Sente, gote and endgame plays

[RobertJasiek]

In an environment with move values Tpq, suppose only the starting player has local sentes with follow-up move values Fij occurring in blocks so that T01 >= T02 >=... > F01 >= F02 >=... >= T11 >= T12 >=... > F11 >= F12 >=... >=...

Can we at least prove that playing any Fkl cannot be better than playing some still available Fij with k > i? That is, can the local sentes be played in order of their blocks? Or which counter-example can we construct?

We can also pose similar questions for only the opponent's local sentes.

[Bill Spight]

In an environment with move values Tpq, suppose only the starting player has local sentes with follow-up move values Fij occurring in blocks so that T01 >= T02 >=... > F01 >= F02 >=... >= T11 >= T12 >=... > F11 >= F12 >=... >=...

Can we at least prove that playing any Fkl cannot be better than playing some still available Fij with k > i? That is, can the local sentes be played in order of their blocks? Or which counter-example can we construct?

We can also pose similar questions for only the opponent's local sentes.

IIUC, I think that this is a counterexample with F01 = 5, T01 = 4.5, F02 = 4.

T = {9 | 0}
S01 = {10 | 0 || -1}
S02 = {8 | 0 || -3}
Black to play

1) Black plays S02, without sente. Then
1a) White takes T, Black plays S01 with sente and then plays the threat of S02. Result: +8
1b) White takes S01 in reverse sente, Black plays T and then White plays the threat of S02. Result: 9 - 1 = 8.

2) Black plays S01, White takes S02 in reverse sente, then Black takes the threat of S01 and White takes T. Result: 10 - 3 = 7.

3) Black takes T, White takes S02 in reverse sente, then Black takes S01 in sente. Result: 9 - 3 = 6.

Black should choose option 1) and play S02.

[RobertJasiek]

Indices simplified, your example in game tree annotation is

T = {9|0}, S1 = {10|0||-1}, S2 = {8|0||-3}

or in move value annotation

T = 4.5, S1 = M1|F1 = 1|5, S2 = M2|F2 = 3|4

with T and follow-up move values

F1 = 5, T = 4.5, F2 = 4

and therefore two blocks of local sentes (each block consisting of one local sente) separated by the high temperature.


I complete the missing sequences and calculate resulting counts (from stated local endgames in which a move occurs) and net profits (stating the profit values of each move), of which either can be used. Sente sequences are ignored in the number value calculations. Rx stands for reverse sente in the local endgame Sx. * means relevant mistake.


1a)

S2 T S1 S1 S2 = 8 - 0 = 8.

F2 - T + F1 - F1 + F2 = 4 - 4.5 + 4 = 3.5.


1b)

S2 R1 T S2 = 0 - 1 + 9 = 8.

F2 - M1 + T - F2 = 4 - 1 + 4.5 - 4 = 3.5.

On move 2, White chooses (1a) or (1b).


2a)

S1* R2 S1 T = 10 - 3 - 0 = 7.

F1* - M2 + F1 - T = 5 - 3 + 5 - 4.5 = 2.5.


2b)

S1 S1* S2 T = 8 - 0 = 8.

F1 - F1* + F2 - T + F2 = 4 - 4.5 + 4 = 3.5.


2c)

S1 S1 T R2 = 9 - 3 = 6.

F1 - F1 + T* - M2 = 4.5 - 3 = 1.5.

On move 3, Black chooses (2b) and discards (2c).


2d)

S1 T S1 R2 = 10 - 0 - 3 = 7.

F1 - T + F1* - M2 = 5 - 4.5 + 5 - 3 = 2.5.


2e)

S1 T S2 S1 S2 = 0 - 0 + 8 = 8.

F1 - T* + F2 - F1 + F2 = 5 - 4.5 + 4 - 5 + 4 = 3.5.

On move 3, Black chooses (2e) and discards (2d).

On move 2, White chooses (2a) and discards (2b) + (2e).


