Here's an updated version using lookaheads to greatly shorten things. Matching time shouldn't change much.
my @E = ([1,3],[1,5],[2,3],[2,4],[2,5],[4,5]);
my $V = 5;
my $verbose = 1;
my @all_edges = map { my $x = $_; map { [$x, $_] } $x+1 .. $V } 1 .. $
+V-1;
my $string = (join(' ', 1 .. $V) . "\n") x $V
. join(' ', map { join "-", @$_ } @all_edges ) . "\n"
. join(' ', map { join "-", @$_ } @E );
my $regex = "^\n"
. ".* \\b (\\d+) \\b .* \\n\n" x $V
. join("", map { my ($x, $y) = @$_;
"(?= .* \\b (?: \\$x-\\$y | \\$y-\\$x ) \\b
+ ) \n"
} @all_edges) . ".*\\n\n"
. join("", map { my ($x, $y) = ($_, $_+1);
"(?= .* \\b (?: \\$x-\\$y | \\$y-\\$x ) \\b
+ )\n"
} 1 .. ($V-1))
. "(?= .* \\b (?: \\$V-\\1 | \\1-\\$V ) \\b )\n";
print "'$string' =~ /\n$regex\n/x\n" if $verbose;
if (my @c = $string =~ /$regex/x) {
local $" = " -> ";
print "Hamiltonian circuit: [ @c -> $1 ]\n";
} else {
print "No Hamiltonian circuit\n";
}
__END__
$string = q[
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-3 1-5 2-3 2-4 2-5 4-5
];
$regex = q[
^ .* \b (\d+) \b .* \n
.* \b (\d+) \b .* \n
.* \b (\d+) \b .* \n
.* \b (\d+) \b .* \n
.* \b (\d+) \b .* \n
(?= .* \b (?: \1-\2 | \2-\1 ) \b )
(?= .* \b (?: \1-\3 | \3-\1 ) \b )
(?= .* \b (?: \1-\4 | \4-\1 ) \b )
(?= .* \b (?: \1-\5 | \5-\1 ) \b )
(?= .* \b (?: \2-\3 | \3-\2 ) \b )
(?= .* \b (?: \2-\4 | \4-\2 ) \b )
(?= .* \b (?: \2-\5 | \5-\2 ) \b )
(?= .* \b (?: \3-\4 | \4-\3 ) \b )
(?= .* \b (?: \3-\5 | \5-\3 ) \b )
(?= .* \b (?: \4-\5 | \5-\4 ) \b )
.*\n
(?= .* \b (?: \1-\2 | \2-\1 ) \b )
(?= .* \b (?: \2-\3 | \3-\2 ) \b )
(?= .* \b (?: \3-\4 | \4-\3 ) \b )
(?= .* \b (?: \4-\5 | \5-\4 ) \b )
(?= .* \b (?: \5-\1 | \1-\5 ) \b )
];
There's no more need for repeating the listing of
@all_edges so many times. With lookaheads, it only needs to be there once. Same with the listing of
@E. This reduces
$string to O(V^2).
Because backtracking doesn't happen within lookaheads, I couldn't use lookaheads to select the captured vertices. So the listing of vertices 1 to $V still has to be there $V times.