# Chinese counting

Someone asked me a tutorial about how we count a game in Chinese rule. (He’s going to have to IRL.)

The very first thing is to understand what we want to do.

We are going to count all together, meaning the stones and the emptyness in the middle of the stones.
We count only for one color and then we can see if this color occupies more as half of the board, or not.

Here the game is finished. We remove the prisoners to the stock of stones (bowls), prisoners are of no interest in Chinese rules.

Let’s count black. I marked all black points.

Is black winning ?

I count 86 one by one.
Half the board is 13x13/2 so 84,5
With a komi of 6.5, black need to substract half of it (3.25)
Black lost.

In next post, i will describe how to count more efficiently as one by one.

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Normally people group stones up to make it easier to count or replace territory with stones(or vice versa) as territory and stones are worth the same amount of points normally you make it so that you count in groups of ten as it is easier to count like that

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Patience.
Welcome back to the forum by the way.

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So we can organize us to make it more easy.
First important thing is to understand that an empty intersection has the same value as one with a stone

Meaning as long as i don’t change the boundaries you can add or remove stones at will.

Here the same game where i arranged the emptyness to be easier to count. I didn’t modify the boundary.

We have 4x10 empty intersections.
We still need to add the remaining stones but now that we know the number of emptyness we can do as we want (we only need to know how many black stones to add to the score )

So we have 40+46=86 (same as one by one)

All good.

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I did not understand this method before, but now I get it.

Thanks @Groin !

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How is seki usually handled? If there is an even number of dame, it would be convinient to fill half of them black and half of them white. But if there is an odd number, do you leave one empty and adjust the total number of points?

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Any unfilled dame, whether from seki or otherwise, count as half a point for each player, so yes, you could just fill in half of the dame for each player, and manually count the half point if there are an odd number.

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Understanding what is underlying helps to integrate sometimes.
Japanese rules as we call them are elaborated from the Chinese in a try to count smaller numbers. We count only the emptyness but then to get a result almost equivalent we have to put back the prisoners on the board.
In this way we ensure that both players play the same quantity of stones, so we can avoid to count them.

Note that generally Chinese players use the Japanese counting during the game to assess the balance of territories.

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I remember this tends to confuse me when I have to think about it

It’s like either black has 86 and white has 169-86+6.5=89.5 so that W+3.5 points or the other way B: 86-3.25=82.75. Then 82.75-84.5=-1.75 (which I’ve heard referred to as Zi at Sensei's Library ). W is ahead (minus sign) by twice this amount which is 3.5 points.

Edit:

Explanation without any seki on the board like the example.

The score (difference), in this game W+3.5 would be calculated like

`Score = (Black score)-(White score)=Black points - (White points +komi)`.

`-3.5=86-(83+6.5)`.

If `Score>0` Black has more than White so Black wins by Score or `B+Score`

If `Score<0` White has more than Black so White wins by |Score| or `W+|Score|`

In this case `W+3.5`.

Now also when there’s no seki in area scoring,
`White points = (Points on board) - (Black points)`, every point should be owned by Black or White

so the above formula could be rewritten

`Score = (Black points) - Komi - (White points) `
`=2*(Black points) - Komi - (Points on board)`

With the same idea if `Score>0` then `0.5*Score>0 ` also, similarly with `Score<0` so one could choose instead of the Score the quantity `0.5*Score` to call the result.

`Zi score:=0.5*Score=(Black points) - 0.5*Komi - 0.5*(Points on board)`

and this is @Groin 's calculation `0.5*(Points on board)=0.5*169=84.5`

`86-3.25-84.5=-1.75` so White wins, and to get the score double it again, by `2*1.75=3.5 ` points.

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Interesting!

I’m no pro at counting, but I’ve noticed that for me Chinese counting can often be easier during a game since I can pick which color to count - the side with fewer, more rectangular areas.

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Well no problem if you can do Chinese way, i think It’s all about the quickest and most confortable for each of us. Your reasons are quite Interesting i would say.

An extreme example:

Pretty easy to see that Black has 36+4 = 40 points here, under Chinese rules. So, with no komi, White wins by 1.

Since White moved last and there’s nothing strange like seki going on, the Japanese result should be the same. Slowly counting that up:

• White has 14 points of territory + 3 prisoners = 17.
• Black has 15 points of territory + 1 prisoner = 16.
• (Don’t forget any already-captured prisoners! None in this case.)
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Before territories are decided, it’s often convenient to not count undecided areas as points for either player. Then it’s no longer possible to use the trick of counting only one side and automatically know the score difference (because you don’t know what the total will be). I think this is the main reason why most players count territory instead of area during the game

It can still be useful though, let’s say for instance you count your own solid area and notice that you need only 10 more points to have half the board. I’ve never really used this in Go except on 9x9, but I’ve used it quite a bit for Tumbleweed (another area control game, which does not have seki, so the total score is always equal to the size of the board).

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Using Japanese counting, it is obvious that white is ahead. I don’t understand how you can see this using Chinese counting.

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I don’t say It’s the most adapted but you can. Half of not yet defined area + white area is bigger as half of the board

Nice exsmple but slightly OT as it is not during but at the end of the game.
Now it can inspire for situation during a game. As long as you can add areas together and see that they are or they will be bigger as the half of the board.

Black score+white score = 361+komi. To know if black won you compare black’s score with 180.5 + half komi.

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I did go into a little more detail above, but essentially it’s the same point. I also think it is very unintuitive but I suppose it’s something one could learn as something that just works. (I believe one could also instead just add half the Komi to half the total points on the board instead – one can check that calculation I suppose.)

I think though dividing by two is also completely arbitrary the way I’ve laid it out.

Deriving in a slightly different (but equivalent) way from Jlt's starting point

`B+W = 361+komi` where B=Blacks score and W=White’s score

`B-W=S` the score needs to be computed.

You know Blacks score `B` by manually counting on the board so adding the equations gives

`2*B=S+361+komi` and rearranging to get `S=2*B-komi-361`.

Again here diving by two feels arbitrary, (one could divide by four or ten etc), but convenient so one doesn’t have to multiply Blacks score by anything.

`Zi score=S/2=B-0.5*komi-(points on board)/2`.

Asking if the `Zi score>0` is the same as asking if `B-0.5*komi>(points on board)/2=180.5` on 19x19. That is if Blacks points minus half the Komi is bigger than half the total points on the board.

I think knowing whether to add half a point or subtract half a point for odd number of points in seki to Black’s score is also unintuitive (I believe it’s adding 0.5 when I look at the calculation)

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Ah, now I get it. It’s as if the board is actually 368.5 intersections but 7.5 of them are always white. And black needs more than half of 368.5 board.

Yeah! That actually helps!

Just imagine the board with komi.

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Suppose there is one neutral point. Then black’s score+white’s score = 360, so to know if black won you need to compare black’s score with 180 instead of 180.5 or, equivalently, add 0.5 to black’s score and compare it to 180.5.

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