3a)

T R2 S1 S1 = 9 - 3 = 6.

T* - M2 + F1 - F1 = 4.5 - 3 = 1.5.


3b)

T R1 S2 S2 = 9 - 1 = 8.

T - M1* + F2 - F2 = 4.5 - 1 = 3.5.

On move 2, White chooses (3a) and discards (3b).

On move 1, Black chooses (1a) + (1b) and discards (2a) + (3a).


Black's correct start is in the local sente S2 taking the follow-up move value F2 in the second block of local sentes.

Many thanks for constructing this counter-example to the conjecture that blocks of local sentes would be played in order of follow-up move values!

[RobertJasiek]

Now for the rules for comparing two sente. Let A = {2b | 0 || -a} and S = {2s | 0 || -r} be two sente which are not equal. Black should prefer A to S, for certain,

1) if a > r and 2b >= 2s or
2) if a = r and 2b > 2s or
3) if a < r but 2b > r + 2s .

Let me try to understand this.

The possible, relevant profit value sequences and their net profits are:

1) b - b + s - s = 0

2) b - r + b = 2b - r

3) s - s + b - b = 0

4) s - a + s = 2s - a

These comparisons occur for choices between the sequences:

(1) ? (2) <=> 0 ? 2b - r.

(3) ? (4) <=> 0 ? 2s - a.

(2) ? (4) <=> 2b - r ? 2s - a <=> a - r ? 2s - 2b.

The first two comparisons are also summarised in the third comparison. There is also the implicit interpretation of the count differences in A between 2b and -a and in S between 2s and -r.

Now, for starting in A, you write the conditions

1) if a > r and 2b >= 2s or
2) if a = r and 2b > 2s or
3) if a < r but 2b > r + 2s.

Let me try to explain this in terms of the comparison (2) ? (4) <=> a - r ? 2s - 2b.

Case a > r:

a - r > 0 so with 2b >= 2s we have 2s - 2b <= 0, (2) > (4) <=> a - r > 2s - 2b fulfilled and therefore Black can start in (2), which means starting in A.

Case a = r:

a - r = 0 so with 2b > 2s (equality excluded by your assumption of different A and S) we have 2s - 2b < 0, (2) > (4) <=> a - r > 2s - 2b fulfilled and therefore Black can start in (2), which means starting in A.

Case a < r:

If a is infinitesimal positive, a - r is, at the mimimum of a - r, slightly less than r smaller than 0. Therefore, adding r compensates the excess of how much a - r is smaller than 0. For a - r = 0, we have 2b > 2s so, with a - r < 0, also 2b > r + 2s is fulfilled. As a consequence and since a > 0, we have (2) > (4) <=> a - r > 2s - 2b <=> a + 2b > r + 2s fulfilled and therefore Black can start in (2), which means starting in A.

This confirms your conditions.

***

How to interpret all this?

Can there be further cases when starting in A is correct?

I think that starting in A being correct does not always exclude the alternative of starting in S also being correct, right?

I mentioned for an environment with low temperature that the sente player could play all local sentes in sente in order of decreasing sente move values. Do your conditions contradict this or are they just another study aspect?

[Bill Spight]

Now for the rules for comparing two sente. Let A = {2b | 0 || -a} and S = {2s | 0 || -r} be two sente which are not equal. Black should prefer A to S, for certain,

1) if a > r and 2b >= 2s or
2) if a = r and 2b > 2s or
3) if a < r but 2b > r + 2s .

Let me try to understand this.

The possible, relevant profit value sequences and their net profits are:

1) b - b + s - s = 0

2) b - r + b = 2b - r

3) s - s + b - b = 0

4) s - a + s = 2s - a

That may not be so when A and S are not the only plays on the board. The above conditions hold when there are other plays on the board — sans ko, now or later.

If those were the only four sequences, then the comparison would be

a+2b >?< r+2s

The above conditions are more stringent, as there are cases where the play in A is correct but those conditions are not met. For instance, when the only two plays on the board are A and S and a+2b > r+2s but a > r and 2s > 2b.

