So, can I have a counterexample to rule 2?
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You can, but it’s not really a complete rule, so I thought you’d reformulate it for the actual guess. What if there’s not at least two white groups?
In any case, I’ll stick with modifying my initial koans for the counter examples, which have more than two white chains:
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It is a complete rule. If there are not at least two white groups, the koan is green. I’m assuming “If A, then B” is interpreted as a material conditional that is, the statement “if A, then B” is true whenever B is true or A is false.
So, in essence, my rule implied that all red koans have at least two white groups.
My new rule guess:
The graph obtained by letting each Black group be a vertex, and putting an edge between two Black groups when they “see” each other, is an acyclic path-connected graph. The same holds for the White groups.
This rule expects the following to be green:
With the graphs:
and the following to be red, because the Black graph has a cycle:
and the following to be red, because the White graph is not connected:
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Do the revealed koans suggest that acyclic is necessary?
Are these koans green?
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I don’t think they do. In case cyclic graphs are fine, my koans would be green, green, red.
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Yup, green green red and that is the rule 
. Two stones of the same color belong to the same group if they see each other. All the stones of the same color belong to the same group.
I was worried that that rule is too complex, but you got there rather fast, it’s impressive.
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I have an idea for a rule I’d like to pose if no one objects 
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My rule is rotation-invariant, reflection-invariant and colour-symmetry-invariant, and defined for rectangular boards with at least side length 2. Koans don’t have to be legal go positions.
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Might be something simple like same number of stones, or something related to symmetry maybe.
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Okay I don’t really have enough data, but if the following are green and red, I’d like to guess the rule that reversing the colors combined with a rotation is a symmetry.
green:
red:
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