one funster emailed me privately:
Well, I would start with a grid "clued" with your diagonal. That should get you there pretty fast if one exists.
And I can't imagine that I missed that approach! I tried to start with a grid and manipulate it to have the desired diagonal. By just starting with the desired diagonal, I could then add "possible" cells until the solver fills in the rest of the grid. Very neat. I easily banged out several more with little trouble. THANKS! Given this approach, it makes me think that every legal diagonal can be extended to a full grid [where I think the only 'legality' constraint is that none of the three 3x3's on the diagonal contain a repeated #]. Just have the solver show you all the remaining possibilities for each unfilled and un-deduced cell, and fill in the unfilled-ones at random... Thanks all for both the sample-grids for me to use and suggestions for how I could generate more! /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--