Let B = {2b | 0} and A = {B | -a}, b > a > 0.
Let T = {2s | 0} and S = {T | -r}, s > r > 0.
Let G = A + S.
Let E = the environment of other plays, with no ko, now or later.

When Black (Left) plays first in G, the result is B + S + E when Black plays from A to B, or T + A + E when Black plays from S to T. Which, if either, does Black prefer? When we compare them, E drops out (because of no ko), so we only have to compare S + B and A + T. We can do that in the game, D = S - T - A + B, or {2s | 0 || -r} + {0 | -2s} + {a || 0 | -2b} + {2b | 0}. If Black wins or ties even when White plays first, for Black to play A is at least as good as for Black to play S.

Can there be further cases when starting in A is correct?

Sure, depending on the rest of the board.

I think that starting in A being correct does not always exclude the alternative of starting in S also being correct, right?

If the above conditions are met and there is no ko in the game tree, Black can play in A instead of S with no reading.

I mentioned for an environment with low temperature that the sente player could play all local sentes in sente in order of decreasing sente move values. Do your conditions contradict this or are they just another study aspect?

In all three conditions, 2b >= 2s, so no, they do not contradict playing the sente in descending order of threats. Note, however, that when a < r but r + 2s >= 2b > 2s, the conditions are not met, and the play in S may be correct, depending on the rest of the board.

[RobertJasiek]

I need to read your message more carefully. For now, just one note: I do not mean "playing the sente in descending order of threats" but mean "playing the sente in descending order of sente move values" (the profit values of the opponent playing reverse sente).

With move value | follow-up move value annotation and the two local sentes 1|6 and 2|3, the sente player starts with 2|3 because 2>1. Sequences:

1) 3 - 3 + 6 - 6 = 0

2) 3 - 1 + 3 = 5

3) 6 - 6 + 3 - 3 = 0

4) 6 - 2 + 6 = 10

[Bill Spight]

I need to read your message more carefully. For now, just one note: I do not mean "playing the sente in descending order of threats" but mean "playing the sente in descending order of sente move values" (the profit values of the opponent playing reverse sente).

Sorry. Condition 3 contradicts that. Given no kos, you should play the smaller sente if its threat is large enough.

[RobertJasiek]

For local sentes a|b, r|s with a+2b > r+2s, a > r, 2s > 2b, correct start with a|b but outside your no-reading conditions, here is an example:

4|5, 1|6

1) 5 - 5 + 6 - 6 = 0

2) 5 - 1 + 5 = 9 mistake on move 2

3) 6 - 6 + 5 - 5 = 0

4) 6 - 4 + 6 = 8 mistake on move 2

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You claim the following conjecture 0:

Suppose no kos in the game tree. There are the non-identical local sentes a|b, r|s of the starting sente player, whose correct start in a|b is at least as good as his start in r|s in the following non-exhaustive cases, without having to read, in a (possibly empty) environment T >= T1 >=...:

a) a > r AND 2b >= 2s

b) a = r AND 2b > 2s

c) a < r AND 2b > r + 2s


What is a proof, if you have already done it?

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Let there be the environment T >= T1 >=...> 0 (or T = 0 if the environment is empty).
The sente player has the local sentes M0|F0, M1|F1,... (Mi are the sente move values, Fi are the gote follow-up move values, Mi < Fi is the local sente condition) with M0 >= M1 >=... The temperature is low: F0, F1,... >= T.

I made the following conjectures:

1) The starting sente player can, e.g., play all his local sentes in sente in the order M0|F0, M1|F1,...

2) The reverse sente player's start is determined by the comparison M0 to twice the alternating sum T - T1 +...

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Counter-example for conjecture 1:

Local sentes 2|8, 3|4. Environment 3.5.

