TABLE OF CONTENTS
Exercises on Flow Controls
Exercises on Conditional (Decision)
(if-else): Write a program called CheckPassFail which prints &PASS& if the int variable &mark& is more than or equal to 50; or prints &FAIL& otherwise.
public class CheckPassFail {
public static void main(String[] args) {
int mark = 49;
System.out.println(&The mark is & + mark);
if ( ...... ) {
System.out.println( ...... );
System.out.println( ...... );
(if-else): Write a program called CheckOddEven which prints &Odd Number&
if the int variable
“number” is odd, or “Even Number”
otherwise.
Hints: n is an even number if (n % 2) is 0.
public class CheckOddEven {
public static void main(String[] args) {
int number = 49;
System.out.println(&The number is & + number);
if ( ...... ) {
System.out.println( ...... );
System.out.println( ...... );
(nested-if, switch-case): Write a program called PrintNumberInWord which prints &ONE&, &TWO&,... , &NINE&, &OTHER& if the int variable &number& is 1, 2,... , 9, or other, respectively.
Use (a) a &nested-if& (b) a &switch-case& statement.
public class PrintNumberInWord {
public static void main(String[] args) {
int number = 5;
if (number == 1) {
System.out.println(&ONE&);
} else if (......) {
} else if (......) {
switch(number) {
case 1: System.out.println(&ONE&);
case 2: ......
default: System.out.println(&OTHER&);
Similarly, write a program called PrintDayInWord, which prints &Sunday&, &Monday&, ... &Saturday& if
the int variable &day& is
0, 1, ..., 6, respectively.
Otherwise, it shall print
&Not a valid day&.
Exercises on Loop (Iteration)
(Loop): Write a program called SumAndAverage to produce the sum of 1, 2, 3, ..., to an upperbound (e.g., 100).
Also compute and display the average.
The output shall look like:
The sum is 5050
The average is 50.5
public class SumAndAverage {
public static void main (String[] args) {
int sum = 0;
int lowerbound = 1;
int upperbound = 100;
for (int number = number &= ++number) {
Modify the program to use a &while-do& loop instead of &for& loop.
int number =
int sum = 0;
while (number &= upperbound) {
Modify the program to use a &do-while& loop.
int number =
int sum = 0;
} while (number &= upperbound);
What is the difference between &for& and &while-do& loops? What is the difference between &while-do& and &do-while& loops?
Modify the program to sum from 111 to 8899, and compute the average. Introduce an int variable called count to count the numbers in the specified range.
int count = 0;
for (...; ...; ...) {
Modify the program to sum only the odd numbers from 1 to 100, and compute the average. (Hint: n is an odd number if n % 2 is not 0.)
Modify the program to sum those numbers from 1 to 100 that is divisible by 7, and compute the average.
Modify the program to find the &sum of the squares& of all the numbers from 1 to 100, i.e. 1*1 + 2*2 + 3*3 + ... + 100*100.
(Loop): Write a program called Product1ToN to compute the product of integers 1 to 10 (i.e., 1&2&3&...&10).
Try computing the product from 1 to 11, 1 to 12, 1 to 13 and 1 to 14. Write down the product obtained and explain the results.
Hints: Declares an int variable called product (to accumulate the product) and initialize to 1.
(Loop): Write a program called HarmonicSum to compute the sum of a harmonic series, as shown below, where n=50000.
The program shall compute the sum from left-to-right as well as from the right-to-left.
Obtain the difference between these two sums and explain the difference.
Which sum is more accurate?
public class HarmonicSum {
public static void main (String[] args) {
int maxDenominator = 50000;
double sumL2R = 0.0;
double sumR2L = 0.0;
for (int denominator = 1; denominator &= maxD ++denominator) {
(Loop & Condition): Write a program called ComputePI to compute the value of &, using the following series expansion. You have to decide on the termination criterion used in the computation (such as the number of terms used or the magnitude of an additional term). Is this series suitable for computing &?
JDK maintains the value of & in a double constant called Math.PI. Compare the values obtained and the Math.PI, in percents of Math.PI.
