Simultaneous Fractional Go Game 1

The reason why I changed my mind:

image

… but it seems I saved black… sorry…

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If there was a stone at T5, would black+green R5 be suicide? (Also, is suicide legal in this variant?)

I think this is the most fun Ive ever had playing go :slight_smile: (Keep in mind 99 % of my time with go is spent watching AGA and dwryin videos, and watching Kata play itself, lol)

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That’s what I was thinking and it would have been good for black, the current position is okay for white I think.

No, captures of “old” groups have priority over captures of “new” groups, so R5 would be captured first, giving Black|Green a liberty

Are you sure?

Yes:

Exactly, R5 would be new, so the four green+black are a green chain with a “recently placed stone”. They would be removed in step 2. Unless T5 would be green too, then T6 would be a liberty for green.

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Ah, wait, yes, I had S5 in my mind, and that it was about last move, never mind.

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Yes it is allowed. This can be very important because chains can include stones of both sides, so tactical suicide is a thing.

A more profound reason is, I do not know how to formulate a rule disallowing suicide in this variant :sweat_smile: Because of the simultaneous moves, before placement of all stones is revealed, it is often unsure if a move will result in one of the connected chains to be captured.

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I’m thinking that @terrific and @yebellz can handle the green threat in the lower-right. So I’ll work on getting my four eyes by playing S17.

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I don’t have the time to do the update right now, but I’ll do it shortly.

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Sorry for the delay.

Round 9

Team Blue Green Pink
Black @Maharani: K15 @Jon_Ko: O3 @Vsotvep: S15
White @yebellz: T5 @Feijoa: P5 @terrific: N3

New Position:

fractional_game2_Round9

Deadline for Round 10: 2021-11-28T09:00:00Z

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In my role as asker of stupid questions, let me enquire:

If white+green plays at Q6 and black+green plays at R5, and no one else plays locally, which (if any) stones are removed from the board?

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This seems like a question to understand the rules better, so I’ll answer: That would result in two chains having no liberties: The light-blue chain containing the stones Q4 and Q5, and the green chain containing the stones R4, R5, S4 and S5. Of these two, the light-blue chain contains no stones that were just placed, so it gets removed first, which frees up liberties of the green chain, which is thus not removed.

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Let’s say that

a) White+green plays at Q6,
b) White+red plays at R7, and
c) Black+green and black+blue collide at R5.

What happens…?

(Edited)

The same green chain and a light-blue chain that is slightly larger than in the last example both have no liberties and are simultaneously removed, because both contain a “recently placed stone”.

fractional_game2_example1

fractional_game2_example2

I really must go to sleep now, good night!

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Would love to hear everyone’s thoughts on the bottom-right situation!

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I was thinking team white should play at R6 and Q5 Q6 and R5 to kill four stones, should I play one of those?

I definitely approve of team white submitting two illegal moves :stuck_out_tongue_winking_eye:

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To me it seems like white was really successful there, although we started with a good position.

The corner is gone IMO, the right side needs fortification. I wonder if I can help with the attack on the upper right.

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