CGT values of connections

0  

• 0
  

Two groups are cut apart

1  

• 1
  

Unbreakable connection

1-ish  

• {1 || 1 | 0} = 1 + tiny-1, with a mean value of 1
  

Two groups are locally connected, but White has a threat to disconnect

mean = 1/2  

• {1 | 0}, with a mean value of 1/2
  

Groups can be cut or connected

mean = 1/3  

• mean value of 1/3
  

Two groups can be connected by taking and connecting a ko

mean = 2/3  

• mean value of 2/3
  

Two groups can be connected by connecting a ko

Charles Matthews I hadn't thought of this before. The idea would be that to be connected, you want independent ways to connect, which add up to at least one.

KarlKnechtel: My understanding of CGT is rather limited, but this is an interesting concept. I think I was trying to get at something similar a long time back with AdjustedLengthAnalysis - to refine my original idea, we might say that the expected territory within a moyo is the area marked out, times some factor depending on the CGT values of the connections around the outside. (The probability of connecting everything up, and thus claiming the whole area, would seem to be the product of all the CGT values; but I think it's more complicated than that - it's not right to assume independance.)

By the way, while we usually think of a kosumi as a solid connection, the difference game suggests its CGT connection value is only 7/8:

The two games. Black to play.  

If Black plays in game 'a', both connections are solid. Similarly if White plays first in game 'b'. But if Black opts to play in game 'b' first and White responds in game 'a', Black has the option to crosscut game 'b'. Assuming that the cutting stones cannot be captured, and that 'points' are only awarded for staying connected, White's only logical response is to return the favour in game 'a' to draw the difference game. Thus 1/8 of the time, the kosumis end up cut.

Do I understand this analysis technique correctly? o.O

White connects  

The original position has a value of {0 | -1 || -1}, where an unbreakable white connection is worth -1. After Black 1 the value is {0 | -1}, a hot position with a mean value of -1/2. White 2 goes to -1.

The two games.  

The original value of the sum of 'a' and 'b' is {1 || 1 | 0} + {0 | -1 || -1} = 0. After Black 1 - White 2 the sum is still 0 ( {1 | 0} + {0 | -1}). Then after Black 3 - White 4 the sum is still 0.