Endgame Strategy (06)
Robert Jasiek
- Ikeda Territory I Scoring.
- A succesion of 2 plays may not recreate the position. (Intervening
passes are ko threats.)
- Fixed Ko Rule.
- The alternation stops with 3 successive passes.
- The Playout ends with 3 successive passes.
CLASS OF POSITIONS I1
Either player to move during the alternation
D >= 0
K > 0
Kb >= Kw >= 0
Tb >= 0
Tw >= 0
Case 1: Either player to move && Kd MOD 3 = 0
---------------------------------------------
Algorithm:
1. Repeat 2 * Kw times: connect ko
2. Choose possible variant A or B and repeat Kd/3 times:
2A.1 Black connects ko 2B.1 White captures ko
2A.2 White captures ko 2B.2 Black connects ko
2A.3 Black connects ko 2B.3 White connects ko
2A.4 White connects ko 2B.4 Black connects ko
3. Repeat D times: play dame
4. Passes
Proposition I1.1.1: The score due to the algorithm is: -Kd/3
Proof: The score due to (1) is 0 in the affected kos. During each of
the Kd/3 repetitions of (2), White gets 1 points more than Black.
During (3), the score does not change. QED.
Remarks: Each ko's miai value is 1 less than under Area Scoring. 2/3
under Area Scoring becomes -1/3 under Ikeda Territory I Scoring. If we
include White's Kd excess stones played elsewhere in alternation, the
Area Score altogether becomes -Kd/3 while under Ikeda Territory I
Scoring nothing changes and the sum over all kos remains -Kd/3. So in
the global view, both scoring methods are in agreement while in the
per ko view they differ. Strategically the latter fact creates
differences.
Lemma I1.1.2: The score is optimal for either player until the end of
the alternation.
Proof: The proof is similar to Kd MOD 3 = 0 cases under Area Scoring,
see there. Now playing a dame is worth 0, whenever it is played during
the alternation. QED.
Proposition I1.1.3: The score is optimal for either player.
Proof: For the score until the end of the alternation, see lemma
I1.1.2. Can either player force his opponent to leave some ko or dame
unfilled until the end of the alternation to improve on the score
during the playout due to its pass stones? No, because during the
alternation either player can follow the strategy 1. answer non-dame
ko threat, 2. connect ko, 3. capture ko if legal, 4. play dame, 5.
pass. If one player follows this strategy, then, regardless of the
opponent's strategy, all kos and all dame are filled at the end of the
alternation. For the answer "no" and filling kos, the details of the
rules do not become relevant in this Kd case, unless some player play
imperfectly. For the value of the playout's passes themselves, see
proposition 13 of the threads Proofs about Pass Fights. QED.
Proposition I1.1.4: A pass-fight-6 does not occur.
Proof: Since all kos and all dame are filled at the end of the
alternation, the position is divided then and propositions 11, 13, and
14 of the threads Proofs about Pass Fights apply. During the
alternation, each (reasonable, i.e. not self-killing) non-ko-threat
tenuki and each non-dame ko threat and answer do not change the state
at the end of the alternation of all kos and all dame being filled.
QED.
Case 2: Black to move && Kd MOD 3 = 1
-------------------------------------
Algorithm:
1. Black connects ko
Then see under case 1 with variant B repeated (Kd - 1)/3 times.
Proposition I1.2.1: The score due to the algorithm is: -(Kd - 1)/3
Proof: See case 1 and propositions under Area Scoring. Connecting the
ko in (1) provides 0 points but reduces the number of remaining kos to
(Kd - 1). QED.
Proposition I1.2.2: The score is optimal for either player.
Proof: See case 1. However, now the rules details have to be studied
more carefully as to whether Black can leave the last ko open until
the playout to fill it then and thereby gain 1 extra point in the form
of an extra opposing pass stone. This is interesting only if Td > 0.
However, even after White would have used up all his ko threat plays,
White retains an essentially infinite number of pass ko threats
because if White passes to use one of them, then Black's pass as an
answer does not stop the alternation yet and White gets his chance
to execute yet another ko capture. Because Td is only finite, after a
finite time, Black is forced already during the alternation to fill
the last ko. QED.
