Recursion Liang Introduction to Java Programming Eighth Edition

Motivations Suppose you want to find all the files under a directory that contains

Motivations The Eight Queens puzzle is to place eight queens on a chessboard such

What is recursive method? FA method that calls itself F With each method call

Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); 0!=1 1!=1 2!=2*1!=2 3!=3*2=6 …. n!

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) Liang, Introduction to Java

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2)

animation Trace Recursive factorial Executes factorial(4) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial Executes factorial(3) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial Executes factorial(2) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial Executes factorial(1) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial Executes factorial(0) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial returns 1 Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial returns factorial(0) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial returns factorial(1) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial returns factorial(2) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial returns factorial(3) Liang, Introduction to Java Programming, Eighth Edition, (c)

animation Trace Recursive factorial returns factorial(4) Liang, Introduction to Java Programming, Eighth Edition, (c)

factorial(4) Stack Trace Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education,

Factorial public static long factorial(int n) { System. out. println( "Method" + n );

Infinite Recursion public static long factorial(int n) { return result = n * factorial(n

Fibonacci Numbers Fibonacci series: 0 1 1 2 3 5 8 13 21 34

Fibonnaci Numbers, cont. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education,

Fibonnaci Numbers, cont. static int count; /** The method for finding the Fibonacci number

Case Study: Non-recursive version of Fibonacci public long static long fib(long n) { f

Dynamic programming F Process of solving subproblems, then combine the solutions to obtain an

f 0 f 1 f 2 Fibonacci series: 0 1 1 2 3 5

Characteristics of Recursion All recursive methods have the following characteristics: – One or more

Problem Solving Using Recursion Let us consider a simple problem of printing a message

Output for n. Println(“hello”, 3) 3>1 => hello n. Println(“hello”, 2) 2>1 => hello

Think Recursively The is. Palindrome problem approach 1 public static boolean is. Palindrome(String s)

Think Recursively Many problems can be solved using recursion if you think recursively. For

Recursive Helper Methods The preceding recursive is. Palindrome method is not efficient, because it

Binary Search public static int binary. Search(int[] list, int key) { int low =

Recursive Binary Search 1. 2. 3. Case 1: If the key is less than

Recursive Implementation /** Use binary search to find the key in the list */

GCD: Greatest Common Divisor gcd(2, 3) = 1 gcd(2, 10) = 2 gcd(25, 35)

Case Study: GCD Algorithms Version 1 public static int gcd(int m, int n) {

Case Study: GCD Algorithms Version 2 for (int k = n; k >= 1;

Case Study: GCD Algorithms Version 3 public static int gcd(int m, int n) {

Euclid’s algorithm Let gcd(m, n) denote the gcd for integers m and n: FIf

Euclid’s Algorithm Implementation public static int gcd(int m, int n) { if (m %

Directory Size The preceding examples can easily be solved without using recursion. This section

Directory Size The size of the directory can be defined recursively as follows: Liang,

Towers of Hanoi F There are n disks labeled 1, 2, 3, . .

Towers of Hanoi, cont. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson

Solution to Towers of Hanoi The Towers of Hanoi problem can be decomposed into

Solution to Towers of Hanoi F F F Move the first n - 1

$Fractals? A fractal is a geometrical figure just like triangles, circles, and rectangles, but$

Sierpinski Triangle 1. 2. 3. 4. It begins with an equilateral triangle, which is

Sierpinski Triangle Solution Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education,

Eight Queens (Homework 3) The Eight Queens puzzle is to place eight queens on

Eight Queens (Homework 3) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson

Questions? Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All

Case Study: Fibonacci Numbers /** The method for finding the Fibonacci number */ public

Complexity for Recursive Fibonacci Numbers Since and Therefore, the recursive Fibonacci method takes. Liang,

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Motivations Suppose you want to find all the files under a directory that contains a particular word. How do you solve this problem? There are several ways to solve this problem. An intuitive solution is to use recursion by searching the files in the subdirectories recursively. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 2

Motivations The Eight Queens puzzle is to place eight queens on a chessboard such that no two queens are on the same row, same column, or same diagonal. How do you write a program to solve this problem? A good approach to solve this problem is to use recursion. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 3

