View Single Post
Old 05-30-2013, 06:41 PM
BillsBayou's Avatar
BillsBayou BillsBayou is offline
Join Date: Mar 2013
Location: New Orleans
Posts: 77

Nailed it, guys, thanks. I need to add that one to my toolbox. I've seen it before.

Got to give it a name of some sort. The chains have the same spacing, but because values in the chain are in the same category, we can eliminate the equality of their related links. Both chains begin from different points in the same category and head in the same direction at the same pace. Uneven Parallel Chains?

Ax + c = Jm
Ay + c = Kn
Jm <> Kn

Category-A-index-x, plus some constant, c, equals Category-J-index-m
Category-A-index-y, plus some constant, c, equals Cateogry-K-index-n
Implies that Category-J-index-m cannot be Category-K-index-n
Where c is the same in both equations
Where Category-J or Category-K can equal Category-A (but not both because the problem would be easy to solve)
Where all four values Ax, Ay, Jm, Kn represent four distinct values.
Reply With Quote