Net Go / Discussion

Sub-page of NetGo

Discussion of the NetGo rules

Article 1

Ellbur: If the object of an alliance is to occupy the greatest territory:

Does this mean more territoy than the other alliances, or just the greatest total territory for the one alliance?

What is the object of the players? Do they want one of their alliances to win? Do they want all of their alliances to be superior to others?

When does an alliance 'win'? When does a player 'win'?

How should the players consider their object relative to the success of the other players? Should they try to do better than others or best for themselves?

The rules don't answer any of these questions.

doraguma: Good points! I think, if for example A has 50, B has 40, and C has 60, the alliance C should be first, A second, and B third winner. There is no rule, if a player has won, or not. Only the alliances count. If one of the players in an alliance is not so strong, (s)he reduces the ability of the alliance.

If there are no non-color alliances, the player is supposed to play in favour for the alliance. (If not, he/she will not be invited often to play ;-) ) Of course all playing alliances can consist of only of one player.

I had no ideas for non-color alliance rules, yet.

Article 6 (ko)

doraguma: I am not sure, if this ko rule is the best extension for a multi player Go. What do you think?

David: How do the one-way connections (the ones marked with an arrowhead) work in practice? This breaks the commutativity of connections; e.g., a stone A can be connected to B but B is not connected to A.


Say that the connections in this diagram go from left to right, so stones get connections and liberties only from the right. Before B1, white+square has one liberty and the other white stone has one liberty through white+square. What happens after B1? Clearly, white+square will be captured, because its only liberty has been occupied. Is the other white stone captured too? Yes, because white+square is in its chain, or no, because it is not in white+square's chain?

doraguma: In this case the white stone on the left has no liberties (also before 1). If white played first, then red, the red stone captures already the white stone. (If red played in the middle first, white could not place a stone there.)

But it is possible to capture more than one color of stones at the same time, like in standard go it is possible to capture two stones, which are not connected.

David: I'm sorry; my example wasn't clear. white+square in the diagram is a not a red stone, but a white stone with a mark. There are only two colors in the diagram. The two white stones are present before B1.

So my question remains: which white stones are captured? Both of them, or just white+square?

doraguma: Sorry, my misunderstanding. The two white stones are connected and build a chain with one liberty (before B1). Each of the white stones is connected to the other. It would not make sense to say, O is connected to white+square, but white+square is not connected to O. So B1 takes the last liberty and the white stones are captured.

axd: Note: commutativity is no longer applicable in this variant; this asymmetry however introduces interesting complications. I think it does make sense to say that "connectedness" is no longer commutative.

Net Go / Discussion last edited by axd on February 8, 2005 - 12:37
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