[Welcome to Sensei's Library!]

StartingPoints
ReferenceSection
About


Paths
SecondCourseOnKo

Referenced by
Ko
TerritoryAndAreaS...
RealHalfPointKo
GTLReview1558
Fujitsu3Game5
EndgameTestTwoWhi...
MiaiValuesList

 

Half-Point Ko
Path: SecondCourseOnKo   · Prev: KoThreatAmplification   · Next: RealHalfPointKo
    Keywords: Ko

A half-point ko is a ko in which nothing is at stake beyond the fate of the two points inside the ko.

Despite the name, the miai value of connecting or capturing in a half-point ko is only one third of a point.


This name is only there because it translates the Japanese hanko. I propose minimal ko, since that is what it is.

Charles Matthews

Half-point ko also translates the Chinese term 半劫 directly.

--unkx80


Bill: I just found a Real Half Point Ko.


Arno: could someone please explain me, why ordinary minimal kos have a value of 1/3? I haven't got the knack of miai counting yet.

Bill: Hi, Arno! :-)

The difference between winning and losing the ko is 1 point. There are 3 moves between the winning positions, 1 move for each player to win the ko and 1 move to take the ko. So each move is worth 1/3 point, on average. Es claro?

Arno: while formulating the next question I guess I got the solution, which I write down for other people like me: Bill counts three moves, because in the following position ...

[Diagram]
minimal ko

... if white plays, she has invested one move more than black (i.e. 1 white move, 0 black moves). If black plays, he has to capture and connect (=2 moves) while white does not play there (=0 moves). Therefore, black played two more moves than white. A possible ko-threat play by White is not counted, because it is outside this position.

Thus the difference of the possible outcomes is 3 moves and 1 point value, thus a miai value of 1/3.

Did I get it right? :o)

Bill: Well put, Arno! :-)


KarlKnechtel: Interesting... I came up with the 1/3 value in an entirely different way. First, let's see if I've got the basics right:

[Diagram]
typical half-point play

In this diagram, white has half a point, under the assumption that it's even probability that each player will be the next to play locally (in turn based on the assumption that there are roughly equal-value plays elsewhere on the board). Therefore, whoever plays at 'a' gains half a point for that move. That's about the limit of my confident understanding of miai counting ^^; But assuming I have the basic idea right, I apply it to the minimal ko as follows:


[Diagram]
minimal ko

Now there is an even chance that white will play next. Define the result to be net 0 points difference. The alternative is that black eventually wins the ko, and gets +1 point as a result.

If, at each move, it is an even chance that either white or black will play again locally, the chance that black wins the ko immediately is 1/4; that white wins immediately is 1/2; the remaining 1/4 of the time, black plays and then white plays, so that the original situation is restored. Thus, when we sum the infinite series, we get a total chance of 2/3 that white wins, and 1/3 that black wins.

Thus the current result gives white a disadvantage of 1/3 point from the 'reference' position where white has won the ko. Thus, playing locally is worth 1/3 point for white. Symmetrically, it is worth 1/3 point for black, since the black play would shift the value from 1/3 to 2/3 in black's favour (the value would become 2/3 since it's 1/3 less than the +1 value for black winning the ko - 1/3 less by the same argument as above).

This way I don't have to think about counting moves or amortizing the value of a play over the moves in the sequence (which I find a bit counterintuitive). But I'm not sure it's any easier to understand ;) (Actually, I just make it sound complicated, I think.)



Path: SecondCourseOnKo   · Prev: KoThreatAmplification   · Next: RealHalfPointKo
This is a copy of the living page "Half-Point Ko" at Sensei's Library.
(OC) 2003 the Authors, published under the OpenContent License V1.0.