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 )
];

[download]
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.

blokhead

In reply to Re: Pure regex Hamiltonian Circuit solution by blokhead
in thread Pure regex Hamiltonian Circuit solution by blokhead

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