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Subject: Considering "Akani Go" rss

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Steven Metzger
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A while back, I posted a comment suggesting a simple merger of the rules of TriGo and Redstone.

Part of the goal with TriGo was to make a fair game and eliminate komi. Redstone eliminates ko, komi, and endgame scoring, but isn't necessarily fair. Merging these two mechanisms together might create a fair, simple, and interesting game, and it feels only natural to give this idea some thought. After spending 3 minutes with Google Translate, I think it should be called Akani Go.

Without spending a lot of time on hit, the rules would be like this:

A) Play on a hexhex board. TriGo is hexhex8 (169 spaces), but with an annihilation game I can see a board as small as hexhex5 (61 spaces) being reasonable.
B) Every turn, place two stones simultaneously (except for the first turn, where black places one stone).
C) If a placed stone results in a capture, immediately replace it with a red stone (which cannot be captured or moved from that point forward).
D) You win if all placed stones of your opponent's color have been captured (your turn or theirs).

So I've got some questions about this that are potential snags in the idea:

1) With annihilation as the goal, how excessively long could the game be? I'd almost be more inclined to play this on a hexhex5 as the final board.
2) What happens if you cannot place one or two stones? Is there a situation where this might even occur?
 
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Interesting idea... If both stones played are used in a capture, only the last one is converted to red?

metzgerism wrote:
2) What happens if you cannot place one or two stones? Is there a situation where this might even occur?

If, for instance, Black has nowhere to legally place one stone and if the game is not over yet, there must be a black group with only one liberty L where playing would result in a suicide. Then I would suggest that Black passes (passing without playing any stone being legal only in this case); White can then play at L to kill the black group.
To sum it up: at least one player will always have an available intersection to place a stone, and if a player has no move available they must pass their turn.
Can that situation really arise from an empty board? There is also the necessary condition that every white group must have at least three liberties each completely surrounded by white and red stones. In any case, I obtained the following board through legal (though stupid) play, where it is Black's turn; thus the answer is yes. ninja
 
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Another (simpler?) solution: just authorise suicide (in which case the suicidal stone is converted to red instead of disappearing)...
 
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Steven Metzger
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eobllor wrote:
Another (simpler?) solution: just authorise suicide (in which case the suicidal stone is converted to red instead of disappearing)...
I spent a little time thinking about this, and the conclusion that I came to WAS that you must suicide your own groups if that's your only move. There are some issues with this, though:

A) If suicide is legal, couldn't you simply suicide your own pieces into an opponent's eye? There would have to be something in the rules clarifying that any capturing placement is marked with a red stone, whether it replaces a black or white stone - or none at all. You'd effectively be filling up eyes with reds this way, meaning that eyes no longer exist.

B) I don't think having simultaneous placement will work - if there's only one space available, you can't play and your tactics could be compromised even if there's some additional space to be had. I think the placements have to be sequential.

C) When does the game truly end? If you have to suicide all of your stones on placement 1 of your turn, do you get to finish the turn? With a small board, I think maybe it should - all captures will result in filling the board with reds, which progress the game to its conclusion. There's probably quite a bit of counting that will go on (and maybe a fair degree of tactics trumping strategy).
 
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metzgerism wrote:
A) If suicide is legal, couldn't you simply suicide your own pieces into an opponent's eye? There would have to be something in the rules clarifying that any capturing placement is marked with a red stone, whether it replaces a black or white stone - or none at all. You'd effectively be filling up eyes with reds this way, meaning that eyes no longer exist.

B) I don't think having simultaneous placement will work - if there's only one space available, you can't play and your tactics could be compromised even if there's some additional space to be had. I think the placements have to be sequential.

C) When does the game truly end? If you have to suicide all of your stones on placement 1 of your turn, do you get to finish the turn? With a small board, I think maybe it should - all captures will result in filling the board with reds, which progress the game to its conclusion. There's probably quite a bit of counting that will go on (and maybe a fair degree of tactics trumping strategy).

A) Maybe we could define a kill by the removal of the last liberty of a group present on the board before the placement. Thus, one would be allowed to fill only their own eyes.

B) Wouldn't it work with a turn consisting in placing one or two stones simultaneously (as in TriGo)? I'd have thought of replacing both stones by red ones in case both had removed a liberty of the group being killed.

C) That's a good question... Of course, in case of simultaneous placements, the issue doesn't arise.
 
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