Normal Progressive Go sounds boring. No capturing newly-atari’d stones? No taking and winning a ko the same turn!? No capturing 50-eye groups in one turn!!? What is with these weak hand-holding milquetoast babying condescending anti-fun guardrails!!!? We don’t want hand talk when we play Progressive Go, we want BLOOD!!!!
I propose Unhinged Progressive Go:
Standard robust area-scoring rules apply (0.0 komi)
On turn n, play up to n stones (only if you play exactly 0 stones does it count as a pass)
After and only after you have placed all the stones you wish to that turn, are opposing and then friendly colors cleared (strings with 0 liberties removed) in that order
When one player passes, their opponent gets one more turn (n still increases even during passes, so they can place up to 2 more stones than they were allowed to place on their previous turn), and then the game is scored as is according to the Tromp-Taylor strict scoring rules, with the exception that all points which reach stones of both colors (aka dame) are counted as points for the player who did not pass
I think that @Samraku is specifically referring to shapes where one could take a ko, and then play another nearby move that captures to end the ko fight, rather than situations where the ko fight is resolved by filling.
If I’m interpreting these rules correctly, it seems that this game may have a quite pathological endgame. By simply playing on until n is large enough, it seems that no group is ever truly safe. So, eventually each player can simply capture any/all of their opponent’s chains on each turn.
Superko would theoretically prevent endless cycling, but players could practically cause the game to continue for a very, very long time (practically indefinitely), by just adjusting the position slightly, between cycles of capturing everything (or even just capturing nearly everything, instead of all).
I hadn’t realized that continue almost forever part, but it does
support the way I was thinking Progressive Go might work:
Have the stones get placed 1 at a time, rather than all up-to-n at once.
Yeah, resolving the captures after each individual stone placement would be one way to prevent such late-game cycling, since one would then be able to make safe groups with just two “small” eyes. However, it seems that the spirit of this variant to allow very large captures.
Another way to potentially address the late-game cycling issue is to introduce some other victory condition:
At any point (after the first two turns), if one player has no stones on the board, then the other player is declared the winner.
This would essentially make the victory condition about who first obtains the advantage to land the first “knockout punch”.
My goal with this rule is to hope games finish before n gets large enough to do exactly that, but I’m open to better ways to do it, especially as my idea might just straight up fail at what it wants to do
I thought of that, but abandoned it when I realized it didn’t apply the first two turns. I suppose just not applying it the first two turns solves the problem. I’m inclined to prefer this way of ending the game over the original proposal
In fact, with your proposal, I think we can modify scoring rules to award the win to the player who would have captured all opposing stones at once first, with a little more thought than I have time for at the moment, as I’m about to clock back into work
Infinite games are allowed, you score at each move how many stones were placed in that move. Let’s assume these scores are B0, W0, B1, W1, B2, W2, then White wins if Wn > Bn for all large enough n, Black wins otherwise.
More or less equivalently, the winner is the player after move 180 that gets to a position where they have at least 181 stones on the board.
It seems that this has to assume ideal, hypothetical play all the way out to infinity…
You’re not including a superko rule, right? And I guess the basic ko rule is limited to just the single stone basic kos, right? And other two-turn cycles are allowed?
I think a proof that you have a winning strategy from a certain position onwards should suffice to claim the win. But it’s a fairly convincing and easy proof.