Hint: Add to sum if the denominator modulus 4 is 1, and subtract from sum if it is 3.
double sum = 0;
int maxDenom = ;
for (int denom = 1; ..... ; denom = denom + 2) {
if (denom % 4 == 1) {
sum += ......;
} else if (denom % 4 == 3) {
sum -= ......;
System.out.println(&The computer has gone crazy?!&);
(Loop & Condition): Write a program called CozaLozaWoza which prints the numbers 1 to 110, 11 numbers per line. The program shall print &Coza& in place of the numbers which are multiples of 3, &Loza& for multiples of 5, &Woza& for multiples of 7, &CozaLoza& for multiples of 3 and 5, and so on.
The output shall look like:
1 2 Coza 4 Loza Coza Woza 8 Coza Loza 11
Coza 13 Woza CozaLoza 16 17 Coza 19 Loza CozaWoza 22
23 Coza Loza 26 Coza Woza 29 CozaLoza 31 32 Coza
public class CozaLozaWoza {
public static void main(String[] args) {
int lowerbound = 1;
int upperbound = 110;
for (int number = number &= ++number) {
if (......) {
System.out.print(&Coza&);
if (......) {
System.out.print(.....);
if (......) {
if (......) {
System.out.println();
TRY: Modify the program to use nested-if (if ... else if ... else if ... else) instead.
Write a program called Fibonacci to display the first 20 Fibonacci numbers F(n), where F(n)=F(n–1)+F(n–2) and F(1)=F(2)=1. Also compute their average. The output shall look like:
The first 20 Fibonacci numbers are:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 81 6765
The average is 885.5
public class Fibonacci {
public static void main (String args[]) {
int n = 3;
int fnMinus1 = 1;
int fnMinus2 = 1;
int nMax = 20;
int sum = fnMinus1 + fnMinus2;
System.out.println(&The first & + nMax + & Fibonacci numbers are:&);
while (n &= nMax) {
Tribonacci numbers are a sequence of numbers T(n) similar to Fibonacci numbers, except that a number is formed by adding the three previous numbers, i.e., T(n)=T(n-1)+T(n-2)+T(n-3), T(1)=T(2)=1, and T(3)=2.
Write a program called Tribonacci to produce the first twenty Tribonacci numbers.
: Write a program to extract each digit from an int, in the reverse order. For example, if the int is 1542, the output shall be &2,4,5,1&, with a comma separating the digits.
Hints: Use n % 10 and n = n / 10 to discard the last digit.
int n = ....;
while (n > 0) {
int digit = n % 10;
n = n / 10;
Exercises on Nested-Loop
(nested-loop): Write a program called SquareBoard that displays the following n×n (n=5) pattern using two nested for-loops.
Your program should use only two output statements, one EACH of the followings:
System.out.print(&# &);
System.out.println();
public class SquareBoard {
public static void main (String[] args) {
int size = 5;
for (int row = 1; ......; ......) {
for (int col = 1; ......; ......) {
(nested-loop): Write a program called CheckerBoard that displays the following n×n (n=7) checkerboard pattern using two nested for-loops.
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
Your program should use only three output statements, one EACH of the followings:
System.out.print(&# &);
System.out.print(& &);
System.out.println();
public class CheckerBoard {
public static void main (String[] args) {
int size = 7;
// size of the board
for (int row = 1; ......; ......) {
if ((row % 2) == 0) {
for (int col = 1; ......; ......) {
(nested-loop): Write a program called TimeTable to produce the multiplication table of 1 to 9 as shown using two nested for-loops:
-------------------------------
8 10 12 14 16 18
9 12 15 18 21 24 27
8 12 16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18 24 30 36 42 48 54
7 14 21 28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27 36 45 54 63 72 81
Modify the program to print the multiplication table of 1 to 12.
Exercises on Keyboard and File Input
(Keyboard Input):
Write a program called KeyboardScanner to prompt user for an int, a double, and a String. The output shall look like (the inputs are shown in bold):
Enter an integer: 12
Enter a floating point number: 33.44
Enter your name: Peter
Hi! Peter, the sum of 12 and 33.44 is 45.44
import java.util.S
public class KeyboardScanner {
public static void main(String[] args) {
double num2;
Scanner in = new Scanner(System.in);
System.out.print(&Enter an integer: &);
num1 = in.nextInt();
System.out.print(&Enter a floating point number: &);
num2 = in.nextDouble();
System.out.print(&Enter your name: &);
name = in.next();
(File Input): Write a program called FileScanner to read an int, a double, and a String form a text file called &in.txt&, and produce the following output:
The integer read is 12
The floating point number read is 33.44
The String read is &Peter&
Hi! Peter, the sum of 12 and 33.44 is 45.44
You need to create a text file called &in.txt& (in Eclipse, right-click on the &project& &rA &New& &rA &File&) with the following contents:
import java.util.S
import java.io.F
import java.io.FileNotFoundE
public class FileScanner {
public static void main(String[] args)
throws FileNotFoundException {
double num2;
Scanner in = new Scanner(new File(&in.txt&));
num1 = in.nextInt();
num2 = in.nextDouble();
name = in.next();
(User Input): Write a program called CircleComputation, which prompts user for a radius (of double) and compute the area and perimeter of a circle.