Proposition I1.2.3: A pass-fight-6 does not occur.
Proof: See case 1. Even an attempt by Black to leave the last ko open
until the playout is futile because also that ko will be conected
until the end of the alternation. QED.
Case 3: White to move && Kd MOD 3 = 1 && Td > 0
-----------------------------------------------
Algorithm:
1. Repeat 2 * Kw times: connect ko
2. Repeat (Kd - 1)/3 times:
2.1 White captures ko
2.2 Black connects ko
2.3 White connects ko
2.4 Black connects ko
3. Repeat 2 * Tw + 1 times:
3.1 capture ko
3.2 non-dame ko threat
3.3 answer ko threat
4. Black captures ko
5. Choose variant A or else B:
5A. play dame 5B. pass
6. Black connects ko
7. Choose the same variant letter as in (5):
7A. Repeat (D - 1) times: play dame 7B. Repeat D times: play dame
8. Passes
Proposition I1.3.1: The score due to the algorithm is: -(Kd - 1)/3
Proof: See earlier proofs. During each pair of the first 2 * Tw
repetitions of (3), the score does not change. During the last
repetition, White captures the ko and gets 1 point more than Black.
During (4), Black gets 1 point more than White. During (5) - (7), the
score does not change, if we observe that connecting the final ko
provides 0 points. QED.
Proposition I1.3.2: The score is optimal for either player.
Proof: See the proof for proposition I1.1.3. With reagrds to
attempting to leave the last ko open until the playout, see the proof
for proposition I1.2.2. QED.
Proposition I1.3.3: A pass-fight-6 does not occur.
Proof: See the proof for proposition I1.2.3.
Case 4: White to move && Kd MOD 3 = 1 && Td <= 0
------------------------------------------------
Algorithm:
1. Repeat 2 * Kw times: connect ko
2. Repeat (Kd - 1)/3 times:
2.1 White captures ko
2.2 Black connects ko
2.3 White connects ko
2.4 Black connects ko
3. Repeat 2 * Tb times:
3.1 capture ko
3.2 non-dame ko threat
3.3 answer ko threat
4. White captures ko
5. Choose variant A or else B:
5A. play dame 5B. pass
6. White connects ko
7. Choose the same variant letter as in (5):
7A. Repeat (D - 1) times: play dame 7B. Repeat D times: play dame
8. Passes
Proposition I1.4.1: The score due to the algorithm is: -(Kd - 1)/3 -
1
Proof: See earlier proofs. During (3), the score does not change.
During (4), White gets 1 point more than Black. During (5) - (7), the
score does not change, if we observe that connecting the final ko
provides 0 points. QED.
Proposition I1.4.2: The score is optimal for either player.
Proof: See the proof for proposition I1.1.3. With reagrds to
attempting to leave the last ko open until the playout, see the proof
for proposition I1.2.2, now adopted for White trying to leave that ko
open. QED.
Proposition I1.4.3: A pass-fight-6 does not occur.
Proof: See the proof for proposition I1.2.3, now adopted for White
trying to leave the last ko open. QED.
Case 5: Black to move && Kd MOD 3 = 2 && Td > 0
-----------------------------------------------
Algorithm:
1. Repeat 2 * Kw times: connect 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 non-dame ko threat
4.3 answer ko threat
5. Black captures ko
6. Choose variant A or else B:
6A. play dame 6B. pass
7. Black connects ko
8. Choose the same variant letter as in (6):
8A. Repeat (D - 1) times: play dame 8B. Repeat D times: play dame
9. Passes
Proposition I1.5.1: The score due to the algorithm is: -(Kd - 2)/3
Proof: See I1.3.1, except that now (2) has a different internal order
and is repeated (Kd - 2)/3 times and the inserted (3) provides 0
points. QED.
Proposition I1.5.2: The score is optimal for either player.
Proof: See case 3.
Proposition I1.5.3: A pass-fight-6 does not occur.
Proof: See case 3.
