Solution for Programming Exercise 3.1

---

This page contains a sample solution to one of the exercises from Introduction to Programming Using Java.

---

Exercise 3.1:

How many times do you have to roll a pair of dice before they come up snake eyes? You could do the experiment by rolling the dice by hand. Write a computer program that simulates the experiment. The program should report the number of rolls that it makes before the dice come up snake eyes. (Note: "Snake eyes" means that both dice show a value of 1.) Exercise 2.2 explained how to simulate rolling a pair of dice.

---

---

Since we want to roll the dice at least once, a do..while is appropriate. A pseudocode algorithm for the program is

Let countRolls = 0
do:
    roll the dice
    count this roll by adding 1 to countRolls
while the roll is not snake eyes
Output the value of countRolls

As in Exercise 2.2, we can simulate rolling one die by computing (int)(Math.random()*6) + 1.

We want to stop rolling the dice when the roll is a double 1. We want to continue rolling the dice while the roll is not a double 1. If die1 and die2 are variables representing the values of the dice, the condition for continuing to roll can be expressed as

while ( ! (die1 == 1 && die2 == 1) )

The exclamation point means "not", so the condition says that it is not the case that (both die1 is 1 and die2 is 1). That is, it is not the case that the dice came up snake eyes. Another way to express the same condition is that at least one of the dice is not 1, that is, that either die1 is not 1 or die2 is not 1. In Java code, this is written:

while ( die1 != 1  ||  die2 != 1 )

This is the test that I use in my program. Students often get the && and || operators mixed up, especially when negation is involved. (In this case, we could have avoided the problem by testing while (die1+die2 != 2).)

Filling in some details gives an algorithm that can be easily converted into a program, which is shown below:

Let countRolls = 0
do:
    die1 = (int)(Math.random()*6) + 1
    die2 = (int)(Math.random()*6) + 1
    count this roll by adding 1 to countRolls
while die1 is not 1 or die2 is not 1
Output the value of countRolls

You could of course write the program using a regular while loop instead of do..while. But if you move the test to the start of the loop, you have to make sure that the variables are given values before you try to test those values. You need to "prime" the loop that exist only to ensure that it runs the first time. The pseudocode then becomes

Let countRolls = 0
Let die1 = 0 // Prime the loop with any value except 1
Let die2 = 0
while die1 is not 1 or die2 is not 1:
    die1 = (int)(Math.random()*6) + 1
    die2 = (int)(Math.random()*6) + 1
    count this roll by adding 1 to countRolls
Output the value of countRolls

Another option would be to use a "while(true)" loop. In that case, the test that is needed is the condition for stopping the loop, which is often easier to come up with than the test for continuing the loop, and no priming is necessary:

Let countRolls = 0
while true:
    die1 = (int)(Math.random()*6) + 1
    die2 = (int)(Math.random()*6) + 1
    count this roll by adding 1 to countRolls
    if die1 is 1 and die2 is 1
        break
Output the value of countRolls

---

---

/**  
 * This program simulates rolling a pair of dice until they
 * come up snake eyes.  It reports how many rolls were needed.
 */

public class SnakeEyes {
  
   public static void main(String[] args) {
   
       int die1, die2;   // The values rolled on the two dice.
       
       int countRolls;   // Used to count the number of rolls.
       
       countRolls = 0;
       
       do {
          die1 = (int)(Math.random()*6) + 1;   // roll the dice
          die2 = (int)(Math.random()*6) + 1;
          countRolls++;                        // and count this roll
       } while ( die1 != 1 || die2 != 1 );
       
       System.out.println("It took " + countRolls + " rolls to get snake eyes.");
   
   }  // end main()

}  // end class