The output shall look like:
Enter the radius: 1.2
The area is 4.5239
The perimeter is 7.5035
Hints: & is kept in a constant called Math.PI.
Exercises on User Input and String Operations
Write a program called ReverseString, which prompts user for a String, and prints the reverse of the String. The output shall look like:
Enter a String: abcdef
The reverse of String &abcdef& is &fedcba&.
import java.util.S
public class ReverseString {
public static void main(String[] args) {
String inS
int inStrL
Scanner in = new Scanner(System.in);
System.out.print(&Enter a String: &);
inStr = in.next();
inStrLen = inStr.length();
For a String called inStr, you can use inStr.length() to get the length of the String; and inStr.charAt(index) to retrieve the char at the index position, where index begins with 0.
: On your phone keypad, the alphabets are mapped to digits as follows: ABC(2), DEF(3), GHI(4), JKL(5), MNO(6), PQRS(7), TUV(8), WXYZ(9).
Write a program called PhoneKeyPad, which prompts user for a String (case insensitive), and converts to a sequence of Keypad digits. Use a nested-if (or switch-case) in this exercise. Modify your program to use an array for table look-up later.
Hints: You can use in.next().toLowerCase() to read a string and convert it to lowercase to reduce your cases.
: A word that reads the same backward as forward is called a palindrome, e.g., &mom&, &dad&, &racecar&, &madam&, and &Radar& (case-insensitive). Write a program called TestPalindromicWord, that prompts user for a word and prints &&xxx& is|is not a palindrome&.
Hints: Read in a word and convert to lowercase via in.next().toLowercase().
A phrase that reads the same backward as forward is also called a palindrome, e.g., &Madam, I'm Adam&, &A man, a plan, a canal - Panama!& (ignoring punctuation and capitalization). Modify your program (called TestPalindromicPhrase) to test palindromic phrase.
Hints: Read in the lowercase phrase via in.nextLine().toLowercase(). Maintain two indexes, forwardIndex and backwardIndex, used to scan the phrase forward and backward.
: Write a program called Bin2Dec to convert an input binary string into its equivalent decimal number.
Your output shall look like:
Enter a Binary string: 1011
The equivalent decimal number for binary &1011& is 11
Enter a Binary string: 1234
Error: Invalid Binary String &1234&
Hints: For a n-bit binary number bn-1bn-2...b1b0, bi∈{0,1}, the equivalent decimal number is bn-1×2n-1+bn-2×2n-2+ ...+b1×21+b0×20.
import java.util.S
public class Bin2Dec {
public static void main(String[] args) {
String binS
int binStrL
int dec = 0;
Scanner in = new Scanner(System.in);
: 1 0 1 1 1 0 0 1
charAt(idx)
: 0 1 2 3 4 5 6 7
Math.pow(2, order) : 7 6 5 4 3 2 1 0
binStr.length() = 8
idx + order = binStr.length() - 1
You can use JDK method Math.pow(x, y) to compute the x raises to the power of y.
This method takes two doubles as argument and returns a double.
You may have to cast the result back to int.
To convert a char (of digit '0' to '9') to int (0 to 9), simply subtract by char '0', e.g., '5'-'0' gives int 5.
: Write a program called Hex2Dec to convert an input hexadecimal string into its equivalent decimal number.
Your output shall look like:
Enter a Hexadecimal string: 1a
The equivalent decimal number for hexadecimal &1a& is 26
Enter a Hexadecimal string: 1y3
Error: Invalid Hexadecimal String &1y3&
For a n-digit hexadecimal number hn-1hn-2...h1h0, hi∈{0,…,9,A,…,F}, the equivalent decimal number is hn-1×16n-1+hn-2×16n-2+ ...+h1×161+h0×160.
