in reply to Re: Algorithm for cancelling common factors between two lists of multiplicands

in thread Algorithm for cancelling common factors between two lists of multiplicands

Here is my shot at a pseudo-code

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 ;)

Replies are listed 'Best First'. | |
---|---|

Re^3: Algorithm for cancelling common factors between two lists of multiplicands
by BrowserUk (Pope) on Aug 08, 2005 at 20:50 UTC |

In Section
Seekers of Perl Wisdom

Comment onRe^2: Algorithm for cancelling common factors between two lists of multiplicands