1. Intro to OOP with Java, C. Thomas Wu
Chapter 6
Repetition Statements
(powerpoint slides from the publisher, tantc, and leeml)
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2. Intro to OOP with Java, C. Thomas Wu
Objectives
After you have read and studied this chapter, you should be able to
Implement repetition control in a program using while statements.
Implement repetition control in a program using do-while statements.
Implement a generic loop-and-a-half repetition control statement
Implement repetition control in a program using for statements.
Nest a loop repetition statement inside another repetition statement.
Choose the appropriate repetition control statement for a given task
Prompt the user for a yes-no reply using the showConfirmDialog method of JOptionPane.
(Optional) Write simple recursive methods
We will study two forms of repetition statements in this lesson. They are while and do-while statement.
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3. Intro to OOP with Java, C. Thomas Wu
Definition
Repetition statements control a block of code to be executed for a fixed number of times or until a certain condition is met.
Count-controlled repetitions terminate the execution of the block after it is executed for a fixed number of times.
Sentinel-controlled repetitions terminate the execution of the block after one of the designated values called a sentinel is encountered.
Repetition statements are called loop statements also.
In Chapter 5, we studied selection control statements. We will study in this chapter the second type of control statement, a repetition statement, that alters the sequential control flow. It controls the number of times a block of code is executed. In other words, a block of code is executed repeatedly until some condition occurs to stop the repetition. There are fundamentally two ways to stop the repetition—count-controlled and sentinel-controlled.
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4. Intro to OOP with Java, C. Thomas Wu II. while Statement
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5. Intro to OOP with Java, C. Thomas Wu
The while Statement
int sum = 0, number = 1;
while ( number <= 100 ) {
sum = sum + number;
number = number + 1; }
These statements are executed as long as number is less than or equal to 100.
The first repetition control we will study is the while statement. Here’s an example that computes the sum of integers from 1 to 100, inclusively. Note: there’s a closed form to compute the sum of 1 to 100, which is (100 * 101) / 2, so this repetition statement is a illustration purpose only.
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6. Syntax for the while Statement
Intro to OOP with Java, C. Thomas Wu
Syntax for the while Statement
while ( <boolean expression> )
<statement>
Boolean Expression
while ( number <= ) {
sum = sum + number;
number = number + 1; }
Statement (loop body)
Here’s the general syntax of a while statement. As long as the <boolean expression> is true, the loop body is executed. Notice that the loop body may not be executed at all.
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7. Intro to OOP with Java, C. Thomas Wu
Control Flow of while
int sum = 0, number = 1
number <= 100 ?
sum = sum + number;
number = number + 1;
true
false
This flowchart shows the control flow of the while statement. If the <boolean expression> is true, the loop body is executed and the control returns to the top. If the <boolean expression> is false, then the control flows to the next statement that follows this while statement.
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8. Intro to OOP with Java, C. Thomas Wu
More Examples
int sum = 0, number = 1;
while ( sum <= ) {
sum = sum + number;
number = number + 1; } 1
Keeps adding the numbers 1, 2, 3, … until the sum becomes larger than 1,000,000.
int product = 1, number = 1, count = 20, lastNumber;
lastNumber = 2 * count - 1;
while (number <= lastNumber) {
product = product * number;
number = number + 2; } 2
Computes the product of the first 20 odd integers.
Variation on computing the product of the first 20 odd integers:
int product = 1, number = 1,
lastTerm = 20, count = 1;
while ( count <= lastTerm ) {
product = product * (2 * number – 1);
count = count + 1; }
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9. Example: Testing Input Data
Intro to OOP with Java, C. Thomas Wu
Example: Testing Input Data
String inputStr;
int age;
inputStr = JOptionPane.showInputDialog(null,
"Your Age (between 0 and 130):");
age = Integer.parseInt(inputStr);
while (age < 0 || age > 130) {
JOptionPane.showMessageDialog(null,
"An invalid age was entered. Please try again.");
age = Integer.parseInt(inputStr); }
Priming Read
Here's a more practical example of using a repetition statement. This code will only accept a value greater than 0 but less than 130. If an input value is invalid, then the code will repeat until the valid input is read.
