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Chapter 20 Recursion
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
1
Recursion
The idea of calling one function from another
immediately suggests the possibility of a function
calling itself.
The function-call mechanism in Java supports this
possibility, which is known as recursion.
Recursion is a powerful general-purpose
programming technique, and is the key to
numerous critically important computational
applications, ranging from combinatorial search
and sorting methods methods that provide basic
support for information processing
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
2
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.
Your first recursive program.
The HelloWorld for recursion is to implement the factorial
function, which is defined for positive integers N by the equation
public static int factorial(int N) {
if (N == 1) return 1;
return N * factorial(N-1);
}
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
3
// Find the GCD of two integers
import java.util.Scanner;
public class FindGCDInRecursion {
public static void main (String args[]) {
Scanner input = new Scanner(System.in);
GCD using
Recursion.
// Enter the first number
System.out.print("Enter the first number: ");
int m = input.nextInt();
// Enter the second number
System.out.print("Enter the second number: ");
int n = input.nextInt();
System.out.println("The GCD of " + m + " and " + n + " is " +
gcd(m, n));
}
}
----jGRASP exec: java FindGCDInRecursion
public static int gcd(int m, int n) {
if (m % n == 0)
return n;
else
return gcd(n, m % n);
}
Enter the first number: 45
Enter the second number: 33
The GCD of 45 and 33 is 3
----jGRASP: operation complete.
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
4
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, as shown in Figure 20.1.
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
5
import java.awt.*;
import javax.swing.*;
// Create a maze
public class Exercise21_36 extends JApplet {
public static int SIZE = 8;
private int queens[] = new int[SIZE];
private JPanel solutionPanel = new JPanel();
private int count = 0;
public static void main(String[] args) {
JFrame frame = new JFrame("Exercise20_37");
Exercise20_37 applet = new Exercise20_37();
frame.add(applet, BorderLayout.CENTER);
applet.init();
applet.start(); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setSize(400, 400);
frame.setLocationRelativeTo(null); // Center the frame
frame.setVisible(true);
}
public Exercise20_37() {
this.setLayout(new BorderLayout());
add(new JScrollPane(solutionPanel), BorderLayout.CENTER);
search(0);
}
private void search(int row) {
for (int column = 0; column < SIZE; column++) {
queens[row] = column;
if (isValid(row, column))
if (row < SIZE - 1)
search(row + 1);
else {
count++;
solutionPanel.add(new ChessBoardWithLabel("Solution " + count, queens));
}
}
}
public boolean isValid(int row, int column) {
for (int i = 1; i <= row; i++)
if (queens[row - i] == column // Check column
|| queens[row - i] == column - i // diagonal to upper left
|| queens[row - i] == column + i) // diagonal to upper right
return false;
return true;
}
private class ChessBoardWithLabel extends JPanel {
private JLabel jlblCount = new JLabel("", JLabel.CENTER);
private int[] queens;
private Image queenImage = new ImageIcon("image/queen.jpg").getImage();
public ChessBoardWithLabel(String count, int[] queens) {
this.setBorder(BorderFactory.createLineBorder(Color.BLACK, 2));
this.queens = queens.clone();
setLayout(new BorderLayout());
jlblCount.setText(count);
add(jlblCount, BorderLayout.NORTH);
add(new ChessBoard(), BorderLayout.CENTER);
}
private class ChessBoard extends JPanel {
ChessBoard() {
this.setBackground(Color.WHITE);
this.setBorder(BorderFactory.createLineBorder(Color.BLACK, 2));
}
protected void paintComponent(Graphics g) {
//g.clearRect(0, 0, getWidth(), getHeight());
//
g.setColor(Color.WHITE);
//
g.fillRect(0, 0, getWidth(), getHeight());
g.setColor(Color.BLACK);
// Add the Queen image
for (int i = 0; i < SIZE; i++) {
int j = queens[i];
g.drawImage(queenImage, j * getWidth() / SIZE, i * getHeight()
/ SIZE, getWidth() / SIZE, getHeight() / SIZE, this);
}
// Paint the lines
for (int i = 0; i < SIZE; i++) {
g.drawLine(0, i * getHeight() / SIZE, getWidth(), i * getHeight()
/ SIZE);
g.drawLine(i * getWidth() / SIZE, 0, i * getWidth() / SIZE,
getHeight());
}
}
/** Specify preferred size */
public Dimension getPreferredSize() {
return new Dimension(200, 200);
}
}
}
}
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
6
Objectives
F
F
F
F
F
F
F
F
F
To know what a recursive method is and the benefits of using
recursive methods (§20.1).
