Half-point ko

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Chinese: 半劫 (bànjié or ban4 jie2); 单劫 (dānjié or dan1 jie2);
Japanese: (hankō)
Korean: 반패

A half-point ko is the common term for a ko in which nothing is at stake beyond the fate of the stone 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.


      • WME***

See /WME Discussion.


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! :-)

Alistair Turnbull: Furthermore, it doesn't matter how long the ko fight continues, because in each capture-threat-answer-capture-threat-answer cycle the two players play the same number of moves.


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.)

MK In fact this reasoning nearly convinced me. Could you please make the 'infinite series' part clearer?

victim 1/2 - 1/4 + 1/8 - 1/16 + ... = 1/3.

Another way to come to the same result: Let's call the position where Black can capture or White can connect, B. If White connects, we get zero. If Black captures, we get a position we call W, where Black can connect or White can capture. So B = 0 * 1/2 + W * 1/2, if the probability is 1/2 for each outcome.

Similarly, for the next move, W = B * 1/2 + 1 * 1/2. Now we have two equations with two variables.

B = W/2

W = B/2 + 1/2 = W/4 + 1/2

W*3/4 = 1/2

W = 2/3

B = W/2 = 1/3

By the way, this is not as useless as it seems to be at first sight. The computation works the same way for a bigger ko. So, as a first estimate of how big capturing a ko is, compute the difference between "White wins" and "Black wins" and multiply by 2/3. If you don't have much time, play a gote move bigger than that rather than playing the ko.

[Diagram]

Two-stage ko

Of course the number changes if it's a two-stage ko.

Capturing or connecting here is worth the same as two points in gote, as an approximation. Use the same methods as above to calculate that.


RafaelCaetano Karl, I agree that you make it seem more complicated than it is. :-) I don't see why introduce probabilities and infinite series here.

I find Bill and Arno's reasoning convincing enough. But since some people don't, I'll try a simpler approach. Actually I read it somewhere else a long time ago, probably r.g.g (maybe by Bill?). Consider the following position.

[Diagram]

1/3 * 3 = 1

[Diagram]

Score is 1

Black plays first. White connects, since it's no use to fight the ko. Black 3 also connects (below 1), since again it's useless to try to win both kos. Finally White connects. Black got a stone, that is, 1 point.

[Diagram]

Score is 1

White plays first. Again, after Black captures with 2, White simply connects the other ko. Black 4 below 2. Again Black got 1 point.





So the value of the position above is 1. Therefore, the value of each of the 3 kos is 1/3.

Is this reasoning correct? It looks so simple that I feel something is missing.

victim It's fine. Of course one has to remember that it's still an approximation.

Bill: It's not just an approximation, as Rafael's example shows. If there is one such ko, the final territory result will differ by one point depending on who wins the ko. But when there are three kos, as in Rafael's diagram, the result is the same, no matter who plays first or how many ko threats any player has. If the value were simply an approximation, there would be a range of results with three kos.

Alistair Turnbull: Well, Rafael's example shows that the exact value is some game G such that G+G+G = 1. It doesn't show that G is a number. Or am I missing the point?

Karl Knechtel: Is it possible for a scalar multiple of a game that's not a number, to be a number? If not, there would seem to be no problem.

Bill: No, it does not show that G is a number, but that is not the claim. We are finding the average value of G.


TriumShockwave? - Frequently in pro games I see these kos left unresolved at the end of the game. Are they counted as a point for the player who currently has them surrounded, or are they uncounted as a dame like in seki?

Mef: The point from these kos comes from capturing the opponent's stone for the last time, so it doesn't really matter if it's filled or not, just so long as they know who captured. last. I guess to answer you question, it would be considered dame.

Bill: They are not left unresolved at the end of the game, it is just that the final ko fight is not shown in the published game record. The text tells us who wins the ko. If that information is not in a comment, it should be decipherable from the score.


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