Sub-page of AmbiguousPosition

This is what Ricky felt had to be purged from the main page, quote [0] *the hane was strictly better than the descent*.

There seems to be a temptation to underestimate simple and small examples, let me therefore explain.

Let’s draw the tree belonging to this 4×3 board position. The page count nicely explains that. Black goes down to the left, Right to the white, oops. Counts (from Black’s perspective) are given at the nodes.

0 0 / \ / \ / \ / \ / \ b / \ a a / \ b / \ / \ 5 -5 5 -5 a / \ / \ b b / \ / \ a / \ / \ / \ / \ 10 0 0 -10 10 0 0 -10 c/ \d c/ \d 1 -1 1 -1

Actually this is *one* tree, roots joined. The counts above the leafs were made by averaging. This only was a first guess, but luckily the temperature (distance between mother and daughter) going down never rises, sparing us a recalculation.

Conclusion:

Descent as well as hane share the same miai value of 5 points, and since both do not raise the temperature (it’s 5 before and after), they both are ambiguous moves.

Don’t confuse the situation above with the 4×5 board one below.

Its count is 0 too

(due to symmetry, of course),

but hane’s miai value only is 1,

setting the temperature, and

descent’s miai value only is ½.

I don’t know about the temptation you mentioned, but I know that I fell for the temptation of assuming that you would **not** do

- a generally-complicated calculation that yields bounds and generally-good play

rather than

- a generally-simpler calculation that yields perfect play,

even though there’s barely any difference in complexity for your small example.

(I was assuming that your diagram gave a region^{[1]} in which the already-present stones are immortal and the left will be white territory to the extent that Black doesn't penetrate underneath from the right and similarly for the right being black territory.

For such a region, the comment^{[0]} I left with my revert would've been correct.)

Also, since your calculation came up with a non-zero temperature (namely, 5) such that
under alternating play, neither player can get more moves at that temperature,

what your calculation found is the temperature at which the mast changes from purple (below) to something else (above).

Thus the mast temperature is strictly less than what your calculation came up with.^{[2]}
(For your example, the mast temperature is 1.)

In your example, I agree that play here should be at temperature 5, since whoever begins here chooses what the resulting infinitesimal will be if the opponent gets the next play here, so I agree that your example is an Ambiguous Position and that in it, the descent and the hane are both ambiguous.

^{[3]}

However, for other (loopless) positions, I believe your method can get a temperature which is higher than the temperature at which the position should be played.

Let UP be

/\ / \ 1 / \ / \ 0 -2

and let DOWN be the negative of that.

The result of joining the roots of

/\ /\ /\ / \ / \ / \ / \ 2 -2 / \ / \ / \ / \ / \ / \ / \ /\ / \ /\ / \ / \ / \ / \ / \ 10 DOWN UP -10 8 UP DOWN -8

should be played at temperature 4, even though I believe your method would conclude its temperature is 5.

Robert Pauli: Hi, Ricky, nice to hear from you!

[1] That's why I once suggested to have *round* corners. They are! Zoom in. It's a *full-board* example.

[2] Sorry, Ricky, I'm not yet into thermography. Maybe later . . .

[3] Great that we agree!