Notice that the 'age' variable must have a value before the boolean expression of the while statement can be evaluated. We therefore read the input value before the while test. This reading of input values before the test is called priming read.
The loop body of this while statement is executed zero times if the input is valid the first time.
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10. Useful Shorthand Operators
Intro to OOP with Java, C. Thomas Wu
Useful Shorthand Operators
sum = sum + number;
sum += number;
is equivalent to
Operator
Usage
Meaning +=
a += b;
a = a + b; -=
a -= b;
a = a – b; *=
a *= b;
a = a * b; /=
a /= b;
a = a / b; %=
a %= b;
a = a % b;
When writing a repetition statement, we often see the statement that modifies the value of a variable in the form such as
sum = sum + number. Because of a high occurrence of such statement, we can use shorthand operators.
These shorthand assignment operators have precedence lower than any other arithmetic operators, so, for example, the statement
sum *= a + b;
is equivalent to
sum = sum * (a + b);
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11. Intro to OOP with Java, C. Thomas Wu
Confirmation Dialog
A confirmation dialog can be used to prompt the user to determine whether to continue a repetition or not.
JOptionPane.showConfirmDialog(null,
/*prompt*/ "Play Another Game?",
/*dialog title*/ "Confirmation",
/*button options*/ JOptionPane.YES_NO_OPTION);
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12. Example: Confirmation Dialog
Intro to OOP with Java, C. Thomas Wu
Example: Confirmation Dialog
boolean keepPlaying = true;
int selection;
while (keepPlaying){
//code to play one game comes here
// . . .
selection = JOptionPane.showConfirmDialog(null,
"Play Another Game?",
"Confirmation",
JOptionPane.YES_NO_OPTION);
keepPlaying = (selection == JOptionPane.YES_OPTION); }
Here's an example of how we can use a confirmation dialog in a loop. At the end of the loop, we prompt the user to repeat the loop again or not.
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13. Intro to OOP with Java, C. Thomas Wu
Watch Out for Pitfalls
Watch out for the off-by-one error (OBOE).
Make sure the loop body contains a statement that will eventually cause the loop to terminate.
Make sure the loop repeats exactly the correct number of times.
If you want to execute the loop body N times, then initialize the counter to 0 and use the test condition counter < N or initialize the counter to 1 and use the test condition counter <= N.
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14. Intro to OOP with Java, C. Thomas Wu
Loop Pitfall - 1
int product = 0;
while ( product < ) {
product = product * 5; } 1
Infinite Loops
Both loops will not terminate because the boolean expressions will never become false.
int count = 1;
while ( count != 10 ) {
count = count + 2; } 2
If you are not careful, you can easily end up writing an infinite loop. Make sure the test is written in such a way that the loop will terminate eventually.
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15. Intro to OOP with Java, C. Thomas Wu
Overflow
An infinite loop often results in an overflow error.
An overflow error occurs when you attempt to assign a value larger than the maximum value the variable can hold.
In Java, an overflow does not cause program termination. With types float and double, a value that represents infinity is assigned to the variable. With type int, the value “wraps around” and becomes a negative value.
When an overflow error occurs, the execution of the program is terminated in almost all programming languages. When an overflow occurs in Java, a value that represents infinity (IEEE 754 infinity, to be precise) is assigned to a variable and no abnormal termination of a program will happen. Also, in Java an overflow occurs only with float and double variables; no overflow will happen with int variables. When you try to assign a value larger than the maximum possible integer an int variable can hold, the value “wraps around” and becomes a negative value.
Whether the loop terminates or not because of an overflow error, the logic of the loop is still an infinite loop, and we must watch out for it. When you write a loop, you must make sure that the boolean expression of the loop will eventually become false.
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16. Intro to OOP with Java, C. Thomas Wu
Loop Pitfall - 2
float count = 0.0f;
while ( count != 1.0f ) {
count = count f;
} //seven 3s 1
Using Real Numbers
Loop 2 terminates, but Loop 1 does not because only an approximation of a real number can be stored in a computer memory.
float count = 0.0f;
while ( count != 1.0f ) {
count = count f;
} //eight 3s 2
Although 1/3 + 1/3 + 1/3 == 1 is mathematically true, the expression
1.0/ / /3.0
in computer language may or may not get evaluated to 1.0 depending on how precise the approximation is. In general, avoid using real numbers as counter variables because of this imprecision.
