Re: Endgame Strategy (07)

[Robert Jasiek]

>>
Case 4: D EVEN && -D/2 < Td <= D/2
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[...]
Proposition 5.4.1: The score due to the algorithm is: 2/3 * (Kd - 1)
[...]
Proposition 5.4.2: The score is optimal for Black.

Proof: As is seen by the cases 3 and 5, neither player can both win
the one excess ko and get exactly half of the dame. (Therefore using
ko threats would be superfluous.) The opponent of the excess ko winner
fills two dame in a row exactly once; it does not matter which player
does which. QED.

Proposition 5.4.3: The score is optimal for White.

Proof: Likewise. QED.
<<

I have made mistakes in this case. First of all, the case should be
split into two cases. Then the proofs for the propositions 5.4.2 and
5.4.3 are wrong. Proposition 5.4.1 is still correct though. Let me
correct now:

Case 4A: D EVEN && 0 < Td <= D/2
--------------------------------

Proposition 5.4A.2: The score is optimal for Black.

Proof: As is seen by the cases 3 and 5, neither player can both win
the one excess ko and get exactly half of the dame. The opponent of
the excess ko winner fills two dame in a row exactly once; it does not
matter which player does which. Black can choose whether to win or
lose the one excess ko. Winning the ko means that Black gets 2 points
there while White gets 2 points for filling two dame in a row exactly
once. Losing the ko means that White gets 2 points there while Black
gets 2 points for filling two dame in a row exactly once. The
remaining dame are shared. - The lower end of the condition 0 < Td is
just enough for Black to win the ko if he chooses so because, even
though White can capture the ko first, Black has at least one ko
threat more. QED.

Proposition 5.4A.3: The score is optimal for White.

Proof: See the proof for 5.4A.2. Besides note that White at some time
has to start capturing the ko if Black should have started by filling
dame. Otherwise Td would become greater than half the number of
remaining dame. I.e., failing to start capturing the ko as White in
time does not improve on the score for White but makes it even worse.
QED.

Case 4B: D EVEN && -D/2 < Td <= 0
---------------------------------

Proposition 5.4B.2: The score is optimal for Black.

Proof: Like 5.4A.2, except that now White is the one who can choose to
win or lose the ko. QED.

Proposition 5.4B.3: The score is optimal for White.

Proof: Like 5.4A.3, except that now Black is the one who has to
connect the ko in time and a "greater than half" becomes "smaller than
or equal to half". QED.