Go endgame values with area counting: part 3, ko

Ko

By now we have enough tools that ko is a straightforward extension of our previous work.

If you’ve evaluated kos the normal way, you already know that this is “worth a third of a point”. Now we really have to use miai counting to make sense of the numbers. By territory, the difference between one side or the other winning the ko is one prisoner, and the local tally difference is $3$, so the miai move value is $\frac13$.

By area: two intersections are at stake, so the move value is $\frac{2\cdot 2 - 3}3 = \frac13$. As noted last time, you could also calculate this as $\frac{2 \cdot 2} 3 - 1$, and that’s the way I’ll generally do it in the future.

To me it’s actually a little nicer to just say “two intersections are at stake” rather than “the difference between the difference between the prisoner counts is one” (that’s not a typo), as you would calculating this the territory way. We don’t have to cancel out, even theoretically, all the stones captured in the ko fight, just eyeball the final state.

Let’s do a multi-stage ko, for fun:

Four intersections are at stake, and the local tally difference is $4$ (White can finish by connecting, and Black can finish by taking two kos and connecting), so the value of a move here is $\frac{2\cdot4} 4 - 1 = 1$. With territory counting we’d have to calculate that the end states are $3$ points for Black (two extra captures and one eye) vs $1$ point for White (one eye).

Comparing work

OK, that’s it for the description. How much work does this really save (or cost) us, compared to normal territory counting? I’ve spent so much time describing it that it may look like much more of a bother than it actually is.

The main work-saving idea is that with area counting every point ultimately belongs to one player or the other. This is why you only need to count one side when computing the winner of the entire game, and is also why you only need to count one side when computing the size of a move. We do have to multiply by two to get the full swing, the same way that the difference between $3$ and $-3$ is $6$.

Going back to our formulas from part 1:

$$ \begin{aligned} V_T &= (B_T^1 - B_T^2) - (W_T^1 - W_T^2) \\ V_A &= 2 \cdot (B_A^1 - B_A^2) - \Delta T \end{aligned} $$

Before we had to juggle four territories and now we just have to juggle two areas. We do also need the local tally difference, but we can get that without thinking just by considering whether the sequence is gote or sente.

More examples

To recap everything we’ve learned, let’s do a few problems from O Meien’s book Absolute Endgame. In each case we’re going to calculate the miai value of a move in the local area.

Gote

Just looking at the $2\times5$ rectangle in the corner, Black will own all $10$ points if they go first, and will own either $6$ or $3$ if White goes first, depending on who plays the gote followup afterwards. The difference is $10 - 4\frac12 = 5\frac12$, so the value of the gote move here is one less than that, or $4\frac12$. (Technically we should double the difference to get a swing and then immediately halve it for being gote.)

By territory counting, we’d say that Black would locally lead by $10$ ($3$ prisoners, $1$ dead stone, and $5$ spaces vs. nothing) if they went first, and lead by either $3$ ($1$ prisoner and $2$ spaces vs. nothing) or $-1$ (nothing vs $1$ prisoner) if White went first, which averages to $1$, so the total difference is $10 - 1 = 9$, halved for being gote to give us a miai move value of $4\frac12$.

(Before you wonder why the answer in the book is $5\frac12$, that’s because it’s asking a different question: how many points can we consider Black as having right now? That’s halfway between $10$ and $1$.)

Sente

Four points are at stake (a $2\times2$ rectangle with the second-line Black stone in its northwest corner), so the miai value of a move is $2\cdot4 - 1 = 7$. (It’s sente because Black has to play a move to live if White goes first.) I will leave the territory calculations as an exercise for the reader (or go read O Meien’s explanation).

Ko

$10$ points are at stake, and it’s a direct ko, so the miai value of a move is $\frac{2 \cdot 10}3 - 1 = 5 \frac23$. To do this the territory way you’ll need to carefully count up Black’s bounty if they win and connect the ko: $6$ dead stones, $2$ empty points, and $1$ prisoner, for a total of $15$ points. (Then you compare it with White’s two points if they win, and divide the whole thing by three.)

Two-stage ko

For the grand finale, we’ll take the most complicated position yet. White has played the tesuji of 1 to start a two-stage ko for life. If they win the ko by making three captures from the position after Black 4, it’s their privilege to make the square-triangle exchanges on the right side. If Black wins it by making one capture from the same position, it’s their privilege to make the triangle-square exchanges on the bottom side.

This means that every circled point in Diag. 2e is at stake. There are $20$ of them, so each move is worth $\frac{2 \cdot 20}4 - 1 = 9$ points. O Meien arrives at the same conclusion, declaring that between the two variations Black’s difference in territory is 17 and White’s is 19, giving a move value of $\frac{17+19}{4} = 9$. Perhaps you’d like to check his counts; the whole reason I did this research is so that I don’t have to.

Creating the ko

By the way, how much is White 1 in Diag. 2d worth? If Black had gone first instead, then every one of the $20$ marked points in Diag. 2e would be Black. As it is, after White 3, White has half equity of all those marked points. So it was a gote sequence that effectively changed the ownership of $10$ points, therefore it’s worth $\frac{2\cdot10}2 - 1 = 9$ points, just like the moves in the ko! (Since the values of the ko moves are all $9$, we’d get the same value if we treated 3 as sente.) This shouldn’t really be a surprise; basically, the original position was exactly equivalent to the situation after Black 4, where Black can win the ko in one move or White can play a move to bring it to a perfectly intermediate state.

References

Unsurprisingly, I’ve been told that this way of calculating endgame move values is not original, but I’ve never seen it presented anywhere. (Edit: that's because I hadn't made it all the way through Jasiek's Endgame 2 – Values; see below.) Hopefully these posts are enough to make it clear, since I don’t have anywhere else to point you (now I do!).

If these concepts seem cool but your endgame-counting foundation still feels shaky, here are your main written options. They all cover miai counting thoroughly.

• Sensei’s Library page on Basic Endgame Theory.

  • Pros: it’s free! And it’s quite up-to-date with theory.

  • Cons: it’s written by committee, and is a bit more of a reference than a textbook. Of course it is written by amateurs, but some of them are as much experts in the theory as anyone.

• O Meien, Absolute Counting.

  • Pros: written conversationally, covers more topics than just move values, tons of examples.

  • Cons: written conversationally. Sometimes I want a proof and instead I just get an exhortation that the point is very important and I must remember it.

• Antti Törmänen, Rational Endgame.

  • Pros: written like a math textbook. He tends not to repeat himself much, so you really have to make sure you understand every sentence, but it’s all there. Also, it’s a very nice-looking book.

  • Cons: written like a math textbook, fewer examples.

• Robert Jasiek, Endgame 1–5, Basic Endgame Problems 1–2.

  • Pros: Exhaustive (seven books!), extremely precise, written by one of the principal researchers of modern endgame theory, zillions of problems. Area counting is discussed at length in Endgame 2 – Values (I didn't realize this until after writing these posts).

  • Cons: Seven books, dense and stilted writing style.

Bonus content

It turns out there was still some more to say, so now there's a Part 4.