(At last, some of these koans would have been red under the 40-word version of the rule, but I will not reveal which ones.)
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Ok one more. At some point one of them has to be red 
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Ok so I looks like I have been on the wrong track here since 640 (only the four white stones) is green, and the empty board is green. But adding a single black stone to the empty board makes it red, while adding a black stone to the four stones stays green. That is not possible if there is some property of the black and white stones independently that has to be the same. What we have not checked so far is whether the color of the stones makes any difference at all. With the information we have it could be that the rule is colorblind. So let me check some green koans where I replace all stones by black stones
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If yāall get tired of expert mode, I could recolour those koans grey that would be green under one version of the rule and red under the other. Then youād have three datasets to compareā¦ ~ 24-hour poll:
Keep going in expert mode?
- Yes, weāve got this 8)
- No, we need help :,(
0 voters
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So far, if all stones are within the center 3x3 area, the koans are green if there are an equal number of black and white stones, and red otherwise.
I wonder if the rule might demand that there is some sort of equilibrium between Black and White stones, where stones receive different weight depending on their position (maybe with an amount of tolerated difference). For example, this Koan may mean that the center is worth two stones on the middle-line between center and edge:
Iāll say that it is a far fetched theory at this point, but so far I havenāt had a better idea.
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The green ones do look more balanced than the red ones.
But it seems to break down with stones on the edge of the board. What about a single stone on the edge?
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Second line?
570 is a single stone on the third line and is red
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Guess:
green:
red:
If that is correct Iād like to guess the rule that the imbalance (defined as abs(white stones - black stones)) is greater on the rim than in the center, where the center is 3x3 around tengen and the rest is the rim. On a 7x7 board at least, I donāt know how the rim and center generalize, so Iām only guessing the rule for 7x7.
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If I were to generalize ācenterā and ārimā for rectangular boards, I might define the ācenterā as that central rectangle of the board whose number of intersections is closest to the number of intersections of the remaining board (ārimā). On a 7 x 7 board, that center would be 5 x 5 (25 intersections to the rimās 24 intersections). On a 13 x 8 board, I think the center according to this definition would be 9 x 6 (54 intersections to the rimās 50):
(I hereby neither affirm nor deny that this ācenter/rimā definition is(nāt) included in my rule.)
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