Shuffles Random Numbers with no repetition in Javascript/PHP

Question

I've searched through some of the answers here but it doesn't seem the thing that I needed or I just don't know how to apply it though.

I haven't started any codes and I'm only thinking on how to do it and I have no idea how to do it. I need your help guys.

Let's assume that I have an array which consists of these values below

[1,2,3,4,5,6,7,8,9]

And I need to shuffle it without repeating the position of each numbers of the last result. so it would probably like

[5,3,9,6,2,8,1,4,7]

if I shuffle it again it would be like

[4,7,2,1,8,3,6,9,5]

And so on.

Well I don't know if there' any relevance to it but, would rather not to use rand() though. Any solution for this stuff?

Answer 1

Try this,

$count = 15;
$values = range(1, $count);
shuffle($values);
$values = array_slice($values, 0, 15);

OR

$numbers = array();
do {
   $possible = rand(1,15);
   if (!isset($numbers[$possible])) {
      $numbers[$possible] = true;
   }
} while (count($numbers) < 15);
print_r(array_keys($numbers));

may this help you.

Answer 2

What you want to do is to add elements from your array to another array, randomly, but to ensure that elements are not in the same indexed position. Try this:

$array = [1,2,3,4,5,6,7,8,9];
$new = array();
for($i = 0; $i < $array.length; $i++){
  $rand = $i;
  do {
    $rand = Math.floor( Math.random() * ( $array.length + 1 ) );
  } while ($rand == $i || array_key_exists($rand, $new))
  // Check that new position is not equal to current index
  // and that it doesnt contain another element

  $new[$rand] = $array[i];
}

Not the most efficient, but guaranteed to put elements in non same indices.