I was just thinking: 3-3 invasions must generally be bad (for the invader) since in the double hane variation, White captures one stone and Black captures two?
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Here is an example where life and death does change compared to regular go:
White would need to sacrifice 6+5+4+3+2+1 = 21 stones to capture 18 black stones, so white will just leave it as seki.
This would be quite unlikely to come up in a real game! 6-stone nakade is rare by itself, but then you also need the surrounding group to be almost as small as possible to make it work.
You can make the black group smaller by putting it in the corner, but then white also needs to sacrifice fewer stones because of the special properties of the corner:
In this case white would have to sacrifice 6+4+3+2+1 = 16 stones to capture 14 black stones (still not worth it, but just like with the side group above there’s not much room to make the black group bigger before it starts being worth it for white to capture).
There’s an interesting asymmetry with life and death. Sacrificing stones to kill is usually good (except in the rare extreme cases above). But sacrificing stones to live doesn’t work!
This group can live in regular go, but is already dead in the variant.
A more mundane example:
(black is alive in regular go, but in the variant if white plays first black is dead)
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For area scoring rules, where I think the variant should require playing out life-and-death disputes, I fully agree with this analysis.
However, for applying this variant on top of Japanese rules, I think that one would interpret this as still Black being dead, with zero White stones being captured, as the judgement phase would be considering hypothetical play.
Well, technically Black can live while letting White gain a capture. While that might often lead to White winning, it is still optimal play (locally) for Black to live while sacrificing one stone, and hope to turn the game around by later making a capture elsewhere, if possible.
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Indeed, it seems we need new terminology along the lines of “Black lives with -1 capture”!
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meanwhile, simple stone counting never has such problems
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We could generalize this: for all (a,b)≠(0,0), define the variant (a,b)-go so that prisoners are worth a points and stones on the board at the end of the game are worth b points. If t>0 then (a,b)-go is equivalent to (ta,tb)-go.
Stone defender’s variant is (a,1)-go for a large enough.
(0,1)-go is go with stone counting.
There could be many different variants, but probably finitely many of them since (a,1)-go and (a’,1)-go should be equivalent if 0<a<a’<a+1/361.
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More generally, I think we could use three parameters to describe (a, b, c)-Go:
- a = captured stones
- b = living stones
- c = territory
Then, we have:
- Stone counting is (0, 1, 0)-Go
- Territory counting is (1, 0, 1)-Go
- Area counting is (0, 1, 1)-Go
I would say that @stone.defender’s variant could be flexibly combined with either of these, by changing a to be a very large value.
I’m not sure whether (on a given finite-sized board) there would be a finite or infinite number of strategically distinct variants that could be created by playing with non-negative choices for these parameters. I guess that if we consider Go to be a game that could be play on any arbitrary board size, then there would be an infinite number of strategically distinct variants across these choices. In that case, perhaps we would have to define a = ω, if we don’t know ahead of time what finite board size the players would select. And then, what if the players decided to play on an infinite board? a = ω * ω?
Certainly, many highly pathological cases (in the sense that the best strategy would look very different than normal Go) exist across the parameter space.
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