What is recursive method? FA method that calls itself F With each method call the problem becomes simpler F Must have a condition that leads to the method no longer making another call on itself Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 4

Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); 0!=1 1!=1 2!=2*1!=2 3!=3*2=6 …. n! = n * (n-1)! Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 5

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 6

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 7

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 8

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) = 3 * ( 2 * (1 * factorial(0))) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 9

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) = 3 * ( 2 * (1 * factorial(0))) = 3 * ( 2 * ( 1 * 1))) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 10

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) = 3 * ( 2 * (1 * factorial(0))) = 3 * ( 2 * ( 1 * 1))) = 3 * ( 2 * 1) Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 11

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) = 3 * ( 2 * (1 * factorial(0))) = 3 * ( 2 * ( 1 * 1))) = 3 * ( 2 * 1) =3*2 Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 12

animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) = 3 * ( 2 * (1 * factorial(0))) = 3 * ( 2 * ( 1 * 1))) = 3 * ( 2 * 1) =3*2 =6 Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 13

Factorial public static long factorial(int n) { System. out. println( "Method" + n ); if (n == 0){ // Base case System. out. println( "Returned 1" ); return 1; }else{ long result = n * factorial(n - 1); // Recursive call System. out. print( "Returned " + result); System. out. println( " : " + n + " * factorial("+ n + "-1)" ); return result; } } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 26

Infinite Recursion public static long factorial(int n) { return result = n * factorial(n - 1); // Recursive call } Stack over flow error Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 27

Fibonacci Numbers Fibonacci series: 0 1 1 2 3 5 8 13 21 34 55 89… indices: 0 1 2 3 4 5 6 7 8 9 10 11 fib(0) = 0; fib(1) = 1; fib(index) = fib(index -1) + fib(index -2); index >=2 fib(3) = fib(2) + fib(1) = (fib(1) + fib(0)) + fib(1) = (1 + 0) +fib(1) = 1 + 1 = 2 A model to show the growth of rabbit population Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 28

Fibonnaci Numbers, cont. static int count; /** The method for finding the Fibonacci number */ public static long fib(long index) { System. out. print(count++); System. out. println(": call fib( " + index +")" ); if (index == 0){ // Base case System. out. print(count++); System. out. println(" : Returned 0 "); return 0; }else if (index == 1){ // Base case System. out. print(count++); System. out. println(" : Returned 1 "); return 1; }else{ // Reduction and recursive calls long result = fib(index - 1) + fib(index - 2); System. out. print(count++); System. out. println( ": Returned fib(" + index-- + ")"); return result; } } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 30

Case Study: Non-recursive version of Fibonacci public long static long fib(long n) { f 0 = 0; // For fib(0) f 1 = 1; // For fib(1) f 2 = 1; // For fib(2) if (n == 0) return f 0; else if (n == 1) return f 1; else if (n == 2) return f 2; Obviously, the complexity of this new algorithm is. This is a tremendous improvement over the recursive algorithm. This is known as dynamic programming. for (int i = 3; i <= n; i++) { f 0 = f 1; f 1 = f 2; f 2 = f 0 + f 1; } return f 2; } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 31

Dynamic programming F Process of solving subproblems, then combine the solutions to obtain an overall result. F This naturally leads to a recursive solution. – Sometimes it is not efficient to use recursions, because subproblems overlap. – The key idea behind dynamic programming is to solve each subproblem only once and store the results for later use and avoid redundant computing. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 32

f 0 f 1 f 2 Fibonacci series: 0 1 1 2 3 5 8 13 21 34 55 89… indices: 0 1 2 3 4 5 6 7 8 9 10 11 f 0 f 1 f 2 Fibonacci series: 0 1 1 2 3 5 8 13 21 34 55 89… indices: 0 4 5 6 7 8 f 0 f 1 f 2 1 1 2 3 5 8 13 21 34 55 89… 1 5 6 7 8 9 10 11 Fibonacci series: 0 indices: 0 1 2 2 3 3 4 9 10 11 Fibonacci series: 0 1 1 2 3 5 8 f 0 f 1 f 2 13 21 34 55 89… indices: 0 1 2 3 4 5 6 7 8 9 Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 10 11 33

