CODEWITHSUNDEEP
javarandom
user2900748
user2900748

Reputation: 43

Random numbers within a range excluding zero

I'm trying to create a Random number generator using, java.util.Random. I need to generate numbers between -5 and +5 excluding zero. This is for bouncingbox app for one of my labs. The random number is the direction of the velocity of the box.

Random v = new Random();
       int deltaX = -5 + v.nextInt(10)  ;
       for(; deltaX>0 && deltaX<0;){

           System.out.println(deltaX);
       }

I've tried this, but it doesn't exclude zero. Any help would be appreciated.

Upvotes: 4

Views: 7389

Answers (7)

pjs
pjs

Reputation: 19855

Yet another way:

int[] outcomeSet = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5};
int deltaX = outcomeSet[v.nextInt(10)];

If the outcomeSet is made static this should be the most computationally efficient.

Upvotes: 4

Langusten Gustel
Langusten Gustel

Reputation: 11002

You could also do this:

Random random = new Random();
int random = random.nextInt(5) + 1;
int result = random.nextBoolean() ? random : -random;

Yes its less effective because it requires two random values. The resulting intervals are [-5 - 0) and [1 - 5]

Upvotes: 1

user000001
user000001

Reputation: 33327

Here is one way:

int deltaX = -5 + v.nextInt(10);
if (deltaX >= 0) deltaX++;

Upvotes: 5

Rahul Tripathi
Rahul Tripathi

Reputation: 172448

You may try like this:-

int deltaX = -5 + v.nextInt(10); 
if (deltax >= 0) deltax++;

Upvotes: 1

Pshemo
Pshemo

Reputation: 124225

I like other answers which random in range [-5,4] and if value is 0 or grater they add 1 making real range [-5; -1] [1; 5].

But here is some easier approach, although not shorter.

Random v = new Random();

boolean makeNegative = v.nextBoolean();
int value = v.nextInt(5) + 1; // range [1; 5]

if (makeNegative)
    value = -value;

It is less effective because it requires two random values (boolean and int) but code is easier to read.

Upvotes: 2

user2738336
user2738336

Reputation:

One possible solution without regeneration the random each time is to use the following algorithm:

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;
}

The following can be either called with an array reference:

int[] ex = { 2, 5, 6 };
val = getRandomWithExclusion(rnd, 1, 10, ex)

or by directly inserting the numbers into the call:

val = getRandomWithExclusion(rnd, 1, 10, 2, 5, 6)

It generates a random number (int) between start and end (both inclusive) and does not give you any number which is contained in the array exclude. All other numbers occur with equal probability. Note, that the following constrains must hold: exclude is sorted ascendingly and all numbers are within the range provided and all of them are mutually different.

Upvotes: 3

Silviu Burcea
Silviu Burcea

Reputation: 5348

do {
  deltaX = -5 + v.nextInt(11);
} while (deltaX == 0)

The easiest way, still random(no +1).

Upvotes: 1

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