You do not need a big nested-if statement of 16 cases (or 22 considering the upper and lower letters).
Extract the individual character from the hexadecimal string, says c. If char c is between '0' to '9', you can get the integer offset via c-'0'.
If c is between 'a' to 'f' or 'A' to 'F', the integer offset is c-'a'+10 or c-'A'+10.
String hexS
hexChar = hexStr.charAt(i);
if (hexChar &= '0' && hexChar &= '9') {
... (hexChar-'0') ...
} else if (hexChar &= 'a' && hexChar &= 'f') {
... (hexChar-'a'+10) ...
} else if (hexChar &= 'A' && hexChar &= 'F') {
... (hexChar-'A'+10) ...
System.out.println(&Error: Invalid hexadecimal string&);
System.exit(1);
Exercises on Array
(Array): Write a program called GradesAverage, which prompts user for the number of students, reads
it from the keyboard, and saves it in an int variable called numStudents.
It then prompts user for the grades of each of the students and saves them in an
int array called grades.
Your program shall check that the grade is
between 0 and 100. A sample session is as follow:
Enter the number of students: 3
Enter the grade for student 1: 55
Enter the grade for student 2: 108
Invalid grade, try again...
Enter the grade for student 2: 56
Enter the grade for student 3: 57
The average is 56.0
(Array and Table Lookup): Write a program called Hex2Bin to convert a hexadecimal string into its equivalent binary string.
The output shall look like:
Enter a Hexadecimal string: 1abc
The equivalent binary for hexadecimal &1abc& is 11 1100
Hints: Use an array of 16 binary Strings corresponding to hexadecimal number '0' to 'F' (or 'f'), as follows:
String[] hexBits = {&0000&, &0001&, &0010&, &0011&,
&0100&, &0101&, &0110&, &0111&,
&1000&, &1001&, &1010&, &1011&,
&1100&, &1101&, &1110&, &1111&};
Exercises on Command-line Arguments
(Command-line arguments): Write a program called Arithmetic that takes three command-line arguments: two integers followed by an arithmetic operator (+, -, * or /).
The program shall perform the corresponding operation on the two integers and print the result.
For example:
& java Arithmetic 3 2 +
& java Arithmetic 3 2 -
& java Arithmetic 3 2 /
The method main(String[] args) takes an argument: &an array of String&, which is often (but not necessary) named args.
This parameter captures the command-line arguments supplied by the user when the program is invoked.
For example, if a user invokes:
& java Arithmetic
The three command-line arguments &12345&, &4567& and &+& will be captured in a String array {&12345&, &4567&, &+&} and passed into the main() method as the argument args. That is,
args is {&12345&, &4567&, &+&};
args.length is 3;
args[0] is &12345&;
args[1] is &4567&;
args[2] is &+&;
args[0].length() is 5;
args[1].length() is 4;
args[2].length() is 1;
public class Arithmetic {
public static void main (String[] args) {
int operand1, operand2;
if (args.length != 3) {
System.err.println(&Usage: java Arithmetic int1 int2 op&);
operand1 = Integer.parseInt(args[0]);
operand2 = ......
theOperator = args[2].charAt(0);
System.out.print(args[0] + args[2] + args[1] + &=&);
switch(theOperator) {
case ('-'): System.out.println(operand1 - operand2);
case ('+'): ......
case ('*'): ......
case ('/'): ......
System.err.println(&Error: invalid operator!&);
To provide command-line arguments, use the &cmd& shell to run your program in the form &java ClassName arg1 arg2 ....&.
To provide command-line arguments in Eclipse, right click the source code &rA &Run As& &rA &Run Configurations...& &rA Select &Main& and choose the proper main class &rA Select &Arguments& &rA Enter the command-line arguments, e.g., &3 2 +& in &Program Arguments&.
To provide command-line arguments in NetBeans, right click the &Project& name &rA &Set Configuration& &rA &Customize...& &rA Select categories &Run& &rA Enter the command-line arguments, e.g., &3 2 +& in the &Arguments& box (but make sure you select the proper Main class).
Try &java Arithmetic 2 4 *& (in CMD shell and Eclipse/NetBeans) and explain the result obtained. How to resolve this problem?
In Windows' CMD shell, * is known as a wildcard character, that expands to give the list of file in the directory (called Shell Expansion). For example, &dir *.java& lists all the file with extension of &.java&. You could double-quote the * to prevent shell expansion. Eclipse has a bug in handling this, even * is double-quoted. NetBeans??
