To see if I have understood simple sente and reverse sente calculation, let me test my current understanding in the following examples. Are they correct?

EXAMPLE 1:

GOTE ANALYSIS:

Given tree:

x >= 1 (otherwise it would not be a meaningful go tree)



Gote counts:



Gote move values:



CASE ANALYSIS:

gote if (x+1)/4 > (x-1)/2 <=> x < 3
ambiguous if (x+1)/4 = (x-1)/2 <=> x = 3
sente if (x+1)/4 < (x-1)/2 <=> x > 3

Reverse sente applies when sente applies. Gote, sente and reverse sente apply also in the ambiguous case.

SENTE ANALYSIS FOR x>3:

Kept gote move values:



Inherited sente count:



Inherited sente move value:



It is Black's sente, sente move count and sente move value.

REVERSE SENTE ANALYSIS FOR x>3:

Kept gote move values:



Inherited reverse sente count:



Reverse sente move value:



The reverse sente move value is calculated from the leaves 1 and 0:
swing = 1 - 0 = 1
tally = (1 - 1) - (-1) = 0 + 1 = 1
reverse sente move value = swing / tally = 1 / 1 = 1

It is White's reverse sente, reverse sente move count and reverse sente move value. When White plays the reverse sente from A with the reverse sente count 1 to the leaf 0, Black loses the reverse sente move value 1 (because White gains it): 1 - 1 = 0.

SUMMARY OF ALL NON-AMBIGUOUS CASES AT A:

gote (x<3), gote count (x+1)/4, gote move value (x+1)/4
sente for Black (x>3), sente count 1, sente move value (x-1)/2
reverse sente for White (x>3), reverse sente count 1, reverse sente move value 1

How, in general, is the temperature derived from this? Must the min or max of two values be taken?

*******************************************************

EXAMPLE 2:

GOTE ANALYSIS:

Given tree:

x >= 1 (otherwise it would not be a meaningful go tree)



Gote counts:



Gote move values:



CASE ANALYSIS:

gote if (1+x)/4 > (-1+x)/2 <=> x < 3
ambiguous if (1+x)/4 = (-1+x)/2 <=> x = 3
sente if (1+x)/4 < (-1+x)/2 <=> x > 3

Reverse sente applies when sente applies. Gote, sente and reverse sente apply also in the ambiguous case.

SENTE ANALYSIS FOR x>3:

Kept gote move values:



Inherited sente count:



Inherited sente move value:



It is White's sente, sente move count and sente move value.

REVERSE SENTE ANALYSIS FOR x>3:

Kept gote move values:



Inherited reverse sente count:



Reverse sente move value:



The reverse sente move value is calculated from the leaves 0 and -1:
swing = 0 - (-1) = 1
tally = 1 - (-1 + 1) = 1 - 0 = 1
reverse sente move value = swing / tally = 1 / 1 = 1

It is Black's reverse sente, reverse sente move count and reverse sente move value. When Black plays the reverse sente from A with the reverse sente count -1 to the leaf 0, Black gains the reverse sente move value 1, i.e., -1 + 1 = 0.

SUMMARY OF ALL NON-AMBIGUOUS CASES AT A:

gote (x<3), gote count -(1+x)/4, gote move value (1+x)/4
sente for Black (x>3), sente count -1, sente move value (-1+x)/2
reverse sente for White (x>3), reverse sente count -1, reverse sente move value 1

How, in general, is the temperature derived from this? Must the min or max of two values be taken?

*******************************************************

COMPARISON OF EXAMPLES 1+2:

Because of symmetry, the counts are negated and the move values are equal.

The local temperature for the gote is the gote move value. The local temperature for the sente/reverse sente is the reverse sente move value. Note that for the gote, the temperature (the gote move value) is less than the reverse sente move value. For the sente/reverse sente, the temperature (the reverse sente value) is less than the gote value.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Let me try to understand this. I forgo the ambiguous gote = sente cases.

VALUE CONDITION:

Given the game, {a | b || c | d}, a > b > c > d.

