Endgame Strategy (01)
Robert Jasiek
positions under the given rules and assumptions.
The studied positions shall have two-sided dame, basic endgame kos, ko
threats, an arbitrarily large number of tenukis, and otherwise be
regularly divided.
The rules use area scoring, positional superko, 2 passes.
Without loss of generality, Black shall have equally many open kos as
or more open kos than White. The numbers of ko threats are finite, all
ko threats are independent from each other, and all ko threat and
answer sequences shall consist of exactly 2 plays.
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# := number of
D := #two-sided dame
K := #kos = Kb + Kw
Kb := #kos open for Black
Kw := #kos open for White
Kd := Kb - Kw
Tb := #pairs of black ko threats and white ko threat answers
Tw := #pairs of white ko threats and black ko threat answers
Td := Tb - Tw
"ko" := "basic endgame ko"
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CLASS OF POSITIONS 1
Black to move
D = 0
K > 0
Kb >= Kw >= 0
Tb >= 0
Tw >= 0
STRATEGY FOR EACH PLAYER
1. answer ko threat
2. connect ko
3. capture ko if legal
4. ko threat if there is some ko
5. pass
ANALYSIS
Case 1: Kd MOD 3 = 0
----------------------
Algorithm:
1 Repeat Kw times:
1.1 Black connects ko
1.2 White connects ko
2 Repeat Kd/3 times:
2.1 Black connects ko
2.2 White captures ko
2.3 Black connects ko
2.4 White connects ko
3 Passes
Proposition 1.1.1: The score due to the algorithm is: 2/3 * Kd
Proof: During (1), the score is 0. During each of the Kd/3 repetitions
of (2), Black gets 2 points more than White.
Proposition 1.1.2: The score is optimal for Black.
Proof: In the algorithm, Black wins Kw + 2/3 * Kd kos. If Black wants
to improve on the score, he has to win more kos. To win one further
ko, Black has to replace once a "connect ko" by a "capture ko"
(provided this is legal). For the two involved kos, this replaces "one
ko is connected by Black and one ko is open for White and White moves
next" by "two kos are open for Black and White moves next". The former
results in "one ko is connected by Black and one ko is connected by
White". The latter first becomes "one ko is open for Black and one ko
is open for White and Black moves next" and later "one ko is connected
by Black and one ko is connected by White". Thereby Black does not
improve on the score. For the final score within the two affected kos,
it does not matter when the plays in them are made. Since conections
are not prohibited by the positional superko rule, they are possible.
In the two affected kos, immediate recapture is prohibited by the
positional superko rule but such recaptures are not necessary to
produce the mentioned results for these two affected kos. Other
mentioned captures in them are possible because the worst thing that
could happen is a positional change elsewhere and this does not cause
a prohibition for a capture in either of the two affected kos. If
Black wants to win more than one further ko, then for each of them
Black faces the problem of trying to win exactly one further ko. By
equality of all the kos' shapes, it does not matter which of them
Black captures for an attempt to win yet another ko (provided the
capture is legal). QED.
Proposition 1.1.3: The score is optimal for White.
Proof: Like for 1.1.2, except that White wins Kw + 1/3 * Kd kos.
Remark: In the other cases, the analogous propositions and proofs are
possible.
Proposition 1.1.4: The algorithm is well-defined and lets each player
follow the strategy.
Proof: "answer ko threat" is not available. During (2.1), "connect ko"
applies the next strategic option. During (2.2), "connect ko" is not
available for White because all remaining open kos are open for Black.
During (2.2), the next strategic option "capture ko if legal" is
considered for legality, is legal because Black does not make any ko
captures, and is applied. During (2.3), "connect ko" applies the first
available strategic option. During (2.4), "connect ko" applies the
first available strategic option and a ko connection is available for
White due to White's previous ko capture in (2.2). By the assumption
that Black has at least as many open kos as White, each player has at
least Kw open kos for connection; this is the number of repetitions in
(1). After (1), all kos initially open for White are connected; (2)
starts with only kos open for Black left, namely Kd kos. During each
repetition of (2), 3 of those kos are connected. Since (2) is repeated
Kd/3 times, altogether during (2) 3 * Kd/3 = Kd kos are connected,
i.e. exactly all those available at the start of (2). Hence just since
the start of (3), also the strategic options "connect ko", "capture ko
if legal", and "ko threat if there is some ko" are not available and
this makes passes available. QED.
Remark: Similar propositions and proofs should be made for the other
cases, too.
Case 2: Kd MOD 3 = 1
----------------------
Algorithm:
1 Repeat Kw times:
1.1 Black connects ko
1.2 White connects ko
2 Repeat (Kd - 1) / 3 times:
2.1 Black connects ko
2.2 White captures ko
2.3 Black connects ko
2.4 White connects ko
3 Black connects ko
4 Passes
Proposition 1.2: The score due to the algorithm is: 2/3 * (Kd - 1) +
2
Proof: During (1), the score is 0. During each of the (Kd - 1) / 3
repetitions of (2), Black gets 2 points more than White. During (3),
Black gets 2 points more than White.
Case 3: Kd MOD 3 = 2 && Td > 0
----------------------------------
Algorithm:
1 Repeat Kw times:
1.1 Black connects ko
1.2 White connects ko
2 Repeat (Kd - 2) / 3 times:
2.1 Black connects ko
2.2 White captures ko
2.3 Black connects ko
2.4 White connects ko
3 Black connects ko
4 Repeat 2 * Tw + 1 times:
4.1 capture ko
4.2 ko threat
4.3 answer ko threat
5 Black captures ko
6 White passes
7 Black connects ko
8 Passes
Proposition 1.3: The score due to the algorithm is: 2/3 * (Kd - 2) +
4
Proof: During (1), the score is 0. During each of the (Kd - 2) / 3
repetitions of (2), Black gets 2 points more than White. During (3),
Black gets 2 points more than White. During (4.1) and (5), the final
ko is not yet fixed. During (7), the final ko is fixed and Black gets
2 points more than White.
Case 4: Kd MOD 3 = 2 && Td <= 0
-----------------------------------
Algorithm:
1 Repeat Kw times:
1.1 Black connects ko
1.2 White connects ko
2 Repeat (Kd - 2) / 3 times:
2.1 Black connects ko
2.2 White captures ko
2.3 Black connects ko
2.4 White connects ko
3 Black connects ko
4 Repeat 2 * Tb times:
4.1 capture ko
4.2 ko threat
4.3 answer ko threat
5 White connects ko
6 Passes
Proposition 1.4: The score due to the algorithm is: 2/3 * (Kd - 2)
Proof: During (1), the score is 0. During each of the (Kd - 2) / 3
repetitions of (2), Black gets 2 points more than White. During (3),
Black gets 2 points more than White. During (4.1), the final ko is not
yet fixed. During (5), the final ko is fixed and White gets 2 points
more than Black.
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robert jasiek