NO. (For an explanation, time it with RandomiseArray(1..1000) and RandomiseArray(1..100000) - it has exponential complexity in time)

Here is a better way to do it: 

sub RandomiseArray {
  my @a=@_;
  my @b=();
  while (scalar @a) {
    push(@b,splice(@a,int(rand @a)));
  }
  return(@b);
}

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As the array becomes longer, the probablility of your routine finding an unused space becomes smaller and smaller. With a 1..10 shuffle the probablility of finding the right element for the last space is one in ten - which means you will need 10 tries (on the average). Not too bad. But with a 1..100000 array, you will need 100_000 tries to fill in the last element. and 99_999 to find the next to last, and... you get the idea. It will take a LOOONG time.

Yeah, I could do it inplace (without the second copy of the array), but I'm going to save that for another time.

In reply to Re: Answer: How can I print all the numbers from 1 to n in random order? by matija
in thread How can I print all the numbers from 1 to n in random order? by vroom

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