(Command-line arguments): Write a program called SumDigits to sum up the individual digits of a positive integer, given in the command line.
The output shall look like:
& java SumDigits 12345
The sum of digits = 1 + 2 + 3 + 4 + 5 = 15
Exercises on Method
(Method): Write a program called GradesStatistics, which reads in n grades (of int between 0 and 100, inclusive) and displays the average, minimum, maximum, and standard deviation. Your program shall check for valid input. You should keep the grades in an int[] and use a method for each of the computations.
Your output shall look like:
Enter the number of students: 4
Enter the grade for student 1: 50
Enter the grade for student 2: 51
Enter the grade for student 3: 56
Enter the grade for student 4: 53
The average is 52.5
The minimum is 50
The maximum is 56
The standard deviation is 2.92
Hints: The formula for calculating standard deviation is:
public class GradesStatistics {
public static int[]
public static void main(String[] args) {
readGrades();
System.out.println(&The average is & + average());
System.out.println(&The minimum is & + min());
System.out.println(&The maximum is & + max());
System.out.println(&The standard deviation is & + stdDev());
public static void readGrades() { ....... }
public static double average() { ...... }
public static int max() { ...... }
public static int min() { ....... }
public static double stdDev() { ....... }
(Method): Write a program called GradesHistogram, which reads in n grades (of int between 0 and 100, inclusive) from a text file called &grades.in& and displays the histogram horizontally and vertically. The file has the following format:
numStduents:int
grade1:int grade2:int .... gradeN:int
For example:
49 50 51 59 0 5 9 10 15 19 50 55 89 99 100
The output shall consist of a horizontal histogram and a vertical histogram as follows:
10 - 19: ***
40 - 49: *
50 - 59: *****
80 - 89: *
90 -100: **
10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-100
public class GradesHistogram {
public static int[]
public static int[] bins = new int[10];
public static void main(String[] args) {
readGrades(&grades.in&);
computeHistogram();
printHistogramHorizontal();
printHistogramVertical();
public static void readGrades(String filename) { ...... }
public static void computeHistogram() { ....... }
public static void printHistogramHorizontal() { ...... }
public static void printHistogramVertical() { ...... }
(Method): Write a method called reverseArray() with the following signature:
public static void reverseArray(int[] intArray)
The method accepts an int array, and reverses its orders. For example, if the input array is {12, 56, 34, 79, 26}, the reversal is {26, 79, 34, 56, 12}.
You MUST NOT use another array in your method (but you need a temporary variable to do the swap).
Also write a test class called ReverseArrayTest to test this method.
Take note that the array passed into the method can be modified by the method (this is called &pass by reference&).
On the other hand, primitives passed into a method cannot be modified.
This is because a clone is created and passed into the method instead of the original copy (this is called &pass by value&).
More (Difficult) Exercises
Extract the source code of the class Math from the JDK source code (&$JAVA_HOME& &rA &src.zip& &rA &Math.java& under folder &java.lang&).
Study how constants such as E and PI are defined. Also study how methods such as abs(), max(), min(), toDegree(), etc, are written.
: Similar to Math class, write a Matrix library that supports matrix operations (such as addition, subtraction, multiplication) via 2D arrays. The operations shall support both doubles and ints. Also write a test class to exercise all the operations programmed.
public class Matrix {
public static void printMatrix(int[][] m) { ...... }
public static void printMatrix(double[][] m) { ...... }
public static boolean haveSameDimension(int[][] m1, int[][] m2) { ...... }
public static boolean haveSameDimension(double[][] m1, double[][] m2) { ...... }
public static int[][] add(int[][] m1, int[][] m2) { ...... }
public static double[][] add(double[][] m1, double[][] m2) { ...... }
(Special Characters and Escape Sequences): Write a program called PrintAnimalPattern, which uses println() to produce this pattern:
/========\/
/ || %% ||
Use escape sequence \uhhhh where hhhh are four hex digits to display Unicode characters such as ? and (C).
? is 165 (00A5H) and (C) is 169 (00A9H) in both ISO-8859-1 (Latin-1) and Unicode character sets.
Double-quote (&) and black-slash (\) require escape sign inside a String. Single quote (') does not require escape sign.
