Generate random numbers without the exact same number the second time

Question

So I am using this code to generate a random integer from 1 to 13. This is the code that gets executed, when pressing my UIButton:

currentValue = Int.random(in: 1 ... 13)

This generates a random value, but how can I avoid getting the same number two times in a row?

By "two times in a row" do you mean that a number can come up more than once, just never consecutively, or the same number shouldn't show up ever again? — Tor, Jul 01 '19 at 23:55
@Tor Sorry; The same number can show up another time, but never consecutively. Sorry for by bad English. — Kurt Nobel, Jul 01 '19 at 23:56
Well you can't. Otherwise it wouldn't be random. If you want to apply additional rules to a random function, you will need to write your own 'random' function. If this is the only rule, then you can (a) remember each return value by storing to a variable as `lastRandom`, then use a while loop to generate the next random number until the answer does not equal `lastRandom`. — Chris Shaw, Jul 01 '19 at 23:57
I just created an [`Int.random(in:excluding:)`](https://stackoverflow.com/a/27548748/1630618) — vacawama, Jul 02 '19 at 01:03

score 1 · Answer 1 · answered Jul 02 '19 at 00:00

Well you can't. Otherwise it wouldn't be random. If you want to apply additional rules to a random function, you will need to write your own 'random' function. If this is the only rule, then you can (a) remember each return value by storing to a variable as lastRandom, then use a while loop to generate the next random number until the answer does not equal lastRandom.

var lastRandom: Int = -1
func KurtsRandom() -> Int
{
  var result = lastRandom
  while result == lastRandom
  {
    result = Int.random(in: 1 ... 13)
  }

  lastRandom = result

  return result
}

Generate random numbers without the exact same number the second time

1 Answers1