Waterdancer -

I think that your variant is a viable one, but it would probably take a while to catch on because it is so different. I think that you haven't gotten much response because May is about one of the worst months for people to be busy with finals, graduations, kids sports games, spring cleaning, you name it. It's just not a good month for board games.

I do apprecate you putting up the $100 to get people interested in the variant. I looked at it 2 nights ago for 1-1/2 hours and was close but just didn't have any more time to look at it. I eliminated most of the 'obvious-looking' moves for player 1 and was down to what I thought was 2 or 3 more reasonable possibilities for him.

I echo Dmitri King's question here. How is it possible to draw your variant? I can only think of one possible way:
A player 'poofs' a pair of his own stones while at the same time capturing a pair of his opponents stones (or some variation of that thereof) so that the stone capture count is 10-10.

I have a comment and a couple of questions though:
1. Draws should not be allowed. If the stone capture count is 10-10, then the game should continue until the tied capture count is broken or someone gets a 5 in a row.
2. As you have alluded to, it is possible to poof an odd # of ones own stones (Pairs poofed in 2 different directions that contain a single common stone would be 3 stones poofed). What happens if the stone capture count is 8-8, I poof 3 of my own stones while capturing 2 of my opponent's stones? Then it would be 11-10 in my opponent's favor. I would think that my opponent should win even though in theory, both sides have captured at least 10 stones (5 theoretical pairs).
3. I am assuming that we would always use a stone capture count and not a pair capture count. It could get too confusing if I poof 3 of my own stones so that the stone capture count is 9-8. So my opponent wouldn't win the game even though in logical theory, he now has 5 pairs. It's just that 2 of the 5 pairs contained a common stone.


Gary