RE: finding differences in a list of amounts

> -----Original Message-----
> From: rpg400-l-bounces@xxxxxxxxxxxx / CWilt@xxxxxxxxxxxx
> Sent: Monday, September 20, 2004 10:59 AM
>
> I'm thinking some sort of a shifting algorithm...
<snip>
>
> Again, two bits is easy.  Three+ takes more thought, which I
> don't have time for right now ;-)
>
> But it is an interesting task, I'll be thinking about it at home tonight.

(grin)

> If you have to brute force it, you might consider running the 2-bit before
> spending much time validating the 3+ bit.  You may get lucky and find the
> difference only involves 2 amounts.

Yeah, see, this is what I'm thinking as well.  Finding differences in
reconciliation, in my experience, usually involves a small set of numbers.
Since this seems to be getting into serious bit-twiddling territory, I had
posted this part of the problem to the MI400 list.  A couple of responses
there gave me an idea, and I replied:

I am starting to think that this problem requires another array that I can't
quite put in words (there's a horrendous machine buzzing in my background
all morning - need to get some noise-cancelling headphones).  Essentially,
it would be an array of integers that would have as many elements as there
would be the number amount of combinations as I want to test for.  In my
first iteration, only elements 1 & 2 would be used.  Element1 is set to 1
and element2 is set to 2.  Increment element2 until get to the high limit,
then add 1 to element1 and set element2 to (element1 + 1).

Once I get to the end of 2-bit combos, set e1 to 1, e2 to 2, and e3 to 3,
and repeat the process as described for the 2-bit combos.

Does that make any sense?

tia,
db