use Math::BigInt ();

*gcd = \&Math::BigInt::bgcd;

sub embiggen {
   local $_ = @_ ? $_[0] : $_;
   return Math::BigInt->new($_);
}


my @a = ( 10, 20, 33, 45, 60 );
my @b = ( 2, 5, 10, 12, 16, 23, 45 );


# my %a = map { $_ => 1 } @a;
# my %b = map { $_ => 1 } @b;
#
# my @c = map embiggen,
#         grep { !$b{$_} }
#         @a;
#
# my @d = map embiggen,
#         map { Math::BigInt->new($_) } 
#         grep { !$a{$_} }
#         @b;

my @c = @a;
my @d = @b;

foreach my $c (@c) {
   foreach my $d (@d) {
      my $gcd = gcd($c, $d);
      $c /= $gcd;
      $d /= $gcd;
   }
}

@c = grep { $_ != 1 } @c;
@d = grep { $_ != 1 } @d;
[download]

use Math::BigInt       ();
use Math::Big::Factors ();

*gcd    = \&Math::BigInt::bgcd;
*factor = \&Math::Big::Factors::factors_wheel;

sub embiggen {
   local $_ = @_ ? $_[0] : $_;
   return Math::BigInt->new($_);
}


my @a = ( 10, 20, 33, 45, 60 );
my @b = ( 2, 5, 10, 12, 16, 23, 45 );


# my %a = map { $_ => 1 } @a;
# my %b = map { $_ => 1 } @b;
#
# my @c = map embiggen,
#         grep { !$b{$_} }
#         @a;
#
# my @d = map embiggen,
#         grep { !$a{$_} }
#         @b;

my @c = @a;
my @d = @b;


my %c_factors;

foreach my $c_idx (0..$#c) {
   my @c_factors = factor($c[$c_idx]);
   foreach my $c_factor (@c_factors) {
      push(@{$c_factors{$c_factor}}, $c_idx);
   }
}


foreach my $d (@d) {
   my @d_factors = factor($d);
   foreach my $d_factor (@d_factors) {
      next unless $c_factors{$d};
      next unless @{$c_factors{$d}};

      my $c_idx = shift(@{$c_factors{$d}});
      $c[$c_idx] /= $d_factor;
      $d         /= $d_factor;
   }
}


@c = grep { $_ != 1 } @c;
@d = grep { $_ != 1 } @d;
[download]

Update I wrote this when we were trying to figure out the exact output. Not relevant anymore

ikegami,

Your first should work correctly. Did not read the second one yet. One thing you want to note when you test this code is that the result will not match exactly what BrowserUk had coz the cancellation process is kind of subjective when done by humans as opposed to computers

take for example if i have 15 * 3 * 12 / 9 * 17

then I could either do 5 * 12/17 or 15 * 4/17

I guess the final solution should match product of the elements in the array

Another thing, you might get a tiny little benefit if you check for the elements == 1 inside the loop construct and pop them out. I am not sure if that is even possible.

BTW, i think you can add next if $c == 1 or $d == 1; to avoid computing gcd's on 1's.

-SK

1. Pick the first element of array a

2. Calcuate the GCD(a[0],b[0]) or better yet a[0] = mod (a[0], gcd(a[0],b[0]) and b[0] =mod (a[0], gcd(b[0],b[0]) (just divide elements by gcd, it should be an integer)

3. If anything is 1 pop it out of the list

4. Go through the list of elements for b with a[0] again. Do this until all elements are scanned in b.

5. Take the second element in a and repeat the process.

Does this work?

I shall post the code if i get a chance to implement it

-SK

PS: This reminds of division in high school days ;)

PS: This reminds of division in high school days ;)

I guess that's where the "cancelling out" phrase comes from :)

Maybe the confusion comes from the fact that most people younger than me never learned to cancel out their equations because they all used calculators from an early age.

They probably don't know what "casting out the nines" is either.

---

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?

"Science is about questioning the status quo. Questioning authority".

The "good enough" maybe good enough for the now, and perfection maybe unobtainable, but that should not preclude us from striving for perfection, when time, circumstance or desire allow.

foreach, element in an array, factor to primes (i.e. if starting from 10, then 1*2*5). Create a hash for for the first array where keys are factors of each element and values are the number of times each factor has occurred. Repeat for second array. "Subtract" the two hashes. Multiply the resulting keys by their respecting values --> should give you the elements of the resulting array.