Profit sequences:

1) 8 - 8 + 4 - 4 + 3.5 = 3.5 correct start with 2|8

2) 8 - 8 + 3.5 - 3 = 0.5 mistake on move 3

3) 8 - 3 + 8 - 3.5 = 9.5 mistake on move 2

4) 8 - 3.5 + 8 - 3 = 9.5 mistake on move 2

5) 8 - 3.5 + 4 - 8 + 4 = 4.5 mistake on move 3

6) 4 - 4 + 8 - 8 + 3.5 = 3.5 mistake on move 2

7) 4 - 2 + 4 - 3.5 = 2.5 mistake on move 1

8) 3.5 - 3 + 8 - 8 = 0.5 mistake on move 1

9) 3.5 - 2 + 4 - 4 = 1.5 mistake on move 2

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Nevertheless, it may be instructive to see my failed sketch of a proof of conjecture 1 and suggest where I made mistakes in it:

Iteratively, the sente player can start the next available local sente with the index k or play in the environment with its updated temperature T. At iteration step k for a sequence, if the opponent incurs a loss due to a positive net profit, he never gets a chance to compensate it at a later iteration step because none of the sequences below played to the opponent's turn has a negative net profit favouring the opponent. His playing of M_k+1, M_y or M_z as below prevents the sente player from later playing M_k+1 - M_k+1 = 0, or F_y - F_y = 0 and F_z - F_z = 0; this means that the opponent incurs a earlier loss by playing M_k+1, M_y or M_z instead of later neither gaining compensation nor losing anything from M_k+1 - M_k+1 = 0, or F_y - F_y = 0 and F_z - F_z = 0.


Special case: afterwards, zero local sentes remain:

1) F_k - F_k = 0.

2) F_k - T >(*2) 0 is the minimising opponent's mistake.

3) T... is dominated because of (*2).

(1) is correct, the sente player has started all local sente sequences and the proposition is proven.


Special case: afterwards, only one local sente remains:

1) F_k - F_k = 0.

2) F_k - M_k+1 >(*4) 0 is the minimising opponent's mistake.

3) F_k - T >(*2) 0 is the minimising opponent's mistake.

4) T... is dominated because of (*2).


Else regular case: afterwards, at least two local sentes remain:

Let y ≠ z be indices so that k < y, z.

1) F_k - F_k = 0.

2) F_k - M_y + F_k - M_z >(*4) 0 is the minimising opponent's mistake.

3) F_k - M_y + F_k - T >(*2)(*4) 0 is the minimising opponent's mistake.

4) F_k - T + F_k - M_y >(*2)(*4) 0 is the minimising opponent's mistake.

5) F_k - T + F_k - T1 ≥(*2) 0 is the minimising opponent's mistake or (1) is equally good so, without loss of generality, we assume (1).

6) T - M_k.

If F_k > T, this is the sente player's mistake because he does not follow the simple sente strategy by failing to prevent the opponent from taking M_k. If F_k = T, move 1 in (1) is equally good so, without loss of generality, we assume (1).

7) T - T1. is dominated by (1) to (5) because, after the sente player has started all local sente sequences, he starts playing the alternating sum of the environment.


Footnotes:

(*3) By the definition of local sente, we have M_k < F_k for each local sente with index k = 0, 1, 2...

(*4) F_k >(*3) M_k ≥(*1) M_y ≥(*1) M_z.

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My failing sketch of a proof of conjecture 2 depends on conjecture 1:

Additional Presupposition:

We have the net profit S = 0 (expressed from the opponent's value perspective) of the opponent's remaining local sentes still available to him after any possible reverse sente play of the starting reverse sente player. (I use ∆ for the alternating sum starting at the denoted value.)

Conjecture 2 repeated:

If F0, F1, F2,... ≥ T, the reverse sente player starts
in the environment if 2∆T ≥ M0,
locally if 2∆T ≤ M0.

Failing sketch of a proof:

1) T - S - ∆T1 = ∆T.

2) M0 - S - ∆T = M0 - ∆T.