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17. Intro to OOP with Java, C. Thomas Wu
Loop Pitfall – 2a
int result = 0; double cnt = 1.0;
while (cnt <= 10.0){
cnt += 1.0;
result++; }
System.out.println(result); 1
Using Real Numbers
Loop 1 prints out 10, as expected, but Loop 2 prints out 11. The value 0.1 cannot be stored precisely in computer memory. 10 11
int result = 0; double cnt = 0.0;
while (cnt <= 1.0){
cnt += 0.1;
result++; }
System.out.println(result); 2
Here's another example of using a double variable as a counter. The two loops are identical in concept, and therefore, should output the same result. They would if real numbers are stored precisely in computer memory. Because the value of 0.1 cannot be represented precisely in computer memory, the second one will actually print out 11, while the first one prints out 10, as expected.
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18. Intro to OOP with Java, C. Thomas Wu
Loop Pitfall - 3
Goal: Execute the loop body 10 times.
count = 1;
while ( count < 10 ){
. . .
count++; } 1
count = 1;
while ( count <= 10 ){
. . .
count++; } 2
count = 0;
while ( count <= 10 ){
. . .
count++; } 3
count = 0;
while ( count < 10 ){
. . .
count++; } 4
Yes, you can write the desired loop as
count = 1;
while (count != 10 ) {
...
count++; }
but this condition for stopping the count-controlled loop is dangerous. We already mentioned about the potential trap of an infinite loop. 1 3
and
exhibit off-by-one error.
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19. Loop-and-a-Half Repetition Control
Intro to OOP with Java, C. Thomas Wu
Loop-and-a-Half Repetition Control
Loop-and-a-half repetition control can be used to test a loop’s terminating condition in the middle of the loop body.
It is implemented by using reserved words while, if, and break.
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20. Example: Loop-and-a-Half Control
Intro to OOP with Java, C. Thomas Wu
Example: Loop-and-a-Half Control
String name;
while (true){
name = JOptionPane.showInputDialog(null, "Your name");
if (name.length() > 0) break;
JOptionPane.showMessageDialog(null, "Invalid Entry." +
"You must enter at least one character."); }
This is a simple example of a loop-and-a-half control. Notice the priming read is avoided with this control.
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21. Pitfalls for Loop-and-a-Half Control
Intro to OOP with Java, C. Thomas Wu
Pitfalls for Loop-and-a-Half Control
Be aware of two concerns when using the loop-and-a-half control:
The danger of an infinite loop. The boolean expression of the while statement is true, which will always evaluate to true. If we forget to include an if statement to break out of the loop, it will result in an infinite loop.
Multiple exit points. It is possible, although complex, to write a correct control loop with multiple exit points (breaks). It is good practice to enforce the one-entry one-exit control flow.
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22. Loop: break Statement (1/4)
int i = 1;
int sum = 0;
while (sum < 20) {
sum += i;
++i; }
System.out.println("sum is " + sum);
System.out.println("i is " + i);
sum is 21
i is 7
int i = 1;
int sum = 0;
while (sum < 20) {
sum += i;
if (sum % 2 == 0) break;
++i; }
System.out.println("sum is " + sum);
System.out.println("i is " + i);
sum is 6
i is 3
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23. Loop: break Statement (2/4)
int i = 1;
int sum = 0;
while ( true ) {
sum += i;
if (sum > 10) break;
++i; }
System.out.println("sum is " + sum);
System.out.println("i is " + i);
sum is 15
i is 5
Note: Without the break, this is an infinite loop!
int i = 0;
int sum = 0;
while (sum <= 10) {
++i;
sum += i; }
System.out.println("sum is " + sum);
System.out.println("i is " + i);
Try to avoid using break, unless very necessary.
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24. Loop: break Statement (3/4)
Random number = new Random();
int i;
while (true) {
i = number.nextInt(100);
System.out.println("i is " + i);
if (i >= 80) break; }
i is 43
i is 2
i is 12
i is 95
Random class will be discussed later.