To determine the base cases in a recursive method (§§20.2-20.5).
To understand how recursive method calls are handled in a call
stack (§§20.2-20.3).
To solve problems using recursion (§§20.2-20.5).
To use an overloaded helper method to derive a recursive method
(§20.5).
To get the directory size using recursion (§20.6).
To solve the Towers of Hanoi problem using recursion (§20.7).
To draw fractals using recursion (§20.8).
To understand the relationship and difference between recursion
and iteration (§20.9).
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
7
Computing Factorial
Many mathematical functions
are defined using recursion.
We began using simple
example. The factorial of a
number n can be recursively
define as follow:
factorial(0) = 1;
factorial(n) = n*factorial(n-1);
n! = n * (n-1)!
ComputeFactorial
import java.util.Scanner;
public class ComputeFactorial {
/** Main method */
public static void main(String[] args) {
// Create a Scanner
Scanner input = new Scanner(System.in);
System.out.print("Enter a non-negative integer: ");
int n = input.nextInt();
// Display factorial
System.out.println("Factorial of " + n + " is " +
factorial(n));
}
/** Return the factorial for a specified number */
public static long factorial(int n) {
if (n == 0) // Base case
return 1;
else
return n * factorial(n - 1); // Recursive call
}
}
Run:
Enter a non-negative integer: 4
Factorial of 4 is 24
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)
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)
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))
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)))
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)))
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
13
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
14
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
15
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
16
animation
Trace Recursive factorial
Executes factorial(4)
factorial(4)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Main method
17
animation
Trace Recursive factorial
factorial(4)
Step 0: executes factorial(4)
Step 9: return 24
Executes factorial(3)
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(4)
Main method
18
animation
Trace Recursive factorial
factorial(4)
Executes factorial(2)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
19
animation
Trace Recursive factorial
factorial(4)
Executes factorial(1)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(2)
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
20
animation
Trace Recursive factorial
factorial(4)
Executes factorial(0)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(1)
Space Required
for factorial(2)
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
21
animation
Trace Recursive factorial
factorial(4)
returns 1
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(0)
Space Required
for factorial(1)
Space Required
for factorial(2)
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
22
animation
Trace Recursive factorial
factorial(4)
returns factorial(0)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(1)
Space Required
for factorial(2)
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
23
animation
Trace Recursive factorial
factorial(4)
returns factorial(1)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(2)
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
24
animation
Trace Recursive factorial
factorial(4)
returns factorial(2)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(3)
Space Required
for factorial(4)
Main method
25
animation
Trace Recursive factorial
factorial(4)
returns factorial(3)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Space Required
for factorial(4)
Main method
26
animation
Trace Recursive factorial
returns factorial(4)
factorial(4)
Step 0: executes factorial(4)
Step 9: return 24
return 4 * factorial(3)
Step 1: executes factorial(3)
Step 8: return 6
return 3 * factorial(2)
Step 2: executes factorial(2)
Step 7: return 2
Stack
return 2 * factorial(1)
Step 6: return 1
Step 3: executes factorial(1)
return 1 * factorial(0)
Step 4: executes factorial(0)
Step 5: return 1
return 1
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
Main method
27
factorial(4) Stack Trace
Required
5 Space
for factorial(0)
Required
1 Space
for factorial(4)
Required
4 Space
for factorial(1)
Space Required
for factorial(1)
Required
3 Space
for factorial(2)
Space Required
for factorial(2)
Space Required
for factorial(2)
Required
2 Space
for factorial(3)
Space Required
for factorial(3)
Space Required
for factorial(3)
Space Required
for factorial(3)
Space Required
for factorial(4)
Space Required
for factorial(4)
Space Required
for factorial(4)
Space Required
for factorial(4)
Required
6 Space
for factorial(1)
Space Required
for factorial(2)
Required
7 Space
for factorial(2)
Space Required
for factorial(3)
Space Required
for factorial(3)
Required
8 Space
for factorial(3)
Space Required
for factorial(4)
Space Required
for factorial(4)
Space Required
for factorial(4)
Required
9 Space
for factorial(4)
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
28
Other Examples
f(0) = 0;
f(n) = n + f(n-1);
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
29
Fibonacci Numbers
By definition, the first two numbers in the Fibonacci sequence are 0 and 1,
and each subsequent number is the sum of the previous two.