Equal values simplify the game, unequal values provide a more difficult study case. Except for equalities, a meaningful go game requires a>b, c>d, a>c, b>d, a>d; Black moving to (a|b) then White moving to b must give Black more than letting White move to (c|d) then Black moving to c so that Black's move to (a|b) is meaningful at all, hence a meaningful go game requires also b>c; we get a > b > c > d. Thus, except for equalities, all meaningful go games {a | b || c | d} are represented by the value condition.

GOTE VALUES:

CG := gote count, MG := gote move value.

CG(a|b) = (a+b)/2.
MG(a|b) = (a-b)/2.
CG(c|d) = (c+d)/2.
MG(c|d) = (c-d)/2.
CG(a|b||c|d) = ((a+b)/2 + (c+d)/2)/2 = (a+b+c+d)/4.
MG(a|b||c|d) = ((a+b)/2 - (c+d)/2)/2 = (a+b-c-d)/4.

BLACK SENTE CONDITION:

For the game to be sente for Black, (a+b-c-d)/4 < (a-b)/2 <=>
(a+b-c-d)/4 - (a+b)/2 < (a-b)/2 - (a+b)/2 <=>
-(a+b+c+d)/4 < -b <=>
(a+b+c+d)/4 > b. QED.

WHITE SENTE CONDITION:

For the game to be sente for White, (a+b-c-d)/4 < (c-d)/2 <=>
(a+b-c-d)/4 + (c+d)/2 < (c-d)/2 + (c+d)/2 <=>
(a+b+c+d)/4 < (c-d+c+d)/2 <=>
(a+b+c+d)/4 < c. QED.

GOTE CONDITION:

Gote occurs when neither player has sente, so b > (a+b+c+d)/4 > c. QED.

Note: By this definition of gote, double sente (seen as a value type) is superfluous.

BLACK SENTE:

CS := sente count, MS := sente move value for a particular player.

CS(a|b||c|d) = b. (Inherited from the leaf b.)
MS(a|b||c|d) = (a-b)/2. (Inherited MG(a|b).)

WHITE SENTE:

CS(a|b||c|d) = c. (Inherited from the leaf c.)
MS(a|b||c|d) = (c-d)/2. (Inherited MG(c|d).)

MEAN VALUE:

What is the definition of 'mean value' of a game? Is it, by definition, the middle value of the gote count and the sente counts of the game? This would explain why the mean value of the game is the middle value of b, c and (a+b+c+d)/4. Or is the explanation more difficult than that? What is the purpose of the concept 'mean value'?

REVERSE SENTE FOR BLACK:

Note: Black's reverse sente considers White's sente path.

MRB := Black's reverse sente move value.

MRB(a|b||c|d) = (MG(a|b) - c) / (1 - (-1+1)) = ((a-b)/2 - c) / 1 = (a-b)/2 - c.

REVERSE SENTE FOR WHITE:

Note: White's reverse sente considers Black's sente path.

MRW := White's reverse sente move value.

MRW(a|b||c|d) = (b - MG(c|d)) / ((1-1) - (-1)) = (b - (c-d)/2) / 1 = b - (c-d)/2.

TEMPERATURE:

The local temperature of the gote game is the gote move value or of the sente game for a particular player is the opponent's reverse sente move value.

Gote temperature = (a+b-c-d)/4.
Black sente temperature = b - (c-d)/2.
White sente temperature = (a-b)/2 - c.

Now, the task is to derive the temperature of the game by solving
min((a+b)/2 - t, b) = max(c, (c+d)/2 + t)

One player wants to minimise, the other player wants to maximise - but which is which in this equation? It is an equation because we can view values from either player's perspective. We have the temperatures but how to derive the equation from it?

Please correct any mistakes and help me with my questions.

Given the gote game, G = {a | b}, with a > b, its mean value is (a+b)/2. 2G = a+b , because they are miai. Regardless of who plays first, in one of the games Black plays to a position worth a, and in the other game White plays to a position worth b. So the mean value of G is (a+b)/2.

N multiples of a sente game, G = {a || b | c}, with a > b > c, are not miai, but are equal to Nb if White plays first and (N-1)b + a if Black plays first. the mean value of G converges to b as N approaches infinity.

For the game, G = {a | b || c | b}, with a > b > c > d, the mean value lies between b and c, inclusive, and is (a+b+c+d)/4 if b > (a+b+c+d)/4 > c. The mean value is the middle value of b, c, and (a+b+c+d)/4, by inspection.



