Is this viable? I have no idea. The lack of working ladder doesn't phase me for obvious reasons. I'm beginning to see how lots of living groups could be left at the end of the game if eyespace had a number of red stones in. Effectively, the following shape is alive:
$$Bcm1 Prisoner Count: B-0 W-0 $$ ------------------------- $$ | . . . . . . X O . C . | $$ | . . . . . . X O O O O | $$ | . . X . . . X X O X X | $$ | . . . . . . . X X X . | $$ | . . . . . . . . . . . | $$ | . . . . . . . . X . . | $$ | . . . . . . . . . . . | $$ | . . . . . . . . . . . | $$ | . . O . . . . . O . . | $$ | . . . . . . . . . . . | $$ | . . . . . . . . . . . | $$ -------------------------
[go]$$Bcm1 Prisoner Count: B-0 W-0 $$ ------------------------- $$ | . . . . . . X O . C . | $$ | . . . . . . X O O O O | $$ | . . X . . . X X O X X | $$ | . . . . . . . X X X . | $$ | . . . . . . . . . . . | $$ | . . . . . . . . X . . | $$ | . . . . . . . . . . . | $$ | . . . . . . . . . . . | $$ | . . O . . . . . O . . | $$ | . . . . . . . . . . . | $$ | . . . . . . . . . . . | $$ -------------------------[/go]
I'm not saying how the position arose, but Black can't play in either eye can he ?
Black can't play in either eye. White might eventually be forced to fill one of the eyes though since passing is not allowed. After that, either player could kill the white group with a red stone.
Just to be clear, there's no score. Redstone is a game of annihilation. There will be groups of at most one color left at the end. There could be no groups - just red stones.
My current suspicion is that no matter what your strategy is early midgame, the eventual situation will come down to each player ritually suiciding a few times and the winner being decided on who ended up with the odd or even numbered move - something only really predictable (without an _insane_ move tree) in the last 10-15 moves. As a result, all strategy up to that point seems rather ... pointless is too strong a word, but not by much I suspect?
It might be the case that when it comes down to both players filling their own eyes, whoever has formed more eyes at that point will win. And then if that's true, and if they both have the same number of eyes, then the player on turn (not necessarily Black) will lose.
I believe that the first player to lose a group is more likely, though not certain, to lose the game. You might be able to turn the tables after losing a group, but every time the tables get turned, the advantage gets a little steeper, making future table turnings less and less likely. Even if it got down to the last two groups, one black and one white, it still wouldn't be a game of pure tempo. It would simply be a matter of who had more eyes in their last remaining group, or if equal, who's turn it was.
That endgame implication of continuous play was what I was trying to express (admittedly rather poorly) before. This variant seems quite similar to no-pass Go.
MarkSteere wrote:I believe that the first player to lose a group is more likely, though not certain, to lose the game. You might be able to turn the tables after losing a group, but every time the tables get turned, the advantage gets a little steeper, making future table turnings less and less likely. Even if it got down to the last two groups, one black and one white, it still wouldn't be a game of pure tempo. It would simply be a matter of who had more eyes in their last remaining group, or if equal, who's turn it was.
I'm currently unconvinced that this is the case. Group size and eyespace comes into it, but I actually suspect there'll be a large amount or "remaining moves move parity" issues, and I wonder how many times the game will actually be lost because one side killed himself to remove his final stones. 50% of the time between skilled players maybe?
Note that it wouldn't be most eyes, because of the effective group tax--three eyes for three groups is not worth so much.
I suspect that Mark is right, though, and parity issues won't be the normal case. One player (White) is behind in eyes, so he must either capture his own stones or allow them to be captured first. This creates an empty space on the board, surrounded by Black and Red stones. The only rational play is for White to play in that space. But since that space is surrounded by Black stones with multiple eyes and/or Red stones, these stones are just waiting to be captured. If Black can make even one eye in the space vacated, he is ahead for the next capture. If not, he retains whatever advantage in eyes he had before the first capture, possibly minus one, thanks to parity considerations.
For that reason, I feel confident that parity won't matter unless the "score" difference (importing go terminology) is small once you enter the eye-filling stage. On the other hand, close endgames could be a mess.
