First JAPH - Spell perl in two hundred and eighty five thousand and seventy four easy steps

[Calaban]

#!c:\perl\perl.exe -w
use strict;
my $A="a";for(0..285074){$A++;}print"$A";

[Veachian64]

Corollary 1: $A="a";for(0..176858){$A++;}print"$A";

[blakem]

Combining those I can convert 'japh' into 'perl' in a mere 108216 steps.....

% perl -e '$\="japh";$\++for 0..108215;print'
perl

-Blake

[VSarkiss]

You have a picket-fence problem: it's actually 285,075 iterations. ;-)

[jmcnamara]

Clever idea. :-)

In reply to juerd's golf (28 chars):     perl -e '$\=a;$\++for-74..285e3;print'

--
John.

[Juerd]

Haha! ++! Great one!

Nice piece of code for a game of golf too...

# First attempt
#        1         2         3
#23456789012345678901234567890
$a=a;$a++for 0..285074;print$a

# Second attempt (cheating :P)
print'perl'

2;0 juerd@ouranos:~$ perl -e'undef christmas'
Segmentation fault
2;139 juerd@ouranos:~$

[gmax]

#        1         2        
#234567890123456789012345678
$\=z;$\++for 0..285049;print

Faster, too ;-)

update
Cheating: \:->>


#        1         2        
#2345678901234567890123456
$\=z;$\++for z..perk;print

 _  _ _  _  
(_|| | |(_|><
 _|   

[Anonymous Monk]

So then, if we really want to cheat:

#        1
#234567890123456
print++($_=perk)

[Starky]

So simple it's brilliant! I laughed myself silly.

[pokemonk]

I find this really amazing, but i can't figure out why it happens. can someone please explain?

pokmon

[blakem]

From perlop:
The auto-increment operator has a little extra builtin magic to it. If you increment a variable that is numeric, or that has ever been used in a numeric context, you get a normal increment. If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and matches the pattern /^[a-zA-Z]*[0-9]*$/, the increment is done as a string, preserving each character within its range, with carry:
print ++($foo = '99'); # prints '100'
print ++($foo = 'a0'); # prints 'a1'
print ++($foo = 'Az'); # prints 'Ba'
print ++($foo = 'zz'); # prints 'aaa'


The auto-decrement operator is not magical.

-Blake

[Anonymous Monk]

#!/usr/bin/perl
use strict;
my $theword = @ARGV[0];
my $A='a';
FINDIT: for (0..99999999){
$A++;
if ($A eq "$theword")
    {print "$_\n";
    last FINDIT;
    }
}

[holonaut]

Wow. That can take a really long time to count. This little bit o' code is much faster.

 #!/usr/bin/perl -w

# Convert pseudo-base26 number (digits: a-z) to decimal
# really intended to find the ending number needed in
# the for loop for any given string to be used

my $num = 0;
my $char = "a";
my $answer = 0;

until ($char eq "aa") {
    $equiv{$char}=$num;
    $num++; $char++;
}

print "Enter your lowercase string: ";
chomp($string = <STDIN>);
@strArray=reverse(split(//,$string));

for ($i = 0; $i < @strArray; $i++) {
    $answer += ($equiv{$strArray[($i)]} * (25 ** $i));
}

print --$answer;

[kiat]

hi pokemonk;

In order to get an idea of how the word 'perl' gets printed, you may want to modify the code as follows:



#!c:\perl\perl.exe -w
use strict;
my $A="a";for(0..285074){$A++; print"$A";}

That is, you place the 'print "$A"' statement inside the loop. To see what what happens when, say, 1000 characters are printed, you stop the loop by pressing Ctr 'C' (if you're using Windows).

kiat

[/dev/null]

Great script!! It gave me plenty of pseudo-base 26 fun. The number is incremented exponentially as the number of characters increases. i.e. devnull in base 26 is:
 ((4*26**6) + (5*26**5) + (22*26**4) + (14*26**3) + (21*26**2) + (12*26**1) + (12*26**0))

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