Characteristics of Recursion All recursive methods have the following characteristics: – One or more base cases (the simplest case) are used to stop recursion. – Every recursive call reduces the original problem, bringing it increasingly closer to a base case until it becomes that case. In general, to solve a problem using recursion, you break it into subproblems. If a subproblem resembles the original problem, you can apply the same approach to solve the subproblem recursively. This subproblem is almost the same as the original problem in nature with a smaller size. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 34

Problem Solving Using Recursion Let us consider a simple problem of printing a message for n times. You can break the problem into two subproblems: one is to print the message one time and the other is to print the message for n-1 times. The second problem is the same as the original problem with a smaller size. The base case for the problem is n==0. You can solve this problem using recursion as follows: public static void n. Println(String message, int times) { if (times >= 1) { System. out. println(message); n. Println(message, times - 1); } // The base case is times == 0 } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 35

Output for n. Println(“hello”, 3) 3>1 => hello n. Println(“hello”, 2) 2>1 => hello n. Println(“hello”, 1) 1==1 => hello n. Println(“hello”, 0) 0<1. STOP Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 36

Think Recursively The is. Palindrome problem approach 1 public static boolean is. Palindrome(String s) { int low = 0; int high = s. length() – 1; Boolean is. Palindrome = true; While(low<high){ if(s. char. At(low) != s. char. At(high)){ is. Palindrome=false; break; } low++; high--; } Return is. Palindrome; } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 37

Think Recursively Many problems can be solved using recursion if you think recursively. For example, the palindrome problem approach 2 public static boolean is. Palindrome(String s) { if (s. length() <= 1) // Base case return true; else if (s. char. At(0) != s. char. At(s. length() - 1)) // Base case return false; else return is. Palindrome(s. substring(1, s. length() - 1)); } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 38

Recursive Helper Methods The preceding recursive is. Palindrome method is not efficient, because it creates a new string for every recursive call. To avoid creating new strings, use a helper method: public static boolean is. Palindrome(String s) { return is. Palindrome(s, 0, s. length() - 1); } public static boolean is. Palindrome(String s, int low, int high) { if (high <= low) // Base case return true; else if (s. char. At(low) != s. char. At(high)) // Base case return false; else return is. Palindrome(s, low + 1, high - 1); } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 39

Binary Search public static int binary. Search(int[] list, int key) { int low = 0; int high = list. length - 1; while (high >= low) { int mid = (low + high) / 2; if (key < list[mid]) high = mid - 1; else if (key == list[mid]) return mid; else low = mid + 1; } return -1 - low; } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 40

Recursive Binary Search 1. 2. 3. Case 1: If the key is less than the middle element, recursively search the key in the first half of the array. Case 2: If the key is equal to the middle element, the search ends with a match. Case 3: If the key is greater than the middle element, recursively search the key in the second half of the array. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 41

Recursive Implementation /** Use binary search to find the key in the list */ public static int recursive. Binary. Search(int[] list, int key) { int low = 0; int high = list. length - 1; return recursive. Binary. Search(list, key, low, high); } /** Use binary search to find the key in the list between list[low] list[high] */ public static int recursive. Binary. Search(int[] list, int key, int low, int high) { if (low > high) // The list has been exhausted without a match return -low - 1; int mid = (low + high) / 2; if (key < list[mid]) return recursive. Binary. Search(list, key, low, mid - 1); else if (key == list[mid]) return mid; else return recursive. Binary. Search(list, key, mid + 1, high); } Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 42

GCD: Greatest Common Divisor gcd(2, 3) = 1 gcd(2, 10) = 2 gcd(25, 35) = 5 gcd(205, 301) = 5 gcd(m, n) Approach 1: Brute-force Approach 2: Euclid’s algorithm Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 43

Case Study: GCD Algorithms Version 1 public static int gcd(int m, int n) { int gcd = 1; for (int k = 2; k <= m && k <= n; k++) { if (m % k == 0 && n % k == 0) gcd = k; } return gcd; } Obviously, the complexity of this algorithm is. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 44

