exclude one number from a random range

Question

I want to generate a random number from a specific range in each iteration

for (int i = 0; i < 10; i++)
{
    v1[i] = rand() % n;
}

This code will generate a number between 0 and 9. However, I do not want the selected number to be the same as the index i. For example, if I am in the first iteration (ie: i == 1), I want the random number to be either 0, 2, 3, 4, 5, 6, 7, 8, 9 and not 1.

Can someone help me in this?

fredoverflow · Answer 1 · 2015-07-02T16:46:25.413

8

If you want to exclude one number, just remove the greatest number from the set of possible choices, and if you happen to pick the one you didn't want, choose the greatest number instead.

For example, if you want to pick a number between 0 and 9, but it should not be the number 1, then pick a number between 0 and 8. If you pick 1, choose 9 instead.

edited Jul 02 '15 at 16:46

answered Jul 02 '15 at 11:16

fredoverflow

256,549
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what if the index is 2 and random number generated is 2 also? – rahul tyagi Jul 02 '15 at 11:36
@rahultyagi "if you happen to pick the one you didn't want, choose the greatest number instead", so you would pick 9 in this case. – fredoverflow Jul 02 '15 at 11:37
the range will remain the same 0 to 8 ,this solution seems to work then – rahul tyagi Jul 02 '15 at 11:49

score 3 · Answer 2 · answered Jul 02 '15 at 11:31

3

Try something like:

for (int i = 0; i < 10; i++)
{
  do
  {
    v1[i] = rand() % n;
  } while (v1[i] == i);
}

Given the numbers you are using the do-loop will only cycle more than once about every ten calls.

answered Jul 02 '15 at 11:31

rossum

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1

This solution may distort the statistical properties of the result. – Toby Speight Jul 02 '15 at 14:18
@TobySpeight No it won't, or at least no more so than using % does in general. This is "acceptance/rejection", and is a well-known and mathematically correct technique. – pjs Jul 02 '15 at 15:03

exclude one number from a random range

2 Answers2