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25. Loop: break Statement (4/4)
break causes control to get out of the inner-most loop that contains it.
int a = 1;
while (a < 5) {
int b = 1;
while (b < 5) {
System.out.println("a is " + a
+ ", b is " + b);
if (b == 2) break;
++b; }
++a;
a is 1, b is 1
a is 1, b is 2
a is 2, b is 1
a is 2, b is 2
a is 3, b is 1
a is 3, b is 2
a is 4, b is 1
a is 4, b is 2
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26. Loop: continue Statement (1/2)
Sometimes we may want to immediately go to the next iteration, without executing the rest of the statements in the body of the loop.
This can be done using continue.
Like break, continue only causes control to go to the next iteration of the inner-most loop that contains it.
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27. Loop: continue Statement (2/2)
i is 1
i is 2
i is 4
i is 5
i is 6
i is 7
int i = 0;
while (i <= 6) {
++i;
if (i == 3) continue;
System.out.println("i is " + i); }
int i = 1;
int sum = 0;
while (sum < 20) {
sum += i;
if (sum % 2 == 0) continue;
++i; }
System.out.println("sum is " + sum);
System.out.println("i is " + i);
sum is 23
i is 6
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28. Intro to OOP with Java, C. Thomas Wu III. do while Statement
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29. The do-while Statement
Intro to OOP with Java, C. Thomas Wu
The do-while Statement
int sum = 0, number = 1;
do {
sum += number;
number++;
} while ( sum <= );
These statements are executed as long as sum is less than or equal to 1,000,000.
Here's an example of the second type of repetition statement called do-while. This sample code computes the sum of integers starting from 1 until the sum becomes greater than 1,000,000. The main difference between the while and do-while is the relative placement of the test. The test occurs before the loop body for the while statement, and the text occurs after the loop body for the do-while statement. Because of this characteristic, the loop body of a while statement is executed 0 or more times, while the loop body of the do-while statement is executed 1 or more times.
In general, the while statement is more frequently used than the do–while statement.
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30. Syntax for the do-while Statement
Intro to OOP with Java, C. Thomas Wu
Syntax for the do-while Statement do
<statement>
while ( <boolean expression> ) ;
do {
sum += number;
number++;
} while ( sum <= );
Statement (loop body)
Boolean Expression
Here’s the general syntax of a do-while statement. As long as the <boolean expression> is true, the loop body is executed. Notice that the loop body is executed at least once.
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31. Control Flow of do-while
int sum = 0, number = 1
sum += number;
number++;
true
sum <= ?
false
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32. Intro to OOP with Java, C. Thomas Wu IV. for Statement
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33. Intro to OOP with Java, C. Thomas Wu
The for Statement
int i, sum = 0, number;
for (i = 0; i < 20; i++) {
number = scanner.nextInt( );
sum += number; }
These statements are executed for 20 times ( i = 0, 1, 2, … , 19).
This is a basic example of a for statement. This for statement reads 20 integers and compute their sum.
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34. Syntax for the for Statement
Intro to OOP with Java, C. Thomas Wu
Syntax for the for Statement
for ( i = 0 ; i < ; i ) {
number = scanner.nextInt();
sum += number; }
for ( <initialization>; <boolean expression>; <increment> )
<statement>
Initialization
Boolean Expression
Increment
Statement (loop body)
This shows the general syntax for the for statement.
The <initialization> component also can include a declaration of the control variable. We can do something like this:
for (int i = 0; i < 10; i++)
instead of
int i;
for (i = 0; i < 10; i++)
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35. Intro to OOP with Java, C. Thomas Wu
Control Flow of for
i = 0;
i < 20 ?
false
number = ; sum = number;
true
i ++;
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36. Intro to OOP with Java, C. Thomas Wu
More for Loop Examples
for (int i = 0; i < 100; i += 5) 1
i = 0, 5, 10, … , 95
for (int j = 2; j < 40; j *= 2) 2
j = 2, 4, 8, 16, 32
for (int k = 100; k > 0; k--) ) 3
Here are some more examples of a for loop. Notice how the counting can go up or down by changing the increment expression accordingly.
k = 100, 99, 98, 97, ..., 1
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37. i++ or ++i? Any difference? for ( i = 0; i < 20; i++ ) { ... }
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38. Index Variable Scope
Consider:
int currentTerm = 2;
for (int i = 0; i < 5; ++i) {
System.out.println(CurrentTerm);
CurrentTerm *= 2; }
System.out.println("i is " + i);
Does not compile, because the scope of the index variable i is limited to the for loop.