After move 1 in (1) or (2), apply conjecture 1. Comparison (1) ? (2) <=> ∆T ? M0 - ∆T <=> 2∆T ? M0. This implies conjecture 2 because the reverse sente player starts in the environment in (1) or locally in (2).

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Now, this is really disappointing. We have the comparison in conjecture 2 for one local sente and low temperature. The question remains: Can conjecture 2 be proven differently or what counter-example can be shown?

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My last hope for a general advice for several local sentes of one or both players possibly in an environment possibly with high temperature are these conjectures (no kos now or later):

3) For a player, playing the first local move in his local sente M0|F0 is least as good as playing the first local move in his local sente M1|F1 if M0 >= M1 AND F0 >= F1.

4) For a player, playing reverse sente in the opponent's local sente M0|F0 is least as good as playing reverse sente in the opponent's local sente M1|F1 if M0 >= M1 AND F0 >= F1.

Can they be proven or which counter-examples exist?

[Bill Spight]

For local sentes a|b, r|s with a+2b > r+2s, a > r, 2s > 2b, correct start with a|b but outside your no-reading conditions, here is an example:

4|5, 1|6

1) 5 - 5 + 6 - 6 = 0

2) 5 - 1 + 5 = 9 mistake on move 2

3) 6 - 6 + 5 - 5 = 0

4) 6 - 4 + 6 = 8 mistake on move 2

When those are the only plays on the board, as I said, a+2b >?< r+2s is a good guide, with no reading. 4 + 10 > 1 + 12, so start with A. Correct play gives the same result, but White's mistake is more costly. There is a psychological point, OC. White might be more likely to make the smaller mistake.

You claim the following conjecture 0:

Suppose no kos in the game tree. There are the non-identical local sentes a|b, r|s of the starting sente player, whose correct start in a|b is at least as good as his start in r|s in the following non-exhaustive cases, without having to read, in a (possibly empty) environment T >= T1 >=...:

a) a > r AND 2b >= 2s

b) a = r AND 2b > 2s

c) a < r AND 2b > r + 2s


What is a proof, if you have already done it?

I trust that a) and b) are obvious. Let's try c) with {2b|0} + {a||0|-2b} + {2s|0||-r} + {0|-2s}. White to play cannot win.

1) White plays in A, Black replies in R for jigo.
2) White plays in B, Black replies in R, White plays in A with sente. Result: Jigo.
3) White plays in S. Not! since 2b > 2s.
4) White plays in R. Black replies in B. White plays in A with sente, then plays in S. Result: 2b - r - 2s > 0. Black wins.

My last hope for a general advice for several local sentes of one or both players possibly in an environment possibly with high temperature are these conjectures (no kos now or later)

I applaud your attempts to come up with general advice. I remember struggling with only 3 sente circa 1980. OC, that was long before I learned CGT. The CGT general results with only 2 sente leave a lot unsaid, and there are surely regularities with sente that do not meet the CGT criteria, along with other plays. The question is whether the rules are easy for humans to remember and how often the situations arise in actual games.

[RobertJasiek]

Conjecture 3 (repeated)

In a position with possibly an environment and possibly several local sentes, there are local sentes M0|F0, M1|F1 with M0 > M1 and F0 > F1 of the starting sente player. Starting in M0|F0 is at least as good as starting in M1|F1.

Draft of a Proof

S is the starting sente player's strategy for (1) to (6). The first F0 is the first move. The last three denoted values of a bracket can occur in any order and be preceded, interrupted or succeeded by moves elsewhere, except that the second F1 is a follow-up and can only be played after the first F1 has already been played.

S' for (7) to (12) is S except for the denoted values. The first, second, third and fourth denoted value of S is substituted by the respective denoted value of S'.

We ignore the profit values of the plays elsewhere because replacing S by S' keeps them constant and does not affect the comparisons of net profits below.