TRY: Print the same pattern using printf(). (Hints: Need to use %% to print a % in printf() because % is the suffix for format specifier.)
: Write a method to print each of the followings patterns using nested loops in a class called PrintPatterns. The signatures of the methods are:
public static void printPatternX(int size)
# # # # # # # #
# # # # # # # #
# # # # # # #
# # # # # # #
# # # # # #
# # # # # #
# # # # # #
# # # # # #
# # # # # # #
# # # # # # #
# # # # # # # #
# # # # # # # #
Hints: On the diagonal, row = col.
On the opposite diagonal, row + col = size + 1.
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # # # # # #
# # # # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # # # #
# # # # # # # # #
# # # # # # # # # # #
# # # # # # # # # # #
# # # # # # # # #
# # # # # # #
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
1 2 3 4 5 6 7
7 6 5 4 3 2 1
1 2 3 4 5 6
6 5 4 3 2 1
1 2 3 4 5 6
6 5 4 3 2 1
1 2 3 4 5 6 7
7 6 5 4 3 2 1
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7
7 6 5 4 3 2 1
1 2 3 4 5 6
6 5 4 3 2 1
1 2 3 4 5 6
6 5 4 3 2 1
1 2 3 4 5 6 7
7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
4 5 6 7 6 5 4
5 6 7 8 9 8 7 6 5
6 7 8 9 0 1 0 9 8 7 6
7 8 9 0 1 2 3 2 1 0 9 8 7
8 9 0 1 2 3 4 5 4 3 2 1 0 9 8
: Write a method to print each of the following patterns using nested-loops in a class called PrintTriangles. The signatures of the methods are:
public static void printXxxTriangle(int numRows)
Write the main() which prompts the user for the numRows and prints all the patterns.
(a) PowerOf2Triangle
6 15 20 15
(b) PascalTriangle1
(c) PascalTriangle2
: Write a method to compute sin(x) and cos(x) using the following series expansion, in a class called TrigonometricSeries. The headers of the methods are:
public static double sin(double x, int numTerms)
public static double cos(double x, int numTerms)
Compare the values computed using the series with the JDK methods Math.sin(), Math.cos() at x=0, &/6, &/4, &/3, &/2 using various numbers of terms.
Hints: Avoid generating large numerator and denominator (which may cause arithmetic overflow, e.g., 13! is out of int range). Compute the terms as:
: Write a method to compute the sum of the series in a class called SpecialSeries.
The signature of the method is:
public static double sumOfSeries(double x, int numTerms)
(Overflow) : Write a program called FibonacciInt to list all the Fibonacci numbers, which can be expressed as an int (i.e., 32-bit signed integer in the range of [-, ]). The output shall look like:
F(46) is out of the range of int
Hints: The maximum and minimum values of a 32-bit int are kept in constants Integer.MAX_VALUE and Integer.MIN_VALUE, respectively. Try these statements:
System.out.println(Integer.MAX_VALUE);
System.out.println(Integer.MIN_VALUE);
System.out.println(Integer.MAX_VALUE + 1);
Take note that in the third statement, Java Runtime does not flag out an overflow error, but silently wraps the number around. Hence, you cannot use F(n-1) + F(n-2) & Integer.MAX_VALUE to check for overflow.
Instead, overflow occurs for F(n) if (Integer.MAX_VALUE – F(n-1)) & F(n-2) (i.e., no room for the next Fibonacci number).
Write a similar program for Tribonacci numbers.
(Overflow): Write a program called Factorial1to10, to compute the factorial of n, for 1&n&10.
Your output shall look like:
The factorial of 1 is 1
The factorial of 2 is 2
The factorial of 10 is 3628800
Modify your program (called FactorialInt), to list all the factorials, that can be expressed as an int (i.e., 32-bit signed integer).
Your output shall look like:
The factorial of 1 is 1
The factorial of 2 is 2
The factorial of 12 is
The factorial of 13 is out of range
Hints: Overflow occurs for Factorial(n+1) if (Integer.MAX_VALUE / Factorial(n)) & (n+1).
Modify your program again (called FactorialLong) to list all the factorial that can be expressed as a long (64-bit signed integer).
The maximum value for long is kept in a constant called Long.MAX_VALUE.
: Write a method call toRadix() which converts a positive integer from one radix into another.
