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Perl: the Markov chain saw

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Okay, here's my take.

  • my board is some kind of a 2D array.
  • 'x' are 1.
  • 'o' are -1.
  • empty is 0.

we look for an absolute 3. But I'm sure there is a more mathematical way for this:

my @board; while (<DATA>) { chomp; push @board, [split / /]; } for my $i (0 .. 2) { my ($t1, $t2, $t3, $t4) = (0,0,0,0); for my $j ( 0 .. 2 ) { $t1 += $board[$i][$j]; $t2 += $board[$j][$i]; $t3 += $board[$j][$j]; $t4 += $board[2 - $j][$j]; } if (abs($t1) == 3 || abs($t2) == 3 || abs($t3) == 3 || abs($t4) == 3) { print "Win.\n"; last; } } __DATA__ 1 0 1 -1 1 -1 1 0 1

He who asks will be a fool for five minutes, but he who doesn't ask will remain a fool for life.
Chady |
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In reply to Re: Checking Tic-Tac-Toe Win conditions? by Chady
in thread Checking Tic-Tac-Toe Win conditions? by BUU

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