Random Number Chooser programming exercise

[Venny Tank]

Hi, guys
the task is:

Write a method that returns a random number between 1 and 54, excluding the numbers passed in the argument.
The method header is specified as follows:

public static int getRandom(int... numbers)

The code I've wrote:

public class RandomNumberChooser {

	public static void main(String[] args) {

			// invoke random number method and pass numbers to exclude
		getRandom(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
				18, 19, 20, 21, 22, 23, 24, 25);
	}

	public static void getRandom(int... numbers) {

		int randomNumber = 1 + (int)(Math.random() * 54);		// generate random number in 1 to 54 range
		System.out.println("initial random number: " + randomNumber); // print the first random number (for testing)
		  
			// loop through the array of numbers passed as arguments to be excluded
		for (int i = 0; i < numbers.length; i++) {
		  
			System.out.print(numbers[i] + " ");		// print the numbers from the array (for testing)
		  
				// if the random number is equal to any of the numbers in the array, generate new random
			while (numbers[i] == randomNumber) { 
				randomNumber = 1 + (int)(Math.random() * 54); 
			} 
		}
		  
		 System.out.println("\nfinal random number: " + randomNumber);		// print the final random (for testing)
	}
}

The end result for the random generated number should be a value different from the numbers in the array.
In this code I've managed to generate new random if the generated number is equal to one of the numbers in the array, but can't figure out how to test the new value again.
Could you point me the right direction to find my mistake?

Thanks a lot

[Manoj Kumar Jain]

as I can see in your logic it has a bug. when

 while (numbers[i] == randomNumber) {   
                randomNumber = 1 + (int)(Math.random() * 54);   
            } 

will check that the random no. generated is not equal to the current item in the array numbers[i] and onwards but what about the previous items ??. It doesn't consider the case in which it can be equal to the previous item in the list that is already checked.

for example take the case that you first generated the random no. as 22 in the for loop and then you generated the no. 10 in the while loop

[Manoj Kumar Jain]

I will suggest you to use list and use its contains() method to check if the no. generated should be excluded, if it says yes then generate new no. and check again else you have the no.

[Campbell Ritchie]

Random numbers do not support any relationship between themselves. If you are constrained not to use the same number twice, that is not random numbers. I think it is called random selection from a shrinking population.

[Venny Tank]

Manoj Kumar Jain wrote:as I can see in your logic it has a bug. when

 while (numbers[i] == randomNumber) {   
                randomNumber = 1 + (int)(Math.random() * 54);   
            } 

will check that the random no. generated is not equal to the current item in the array numbers[i] and onwards but what about the previous items ??. It doesn't consider the case in which it can be equal to the previous item in the list that is already checked.

for example take the case that you first generated the random no. as 22 in the for loop and then you generated the no. 10 in the while loop

I'm still on single dimensional arrays chapter of the book that I'm learning from and can't use any more advanced techniques than that.
Is there any way I can check the random number against previous items?

[fred rosenberger]

you would have to keep a list of which ones you found, and each time you want a new one, make sure it isn't already in your list. If it is, you have to select again

Note that the more numbers you select, the slower it will become as you may have to re-try more and more often.

Alternatively, you could move them around...i.e. each time you pick a value, move it to the end of the array, and then next time select a value from 0 to (n-1)...