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 + fib(1) = 1 + 1 = 2
ComputeFibonacci
Run
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
30
import java.util.Scanner;
public class ComputeFibonacci {
/** Main method */
public static void main(String args[]) {
// Create a Scanner
Scanner input = new Scanner(System.in);
System.out.print("Enter an index for the Fibonacci number: ");
int index = input.nextInt();
// Find and display the Fibonacci number
System.out.println(
"Fibonacci number at index " + index + " is " + fib(index));
}
/** 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);
}
}
Run
Enter an index for the Fibonacci number: 8
Fibonacci number at index 8 is 21
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
31
Fibonnaci Numbers, cont.
fib(4)
17: return fib(4)
0: call fib(4)
return fib(3) + fib(2)
11: call fib(2)
10: return fib(3)
1: call fib(3)
16: return fib(2)
return fib(2) + fib(1)
7: return fib(2)
2: call fib(2)
9: return fib(1)
return fib(1) + fib(0)
return fib(1) + fib(0)
8: call fib(1)
return 1
13: return fib(1)
return 1
14: return fib(0)
12: call fib(1)
15: return fib(0)
return 0
5: call fib(0)
4: return fib(1)
3: call fib(1)
return 1
6: return fib(0)
return 0
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
32
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
33
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 nPrintln(String message, int times) {
if (times >= 1) {
System.out.println(message);
nPrintln(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
34
Think Recursively
Many of the problems presented in the early chapters can
be solved using recursion if you think recursively. For
example, the palindrome problem in Listing 7.1 can be
solved recursively as follows:
public static boolean isPalindrome(String s) {
if (s.length() <= 1) // Base case
return true;
else if (s.charAt(0) != s.charAt(s.length() - 1)) // Base case
return false;
else
return isPalindrome(s.substring(1, s.length() - 1));
}
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
35
Recursive Helper Methods
The preceding recursive isPalindrome 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 isPalindrome(String s) {
return isPalindrome(s, 0, s.length() - 1);
}
public static boolean isPalindrome(String s, int low, int high) {
if (high <= low) // Base case
return true;
else if (s.charAt(low) != s.charAt(high)) // Base case
return false;
else
return isPalindrome(s, low + 1, high - 1);
}
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
36
Recursive Selection Sort


Find the largest
number in the list and
swaps it with the last
number.
Ignore the last number
and sort the remaining
smaller list recursively.
public class RecursiveSelectionSort {
public static void sort(double[] list) {
sort(list, 0, list.length - 1); // Sort the entire list
}
public static void sort(double[] list, int low, int high) {
if (low < high) {
// Find the smallest number and its index in list(low .. high)
int indexOfMin = low;
double min = list[low];
for (int i = low + 1; i <= high; i++) {
if (list[i] < min) {
min = list[i];
indexOfMin = i;
}
}
// Swap the smallest in list(low .. high) with list(low)
list[indexOfMin] = list[low];
list[low] = min;
// Sort the remaining list(low+1 .. high)
sort(list, low + 1, high);
}
}
public static void main(String[] args) {
double[] list = {2, 1, 3, 1, 2, 5, 2, -1, 0};
sort(list);
for (int i = 0; i < list.length; i++)
System.out.print(list[i] + " ");
}
-1.0 0.0 1.0 1.0 2.0 2.0 2.0 3.0 5.0
}
Output:
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
37
Recursive Binary Search



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.
public class RecursiveBinarySearch {
public static int recursiveBinarySearch(int[] list, int key) {
int low = 0;
int high = list.length - 1;
return recursiveBinarySearch(list, key, low, high);
}
public static int recursiveBinarySearch(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 recursiveBinarySearch(list, key, low, mid - 1);
else if (key == list[mid])
return mid;
else
return recursiveBinarySearch(list, key, mid + 1, high);
}
}
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
38
Recursive Implementation
/** Use binary search to find the key in the list */
public static int recursiveBinarySearch(int[] list, int key) {
int low = 0;
int high = list.length - 1;
return recursiveBinarySearch(list, key, low, high);
}
/** Use binary search to find the key in the list between
list[low] list[high] */
public static int recursiveBinarySearch(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 recursiveBinarySearch(list, key, low, mid - 1);
else if (key == list[mid])
return mid;
else
return recursiveBinarySearch(list, key, mid + 1, high);
}
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
39
Towers of Hanoi
F
F
F
F
There are n disks labeled 1, 2, 3, . . ., n, and three
towers labeled A, B, and C.
