Just a kind reminder that the deadline is approaching.
Round 12
Team | Blue | Green | Pink |
---|---|---|---|
Black | @Maharani: J14 | @Jon_Ko: C7 | @Vsotvep: R14, Q14 |
White | @yebellz: C12 | @Feijoa: R14 | @terrific: O2 |
New Position:
check in time for Round 13: 2021-12-04T07:00:00Z
Vsotvep and Feijoa collided at R14. Thank you all for voting in the poll. From now on I will extend the round by 24 hours whenever a move submission is missing, and write a reminder post
It’s getting quiet here…but I’ll play P14 unless someone has a better idea.
I have a better idea, but you’re probably not interested in it
I’m very paranoid about mentioning any of my intentions in here. Collisions seem like such a powerful “mwahahahaha you fool, you’ve made me immortal!” instrument…
Round 13
Team | Blue | Green | Pink |
---|---|---|---|
Black | @Maharani: S10 | @Jon_Ko: E7 | @Vsotvep: K17 |
White | @yebellz: D14 | @Feijoa: R7 | @terrific: C8 |
New Position:
Check-in Time for Round 14: 2021-12-06T07:00:00Z
Hmm, looks like I was wrongly expecting to get a capture there
I’ve been struggling to think about how groups without enough eyes will actually get killed:
Black can make a triple attack, but any White stone can collide to save it for another move:
No matter where they collide, both pink and black chains will have a liberty at H9.
Suppose Black-Green tries to kill by playing there. White-Pink can counter:
(before captures)
Since both pink and white chains contain a new stone, they are not removed in the initial phase, and it ends up like this:
On the other hand, if Black-Blue joins the attack:
(before captures)
Then all colors are represented in the new stone, so nothing is removed in the first phase, and everything dies in the second phase.
I have no idea if that’s a reasonable sequence, but this seems much more complicated than I expected. I think we’re in for a crazy endgame.
Anyway, I think I need to penetrate the Black-Blue wall so I’m going to aim for a collision at R10.
Now I’m wondering how counting liberties will work in semeais. You showed that white can play inside to delay being captured, but that means white plays fewer stones on the outside. And in your example black can still kill by playing three more stones than white (which is the case if black+blue does indeed add a stone in the end).
Round 14
Team | Blue | Green | Pink |
---|---|---|---|
Black | @Maharani: R9 | @Jon_Ko: C6 | @Vsotvep: J17 |
White | @yebellz: S7 | @Feijoa: R10 | @terrific: C9 |
New Position:
Check-in Time for Round 15: 2021-12-08T10:00:00Z
It’s getting very exciting in multiple places. I’m still baffled that there were only two collisions so far - are you trying to dodge them on purpose?
PS: I can’t refrain from commenting how (it seems that) Maharani had the perfect read on Feijoa on the right side - well played.
Or was it the other way round? Feijoa announced R10 here. To me it comes as a surprise that Feijoa actually played their announced move .
Right, I forgot about that ^^
I’m not sure if we are any better than a random bot at this game
I sincerely believe that all of you play much better than random-bot.
Well, that’ll be the last time
Round 15
Team | Blue | Green | Pink |
---|---|---|---|
Black | @Maharani: Q7 | @Jon_Ko: B6 | @Vsotvep: R12 |
White | @yebellz: E11 | @Feijoa: R11 | @terrific: E4 |
New Position:
Check-in Time for Round 16: 2021-12-10T07:00:00Z
At the end of the game, only the white and black score is relevant for the result. But we can count points for blue, green and pink too, just for fun Which of the secondary colours do you think will have the most?
Hard to say. The upper left will stay blue, I’d guess. But green and pink are fighting in two places at once, anything can happen there.
The right side looks very interesting.
Here’s a pass-alive structure with only white territory, I think:
And a smaller one, featuring collisions:
@yebellz do you want to work on killing the black group at RS6? I could probably keep my ataried stone there alive a little longer.
We have to clean up all the dead stones before scoring, and much of that could involve arbitrary choices of secondary color, so I think the secondary color scoring will be pretty random.