1) S(F0;-F0,F1,-F1) = 0

2) S(F0;-F0,F1,F1) = 2F1

3) S(F0;-F0,-M1) = -M1

4) S(F0;F0,F1,-F1) = 2F0

5) S(F0;F0,F1,F1) = 2F0 + 2F1

6) S(F0;F0,-M1) = 2F0 - M1

7) S'(F1;-F1,F0,-F0) = 0

8) S'(F1;-F1,F0,F0) = 2F0

9) S'(F1;-F1,-M0) = -M0

10) S'(F1;F1,F0,-F0) = 2F1

11) S'(F1;F1,F0,F0) = 2F0 + 2F1

12) S'(F1;F1,-M0) = 2F1 - M0

These comparisons are constant: (1) = (7), (5) = (11). For the comparisons (3) > (9), (4) > (10), (6) > (12), the starting sente player incurs a loss by changing from S to S', that is, replacing the start with F0 by F1.

The comparison (2) < (8) is the starting sente player's gain. [---] Therefore, it remains to be shown that (2) does not occur due to strategic choices.

Can you either complete the proof or provide a counter-example?

[Bill Spight]

Conjecture 3 (repeated)

In a position with possibly an environment and possibly several local sentes, there are local sentes M0|F0, M1|F1 with M0 > M1 and F0 > F1 of the starting sente player. Starting in M0|F0 is at least as good as starting in M1|F1.

{snip}

Can you either complete the proof or provide a counter-example?

Using CGT notation, we are comparing playing first in {2*F0 | 0 || -M0} with {2*F1 | 0 || - M1}, plus or minus some constant, with F0 > M0 > M1 and F0 > F1 > M1. The no ko conditions in the environment apply.

To prove: Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1| 0} + {2*F0 | 0 || -M0}. That is, D = {2*F0 | 0} + {2*F1 | 0 || -M1} - {2*F1 | 0} - {2*F0 | 0 || -M0} is at least as good as a score of 0. That is, even if White plays first in D, Black can reply to a score >= 0.

Proof:

Let A = {M0 || 0 | -2*F0}
Let B = {2*F0 | 0}
Let Y = {2*F1 | 0 || -M1}
Let Z = {0 | -2*F1}

Because F0 > F1, we already know that White will not start in Z.

Case 1: Let White start in A. Then Black will reply in Y, leaving

B - B + Z - Z = 0.

Case 2: Let White start in B. Then Black will reply in Y Leaving

A + Z - Z = A. We may ignore the miai, Z - Z.

Then White will play in A and Black will reply to 0.

Case 3: Let White start in Y. Then Black will reply in A, leaving

B + Z + M0 - M1.

Since F0 > F1, White will play in B and Black will reply in Z, leaving

M0 - M1 > 0.

QED.

[RobertJasiek]

Wonderful, very nice!

I confirm the proof of your lemma.

IIUC, your proof completes the proof of my conjecture 3, turning it into an important theorem. Right?

The details may be worked out: a careful description of the strategy; mentioning that each of your cases represents a "good enough" choice by Black on move 2 so that we do not need to consider his alternative move 2s.

The constant in your proof accounts the plays elsewhere, which I have assumed to be other local sentes or simple gotes of the environment. However, we may as well also allow local gotes with (even iterative) follow-ups as plays elsewhere and allow the other local sentes to have iterative follow-ups, if only the two speicific local sentes do not have iterative follow-ups (nor a follow-up to the reverse sente), there are no kos and all local endgames are separate to form a CGT sum. Right?

What happens if there are basic endgame kos elsewhere?

What do the theorem and its proof mean for my conjecture 4 of the starting reverse sente player?

[Bill Spight]

Wonderful, very nice!

I confirm the proof of your lemma.

IIUC, your proof completes the proof of my conjecture 3, turning it into an important theorem. Right?

Right.

The details may be worked out: a careful description of the strategy; mentioning that each of your cases represents a "good enough" choice by Black on move 2 so that we do not need to consider his alternative move 2s.

One of the nice things about difference games is that "good enough" is good enough. You don't necessarily have to find best play.

