Size- 1);//Stop is last valid index would result in the recursive trace: // return values: SumArray(X, 0, 4) // == 134 return(x[0]+sumarray(x,1,4)) // == return(x[1]+sumarray(x,2,4)) // == return(x[2]+sumarray(x,3,4) ) // == return(x[3]+sumarray(x,4,4)) // == return X[4] // == 19

15 Recursive Design Problem: Code a function void intcomma( long ) that outputs the integer argument comma separated : Example: the call: intcomma ( ) ; displays: 123,456,789 Solution: in class Example of of Head Recursion

16 Middle Decomposition Problem: Given an array of integers of n+1 elements code a function to return the index of the maximum value in the array. Solution: Check if the middle element is the largest if so return its index otherwise return the index of either the largest element in the lower half or the largest element in the upper half, whichever is the larger of the two.

17 Middle Decomposition Code public int rmax(int ray[], int start, int end ) { final int Unknown = -1; int mid, h1max, h2max; if (end < start) return Unknown; Example of of splitting into into halves recursion mid = (start + end) / 2; h1max = rmax(ray, start, mid-1); //left half if (h1max == Unknown) h1max = start; h2max = rmax(ray, mid+1, end); //right half if (h2max == Unknown) h2max = end; if ( (ray[mid] >= ray[h1max]) && (ray[mid] >= ray[h2max]) ) return mid; else return( (ray[h1max] > ray[h2max])? h1max : h2max ); Unknown checks ensure that indices are within array subscript range

18 Call Tree Traces Given: ray = [0] [1] [2] [3] [4] Call Tree Trace of rmax(ray, 0, 4); Middle decomposition (splitting problem into into halves), recursive functions are are best best traced with with tree tree diagrams h1max=0 0 mid=2 h2max=4 4 h1max=0 h2max=1 h1max=3 h2max=4 rmax(ray, 0, 1) rmax(ray, 3, 4) unknown 1 mid=0 unknown mid=3 4 h2max h1max=1 h2max=1 h1max=4 =4 rmax(ray, 0, -1) rmax(ray, 1, 1) rmax(ray, 3, 2) rmax(ray, 4, 4) unknown unknown unknown mid=1 unknown mid=4 rmax(ray, 1, 0) rmax(ray, 2, 1) rmax(ray, 4, 3) rmax(ray, 5, 4)

19 Edges / Center Decomposition Problem: sort a subset, (m:n), of an array of integers (ascending order) Solution: Find the smallest and largest values in the subset of the array (m:n) and swap the smallest with the mth element and swap the largest with the nth element, (i.e. order the edges). Sort the center of the array (m+1: n-1). Solution Trace: m unsorted array n [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] after call#1 [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] after call#3 [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] Variation of of the the selection sort sort algorithm

20 Recursive Sorting Code public void duplexselection(int ray[], int start, int end ) { Integer mini = new Integer(start), maxi = new Integer(end); if (start < end) { //start==end only 1 element to order findminimaxi( ray, start, end, mini, maxi); swapedges(ray, start, end, mini, maxi); duplexselection( ray, start+1, end-1); public void findminimaxi(int ray[], int start, int end, Integer mini, Integer maxi) { if (start < end) { //subset to search exists if (ray[start] < ray[mini.intvalue()]) mini = new Integer(start); else if (ray[start] > ray[maxi.intvalue()]) maxi = new Integer(start); findminimaxi( ray, start+1, end, mini, maxi);

21 Recursive Sorting Code public void swapedges(int ray[], int start, int end, Integer mini, Integer maxi) { //check for swap interference if ( (mini.intvalue() == end) && (maxi.intvalue() == start) ) { swap( ray, start, end ); //check for low 1/2 interference else if (maxi.intvalue() == start) { swap( ray[maxi.intvalue()], ray[end] ); swap( ray[mini.intvalue()], ray[start] ); // (mini == end) no interference else { swap( ray[mini.intvalue()], ray[start] ); swap(ray[maxi.intvalue()], ray[end] ); private void swap (int ray[], int x, x, int y) { int tmp= ray[x] ; ray[x] = ray[y] ; ray[y] = ray[tmp] ;

22 Maze Can you write a program that traverses a maze from one corner to the opposite? Start Recursive definition: Traverse x 1,y 1 to x 2, y 2 if Path from x i,y i to x i+1,y i and Traverse x i+1,y i to x 2, y 2 What is Path defined as? End

23 Maze in pieces Traverse from (0,0) to (1,1)? First consider neighbors of (0,0) Neighbors: (0,1), (1,0) Path from (0,0) to (1,0)? If yes, then reduce the problems to Traverse from (1,0) to (1,1)? If no, try (0,0) to (1,0)? Start Start End Work bottom maze End