The method has the following header:
public static String toRadix(String in, int inRadix, int outRadix)
Write a program called NumberConversion, which prompts the user for an input number, an input radix, and an output radix, and display the converted number.
The output shall look like:
Enter a number and radix: A1B2
Enter the input radix: 16
Enter the output radix: 2
&A1B2& in radix 16 is &0010& in radix 2.
: Write a program called NumberGuess to play the number guessing game.
The program shall generate a random number between 0 and 99. The player inputs his/her guess, and the program shall response with &Try higher&, &Try lower& or &You got it in n trials& accordingly. For example:
& java NumberGuess
Key in your guess:
Try higher
You got it in 4 trials!
Hints: Use Math.random() to produce a random number in double between 0.0 and (less than) 1.0.
To produce an int between 0 and 99, use:
int secretNumber = (int)(Math.random()*100);
: Write a program called WordGuess to guess a word by trying to guess the individual characters. The word to be guessed shall be provided using the command-line argument.
Your program shall look like:
& java WordGuess testing
Key in one character or your guess word: t
Trail 1: t__t___
Key in one character or your guess word: g
Trail 2: t__t__g
Key in one character or your guess word: e
Trail 3: te_t__g
Key in one character or your guess word: testing
Trail 4: Congratulation!
You got in 4 trials
Set up a boolean array to indicate the positions of the word that have been guessed correctly.
Check the length of the input String to determine whether the player enters a single character or a guessed word. If the player enters a single character, check it against the word to be guessed, and update the boolean array that keeping the result so far.
Try retrieving the word to be guessed from a text file (or a dictionary) randomly.
boolean isLeapYear(int year): returns true if the given year is a leap year. A year is a leap year if it is divisible by 4 but not by 100, or it is divisible by 400.
boolean isValidDate(int year, int month, int day): returns true if the given year, month and day constitute a given date.
Assume that year is between 1 and 9999, month is between 1 (Jan) to 12 (Dec) and day shall be between 1 and 28|29|30|31 depending on the month and whether it is a leap year.
int getDayOfWeek(int year, int month, int day): returns the day of the week, where 0 for SUN, 1 for MON, ..., 6 for SAT, for the given date. Assume that the date is valid.
String toString(int year, int month, int day): prints the given date in the format &xxxday d mmm yyyy&, e.g., &Tuesday 14 Feb 2012&.
Assume that the given date is valid.
To find the day of the week (Reference: Wiki &Determination of the day of the week&):
Based on the first two digit of the year, get the number from the following &century& table.
Take note that the entries 4, 2, 0, 6 repeat.
Add to the last two digit of the year.
Add to &the last two digit of the year divide by 4, truncate the fractional part&.
Add to the number obtained from the following month table:
Non-Leap Year
same as above
Add to the day.
The sum modulus 7 gives the day of the week, where 0 for SUN, 1 for MON, ..., 6 for SAT.
For example: 2012, Feb, 17
(6 + 12 + 12/4 + 2 + 17) % 7 = 5 (Fri)
public class DateUtil {
public static String strMonths[]
= {&Jan&, &Feb&, &Mar&, &Apr&, &May&, &Jun&,
&Jul&, &Aug&, &Sep&, &Oct&, &Nov&, &Dec&};
public static int daysInMonths[]
= {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
public static boolean isLeapYear(int year) { ...... }
public static boolean isValidDate(int year, int month, int day) { ...... }
public static int getDayOfWeek(int year, int month, int day) { ...... }
public static String printDate(int year, int month, int day) { ...... }
public static void main(String[] args) {
System.out.println(isLeapYear(1900));
System.out.println(isLeapYear(2000));
System.out.println(isLeapYear(2011));
System.out.println(isLeapYear(2012));
System.out.println(isValidDate());
System.out.println(isValidDate());
System.out.println(isValidDate());
System.out.println(isValidDate());
System.out.println(getDayOfWeek());
System.out.println(getDayOfWeek());
System.out.println(getDayOfWeek());
System.out.println(getDayOfWeek());
System.out.println(toString());
You can compare the day obtained with the Java's Calendar class as follows:
Calendar cal = new GregorianCalendar(year, month - 1, day);
int dayNumber = cal.get(Calendar.DAY_OF_WEEK);
String[] calendarDays = { &Sunday&, &Monday&, &Tuesday&, &Wednesday&,
&Thursday&, &Friday&, &Saturday& };
System.out.println(&It is & + calendarDays[dayNumber - 1]);
The calendar we used today is known as Gregorian calendar, which came into effect in October 15, 1582 in some countries and later in other countries. It replaces the Julian calendar. 10 days were removed from the calendar, i.e., October 4, 1582 (Julian) was followed by October 15, 1582
(Gregorian). The only difference between the Gregorian
and the Julian calendar is the &leap-year rule&. In Julian calendar, every four years is a leap year. In Gregorian calendar, a leap year is a year that is divisible by 4 but not divisible by 100, or it is divisible by 400, i.e., the Gregorian calendar omits
years which are not divisible by 400. Furthermore, Julian calendar considers the first day of the year as march 25th, instead of January 1st.
