The Original IQ Test

[japhy]

Lots of family restaurants have the "peg game" at their tables. You know, 15 holes, 14 pegs, and you have to jump one peg over another into an empty spot. The object is to leave ONE peg remaining. I've done it before, but I've forgotten it. So I had Perl refresh my memory.

#!/usr/bin/perl -w

use strict;

# $board[$empty_pos]{$jumpable} = $jumper
my @board = (
  { 1 => 3, 2 => 5, },
  { 3 => 6, 4 => 8, },
  { 4 => 7, 5 => 9, },
  { 1 => 0, 4 => 5, 6 => 10, 7 => 12, },
  { 7 => 11, 8 => 13, },
  { 2 => 0, 4 => 3, 8 => 12, 9 => 14, },
  { 3 => 1, 7 => 8, },
  { 4 => 2, 8 => 9, },
  { 4 => 1, 7 => 6, },
  { 5 => 2, 8 => 7, },
  { 6 => 3, 11 => 12, },
  { 7 => 4, 12 => 13, },
  { 7 => 3, 8 => 5, 11 => 10, 13 => 14, },
  { 8 => 4, 12 => 11, },
  { 9 => 5, 13 => 12, },
);

# 0 = empty, 1 = peg
my @state = (1) x 15;


# choosing a blank spot, one by one
# we only do five spots, since the board
# is made up of three sets of these five spots
# merely rotated -- there's no need to do extra work
  
for (0 .. 4) {
  print "Trying to make ", 1 + $_, " the remaining peg...\n";
  
  $state[$_] = 0;
  jump(\@state, [], {}, $_);
  $state[$_] = 1;
} 
  
  
sub jump {
  # state, history, repetitions, original blank spot
  my ($st, $hist, $rep, $orig) = @_;
  my $ok = 0;
  
  # we don't count permutations of movements
  # update:  sort @$hist  =>  sort map "@$_", @$hist
  return if $rep->{ join "\n", sort map "@$_", @$hist }++;

  # for each of the blank positions on the board...
  for my $pos (grep !$st->[$_], 0 .. $#$st) {

    # if there are two available pegs (one to jump, one to be jumped).
+..
    for (grep $st->[$_] && $st->[$board[$pos]{$_}], keys %{ $board[$po
+s] }) {

      # the empty place becomes pegged
      # and the jumper and jumped become empty
      $ok = $st->[$pos] = !($st->[$_] = $st->[$board[$pos]{$_}] = 0);
  
      # add this event to our history
      push @$hist, [ 1 + $board[$pos]{$_}, 1 + $pos ];
  
      # and now jump again!  (the meat of recursion)
      jump($st, $hist, $rep, $orig);
  
      # after we're done, remove the latest addition to the history
      pop @$hist;
  
      # and set the pegged place to empty again
      # and the two empty places back to pegged
      $st->[$pos] = !($st->[$_] = $st->[$board[$pos]{$_}] = 1);
    }
  } 

  # if we have ONE peg left, and it's the where the blank spot was
  # show the history to explain how we got here!
  print map("$_->[0] to $_->[1]\n", @$hist), "\n"
    if grep($_, @$st) == 1 and $st->[$orig] == 1;
}

You can modify the final statement to figure out how to get 8 pegs remaining with no jumps available:

  print map("$_->[0] to $_->[1]\n", @$hist), "\n"
    if grep($_, @$st) == 8 and !$ok;

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

[dmmiller2k]

Cool! It's the old Hi-Q game!

Thanks japhy! Now I have something to do to eat up all those extra CPU cycles while I'm stting around waiting for long-running database-intensive code to complete...

dmm

You can give a man a fish and feed him for a day ...
Or, you can teach him to fish and feed him for a lifetime

[runrig]

Cool! It's the old Hi-Q game!

Actually Hi-Q was a game with 33 holes, 32 pegs. Same concept though. The pattern was this (starting w/the center hole empty):

    o o o
    o o o
o o o o o o o
o o o o o o o
o o o o o o o
    o o o
    o o o

I used to actually know the answer to solving this (ending w/one peg in the center) in the fewest number of moves (multiple jumps w/one peg == 1 move).

[dmmiller2k]

Whoops.

'nuff said. Still cool, though...

dmm

You can give a man a fish and feed him for a day ...
Or, you can teach him to fish and feed him for a lifetime