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39. Intro to OOP with Java, C. Thomas Wu V. Nested Repetition Statements
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40. The Nested-for Statement
Intro to OOP with Java, C. Thomas Wu
The Nested-for Statement
Nesting a for statement inside another for statement is commonly used technique in programming.
Let’s generate the following table using nested-for statement.
Just an if statement can be nested inside another if statement, we often nest for loops. For example, using a nest-for loop is the most appropriate way to generate a table such as the illustration.
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41. Intro to OOP with Java, C. Thomas Wu
Generating the Table
int price;
for (int width = 11; width <=20, width++){
for (int length = 5, length <=25, length+=5){
price = width * length * 19; //$19 per sq. ft.
System.out.print (" " + price); }
//finished one row; move on to next row
System.out.println("");
OUTER
INNER
Here's how the table can be produced by a nested-for loop.
For each value of width, length will range from 5 to 25 with an increment of 5. Here’s how the values for width and length change over the course of execution.
width length 11 5 10 15 20 25 12 13
and so on…
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42. Nested-for Loops (1/2) i is 0 j is 0 j is 1 j is 2 j is 3 i is 1
for (int i = 0; i < 3; ++i) {
System.out.println("i is " + i);
for (int j = 0; j < 4; ++j) {
System.out.println(" j is " + j); }
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43. Nested-for Loops (2/2) What is the output?
for (int i = 0; i < 3; ++i) {
System.out.println("i is " + i);
for (int j = 0; j < i; ++j) {
System.out.println(" j is " + j); }
What is the output?
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44. break and continue in for statement
The break and continue statements work the same way in a for loop as in a while loop.
i is 0
j is 0
j is 1
i is 1
i is 2
for (int i = 0; i < 3; ++i) {
System.out.println("i is " + i);
for (int j = 0; j < 4; ++j) {
if (j == 2) break;
System.out.println(" j is " + j); }
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45. Intro to OOP with Java, C. Thomas Wu VI. Useful Classes
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46. Random Numbers (1/2)
Random number generation is discussed in section 3.8 of your textbook.
We learn another way here, using the Random class.
Notes:
Need to import java.util.*;
Random( ): constructs a new random number generator Random whose seed is based on the current time.
int nextInt(int n): returns the next pseudo-random, from the interval [0, 1, … n-1], uniformly distributed value of a Random object.
Refer to the API specification for the complete description of class Random.
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47. Random Numbers (2/2) import java.util.*; public class Random1 {
public static void main(String[] args) {
Random num = new Random();
for (int i=0; i<5; i++) {
System.out.println("Next random number is "
+ num.nextInt(100)); }
Next random number is 48
Next random number is 14
Next random number is 89
Next random number is 7
Next random number is 44
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48. Intro to OOP with Java, C. Thomas Wu
Formatting Output
We call the space occupied by an output value the field. The number of characters allocated to a field is the field width. The diagram shows the field width of 6.
From Java 5.0, we can use the Formatter class. System.out (PrintStream) also includes the format method.
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49. Intro to OOP with Java, C. Thomas Wu
The Formatter Class
We use the Formatter class to format the output.
First we create an instance of the class
Formatter formatter = new Formatter(System.out);
Then we call its format method
int num = 467;
formatter.format("%6d", num);
This will output the value with the field width of 6.
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50. The format Method of Formatter
Intro to OOP with Java, C. Thomas Wu
The format Method of Formatter
The general syntax is
format(<control string>, <expr1>, <expr2>, )
Example:
int num1 = 34, num2 = 9;
int num3 = num1 + num2;
formatter.format("%3d + %3d = %5d", num1, num2, num3);
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51. The format Method of PrintStream
Intro to OOP with Java, C. Thomas Wu
The format Method of PrintStream
Instead of using the Formatter class directly, we can achieve the same result by using the format method of PrintStream (System.out)
Formatter formatter = new Formatter(System.out);
formatter.format("%6d", 498);
is equivalent to
System.out.format("%6d", 498);
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52. Intro to OOP with Java, C. Thomas Wu
Control Strings
Integers
% <field width> d
Real Numbers
% <field width> . <decimal places> f
Strings
% s
For other data types and more formatting options, please consult the Java API for the Formatter class.
