My boggle solution

I'll follow up the new trend of explaining my solution. It's not as advanced as the others so far, but i think it's quite clever :)

my solution is quite straight forward. the only thing that a few have missed is the way to find if the next letter is "adjacent" to the current one.

if you see the boggle playing field like this :

  0123
  abcd 0
  efgh 1
  ijkl 2
  mnop 3

then two letters with coordinates (x,y) and (X,Y) are "adjacent" if abs(x-X) <= 1 and abs(y-Y) <= 1.

Here is a runthrough of my solution.

sub r{
  my$i;
  map{$_=$l++,r($i-1),$_=$b[--$l]
  if(abs$i-"@_"<6&abs$i++%4-"@_"%4<2|!$l)&$b[$l]eq$_}@ARGV;
  $f|=$l==@b
}
@b=/./g,r&&print,$f=0for<STDIN>

in my mail loop i run through <STDIN>, split $_ into an array (@b), call the sub "r", and if the returnvalue from r is true, i print $_. 
Then i clean $f which i use to mark that a solution have been found for each word.

@b=/./g,r&&print,$f=0for<STDIN>


in the sub "r", i use a counter $i to register the position in @ARGV. I use $l to count my recursive debth, which is also the position in the @b array.

i loop through @ARGV, and for each value in @ARGV ($ARGV[$i] which is infact $_) i test if it matches the current value in @b ($b[$l]) and that it is 
adjacent with the previous letter in @b, or if there is no previous because it's the first letter ($l==0).

if all test tests are true, then i effectively empty $ARGV[$i], increment $l, call "r" with $i as a parameter, decrement $l and then restore $ARGV[$i].
i guess i dont really empty $ARGV[$i] but doing $ARGV[$i]=$l is basicly the same thing since $l is a number and will never match a letter.
the reason that i use r($i-1) is that i increment $i too early, in the comparison, so i have to subtract 1.

map{$_=$l++,r($i-1),$_=$b[--$l]if(abs$i-"@_"<6&abs$i++%4-"@_"%4<2|!$l)&$b[$l]eq$_}@ARGV;


at last i test if my recursive depth matches the number of letters in @b. if it does i have found a solution.
$f|=$l==@b


my "adjacent letter" test have been golfed a bit, so i'll run throught that as well.

since i use $i as a counter i dont have a $x or a $y, so i use $i%4 as $x and $i/4 as $y, and $_[0]%4 as $X and $_[0]/4 as $Y.
as most probably golfers know, when used in an number comparison $_[0] can be written "@_".

abs(int($x/4)-int($X/4)) <= 1  =>  abs(int(($x-$X)/4)) <= 1
i had to try a bit to remove the int but 
abs($x-$X) < 6 worked.

I hope that made sense.

Terje   
-- 
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