The constant in your proof accounts the plays elsewhere, which I have assumed to be other local sentes or simple gotes of the environment. However, we may as well also allow local gotes with (even iterative) follow-ups as plays elsewhere and allow the other local sentes to have iterative follow-ups, if only the two speicific local sentes do not have iterative follow-ups (nor a follow-up to the reverse sente), there are no kos and all local endgames are separate to form a CGT sum. Right?

Right, except by constant I meant a number. I talked about the environment earlier. In this post I did not mention the environment explicitly, but in the difference game you get E - E = 0, where E is the environment. If there are kos in the environment or its tree, then we cannot say that E - E = 0, because kos are not combinatorial games. But if there are no kos, you can have anything in the environment.

What happens if there are basic endgame kos elsewhere?

I suppose that you mean a ko with a swing of 1 pt. by territory scoring, 4 pts. by area scoring. As I have indicated in This 'n' That, we can treat such a ko as having a temperature as high as 0.5 by territory scoring or 2 by area scoring. If M0 > M1 > 0.5 by territory scoring or M0 > M1 > 2 by area scoring, I suppose that those sente will probably disappear before the ko is fought.

What do the theorem and its proof mean for my conjecture 4 of the starting reverse sente player?

Your conjecture:

4) For a player, playing reverse sente in the opponent's local sente M0|F0 is least as good as playing reverse sente in the opponent's local sente M1|F1 if M0 >= M1 AND F0 >= F1.

You can relax the conditions to allow F1 > F0. Playing in the reverse sente, {M0 || 0 | -2*FO} is at least as good as playing in the reverse sente, {M1 || 0 | -2*F1} if

M0 >= M1 AND M0 + 2*F0 >= M1 + 2*F1.

We can write the second condition as M0 - M1 >= 2*(F1 - F0)

So we would choose to play in {5 || 0 | -11} over {3 || 0 | -12}.

[RobertJasiek]

Mathematical Go Endgames discusses move order of some shape classes of local endgames with iterative follow-ups if the initial moves have move values 1 or smaller.

What about positions with local endgames with iterative follow-ups and also larger move values?

1) What, if any, general simplification of reading is known (other than playing the environment of simple gotes in decreasing order)?

2) Can the theory of Mathematical Go Endgames be applied to, say, moves with move value 3 quite like it can be applied to moves with move value 1 (even if there are move values 3 and move values 1)?

[Bill Spight]

Can the theory of Mathematical Go Endgames be applied to, say, moves with move value 3 quite like it can be applied to moves with move value 1 (even if there are move values 3 and move values 1)?

Sure. E. g., {15 | 6 || 3 ||| 0} vs. {6 | 0} is like {5 | 2 || 1 ||| 0} vs. {2 | 0}.

Also, the insights about CGT infinitesimals can help us to realize when non-infinitesimals act like infinitesimals.

What, if any, general simplification of reading is known (other than playing the environment of simple gotes in decreasing order)?

Well, games and, hence, sums of games can be simplified by recognizing dominance and reversals, both of which may be analyzed using difference games.

And I have done a good bit of work comparing games with variables instead of constants. I studied {a | b || c | d}, with numbers a ≥ b ≥ c ≥ d, back in the '70s. I have also done a good bit of work with difference games. Some of what I have found is fairly general. OTOH, some of those generalities strain the ability of humans to apply.

The idea of an environment of simple gote of the form, {a | b}, a ≥ b, is useful because we can compare other plays with simple gote of different sizes, and that comparison tells us, in general, when to play them. I don't know what use there might be for an environment of plays of the form, {a | b || c}, a ≥ b ≥ c, but we may order those plays such that, given two such plays, A = {a | b || c} and D = {d | e || f}, if d - e ≥ a - b and e - f ≥ b - c, then each player does at least as well to play in D as in A, with the usual caveat about no kos. So we could have such a well-ordered environment. (Note that the plays could be sente, gote, or ambiguous.)