24 Maze How to represent a maze? private int[][] grid = { {1,1,1,0,1,1,0,0,0,1,1,1,1, {1,0,1,1,1,0,1,1,1,1,0,0,1, {0,0,0,0,1,0,1,0,1,0,1,0,0, {1,1,1,0,1,1,1,0,1,0,1,1,1, {1,0,1,0,0,0,0,1,1,1,0,0,1, {1,0,1,1,1,1,1,1,0,1,1,1,1, {1,0,0,0,0,0,0,0,0,0,0,0,0, {1,1,1,1,1,1,1,1,1,1,1,1,1 ; Binary 2-dimensional array if grid[i][j] == 1, then block is not available if grid[i][j] == 0, then block is open if grid[i][j] == TRIED, then we tried it and didn t work if grid[i][j] == PATH, then it is part of solution

25 # recursive calls Where can we go from here? Start Up Down Left Right End

26 Maze solution public boolean traverse (int row, int column){ boolean path = false; if (valid (row, column)) { grid[row][column] = TRIED; // we are trying it if (endposition(row, column)) { path = true; // the maze is solved else { return path; path = traverse (row+1, column); // down if (! path ) path = traverse (row, column+1); // right if (! path ) path = traverse (row-1, column); // up if (! path ) path = traverse (row, column-1); // left if (path) // this loc is part of final path grid[row][column] = PATH;

27 valid() private boolean valid (int row, int column) { boolean result = false; // check if cell is in the bounds of the matrix if (row >= 0 && row < grid.length && column >= 0 && column < grid[row].length) // check if cell is not blocked // and not previously tried if (grid[row][column] == 0)&& (grid[row][column]!= TRIED) result = true; return result;

28 endposition() Is (row, column) the end position? In this example, that is the right/bottom corner private boolean endposition (int row, int column) { return row == grid.length-1 && column == grid[0].length-1;

29 Backtracking Knapsack Problem (very weak form) Given an integer total, and an integer array, determine if any collection of array elements within a subset of the array sum up to total. Assume the array contains only positive integers. Special Base Cases total = 0 : solution: the collection of no elements adds up to 0. total < 0 : solution: no collection adds to sum. start of subset index > end of subset index : solution: no such collection can exist.

30 Knapsack (continued) Inductive Step Check if a collection exists containing the first subset element. Does a collection exist for total - ray[ subset start ] from subset start + 1 to end of subset? If no collection exists containing ray[ subset start ] check for a collection for total from subset start + 1 to the end of the subset. Backtracking step. Function searches for alternative solution undoing previous possible solution search work.

31 Knapsack Solution Backtracking Recursive Solution public boolean Knap (int ray[], int total, int start, int end) { if (total == 0) // empty collection adds up to 0 return true; if ( (total < 0) (start > end) ) //no such return false; //collection exists //check for collection containing ray[start] if (Knap(ray, total-ray[start], start+1, end)) return true; // check for collection w/o ray[start] return (Knap(ray, total, start+1, end));

32 Knapsack Trace Trace Tree [0] [1] [2] [3] [4] ray = Knap(ray, 100, 0, 4) TRUE Knap(ray, 50, 1, 4) TRUE Knap(ray, 30, 2, 4) FALSE TRUE Knap(ray, -10, 3, 4) Knap(ray, 30, 3, 4) Backtracking steps FALSE TRUE Knap(ray, -30, 4, 4) Knap(ray, 30, 4, 4) TRUE Knap(ray, 0, 5, 4)

33 Recursion Underpinnings Compiler Support Every instance of a function execution (call) creates an Activation Record, (frame) for the function. Activation records hold required execution information for functions: Return value for the function Pointer to activation record of calling function Return memory address, (calling instruction address) Parameter storage Local variable storage

34 Runtime Stack Runtime Stack Snapshot Activation records are created and stored in an area of memory termed the runtime stack. Fny() activation record Fnx() activation record main() activation record local storage Parameter storage Return address Pointer previous act. rec. Fn return value local storage Parameter storage Return address Pointer previous act. rec. Fn return value local storage Parameter storage Return address Pointer previous act. rec. Fn return value

35 Storage Organization Typical program execution memory model: System privileges (not accessible to the program) Binary Code (text segment) Lowest memory addresss Static Data (data segment) Runtime Stack Function activation record management Dynamic memory structure management Heap Highest memory addresss

Recursion. Slides. Recursion

Recursion. Slides. Recursion 1 Slides 1. Table of Contents 2. Definitions 3. Simple 4. Recursive Execution Trace 5. Attributes 6. Recursive Array Summation 7. Recursive Array Summation Trace 8. Coding Recursively 9. Recursive Design