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53. Intro to OOP with Java, C. Thomas Wu
VII. Computational Time Consideration
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54. Estimating the Execution Time
Intro to OOP with Java, C. Thomas Wu
Estimating the Execution Time
In many situations, we would like to know how long it took to execute a piece of code. For example,
Execution time of a loop statement that finds the greatest common divisor of two very large numbers, or
Execution time of a loop statement to display all prime numbers between 1 and 100 million
Execution time can be measured easily by using the Date class.
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55. Intro to OOP with Java, C. Thomas Wu
Using the Date Class
Here's one way to measure the execution time
Date startTime = new Date();
//code you want to measure the execution time
Date endTime = new Date();
long elapsedTimeInMilliSec =
endTime.getTime() – startTime.getTime();
We can achieve the same result by using the currentTimeMillis method of the System class as
long start = System.currentTimeMillis();
//code to measure
long end = System.currentTimeMillis();
long elapsedTimeInMilliSec = end – start;
To get the elasped time in seconds, we divide the milliseconds by 1000
long elapsedTimeInSec
= elapsedTimeInMilliSec / 1000;
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56. Intro to OOP with Java, C. Thomas Wu
Finding GCD
If we want to reduce a fraction to its simplest form, we need to find the greatest common divisor (GCD) of its numerator and denominator.
Method 1 – Brute force.
Given two positive integers M and N, we find their GCD by dividing M and N with values from 1 to M. The last integer that divided both M and N perfectly (no remainder) is the GCD.
Finally, let us look at a slight more complicated example. In order to reduce a fraction o its simplest form, we need to find the GCD of its numerator and denominator. We define a method that returns the GCD of two given arguments. We will develop the full definition of the Fraction class later.
Two solutions:
The brute force solution applies the definition of GCD directly. Given two positive integers M and N, we find their GCD by dividing M and N with values from 1 to M. The last integer that divided both M and N perfectly (no remainder) is the GCD. For example, 24 and 36. The numbers that divide 24 and 36 perfectly are 1, 2, 3, 4, 6, and 12. So the GCD is 12. The fraction 24/36 is reduced to the simplest form 2/3 by dividing 24 and 36 by 12. Use the modulo arithmetic operator.
A more elegant solution is based on the Euclidean algo. Given 44 and 16, first divie 44 by 16. The remainder is 12. We have the relation 44 = 2* Conclude that the GCD that divides 44 and 16 must also divide 12. Otherwise, we get a contradiction. So we reduce the problem of finding GCD(44,16) to GCD(16,12). Repeat the process 16 = 1* Reduce the problem to GCD(16,4). Since 12 = 3*4 + 0 gives no remainder, we finish the process and return 4 as the GCD. Sequence of reduction is GCD(44,16) = GCD(16,12) = GCD(12,4) = 4.
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57. Intro to OOP with Java, C. Thomas Wu
Finding GCD
Method 2 – Euclidean Algorithm.
Given two positive integers M and N, we divide M by N to get the remainder R.
We have the relation M = N*Q + R.
GCD that divides M and N must also divide R.
Reduce the problem of finding GCD (M,N) to finding GCD (N, R).
Repeat process until R=0. Return N as the GCD.
Finally, let us look at a slight more complicated example. In order to reduce a fraction o its simplest form, we need to find the GCD of its numerator and denominator. We define a method that returns the GCD of two given arguments. We will develop the full definition of the Fraction class later.
Two solutions:
The brute force solution applies the definition of GCD directly. Given two positive integers M and N, we find their GCD by dividing M and N with values from 1 to M. The last integer that divided both M and N perfectly (no remainder) is the GCD. For example, 24 and 36. The numbers that divide 24 and 36 perfectly are 1, 2, 3, 4, 6, and 12. So the GCD is 12. The fraction 24/36 is reduced to the simplest form 2/3 by dividing 24 and 36 by 12. Use the modulo arithmetic operator.
