# CGT Values Of Eyes

__Keywords__: Life & Death

Eyes can also have CGT values, like liberties and connexions The only possible chilled values are: 0, 1/2, 3/4, 1, 1*, 5/4, 6/4 and 2.

- Bill Spight: Inaccurate. See "Eyespace values in go" by Howard Landman, in
*Games of No Chance*(Cambridge University Press).

Groups with >=2 eyes will live. Note that 2* eyes is not necessarily enough.

Eyespace values are calculated for a portion of a group. The assumption is that white's pieces continue off the board where there may be more eyes or possible eyes for white. I adjusted the first one, so Bill's comment below no longer applies.!]

Here are some examples

If black plays on the circled spot, there are no eyes for white. If white plays on the circled spot, she gets one eye.

- blubb: Well, both groups touch the border and may extend beyond, so their state is unknown in the first place. If we apply the common SL diagrams convention, both are alive; if we don't, we can merely state that White is ahead by (a mean value of) half an eye.

If black plays on the circled spot, there are no eyes. If white plays at the circle spot, there are 1 + 1/2 = 3/2 eyes.

- Bill: White is dead.

The values are quoted from 'Mathematical Go, Chilling gets the last point', which says they come from a communication from Howard Landman. I made up the examples, so I accept any mistakes in those. Which values are missing/what am I missing?

I've found a paper by Landman at http://www.polyamory.org/~howard/Go/Eyespace.ps which gives the values I gave above, and some material about sekis. I'll shut up on this subject 'til I've had a chance to understand this!

Bill: The paper above is, I believe, the same as the chapter in *Games of No Chance*. Landman defines *eye* in a clear way. (See other attempts here on SL.) He defines eye values in terms of "Bargo", a go-like game which is scored by the number of Black eyes of a group. The final scores are 0, 1, 2, and S. *S* stands for seki, and an integral number of eyes more than 2 are scored as 2.

I am not entirely convinced by Howard's theory, but nobody has offered an improvement yet.

I am not sure if Howard would regard this as a half eye (after chilling) or not. If Black makes an eye, White can reduce Black to 0 eyes in 2 consecutive plays.

- Howard Landman Again, the Black group would have to be connected out to something else, not completely surrounded. The theory is most useful for adding up the values of multiple eyespaces. For example 3/4 Eye + 3/4 Eye + 1/2 Eye is a perfect miai worth exactly 2 eyes. The advantage is, you don't have to analyze the whole group, but can do each eyespace separately and then just add.

If I understand him correctly, this is **not** a half eye because if Black makes the eye White can capture right away.

For his diagrams, which look like this next one, except that he uses clipped stones instead of empty space,

he says that the White stones are __immortal__?, but the Black stones are simply connected to others (which may have eye-space) such that the liberty count does not affect the analysis. I am not sure that that condition is coherent on a finite board unless Black to play can live or make ko.

KarlKnechtel: Wouldn't this be 2/3 of an eye?

If B connects, there is one eye. If w takes the ko and connects (or takes again), there are no eyes. There are three plays between the two end results.

Howard Landman Yes, this is worth 2/3 Eye. The simple method of counting moves and dividing is not always correct, especially when there is a sente move involved, but in this case it gives the right answer.