Cubic Seki

    Keywords: Theory

The topology of a cubic seki resembles that of a cube. Pressed flat, that's

  X -------- O
  | \      / |
  |  O -- X  |
  |  |    |  |
  |  X -- O  |
  | /      \ |
  O -------- X

Eight corners - the groups - connected by twelve edges - their shared liberties.

Harry Fearnley:You could also think of this arrangement as a regular Octahedron -- a solid with 8 triangular faces. These are arranged to look like 2 (square-based) pyramids, which are glued together at their bases. The Octahedron is the dual of the Cube: it has has 8 faces, and 6 vertices (corners), whereas the Cube has 8 vertices, and 6 faces. Both have 12 edges, which mean essentially the same things -- in the Cube model they represent the fact that 2 groups are connected, whereas in the Octahedron they represent (equivalently) the existence of a shared border. In some ways, this is more natural -- you can think of the 8 groups being the 8 faces, and then imagine these laid out on a sphere ...


Smallest ?

[Diagram]

  • 8 x 10 intersections
  • 34 stones each
  • 3 liberties each
  • no square (2x2 block)



Here's one not depending on edges:

[Diagram]

  • 8 x 12 intersections
  • 42 stones each
  • 3 liberties each
  • no square
  • no edge



Are either of these sekis? It looks to me like Black can profit nicely capturing groups if he has the first move. I've marked "a" as the first move. White can then capture that black group, of course, but that creates a shortage of liberties that allows black to capture a larger group elsewhere. (Or does a cubic seki have nothing to do with an ordinary seki?) ~~~~


Nicest ?

[Diagram]

  • 10 x 10 intersections
  • 44 stones each
  • 3 liberties each
  • no square



Or rather this one?

[Diagram]

  • 10 x 10 intersections
  • 44 stones each
  • 3 liberties each
  • no square




Generator

To increase the number of shared liberties, take the one below (made of the one above) and duplicate its two rows and two columns framing its 2x2 center:

[Diagram]

i = 0



[Diagram]

i = 1



[Diagram]

i = 2



etc., giving

  • (12 + 2i) x (12 + 2i) intersections
  • 66 + 18i + 2ii stones each
  • 3 + 3i liberties each

Terminality

Robert Pauli: I guess nobody can gain anything if he starts - under territory scoring, of course (and under area scoring the gain might be equal). At i = 0 and i = 1 it's even disadvantageous. From i = 2 on, without having a proof, I feel that if both go after the weakest opposing group without touching the second to last liberty between them, nobody will be captured. True?


Authors


This is a copy of the living page "Cubic Seki" at Sensei's Library.
(OC) 2012 the Authors, published under the OpenContent License V1.0.
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