# BQM 127

From Miai Values List / 1.00 to 1.99, is worth 1.47. Could someone go through the calculations? Thanks.

Robert Pauli: Let me have a try.

Let's first switch colors to have no negatives (Black's lead normally is positive) and switch sides to have Black increase his score by going to the right (in tree and position):

We want to know how much at *a* is worth.
Let's look at following tree:

. / \ / a \ . 4 / \ / b \ 0 . / \ / d \ . 3 / \ / e \ . 2 / \ / f \ 0 1

Each node and leaf corresponds to a position
(and the position's value).
The root (top node) corresponds to the position
above (with *a-f* being empty).
The label beneath each node indicates the
intersection either Black (right branch) or
White (left branch) took to reach the next
node (or leaf).
Each leaf is labeled with the (local) score
of its corresponding position (which is in
the "turn doesn't matter" state).

So, if White takes *a* as well as *b*,
for instance, Black gets nothing (0), etc.

Still missing are the node values. To find them,
we **average** daughter values and work our way up:

.------2 + 1/2 + 1/32 / \ / a \ 1 + 1/16----. 4 / \ / b \ 0 .------2 + 1/8 / \ / d \ 1 + 1/4-----. 3 / \ / e \ 1/2-----. 2 / \ / f \ 0 1

White's move in question goes from node

2 + 1/2 + 1/32 to 1 + 1/16 (minus) ---------------- 1 + 1/2 - 1/32 =1.46875

Ok, that was that (except for some ugly rounding).

But let me add a **warning**. It's just an incident
that the branches above are labeled with *one*
move each. In general, sente follow-ups are appended!

For instance, the value of this position

is **not**

.------4 + 3/4 / \ / a \ . 6 / \ / b \ . 5 / \ / c \ 0 4

but rather

.------5a+bc / \ a / \ 4 6 ___ / \ | / \ | was 2 5 | added / \ | by Bill / \ | but...^{[1]}0 4 ___|

because White's *bc* follow-up is sente.

Bill: I edited the tree just above.

If a move is sente or not can be checked with the
method of multiples.
Take, say, four copies of the 5-deep 2-goodies corridor
with a white stone at *a* in each and treat at *b*
as sente, that's, Black answers four times and gets a
total of 4 * 4 = 16 points.

Then treat at *b* as gote, that's, Black tenukis
two times (2 * 5) and saves one copy from being spoiled
completely (1 * 4), for a total of (just) 10 + 4 = 14
points.

Conclusion: Black should treat at *b* as sente
(provided rich environment and good timing by White).

So, in case of sente there is no branching. The value
of the 5-deep 2-goodies corridor with a white stone at
*a*, for example, would simply be 4, as if there's no
alternative.

Hope that carries a while.

Robert Pauli:

Sorry, forgot a black stone below *c*
in my first diagram.
Luckily the value (if that stone is added) doesn't change,
just the right subtree:

. . / \ a-cd / \ a \ or \ 4 . c+de / \ c / \ 3 5

(appending opposing sente follow-ups with minus)

There's a choice because after at *a*
at *c* is sente as well as gote:
4 + 4 = 3 + 5.

In the second case Black treats it as gote.
However, White has a sente follow-up at *d*:
3 + 3 + 3 + 3 > 0 + 3 + 4 + 4.

[1] Robert Pauli: (12 years later :-)

Thanks for mending, Bill, but I intentionally omitted the subtree below a+bc→4!
My idea was to unite move and sente follow-ups
to force a plain averaging binary tree.
The reverse sente is omitted completely because it does not go into the calculation (shown by the tree, it went into the sente-gote analysis).

Here's a non-tail example (which was fun but also hard for me to construct :-)

Starting at the leafs (for the reuse):

- After white a+bc Black counts 4.5
- White
*b*(after white*a*) is sente: 4*4.5 > 0 + 4.5 + 6 + 6 - White
*a*, however, is gote: 2*6 < 4.5 + 8

My *non-standard* tree would now be

6.25a+bc/ \ a / \ 4.5 8 / \ / d \ 4 5