Random int with exclude

Programming Questions

David Wil..

in Programming Questions • 1 year ago

Hi all,

I'll rephrase my post:

I need to call a function that returns a random int between an upper and lower limit but excludes value(s) inbetween?

This code from StackOverflow is apparenly a way of doing just that, but I am unsure how I would get this working in Processing as the random() function works differently?

If anyone can help that would be great...

Dave

public int getRandomWithExclusion(Random rnd, int start, int end, int... exclude) {
    int random = start + rnd.nextInt(end - start + 1 - exclude.length);
    for (int ex : exclude) {
        if (random < ex) {
            break;
        }
        random++;
    }
    return random;
}

Any help would be greatly appreciated

Thanks

Dave

Replies(8)

jbum

Re: Random int with exclude

1 year ago

A few questions.

Are you excluding just 1 number. Are you excluding a range of numbers, as in the above example?

As you continue to generate random numbers, will you continue to exclude more and more, until there is only one random number left? If so, the more naive ways of accomplishing this will be very inefficient. If not, then simpler methods will work.

David Wil..

Re: Random int with exclude

1 year ago

Hi,

It's a game board, there are 4 types of game objects.

At the beginning of each turn, a new game object is spawned at random.

I want to spawn only game object of a type that have less than 3 instances already in play.

So, if there are:

1 x Red [1]

3 x Blue [2]

2 x Green [3]

2 x Yellow [4]

The function returns a random int of either 1,3 or 4.

Hope this helps

Dave

bustup

Re: Random int with exclude

1 year ago

Untested but should work.

public int getRandomWithExclusion(int start, int end, int[] exclude) {
int rand = 0;
do {
rand = (int) random(start, end);
} while(Arrays.asList(exclude).contains(rand));
return rand;
}

David Wil..

Re: Random int with exclude

1 year ago

Thanks for your help, but this doesn't work - posted code below...

Any thoughts?

int[] excluded = {
1, 4
};
public void setup() {
print("Excluded numbers are: ");
for (int i = 0; i < excluded.length; i++) {
print(excluded[i]+" and " );
}
println("won't appear in the following output: ");
for (int i = 0; i < 100; i++) {
print(getRandomWithExclusion(0, 5, excluded));
}
}
public int getRandomWithExclusion(int start, int end, int[] exclude) {
int rand = 0;
do {
rand = (int) random(start, end);
}
while (Arrays.asList (exclude).contains(rand));
return rand;
}

bustup

Re: Random int with exclude

1 year ago

Okay, sorry, was untested.

This one works.

public int getRandomWithExclusion(int start, int end, int[] exclude) {
int rand = 0;
do {
rand = (int) random(start, end);
}
while (Arrays.binarySearch(exclude, rand) >= 0);
return rand;
}

David Wil..

Re: Re: Random int with exclude

1 year ago

Thankyou, that's perfect!

Arrays.binarySearch is not something I'm familiar with, thanks again.

bustup

Re: Random int with exclude

1 year ago

Something to keep in mind though is to have the int[] excluded array sorted, else the binarySearch wont work.

This can be achieved by

public int getRandomWithExclusion(int start, int end, int[] exclude) {
Arrays.sort(excluded);
int rand = 0;
do {
rand = (int) random(start, end);
}
while (Arrays.binarySearch(exclude, rand) >= 0);
return rand;
}

jossgray

Re: Random int with exclude

1 year ago

Arrays.binarySearch is a java function.

Processing is built on top of the java language so you can use methods from java as well.

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