A more elegant solution is based on the Euclidean algo. Given 44 and 16, first divie 44 by 16. The remainder is 12. We have the relation 44 = 2* Conclude that the GCD that divides 44 and 16 must also divide 12. Otherwise, we get a contradiction. So we reduce the problem of finding GCD(44,16) to GCD(16,12). Repeat the process 16 = 1* Reduce the problem to GCD(16,4). Since 12 = 3*4 + 0 gives no remainder, we finish the process and return 4 as the GCD. Sequence of reduction is GCD(44,16) = GCD(16,12) = GCD(12,4) = 4.
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58. Intro to OOP with Java, C. Thomas Wu
Finding GCD
Example
N M R=N%M
GCD(44,16) 44 % 16 = 12
GCD(16,12) 16 % 12 = 4
GCD(12, 4) 12 % 4 = 0
Finally, let us look at a slight more complicated example. In order to reduce a fraction o its simplest form, we need to find the GCD of its numerator and denominator. We define a method that returns the GCD of two given arguments. We will develop the full definition of the Fraction class later.
Two solutions:
The brute force solution applies the definition of GCD directly. Given two positive integers M and N, we find their GCD by dividing M and N with values from 1 to M. The last integer that divided both M and N perfectly (no remainder) is the GCD. For example, 24 and 36. The numbers that divide 24 and 36 perfectly are 1, 2, 3, 4, 6, and 12. So the GCD is 12. The fraction 24/36 is reduced to the simplest form 2/3 by dividing 24 and 36 by 12. Use the modulo arithmetic operator.
A more elegant solution is based on the Euclidean algo. Given 44 and 16, first divie 44 by 16. The remainder is 12. We have the relation 44 = 2* Conclude that the GCD that divides 44 and 16 must also divide 12. Otherwise, we get a contradiction. So we reduce the problem of finding GCD(44,16) to GCD(16,12). Repeat the process 16 = 1* Reduce the problem to GCD(16,4). Since 12 = 3*4 + 0 gives no remainder, we finish the process and return 4 as the GCD. Sequence of reduction is GCD(44,16) = GCD(16,12) = GCD(12,4) = 4.
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59. Finding GCD Direct Approach More Efficient Approach
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60. Intro to OOP with Java, C. Thomas Wu VIII. Sample Development
Animated Version
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61. Intro to OOP with Java, C. Thomas Wu
Problem Statement
Write an application that will play Hi-Lo games with the user. The objective of the game is for the user to guess the computer-generated secret number in the least number of tries. The secret number is an integer between 1 and 100, inclusive. When the user makes a guess, the program replies with HI or LO depending on whether the guess is higher or lower than the secret number. The maximum number of tries allowed for each game is six. The user can play as many games as she wants.
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62. Intro to OOP with Java, C. Thomas Wu
Overall Plan
Tasks:
do {
Task 1: generate a secret number;
Task 2: play one game;
} while ( the user wants to play );
As a part of the overall plan, we begin by identifying the main tasks for the program. Unlike the overall plan for the previous sample developments, we will use a pseudo code to express the top level logic of the program.
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63. Intro to OOP with Java, C. Thomas Wu
Required Classes
JOptionPane
Math
standard classes
Ch6HiLo
main class
The structure of this program is very simple. We will use two standard classes, one for input and output and another for generating random numbers.
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64. Intro to OOP with Java, C. Thomas Wu
Development Steps
We will develop this program in four steps:
Start with a skeleton Ch6HiLo class.
Add code to the Ch6HiLo class to play a game using a dummy secret number.
Add code to the Ch6HiLo class to generate a random number.
Finalize the code by tying up loose ends.
The second and the third steps correspond to the two major tasks identified in the overall plan.
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65. Intro to OOP with Java, C. Thomas Wu
Step 1 Design
The topmost control logic of HiLo
1. describe the game rules;
2. prompt the user to play a game or not;
while ( answer is yes ) {
3. generate the secret number;
4. play one game;
5. prompt the user to play another game or not; }
In the first step, we determine a little more detailed control logic than the one stated in the overall plan. For each of the five identified functions, we will define a method: describeRules, generateSecretNumber, playGame, and prompt.
