Java: Random number in fixed range excluding specified number

Question

In my (Java Android) game, I spawn coins at random positions.

Coins can appear at one of 6 positions horizontally across the screen (2 max per horizontal level).

What I do is create a random number between 0 and 5 for the first coin, then I want generate another random number excluding the position of the first coin.

So, for example, coin 1 is spawned at a random position between 0 and 5 - so let's say 4.

Then the next coin needs to be able to choose between 0-3 or 5. (basically 0-5 excluding 4).

I have managed to do it, but it's not very elegant and I'm sure there must be a better/cleaner way to achieve this, however, it escapes me.

The random(int number) method in the code below simply returns a random int from 0 to number-1 (uses nextInt) and randomBool() just returns a random boolean

Also, please bear in mind that I don't want to use any technique that keeps re-generating a random number if the one it produces is equal to the one we are trying to avoid.

code

 //Return a random number between 0 and 5 excluding the specified number
 private int getRandomExcluding(int excludedNumber){
 //If previous position was 0 then generate a number between 1 and 5
 if (excludedNumber==0){
 return random(5)+1;
 }
 //If position was 5, then generate and return number from 0-4
 else if (excludedNumber==5){
 return random(5);
 }
 //If number isn't 0 or 5 (then it is in range of 1-4 use a random bool to determine
 // if we are going to get a number less than or greater than the number we are excluding
 //True - get number lower than excluded number
 else if(randomBool()){
 //Excluded number is 1
 if (excludedNumber==1){
 return 0; //Only posibility
 }
 //Excluded number is > 1
 else {
 //Return a random number between 0 (inclusive) and the excluded number (exclusive)
 return random(excludedNumber);
 }
 //False - get number higher than the excluded number (between exludedNumber+1 (inclusive) and 6(exlusive))
 else {
 return random(6-(excludedNumber+1))+(excludedNumber+1);
 }
 }

• Dave Newton: Use a set of possible values, removing each as it's selected.
• assylias: One way is to fill a list with the valid values, shuffle it, and return the first two elements.
• Zippy: I like this idea, I'm guessing this would be in the form of an array of length 5 and maybe a switch on excludedNumber, which would populate the array with the relevant values (so, case 0 would populated with 1,2,3,4 & 5, case 1 with 0,2,3,4 & 5 etc), and then just generate a random number between 0 and 5 and go to that index of the array to retrieve the value?! maybe you could post your idea as an answer if there is a better way to put it into action than my interpretation?! Thanks!!!! –

Answer 1

You could populate a list and shuffle it:

List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5);
Collections.shuffle(numbers);
int coin1 = numbers.get(0);
int coin2 = numbers.get(1);

• Maarten Bodewes: You may want to use shuffle(numbers, new SecureRandom) for an optimal distribution, often important for a game.

Answer 2

Try this solution:

private int getRandomExcluding(int excludedNumber){
 int num = random(5);
 return num >= excludedNumber ? num+1 : num;
}

It simply generates the random number from 0 to 4 and if it's great or equal to the excluded number, it adds one. This way all five possible numbers are uniformly distributed (if your RNG produces uniformly distributed numbers).

• Dave Newton: ... This is broken in multiple ways; it can go out of range, and isn't iterative. A better solution would just use a set of possible values.
• Tagir Valeev: Out of which range? And why it should be iterative? There's nothing about iterativeness in the question.
• Dave Newton: I missed the "only two numbers" part, although I dread magic numbers. I think the solution actually works since you're only generating 0-4, so apologies. I don't think it's particularly random, though, since you always increment. I think the canonical solutions (shuffling or a set) are still better, and better suited for generality.

Answer 3

Solution: (you can change List<Integer> excludes to int... excludes if you prefer this type of input)

 /**
 * Get a random number between a range and exclude some numbers
 *
 * @param start start number
 * @param end end number
 * @param excludes list of numbers to be excluded
 * @return value between {@code start} (inclusive) and {@code end} (inclusive)
 */
 private int getRandomWithExclusion(int start, int end, List<Integer> excludes) {
 Collections.sort(excludes); // this method only works with sorted excludes
 int random = start + new Random().nextInt(end - start + 1 - excludes.size());
 for (int exclude : excludes) {
 if (random < exclude) {
 break;
 }
 random++;
 }
 return random;
 }