On Fri, Oct 11, 2002 at 02:54:20PM -0400, Bernie Cosell wrote:
> The NPL puzzle for 6 oct was an interesting little Perl exercise [I'm not
> sure how to solve it analytically --- I played with it some to little
> avail --- but it was certainly subject to brute force, and it turned out
> to be a cute little thing.
>
> Write out the digits from 1-9 in order. Then add some plus (+) signs
> and times (x) signs to the string to make it add up to 2,002. As
> usual in arithmetic, multiplication is done before addition, and you
> don't have to put a sign between every 2 digits. The answer is
> unique.
>
> What's odd is that my little Perl program found *TWO* solutions, but one
> is potentially ambiguous [in particular, given those rules, what should
> the value of "a*b*c" be?-- it doesn't say whether things should be done
> left-to-right or right-to-left, so perhaps that could be used to exclude
> one of the two solutions.
It doesn't matter whether the multiplication is evaluated left-to-right or
right-to-left, because multiplication is associative. The product is the
same either way.
> Anyhow, here's the little program I whipped up for it... the fun part is
> that it is one of the rare times that counting base-3 is useful:
>
> for (my $count = 0; $count < 3**8; $count += 1)
> { my $try = fixstr($count) ;
> print $try,"\n" if eval($try) == 2002 ;
> }
> exit ;
>
> sub fixstr
> { my $key = $_[0] ;
> my $str = "123456789" ;
> for (my $i = 8; $i > 0; $i -= 1)
> { my $next = $key % 3 ;
> $key = int($key/3) ;
> next unless $next ;
> substr ($str, $i, 0, $next == 1? '+': "*") ;
> }
> return $str ;
> }
>
> Obviously I'm not a golfer, but I'm wondering if there are any other
> interesting approaches to the problem... [base-3 and eval seemed pretty
> clean/cute to me]
There was discussion of this puzzle on the Boston.pm mailing list as well.
The thread starts at
http://mail.pm.org/pipermail/boston-pm/2002-October/000160.html
Ronald
