On Fri, 2009-07-10 at 09:26 +0100, Dermot wrote:
> The algorithm works perfectly but my understanding of it's workings is amiss.
>
> When I look at this I see $E initialised and then concatenate with the
> the modulus of
> 37 % 2 =1
> 18 % 2 = 0
> 9 % 2 = 1
> 4 % 2 = 0
> 2 % 2 = 0
>
> That by my reckoning is 10100. Almost the reverse of the answer but I
> am obviously wrong and I can't see how the final expressions:
>
> my $E = binary($k);
> return $E . $b;
>
> work to give the answer.
>
> Can someone more enlightened than me give me some guidence.
Perhaps it would be easier to understand if we look at the counterpart
to this.
#!/usr/bin/perl
my $dstr = decimal( 123456789 );
print "$dstr\n";
sub decimal {
my ($n) = @_;
return $n if $n < 10;
my $k = int($n/10);
my $b = $n % 10;
my $E = decimal($k);
return $E . $b;
}
__END__
When dealing with the number 123456789, the first time through, stopping
just before the sub is called a second time, we have:
$b = 123456789 % 10 = 9;
$k = int( 123456789 / 10 ) = 12345678;
$E will be assigned the string that represents $k. Clearly, this must
be concatenated before $b in the returned string. The binary version
works the same way.
--
Just my 0.00000002 million dollars worth,
Shawn
Programming is as much about organization and communication
as it is about coding.
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