in reply to Re: The N-queens problem using pure regexes
in thread The N-queens problem using pure regexes

N Original Faster Non-pure 4 0.035 0.034 0.035 5 0.045 0.036 0.036 6 0.769 0.041 0.038 7 4.833 0.042 0.038 8 0.082 0.049 9 0.072 0.044 10 0.113 0.056 11 3.504 0.051 12 0.096 13 0.071 14 0.577 15 0.467 16 3.864 17 2.289 18 19.630 19 1.324 20 117.227

I'm curious - have you had a chance to look at why the speeds actually improve when going from 8 to 9 for both Faster and Non-pure and from 10-11 for Non-pure, but slows down 30x for Faster? And, what's with 17, 18, and 19 when it's 2.289 -> 19.630 -> 1.324??

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The idea is a little like C++ templates, except not quite so brain-meltingly complicated. -- TheDamian, Exegesis 6

Please remember that I'm crufty and crochety. All opinions are purely mine and all code is untested, unless otherwise specified.

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Re: The N-queens problem using pure regexes
by Abigail-II (Bishop) on Oct 11, 2003 at 01:34 UTC
    It has to do with how many positions are rejected before a suitable one is found. The solutions found for n = 8 and n = 9 are:
    [a8 b4 c1 d3 e6 f2 g7 h5] [a9 b7 c4 d2 e8 f6 g1 h3 i5]
    As you can see, for n = 8, it never has to backtrack for the first queen (a8 is choosen), but for the seconde queen, b8, b7, b6, and b5 need to be rejected. b8 and b7 will be rejected right away (as they are attacked by a8), but for b6 and b5 to be rejected, lots of other queens will be have to be placed. For n = 9, no backtracking for the first queen is needed, and for the second queen, the positions b9 and b8 are rejected immediately. It's only the third queen were there's some real backtracking going on - c9, c8, c7, and c6 are rejected immediately, and only for c5 more queens will be tried before rejecting it.

    The timings for 'faster' with n >= 10 cannot be trusted, as the program contained a bug for n >= 10 (see elsewhere in this thread - the bug is now fixed). Here's a new table (done on a different computer, and recording user times, not wall clock time), with the fixed programs:

    N Original Faster Non-Pure 4 0.06 0.05 0.04 5 0.07 0.04 0.05 6 1.57 0.07 0.05 7 9.29 0.06 0.05 8 0.23 0.06 9 0.16 0.06 10 0.50 0.07 11 0.41 0.07 12 2.64 0.14 13 1.58 0.10 14 37.23 0.82 15 35.45 0.70 16 5.45 17 3.18 18 27.17 19 1.89 20

    And, in case you are interested, the code that generated the table:

    #!/usr/bin/perl use strict; use warnings; no warnings qw /syntax/; $| = 1; my $width = 15; my $time_out = 120; my @cmds = ("./queens2 -n ", "./queens3 -n ", "./queens1 -f -n "); my $nr_of_commands = @cmds; my $N = 4; print " N"; printf "%${width}s" => $_ for qw /Original Faster Non-Pure/; print "\n"; while ($nr_of_commands) { printf "%3d" => $N; foreach my $cmd (@cmds) { unless (defined $cmd) { print " " x $width; next; } local $SIG {ALRM} = sub {die "Time out!"}; alarm ($time_out); eval { my $time = (`/usr/bin/time -f "%U" $cmd $N 2>&1`) [-1]; alarm (0); chomp $time; printf "%$width.2f" => $time; }; if ($@ && $@ =~ /Time out/) { undef $cmd; $nr_of_commands --; print " " x $width; } } print "\n"; $N ++; }

    Home work question: the code above is lacking something vital. What is it not doing what it should do?

    Abigail

      Home work question: the code above is lacking something vital. What is it not doing what it should do?

      Not clearing $@ ?


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

        Clearing $@ isn't necessary. $@ is guaranteed to be a "null string" (sic) if there's no error in an eval block or string.

        But it has to do something with the eval.

        Abigail

Re: Re: Re: The N-queens problem using pure regexes
by benizi (Hermit) on Oct 10, 2003 at 19:58 UTC

    For any odd N odd N not divisible by 3: $a_solution = [ map { chr(ord('a') + $_ - 1).(((2 * $_) - 1) % $N); } (1..$N) ]

    (e.g., for $N = 11: [ a1, b3, c5, d7, e9, f11, g2, h4, i6, j8, k10 ])

    This is equivalent to starting in the bottom-left corner of the board and moving right one square and up two squares (wrapping when you hit the edge) N-1 times.

    If the regex picks the first space not-already-capturable in each column (From brief inspection, it appears to do so -- It finds an equivalent solution for odd-N), this is the first solution it will find. In the even-N case, this process will not leave any not-already-capturable squares in the last column on the first pass, so it must then backtrack.

    Update: Whoa there, Ben. I spoke way too soon. The above solution only applies when N is odd AND not divisible by 3. (So, for N = (1,5) mod 6)

    More update: I was wrong about most of the analysis, too. This won't be the first solution found.