Random Generator and 2D Arrays

**hiya54** · September 21st, 2014, 04:26 PM

So im making this ghost game where i display an 8x8 filled with 0s and a randomly generator five 1s in there
I can get it to display 0s and add 1s, however sometimes the 1s that are randomly generated sometimes go on the same spot
making it look like there are only four 1s. How would i go about fixing that?

 
package Grade12;
 
import java.util.Random;
 
public class Ghost {
 
	public static void main(String[] args) {
 
		Random generator = new Random();
 
		int gameboard [][] = new int [8][8];
		int randomx, randomy, counter = 0, sum = 0;
 
		for(int row = 0; row < 8; row++){
			for(int col = 0; col < 8; col++){
 
 
			(gameboard[row][col]) = 0;
 
			}
 
		}
 
		for(int row = 0; row < 8; row++){
			for(int col = 0; col < 8; col++){
 
 
				if( counter != 5){
 
 
				randomx = (int)(Math.random()*8);
				randomy = (int)(Math.random()*8);
 
				gameboard[randomx][randomy] = 1;
 
 
				counter++;
 
				if(gameboard[randomx][randomy] == 1){
					sum++;
				}
 
 
 
				}
				System.out.print(gameboard[row][col] + " ");
 
 
 
 
			}
 
 
 
 
 
		}
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
	}
 
}

**camel-man** · September 21st, 2014, 04:36 PM

You could check to see if a 1 is in that "slot" and if there is already a 1 there, then you should redo the random number generator until it finds an open slot.

**hiya54** · September 21st, 2014, 04:49 PM

So do i need a if loop in the nested for loop? I know i had to do something like that i just didnt figure
how to do it yet

**camel-man** · September 21st, 2014, 04:53 PM

if(gameboard[randomx][randomy] == 0)
{
gameboard[randomx][randomy] = 1;
counter++;
}

That should be under your randomx and randomy math generators.

**Cornix** · September 21st, 2014, 05:26 PM

What camel-man describes could work in practice, but in theory it might end in an infinite loop. Most of the time that isnt too bad because its very very very unlikely, but I would personally not like to use a possibly unsafe method.
As an alternative solution that always terminates you could use a list. You fill it with ALL possible positions, then you shuffle it, and then you take the first 5 elements in the list. Its guaranteed to give you 5 different random positions if you have at least 5 positions. If there are no 5 positions (because your array is too small) an appropriate exception will be thrown instead of an infinite loop.

**hiya54** · September 21st, 2014, 09:20 PM

We werent taught array list or shuffle yet. I learned a bit of array list on my own spare time
but yea not really that well, but anyways camel-man's method works with a simple while loop

Thread: Random Generator and 2D Arrays

LinkBack

Thread Tools

Display

Random Generator and 2D Arrays

Related threads:

Re: Random Generator and 2D Arrays

Re: Random Generator and 2D Arrays

Re: Random Generator and 2D Arrays

Re: Random Generator and 2D Arrays

Re: Random Generator and 2D Arrays

Similar Threads

Random Shape Generator

Random Number Generator

Random Number Generator Always gives 0

random number generator

[SOLVED] Random number method implementation