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### Pure regex Hamiltonian Circuit solution

 on Oct 10, 2003 at 17:44 UTC ( #298341=perlmeditation: print w/ replies, xml ) Need Help??

I'd seen Abigail do nasty things with regexes before, but never really understood them at all. However, in the recent N-Queens solver, it was a pure regex (backrefs only), there were no (?{}) constructs, and I was finally able to understand how Abigail does it..

Then, on a regex kick, I discovered Abigail's pure regex 3-SAT reduction. If you haven't already seen it, it's the coolest couple of lines on the planet. With all this regex excitement, I decided to try my hand at solving some NP-complete problem(s) with pure regexes.

I tried a handful of different NP-complete problems, but after a few emails with MJD, I understood why these attempts went wrong. I kept getting hung up on the string and regex size being exponential on the size of input (not meaning my solution was incorrect, but invalidating its use as an NP-completeness reduction proof). Finally, I think I may have gotten it right with Hamiltonian Circuits. Abigail has done this before using extended regexes, but claimed to be stuck on a pure regex solution that wasn't exponential.

So without further ado, here's my solution. Given a (simple, undirected) graph with E edges and V vertices, it finds a Hamiltonian Circuit (a cycle that visits every vertex exactly once, starting and ending in the same place). The size of the string it creates is bounded by O(V^4), and the size of the regex it creates is bounded by O(V^2).

```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
. "\n"
. (join(' ', map { join "-", @\$_ } @all_edges ) . "\n")
x @all_edges
. "\n"
. (join(' ', map { join "-", @\$_ } @E ) . "\n") x \$V;

my \$regex = "^ "
. ".* \\b (\\d+) \\b .* \\n\n" x \$V
. "\\n\n"
. join("", map { my (\$x, \$y) = @\$_;
".* \\b (?: \\\$x-\\\$y | \\\$y-\\\$x ) \\b .*
+\\n\n"
} @all_edges)
. "\\n\n"
. join("", map { my (\$x, \$y) = (\$_, \$_+1);
".* \\b (?: \\\$x-\\\$y | \\\$y-\\\$x ) \\b .*
+\\n\n"
} 1 .. (\$V-1))
. ".* \\b (?: \\\$V-\\1 | \\1-\\\$V ) \\b .* \\n \\$\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";
}
Now the dirty details.. Here's the value of \$string and \$regex for the example graph included:
```\$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
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5
1-2 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 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
1-3 1-5 2-3 2-4 2-5 4-5
1-3 1-5 2-3 2-4 2-5 4-5
1-3 1-5 2-3 2-4 2-5 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
\n
.* \b (?: \1-\2 | \2-\1 ) \b .* \n
.* \b (?: \1-\3 | \3-\1 ) \b .* \n
.* \b (?: \1-\4 | \4-\1 ) \b .* \n
.* \b (?: \1-\5 | \5-\1 ) \b .* \n
.* \b (?: \2-\3 | \3-\2 ) \b .* \n
.* \b (?: \2-\4 | \4-\2 ) \b .* \n
.* \b (?: \2-\5 | \5-\2 ) \b .* \n
.* \b (?: \3-\4 | \4-\3 ) \b .* \n
.* \b (?: \3-\5 | \5-\3 ) \b .* \n
.* \b (?: \4-\5 | \5-\4 ) \b .* \n
\n
.* \b (?: \1-\2 | \2-\1 ) \b .* \n
.* \b (?: \2-\3 | \3-\2 ) \b .* \n
.* \b (?: \3-\4 | \4-\3 ) \b .* \n
.* \b (?: \4-\5 | \5-\4 ) \b .* \n
.* \b (?: \5-\1 | \1-\5 ) \b .* \n \$
];

__OUTPUT__
Hamiltonian circuit: [ 5 -> 4 -> 2 -> 3 -> 1 -> 5 ]
Here's how it works:
• In the first "paragraph", \$1 through \$5 each get a vertex assigned to them. This paragraph of \$string is bounded by O(V^2), and \$regex by O(V).

• The second "paragraph" is the ugly bit. We have to make sure all the vertices chosen are different. This could have been done by originally choosing \$1 through \$5 to be any permutation from an exhaustive list of permutations, but such a list would have been exponential in size.

The way I checked is by picking any 5 vertices, then ensuring that they are all pairwise distinct. In \$string, we have a repeated list of all "valid" (distinct) vertex pairs. In \$regex, we make sure that every pair of two vertices in {\$1,..,\$5} shows up in that list.

This isn't exponential, because there are V(V-1)/2 "valid" (distinct) pairs, and V(V-1)/2 pairs to verify. So this paragraph of \$string is bounded by O(V^4), and \$regex by O(V^2).

• The third "paragraph" is where we verify that our distinct vertex set has edges from \$1 to \$2 ... to \$5 and then back to \$1. This paragraph of \$string is bounded by O(E*V) = O(V^3), and \$regex by O(V).

If all these conditions matched, then the regex matches, and \$1 through \$5 must be the vertices of a Hamiltonian Circuit.

There you have it. Your feedback is welcome. I really hope I haven't made any mistakes in my analysis. This code isn't one-tenth as beautiful and elegant as Abigail's 3-SAT reduction, and perhaps it could be improved upon. But hey, we all have to start somewhere ;)

Update: modified code so that \$string is a little clearer to read. Instead of comma-separating the edge and vertex lists, they are separated by spaces. \$regex is modified accordingly, matching on \b where appropriate, instead of a comma.

Comment on Pure regex Hamiltonian Circuit solution
Replies are listed 'Best First'.
Re: Pure regex Hamiltonian Circuit solution
by Abigail-II (Bishop) on Oct 11, 2003 at 02:05 UTC
Nice. You beat me to it. I was already pretty sure earlier this week that Hamiltonian Circuits/Paths could be done with pure regexes as well, and just tonight I was making some notes on how to do it. It basically came down to the same principles you are using - although I was thinking of mixing the picking of the vertices with the testing for unique picks and valid paths (just to gain speed by rejecting earlier).

Abigail

Re: Pure regex Hamiltonian Circuit solution
by BrowserUk (Pope) on Oct 11, 2003 at 02:26 UTC

++blokhead. My hat is off to you.

I'd really like to see two things.

1. The same algorithm implemented using normal perl. Iterative, recursive, a mixture of the two. Whatever, so long as it solved the same problem (and arrived at the same answer:).
2. A benchmark of a few, well chosen, 'typical cases' using your pure-regex, Abigail's extended regex, and the pure perl solutions.

Throw in a CPAN Perl-over-'generic paths'-C-library, solution and we'd have a very interesting shoot-out:)

Examine what is said, not who speaks.
"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail

Re: Pure regex Hamiltonian Circuit solution
by blokhead (Monsignor) on Oct 13, 2003 at 05:50 UTC
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.

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