Re^2: The 15 Puzzle

Your

my $inversions = 1;
while ($inversions % 2) {
    @board = (0, shuffle(1..15));
    $inversions = 0;
    for my $x (1 .. 15) {
        for my $y ($x + 1 .. 15) {
            $inversions++ if $board[$x] > $board[$y];
        }
    }
}
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bothers me a little, because the code is asserting we have one inversion, and then that we have none, and then counting them. It seems… wrong, somehow, to assert at the start that we have one inversion, beginning counting them — especially because that assertion is followed immediately by a contradictory one. So I switched it to

my $inversions;
do {
  $inversions = 0;
  @board = (0, shuffle(1..15));
  # @board = (1,0,2..15); # for testing
  for my $x(1..15) {
    for my $y($x+1..15) {
      ++$inversions if $board[$x]>$board[$y]
    }
  }
} while $inversions % 2;
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I'd appreciate your letting me know what you think.

$_="msh210";$"=$\;@_=@{[split//,uc]}[2,0];$_="@_$\1";$\=$/;++$_[0]for$...1;print lc substr crypt($_,"@_"),1,6

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Re^3: The 15 Puzzle by hippo (Archbishop) on Jun 10, 2020 at 21:35 UTC
Both versions of the loop are effectively the same as far as the algorithm goes. By moving the conditional to the end you have avoided the need for the initial flag value. I just picked 1 as it makes `$inversions % 2` a true value but it could equally have been -1 or -9999 or any other (odd) value if that were to make it seem less wrong. You could even go so far as to set up a constant called `INITIAL_INVERSIONS` and set that to have the arbitrary flag value - TIMTOWTDI. It's best to code in a way that's right for you. If the semantics of the variables are most important to you then that's clearly the way to go. There's no computational penalty for this and anything that makes the code easier to maintain, whether objectively or for the specific maintainer, has to be beneficial. Just the process of analysing this and re-working the loop will have been, I hope, a worthwhile exercise. Thanks for taking the time to do so.	[reply] [d/l] [select]

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Re^3: The 15 Puzzle
by hippo (Archbishop) on Jun 10, 2020 at 21:35 UTC

Both versions of the loop are effectively the same as far as the algorithm goes. By moving the conditional to the end you have avoided the need for the initial flag value. I just picked 1 as it makes $inversions % 2 a true value but it could equally have been -1 or -9999 or any other (odd) value if that were to make it seem less wrong. You could even go so far as to set up a constant called INITIAL_INVERSIONS and set that to have the arbitrary flag value - TIMTOWTDI.

It's best to code in a way that's right for you. If the semantics of the variables are most important to you then that's clearly the way to go. There's no computational penalty for this and anything that makes the code easier to maintain, whether objectively or for the specific maintainer, has to be beneficial.

Just the process of analysing this and re-working the loop will have been, I hope, a worthwhile exercise. Thanks for taking the time to do so.

[reply]
[d/l]
[select]