No disk can be on top of a smaller disk at any
time.
All the disks are initially placed on tower A.
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
40
Towers of Hanoi, cont.
A
B
C
A
Original position
A
B
B
C
A
B
Step 3: Move disk 1 from B to C
B
C
Step 5: Move disk 1 from C to A
C
A
Step 2: Move disk 2 from A to C
A
C
Step 4: Move disk 3 from A to B
Step 1: Move disk 1 from A to B
A
B
B
C
Step 6: Move disk 2 from C to B
C
A
B
C
Step 7: Mve disk 1 from A to B
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
41
Solution to Towers of Hanoi
The Towers of Hanoi problem can be decomposed into three
subproblems.
n-1 disks
n-1 disks
.
.
.
.
.
.
A
B
C
A
Original position
B
Step2: Move disk n from A to C
n-1 disks
n-1 disks
.
.
.
.
.
.
A
B
C
C
Step 1: Move the first n-1 disks from A to C recursively
A
B
C
Step3: Move n-1 disks from C to B recursively
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
42
Solution to Towers of Hanoi
import java.util.Scanner;
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.
Run
Enter number of disks: 3
The moves are:
Move disk 1 from A to B
Move disk 2 from A to C
Move disk 1 from B to C
Move disk 3 from A to B
Move disk 1 from C to A
Move disk 2 from C to B
Move disk 1 from A to B
TowersOfHanoi
public class TowersOfHanoi {
/** Main method */
public static void main(String[] args) {
// Create a Scanner
Scanner input = new Scanner(System.in);
System.out.print("Enter number of disks: ");
int n = input.nextInt();
// Find the solution recursively
System.out.println("The moves are:");
moveDisks(n, 'A', 'B', 'C');
}
/** The method for finding the solution to move n disks
from fromTower to toTower with auxTower */
public static void moveDisks(int n, char fromTower,
char toTower, char auxTower) {
if (n == 1) // Stopping condition
System.out.println("Move disk " + n + " from " +
fromTower + " to " + toTower);
else {
moveDisks(n - 1, fromTower, auxTower, toTower);
System.out.println("Move disk " + n + " from " +
fromTower + " to " + toTower);
moveDisks(n - 1, auxTower, toTower, fromTower);
}
}
} Edition, (c) 2011 Pearson Education, Inc. All
Liang, Introduction to Java Programming, Eighth
rights reserved. 0132130807
43
Exercise 20.3 GCD
gcd(2, 3) = 1
gcd(2, 10) = 2
gcd(25, 35) = 5
gcd(205, 301) = 5
gcd(m, n)
Approach 1: Brute-force, start from min(n, m) down to 1,
to check if a number is common divisor for both m and
n, if so, it is the greatest common divisor.
Approach 2: Euclid’s algorithm
Approach 3: Recursive method
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
44
Approach 2: Euclid’s algorithm
// Get absolute value of m and n;
t1 = Math.abs(m); t2 = Math.abs(n);
// r is the remainder of t1 divided by t2;
r = t1 % t2;
while (r != 0) {
t1 = t2;
t2 = r;
r = t1 % t2;
}
// When r is 0, t2 is the greatest common
// divisor between t1 and t2
return t2;
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
45
Approach 3: Recursive Method
gcd(m, n) = n if m % n = 0;
gcd(m, n) = gcd(n, m % n); otherwise;
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
46
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
47
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
48
Sierpinski Triangle Solution
p1
midBetweenP1P2
midBetweenP3P1
p2
p3
midBetweenP2P3
SierpinskiTriangle
Run
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
49
Eight Queens
queens[0]
queens[1]
queens[2]
queens[3]
queens[4]
queens[5]
queens[6]
queens[7]
0
4
7
5
2
6
1
3
EightQueens
Run
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
50
Eight Queens
0
0
1
1
check
colum
n
2
3
4
5
6
7
upright diagonal
2
upleft
3
4
5
6
7
Liang, Introduction to Java Programming, Eighth Edition, (c) 2011 Pearson Education, Inc. All
rights reserved. 0132130807
51