Case 6: Black to move && Kd MOD 3 = 2 && Td <= 0
------------------------------------------------
Algorithm:
1. Repeat 2 * Kw times: connect 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 non-dame ko threat
4.3 answer ko threat
5. White captures ko
6. Choose variant A or else B:
6A. play dame 6B. pass
7. White connects ko
8. Choose the same variant letter as in (6):
8A. Repeat (D - 1) times: play dame 8B. Repeat D times: play dame
9. Passes
Proposition I1.6.1: The score due to the algorithm is: -(Kd - 2)/3 -
1
Proof: See I1.4.1, except that now (2) has a different internal order
and is repeated (Kd - 2)/3 times and the inserted (3) provides 0
points. QED.
Proposition I1.6.2: The score is optimal for either player.
Proof: See case 4.
Proposition I1.6.3: A pass-fight-6 does not occur.
Proof: See case 4.
Case 7: White to move && Kd MOD 3 = 2
-------------------------------------
Algorithm:
1. Repeat 2 * Kw times: connect ko
2. Repeat (Kd - 2)/3 times:
2.1 White captures ko
2.2 Black connects ko
2.3 White connects ko
2.4 Black connects ko
3. White captures ko
4. Black connects ko
5. White connects ko
6. Repeat D times: play dame
7. Passes
Proposition I1.7.1: The score due to the algorithm is: -(Kd - 2)/3 -
1
Proof: The score due to (1) is 0 in the affected kos. During each of
the (Kd - 2)/3 repetitions of (2), White gets 1 point more than Black.
During (3), White gets 1 point more than Black. (4)-(6) do not provide
points. QED.
Proposition I1.7.2: The score is optimal for either player.
Proof: See earlier cases.
Proposition I1.7.3: A pass-fight-6 does not occur.
Proof: See earlier cases.
*********************************************************************************************
Additional Remarks:
- Study of the class of positions I1 is complete. Cases 1 to 7 solve
all possible regular positions with basic endgame kos and two-sided
dame, but without teire!
- In the cases 5 to 7, also trying to leave 2 last kos open for the
same player until the playout does not work because the opponent can
force their connection until the end of the alternation.
- Using J2003's long cycle rules instead of the Fixed Ko Rule does not
change the propositions because the proofs do not rely on arguments
about repeated long cycle positions.
- Introducing teire will affect algorithms, numbers of ko threats, and
- if there is a ko fight - the score by either 0 or +-1 point
depending on the winner of the last ko. However, the author expects
that propositions about pass-fights still apply.
- These aspects should be studied more carefully later, of course.
- Changing the number of playout ending passes from 3 to 2 does not
affect the propositions (see propositions 13 and 14), but demands
slightly different proofs for some.
- Using 3 successive passes to stop the alternation is, however,
essential for many of the propositions. Otherwise, there is the
additional strategic element of trying to postpone connection of the
last one or two kos until the playout. This not only might affect the
score, depending on the ko threat difference, and presumably introduce
dame ko threats for one of the players in some cases, but more
importantly requires a fresh approach to many proofs. IOW, if 2
successive passes shall stop the alternation (or if a superko rule
shall be used), then completely new theoretical study is mandatory and
probably will be significantly more complex due to a greater number of
cases to be considered.
- Below you can extract the miai value per ko of -1/3.
**************************************************************************
Summary of strategic cases:
§ Move Case Score Score Ko
Fight
(other style) Win
1 B/W Kd MOD 3 = 0 -Kd/3 -Kd/3 no
2 Black Kd MOD 3 = 1 -(Kd - 1)/3 -Kd/3 + 1/3 no
3 White Kd MOD 3 = 1 && Td > 0 -(Kd - 1)/3 -Kd/3 + 1/3 B
4 White Kd MOD 3 = 1 && Td <= 0 -(Kd - 1)/3 - 1 -Kd/3 - 2/3 W
5 Black Kd MOD 3 = 2 && Td > 0 -(Kd - 2)/3 -Kd/3 + 2/3 B
6 Black Kd MOD 3 = 2 && Td <= 0 -(Kd - 2)/3 - 1 -Kd/3 - 1/3 W
7 White Kd MOD 3 = 2 -(Kd - 2)/3 - 1 -Kd/3 - 1/3 no
--
robert jasiek