Case Study: GCD Algorithms Version 2 for (int k = n; k >= 1; k--) { if (m % k == 0 && n % k == 0) { gcd = k; break; } } Assume m>=n The worst-case time complexity of this algorithm is still. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 45

Case Study: GCD Algorithms Version 3 public static int gcd(int m, int n) { int gcd = 1; if (m == n) return m; for (int k = n / 2; k >= 1; k--) { if (m % k == 0 && n % k == 0) { gcd = k; break; } } It is proved that a divisor for n can not be greater than n/2. so we can start from n/2 rather than n. return gcd; } The worst-case time complexity of this algorithm is still. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 46

Euclid’s algorithm Let gcd(m, n) denote the gcd for integers m and n: FIf m % n is 0, gcd (m, n) is n. FOtherwise, gcd(m, n) is gcd(n, m % n). m = n*k + r if p is divisible by both m and n m / p = n*k/p + r/p Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 47

Euclid’s Algorithm Implementation public static int gcd(int m, int n) { if (m % n == 0) return n; else return gcd(n, m % n); } The time complexity of this algorithm is O(logn). Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 48

Directory Size The preceding examples can easily be solved without using recursion. This section presents a problem that is difficult to solve without using recursion. The problem is to find the size of a directory. The size of a directory is the sum of the sizes of all files in the directory. A directory may contain subdirectories. Suppose a directory contains files , , . . . , , and subdirectories , , . . . , , as shown below. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 49

Directory Size The size of the directory can be defined recursively as follows: Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 50

Towers of Hanoi F There are n disks labeled 1, 2, 3, . . . , n, and three towers labeled A, B, and C. F No disk can be on top of a smaller disk at any time. F All the disks are initially placed on tower A. F Only one disk can be moved at a time, and it must be the top disk on the tower. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 51

Solution to Towers of Hanoi The Towers of Hanoi problem can be decomposed into three subproblems. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 53

Solution to Towers of Hanoi F F F Move the first n - 1 disks from A to C with the assistance of tower B. Move disk n from A to B. Move n - 1 disks from C to B with the assistance of tower A. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 54

$Fractals? A fractal is a geometrical figure just like triangles, circles, and rectangles, but$

Fractals? A fractal is a geometrical figure just like triangles, circles, and rectangles, but fractals can be divided into parts, each of which is a reduced-size copy of the whole. There are many interesting examples of fractals. This section introduces a simple fractal, called Sierpinski triangle, named after a famous Polish mathematician. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 55

Sierpinski Triangle 1. 2. 3. 4. It begins with an equilateral triangle, which is considered to be the Sierpinski fractal of order (or level) 0, as shown in Figure (a). Connect the midpoints of the sides of the triangle of order 0 to create a Sierpinski triangle of order 1, as shown in Figure (b). Leave the center triangle intact. Connect the midpoints of the sides of the three other triangles to create a Sierpinski of order 2, as shown in Figure (c). You can repeat the same process recursively to create a Sierpinski triangle of order 3, 4, . . . , and so on, as shown in Figure (d). Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 56

Eight Queens (Homework 3) The Eight Queens puzzle is to place eight queens on a chessboard such that no two queens are on the same row, same column, or same diagonal. A good approach to solve this problem is to use recursion. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 58

Case Study: Fibonacci Numbers /** The method for finding the Fibonacci number */ public static long fib(long index) { if (index == 0) // Base case return 0; else if (index == 1) // Base case return 1; else // Reduction and recursive calls return fib(index - 1) + fib(index - 2); } Finonacci series: 0 1 1 2 3 5 8 13 21 34 55 89… indices: 0 1 2 3 4 5 6 7 8 9 10 11 fib(0) = 0; fib(1) = 1; fib(index) =Liang, fib(index -1)Programming, + fib(index >=2 Inc. All Introduction to Java Eighth Edition, -2); (c) 2011 index Pearson Education, rights reserved. 0132130807 61

Complexity for Recursive Fibonacci Numbers Since and Therefore, the recursive Fibonacci method takes. Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All rights reserved. 0132130807 62