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66. Intro to OOP with Java, C. Thomas Wu
Step 1 Code
Directory: Chapter6/Step1
Source Files: Ch6HiLo.java
Program source file is too big to list here. From now on, we ask
you to view the source files using your Java IDE.
Please use your Java IDE to view the source files and run the program.
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67. Intro to OOP with Java, C. Thomas Wu
Step 1 Test
In the testing phase, we run the program and verify confirm that the topmost control loop terminates correctly under different conditions.
Play the game
zero times
one time
one or more times
Run the program and verify that the topmost control loop is functioning correctly.
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68. Intro to OOP with Java, C. Thomas Wu
Step 2 Design
Implement the playGame method that plays one game of HiLo.
Use a dummy secret number
By using a fix number such as 45 as a dummy secret number, we will be able to test the correctness of the playGame method
In order to verify whether our code is working correctly or not, we need to know what is the secret number. The easiest way to do this is to use a fixed number, such as 45, make the temporary generateRandomNumber to return this fixed number.
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69. Intro to OOP with Java, C. Thomas Wu
The Logic of playGame
int guessCount = 0;
do {
get next guess;
guessCount++;
if (guess < secretNumber) {
print the hint LO;
} else if (guess > secretNumber) {
print the hint HI; }
} while (guessCount < number of guesses allowed && guess != secretNumber );
if (guess == secretNumber) {
print the winning message;
} else {
print the losing message;
Here's the playGame method expressed as a pseudocode.
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70. Intro to OOP with Java, C. Thomas Wu
Step 2 Code
Directory: Chapter6/Step2
Source Files: Ch6HiLo.java
We implement the playGame and getNextGuess methods in this step.
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71. Intro to OOP with Java, C. Thomas Wu
Step 2 Test
We compile and run the program numerous times
To test getNextGuess, enter
a number less than 1
a number greater than 100
a number between 2 and 99
the number 1 and the number 100
To test playGame, enter
a guess less than 45
a guess greater than 45 45
six wrong guesses
We need to verify the correctness of two methods: playGame and getNextGuess. Try all cases presented here and confirm that you get the expected responses.
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72. Intro to OOP with Java, C. Thomas Wu
Step 3 Design
We complete the generateSecretNumber method.
We want to generate a number between 1 and 100 inclusively.
private void generateSecretNumber( ) {
double X = Math.random();
secretNumber = (int) Math.floor( X * 100 ) + 1;
System.out.println("Secret Number: " secretNumber); // TEMP
return secretNumber; }
Notice that we have one temporary statement to output the value of secretNumber. We include it for the testing purpose, i.e., we need to check the numbers generated are valid.
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73. Intro to OOP with Java, C. Thomas Wu
Step 3 Code
Directory: Chapter6/Step3
Source Files: Ch6HiLo.java
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74. Intro to OOP with Java, C. Thomas Wu
Step 3 Test
We use a separate test driver to generate 1000 secret numbers.
We run the program numerous times with different input values and check the results.
Try both valid and invalid input values and confirm the response is appropriate
As always, we run the final test by running the program numerous times trying out as many variations as possible. Before testing the generateSecretNumber method as a part of the final program, we will use a separate test driver to generate 1000 (or more) secret numbers and verify that they are valid.
class TestRandom {
public static void main (String[] args) {
int N = 1000, count = 0, number;
double X;
do {
count++;
X = Math.random();
number = (int) Math.floor( X * 100 ) + 1;
} while ( count < N && 1 <= number && number <= 100 );
if ( number < 1 || number > 100 ) {
System.out.println("Error: " + number);
} else {
System.out.println("Okay"); }
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75. Intro to OOP with Java, C. Thomas Wu
Step 4: Finalize
Program Completion
Finish the describeRules method
Remove all temporary statements
Possible Extensions
Allow the user to set her desired min and max for secret numbers
Allow the user to set the number of guesses allowed
Keep the score—the number of guesses made —while playing games and display the average score when the user quits the program
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