Chapter 2 Recursion: The Mirrors

© 2005 Pearson Addison-Wesley. All rights reserved2-2 Recursive Solutions Recursion is an extremely powerful problem- solving technique –Breaks a problem in smaller identical problems –An alternative to iteration, which involves loops A sequential search is iterative –Starts at the beginning of the collection –Looks at every item in the collection in order A binary search is recursive –Repeatedly halves the collection and determines which half could contain the item –Uses a divide and conquer strategy

© 2005 Pearson Addison-Wesley. All rights reserved2-3 Recursive Solutions Facts about a recursive solution –A recursive method calls itself –Each recursive call solves an identical, but smaller, problem –A test for the base case enables the recursive calls to stop Base case: a known case in a recursive definition –Eventually, one of the smaller problems must be the base case

© 2005 Pearson Addison-Wesley. All rights reserved2-4 Recursive Solutions Four questions for construction of recursive solutions –How can you define the problem in terms of a smaller problem of the same type? –How does each recursive call diminish the size of the problem? –What instance of the problem can serve as the base case? –As the problem size diminishes, will you reach this base case?

© 2005 Pearson Addison-Wesley. All rights reserved2-5 A Recursive void Method: Writing a String Backward Problem –Given a string of characters, write it in reverse order Recursive solution –Each recursive step of the solution diminishes by 1 the length of the string to be written backward –Base case: write the empty string backward

© 2005 Pearson Addison-Wesley. All rights reserved2-6 A Recursive void Method: Writing a String Backward Figure 2.6 A recursive solution

© 2005 Pearson Addison-Wesley. All rights reserved2-7 Multiplying Rabbits (The Fibonacci Sequence) “Facts” about rabbits –Rabbits never die –A rabbit reaches sexual maturity exactly two months after birth, that is, at the beginning of its third month of life –At the beginning of every month, each sexually mature male-female pair gives birth to exactly one male-female pair

© 2005 Pearson Addison-Wesley. All rights reserved2-8 Multiplying Rabbits (The Fibonacci Sequence) Problem –Given a pair of rabbits in month 1 –How many pairs of rabbits are alive in month n? Recurrence relation rabbit(n) = rabbit(n-1) + rabbit(n-2)

© 2005 Pearson Addison-Wesley. All rights reserved2-9 Multiplying Rabbits (The Fibonacci Sequence) Figure 2.10 Recursive solution to the rabbit problem

© 2005 Pearson Addison-Wesley. All rights reserved2-10 Multiplying Rabbits (The Fibonacci Sequence) Base cases –rabbit(2), rabbit(1) Recursive definition rabbit(n) = 1if n is 1 or 2 rabbit(n-1) + rabbit(n-2)if n > 2 Fibonacci sequence –The series of numbers rabbit(1), rabbit(2), rabbit(3), and so on

© 2005 Pearson Addison-Wesley. All rights reserved2-11 Implementation: Multiplying Rabbits (Recursive) int rabbit(int n){ // Precondition: n is a positive integer. // Postcondition: Returns the nth Fibonacci // number. if (n <= 2) return 1; else // n > 2, so n-1 > 0 and n-2 > 0 return rabbit(n-1) + rabbit(n-2); }

© 2005 Pearson Addison-Wesley. All rights reserved2-12 Implementation: Multiplying Rabbits (Iterative) int iterativeRabbit(int n){ // Iterative solution to the rabbit problem. // initialize base cases: int previous = 1; // initially rabbit(1) int current = 1; // initially rabbit(2) int next = 1; // result when n is 1 or 2 // compute next rabbit values when n >= 3 for (int i = 3; i <= n; ++i){ // current is rabbit(i-1), previous is rabbit(i-2) next = current + previous; // rabbit(i) previous = current; // get ready for current = next; // next iteration } return next; }

© 2005 Pearson Addison-Wesley. All rights reserved2-13 Binary Search A high-level binary search if (anArray is of size 1) { Determine if anArray’s item is equal to value } else { Find the midpoint of anArray Determine which half of anArray contains value if (value is in the first half of anArray) { binarySearch (first half of anArray, value) } else { binarySearch(second half of anArray, value) }

© 2005 Pearson Addison-Wesley. All rights reserved2-14 Binary Search Implementation issues: –How will you pass “half of anArray ” to the recursive calls to binarySearch ? –How do you determine which half of the array contains value? –What should the base case(s) be? –How will binarySearch indicate the result of the search?

