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Using modern regex flavors like PCRE/PHP, Perl, .NET, or similar, it is possible to determine if a vertical line formation of three Xs exists in an ASCII "image."
Consider the following regex expression:
(?xm) # ignore comments and whitespace, ^ matches beginning of line ^ # beginning of line (?: . # any character except \n (?= # lookahead .*+\n # go to next line ( ?+ . ) # add a character to the 1st capturing group .*+\n # next line ( ?+ . ) # add a character to the 2nd capturing group ) )*? # repeat as few times as needed X .*+\n # X on the first line and advance to next line ?+ # if 1st capturing group is defined, use it, consuming exactly the same number of characters as on the first line X .*+\n # X on the 2nd line and advance to next line ?+ # if 2st capturing group is defined, use it, consuming exactly the same number of characters as on the first line X # X on the 3rd line
This expression employs lookaheads with self-referencing capturing groups to count the number of characters on each line before an X. It attempts to match three X characters in a vertical formation by matching the conditions specified in the lookaheads.
Online Demo: https://regex101.com/r/Xb5FXl/2
While it's not directly solvable in regex flavors with limited lookbehinds like Perl and PCRE, it is possible to indirectly determine the count by manipulating the matched string.
The following solution modifies m.buettner's "partial PCRE solution" by matching and counting the number of characters in the portion of the match where the three Xs occur.
^ (?: (?: # match .+? characters . (?= # counting the same number on the following two lines .*+\n ( ?+ . ) .*+\n ( ?+ . ) ) )+? (?<= X ) # till the above consumes an X (?= # that matches the following conditions .*+\n ?+ (?<= X ) .*+\n ?+ (?<= X ) ) (?= # count the number of matches .*+\n ( ?+ . ) # the number of matches = length of ) )* # repeat as long as there are matches on this line .*\n? # remove the rest of the line
This expression matches the same formations as the previous one, but it adds a third capturing group and counts the length of the matched characters within that group. By substituting the matches with just the contents of the third capturing group, the resulting string length represents the number of three-X formations.
Example Usage in Perl:
$in =~ s/regex//gmx; $count = length $in;
Online Demo: https://regex101.com/r/iqxY1a/1
Test Cases:
Test #0: -------------------- X X X result: 1 (X) Test #1: -------------------- ..X.... ..X.... ..X.... result: 1 (.) Test #2: -------------------- ..X.X.. ..X.X.. ....X.. result: 1 (.) Test #3: -------------------- ..X.... ..X.... ...X... result: 0 () Test #4: -------------------- ..X.... ...X... ..X.... result: 0 () Test #5: -------------------- ....X.. .X..X.. .X..... result: 0 () Test #6: -------------------- .X..X.. .X.X... .X.X... result: 1 (.) Test #7: -------------------- .X..X.. .X..X.. .X..X.. result: 2 (.X) Test #8: -------------------- XXX XXX XXX result: 3 (XXX) Test #9: -------------------- X.X.X XXXXX XXXXX .X.X. result: 5 (XXXXX) Test #10: -------------------- 1....X....... 2..X..X...X.... 3X.X...X..X..... 4X....XXXXXX..... 5X..XXX........... 6.....X.......... 7.........X....X 8..X......X....X.... 9..X......X....X....X... A....X..... B.X..X.. C..... XXX XXX XXX . result: 8 (3458.XXX)
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