# Counting Problem 1 / Solution

Sub-page of CountingProblem1

To calculate the size of a yose play, we need to find the results of the main lines of play.

Black fills.

First, suppose that Black fills the ko. The result is +1 (for Black).

White takes and wins.

Black 2 elsewhere.

Now suppose that White takes the ko and then, when Black plays elsewhere, wins it by taking Black's stone with White 3. White can count 1/3 of the marked Black stone, so the result is -4 1/3.

White takes twice and fills.

Black 2, Black 4 elsewhere. White 5 fills.

If White takes the ko twice and then fills, the result is -6.

Let's sketch out the tree for these variations.

```              A -- B -- C
/      \    \
1     -4 1/3 -6
```

(A indicates the original position. / indicates a Black play, \ indicates a White play, and -- indicates a ko play.)

Intuitively, it seems like White should take both legs of the ko and fill, for a score of -6. In that case there are 4 plays between a score of +1 and a score of -6, and each one is worth, on average, 7/4 = 1 3/4.

But what if White should win the ko after taking only once? In that case there are 3 plays between +1 and -4 1/3, and each one is worth, on average, (5 1/3)/3 = 1 7/9.

Well, 1 7/9 > 1 3/4. (Just barely!) Does that mean that White should win the ko after taking only once, instead of taking twice and winning it?

In a real game, either line of play is likely to produce the same result. :-) But, yes, the result that determines the size of the ko is the one that makes the plays bigger, on average. The miai value of the ko is 1 7/9, and the count of the original position is 1 - 1 7/9 = -7/9.

(Note that, since White is not threatening to take and fill the second leg of the ko, Black does not have to fill to prevent that, and we can ignore that line of play for our calculations.)

Original position.

Count: -7/9
Miai value: 1 7/9

What is orthodox play?

Obviously, orthodox play for Black is to fill.

White can get 6 points by taking the ko twice and filling, or 4 1/3 points by taking once and winning that leg. A difference of one play makes a difference of 1 2/3 points. That means that White can take the ko twice to reach this position:

No fill.

(2 captured Black stones on the marked points.)

But now filling the ko is only worth 1 2/3 points. Taking the ko was worth 1 7/9 points, but filling it is worth less. If there is now a play elsewhere worth between 1 7/9 and 1 2/3 points (for instance, a play worth 1 3/4 points), White should normally stop playing locally and take the other play.

Because it is normally preferable to retain the option of filling the 1 2/3 point ko, orthodox play is for White to take the ko twice to reach this position, but not to fill next.

Sente ko.

When the ambient temperature lies between 1 2/3 and 1 7/9, Black 1 can take the ko. Now White has the choice of playing a ko threat and taking the ko back, or playing White 2. In a real game fighting the ko may be best, but for orthodox play we assume that there are plenty of plays on the board in that range, and that White will run out of ko threats before the temperature drops to 1 2/3, allowing White to fill the ko at 1 without loss. That means that Black can take the ko with sente, and it is a sente ko.

dnerra: Now please show me a game with plenty of plays worth between 1 2/3 and 1 7/9 points :))

Bill: ;-)

Counting Problem 1 / Solution last edited by MrTenuki on July 9, 2006 - 02:00