在java中生成随机数对，使得p！= q

Question

I am trying to create pairs of random integers in the range [0,n) . I need to make sure that for any input n , the numbers created ,say p,q are such that p != q

I tried to use java.util.Random with seed sothat I can reproduce the result ..I tried inputs 100,200,400,800 and they all created p,q such that p !=q .But at 1600 two pairs were with p == q

public static void generate(int size){      
    Random ran = new Random();
    ran.setSeed(123456L);       
    for(int i =0;i<size;i++){
        int p = ran.nextInt(size);
        int q = ran.nextInt(size);
        if(p==q)
            System.out.println(p+" equals "+q);
        //else
            //System.out.println(p+" "+q);
    }
}

public static void main(String[] args) {
    generate(1600);

}

this gave

692 equals 692
843 equals 843

I am sure there is some way to make sure that p != q for any input n.. but I cannot recall the math needed

Can someone help?

Answer 1

Just keep picking until they don't match.

int p = ran.nextInt(size);
int q;

do {
    q = ran.nextInt(size);
} while(p==q);

Answer 2

Generate one number in [0,n) and the other one in [0,n-1) If the second one is superior (inclusive) to the first one, add one.

int p = ran.nextInt(size);
int q = ran.nextInt(size-1);

if (q>=p){
    q++;
}

Answer 3

Add 1 to n in a List . Then use Collection.Shuffle to shuffle the whole list . It will shuffle the list with equal likelihood. Then get any 2 from the list

For example

ArrayList a = new ArrayList();
for(int i = 1;i <= n; i++)
    a.add(i);
Collections.shuffle(a);
int first = (int)a.get(0);
int second = (int)a.get(1);

Answer 4

One of the solutions is:

1. Generate first number
2. Generate second number
3. While second number equals to first number return to step 2

In almost 100% step 2 will be executed no more than a couple of times.

But ensure than n is more than 1 because you will have an endless loop in another case (but anyway, you can't get correct results with any algorithm)