This above algorithm work for Gregorian dates only. It is difficult to modify the above algorithm to handle pre-Gregorian dates. A better algorithm is to find the number of days from a known date.
Exercises on Number Theory
A positive integer is called a perfect number if the sum of all its factors (excluding the number itself, i.e., proper divisor) is equal to its value. For example, the number 6 is perfect because its proper divisors are 1, 2, and 3, and 6=1+2+3; but the number 10 is not perfect because its proper divisors are 1, 2, and 5, and 10&1+2+5.
A positive integer is called a deficient number if the sum of all its proper divisors is less than its value. For example, 10 is a deficient number because 1+2+5&10; while 12 is not because 1+2+3+4+6&12.
Write a method called isPerfect(int posInt) that takes a positive integer, and return true if the number is perfect.
Similarly, write a method called isDeficient(int posInt) to check for deficient numbers.
Using the methods, write a program called PerfectNumberList that prompts user for an upper bound (a positive integer), and lists all the perfect numbers less than or equal to this upper bound. It shall also list all the numbers that are neither deficient nor perfect.
The output shall look like:
Enter the upper bound: 1000
These numbers are perfect:
[3 perfect numbers found (0.30%)]
These numbers are neither deficient nor perfect:
12 18 20 24 30 36 40 42 48 54 56 60 66 70 72 78 80 ......
[246 numbers found (24.60%)]
A positive integer is a prime if it is divisible by 1 and itself only. Write a method called isPrime(int posInt) that takes a positive integer and returns true if the number is a prime.
Write a program called PrimeList that prompts the user for an upper bound (a positive integer), and lists all the primes less than or equal to it.
Also display the percentage of prime (up to 2 decimal places).
The output shall look like:
Please enter the upper bound: 10000
[1230 primes found (12.30%)]
Hints: To check if a number n is a prime, the simplest way is try dividing n by 2 to &n.
Write a method isProductOfPrimeFactors(int posInt) that takes a positive integer, and return true if the product of all its prime factors (excluding 1 and the number itself) is equal to its value. For example, the method returns true for 30 (30=2&3&5) and false for 20 (20&2&5).
You may need to use the isPrime() method in the previous exercise.
Write a program called PerfectPrimeFactorList that prompts user for an upper bound. The program shall display all the numbers (less than or equal to the upper bound) that meets the above criteria.
The output shall look like:
Enter the upper bound: 100
These numbers are equal to the product of prime factors:
1 6 10 14 15 21 22 26 30 33 34 35 38 39 42 46 51 55 57 58 62 65 66 69 70 74 77 78 82 85 86 87 91 93 94 95
[36 numbers found (36.00%)]
One of the earlier known algorithms is the Euclid algorithm to find the GCD of two integers (developed by the Greek Mathematician Euclid around 300BC). By definition, GCD(a, b) is the greatest factor that divides both a and b.
Assume that a and b are positive integers, and a&b, the Euclid algorithm is based on these two properties:
GCD(a, 0) = a
GCD(a, b) = GCD(b, a mod b), where (a mod b) denotes the remainder of a divides by b.
For example,
GCD(15, 5) = GCD(5, 0) = 5
GCD(99,88) = GCD(88,11) = GCD(11,0) = 11
GCD() = GCD() = GCD(990,243) = GCD(243,18) = GCD(18,9) = GCD(9,0) = 9
The pseudocode for the Euclid algorithm is as follows:
while (b != 0) {
b & a mod b
Write a method called gcd() with the following signature:
public static int gcd(int a, int b)
Your methods shall handle arbitrary values of a and b, and check for validity.
TRY: Write a recursive version called gcdRecursive() to find the GCD.}