© 2005 Pearson Addison-Wesley. All rights reserved2-15 Organizing Data: The Towers of Hanoi Figure a) The initial state; b) move n - 1 disks from A to C; c) move one disk from A to B; d) move n - 1 disks from C to B

© 2005 Pearson Addison-Wesley. All rights reserved2-16 The Towers of Hanoi Pseudocode solution solveTowers(count, source, destination, spare) if (count is 1) { Move a disk directly from source to destination } else { solveTowers(count-1, source, spare, destination) solveTowers(1, source, destination, spare) solveTowers(count-1, spare, destination, source) }

© 2005 Pearson Addison-Wesley. All rights reserved2-17 Recursion and Efficiency Some recursive solutions are so inefficient that they should not be used Factors that contribute to the inefficiency of some recursive solutions –Overhead associated with method calls –Inherent inefficiency of some recursive algorithms Do not use a recursive solution if it is inefficient and there is a clear, efficient iterative solution

© 2005 Pearson Addison-Wesley. All rights reserved2-18 Fall 2006 – Midterm Question Consider a salesman at a grocery store. The salesman uses a scale with two buckets to measure the weight of the purchased items. The salesman also has a finite number of metal weights (e.g., 1kg, 2kg, 5kg, 8kg, etc.) to be used in the weighing process. Write a recursive function that checks whether these metal weights are sufficient to weigh a purchased item. –Your recursive function should take the available metal weights and the weight of the purchased item as input. The function returns true if the metal weights can be used to weigh the item; returns false otherwise. –For example, if the available metal weights are 1kg, 2kg, 5kg and 8kg, and the purchased item is 13kg, the salesman can put the purchased item into one bucket and the 5kg and 8kg metal weights into the other bucket to measure the weight of the item. If the purchased item is 12kg, the salesman can put the purchased item and the 1kg metal weight into one bucket, and the 5kg and 8kg weights into the other bucket to measure the weight of the item. However, an 18kg item cannot be measured using these metal weights.

© 2005 Pearson Addison-Wesley. All rights reserved2-19 Fall 2006 – Midterm Question bool scale( const int *weights, const int size, const int item, const int i, const int sum ) { if ( sum == item ) return true; if ( i == size ) return false; bool result = false; result = scale( weights, size, item, i + 1, sum + weights[i]); if ( result == false ) result = scale( weights, size, item, i + 1, sum - weights[i]); if ( result == false ) result = scale( weights, size, item, i + 1, sum ); return result; }

© 2005 Pearson Addison-Wesley. All rights reserved2-20 Finding Connected Components Suppose that we want to segment cell nuclei in a tissue image. Suppose that it is a gray level image (its pixels contain intensity values between 0 and 255). For this purpose, we first apply some image processing to obtain its black-and- white form. In this form, each pixel value is either 0 and 1. Then, we are going to find the connected components on the black pixels. So, we write a recursive function for finding the connected components.

© 2005 Pearson Addison-Wesley. All rights reserved2-21 Finding Connected Components Suppose that we want to segment cell nuclei in a tissue image. Suppose that it is a gray level image (its pixels contain intensity values between 0 and 255). For this purpose, we first apply some image processing to obtain its black-and- white form. In this form, each pixel value is either 0 and 1. Then, we are going to find the connected components on the black pixels. So, we write a recursive function for finding the connected components.

Finding Connected Components Similarly, in the image below, we want to identify individual buildings. Connected components analysis can be used after obtaining a black-and-white image of buildings. © 2005 Pearson Addison-Wesley. All rights reserved2-22

Finding Connected Components Similarly, in the image below, we want to identify individual buildings. Connected components analysis can be used after obtaining a black-and-white image of buildings. © 2005 Pearson Addison-Wesley. All rights reserved2-23

© 2005 Pearson Addison-Wesley. All rights reserved2-24 Finding Connected Components int **findConnectedComponents(int **A, int row, int column){ int **labels, i, j, currLabel; labels = new int *[row]; for ( i = 0; i < row; i++ ){ labels[i] = new int[column]; for ( j = 0; j < column; j++ ) labels[i][j] = 0; } currLabel = 1; for ( i = 0; i < row; i++ ) for ( j = 0; j < column; j++ ) if ( A[i][j] && !labels[i][j] ) fourConnectivity( A, labels, row, column, i, j, currLabel++); return labels; }

© 2005 Pearson Addison-Wesley. All rights reserved2-25 Finding Connected Components void fourConnectivity( int **A, int **labels, int row, int column, int i, int j, int currLabel){ if (A[i][j] == 0) return; if (labels[i][j] > 0) return; labels[i][j] = currLabel; if (i-1 >= 0) fourConnectivity(A,labels,row,column,i-1,j,currLabel); if (i+1 < row) fourConnectivity(A,labels,row,column,i+1,j,currLabel); if (j-1 >= 0) fourConnectivity(A,labels,row,column,i,j-1,currLabel); if (j+1 < column) fourConnectivity(A,labels,row,column,i,j+1,currLabel); }