The left side of the equation represents values derived from minimax play at temperature, t, with Black playing first. White chooses how to play in response, and wishes to minimize the Black score, which is why we take the minimum of the different values. Conversely, the right side of the equation is a maximum because White plays first and Black chooses her replies to maximize the score.

An ideal environment of temperature, t, has a mean value of 0 and playing first in the environment gains t/2. We add a game, G, to that environment and find the result of minimax play with Black playing first. Then we subtract t/2 to adjust for the effect of the environment. The result is the left (Black) scaffold of the thermograph of G. It is the left side of the equation. The right side of the equation is found in similar fashion.

Let me illustrate for G = {a | b || c | d}. If Black (correctly) plays in the environment, the result is the mean value of G plus t/2. Since we do not know the mean value, we cannot use that in the equation. If Black plays to {a | b} and then White (correctly) replies in the environment, the result is the mean value of {a | b} - t/2 = (a+b)/2 - t/2. If White replies to b, and the Black plays in the environment, the result is b + t/2. That gives the result of min((a+b)/2 - t/2, b + t/2). Adjusting for the environment gives us min((a+b)/2 - t, b). As advertised.

For more, see On Numbers and Games by Conway, Winning Ways by Berlekamp, Conway, and Guy, and "Extended thermography for multiple kos in go" by yours truly.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Very clear explanation!

Solving min((a+b)/2 - t, b) = max(c, (c+d)/2 + t):

CASE min = b AND max = c:
Solution: b = c.

CASE min = b AND max = (c+d)/2 + t:
Solution: b = (c+d)/2 + t <=> t = b - (c+d)/2.

CASE min = (a+b)/2 - t AND max = c:
Solution: (a+b)/2 - t = c <=> t = (a+b)/2 - c.

CASE min = (a+b)/2 - t AND max = (c+d)/2 + t:
Solution: (a+b)/2 - t = (c+d)/2 + t <=> 2t = (a+b)/2 - (c+d)/2 <=> t = (a+b-c-d)/4.




Apparently, I must have made mistakes in my earlier calculations of the reverse sente move values. So are they not derived from MG(a|b) and MG(c|d) but from CG(a|b) and CG(c|d)?

REVERSE SENTE FOR BLACK:

MRB(a|b||c|d) = (CG(a|b) - c) / (1 - (-1+1)) = ((a+b)/2 - c) / 1 = (a+b)/2 - c.

REVERSE SENTE FOR WHITE:

MRW(a|b||c|d) = (b - CG(c|d)) / ((1-1) - (-1)) = (b - (c+d)/2) / 1 = b - (c+d)/2.

Accordingly, the temperatures are derived.

TEMPERATURE:

The local temperature of the gote game is the gote move value or of the sente game for a particular player is the opponent's reverse sente move value.

Gote temperature = (a+b-c-d)/4.
Black sente temperature = b - (c+d)/2.
White sente temperature = (a+b)/2 - c.

Is now everything correct and are the players (colours) assigned correctly?

C := count, M := move value. Ambiguous gote = sente ignored. No kos. I have not understood reverse sente calculation yet, so I forgo it and the temperatures.

********************************************************************************************

GAME (a|b), a>b

C(a|b) = (a+b)/2
M(a|b) = (a-b)/2
The game is gote because there are no follow-up moves.

********************************************************************************************

GAME (a||b|c), a>b>c

GOTE VALUES:

C(b|c) = (b+c)/2
M(b|c) = (b-c)/2
C(a||b|c) = (a + (b+c)/2) / 2
M(a||b|c) = (a - (b+c)/2) / 2

CASE STUDY:

Gote if (a - (b+c)/2) / 2 > (b-c)/2 <=>
a/2 - b/4 - c/4 > b/2 - c/2 <=>
a - b/2 - c/2 > b - c <=>
a > 3b/2 - c/2.

Sente if (a - (b+c)/2) / 2 < (b-c)/2 <= a < 3b/2 - c/2.

WHITE SENTE VALUES (a < 3b/2 - c/2):

C(a||b|c) = b (inherited from the leaf b)
M(a||b|c) = (b-c)/2 (inherited from (b|c))

********************************************************************************************

GAME (c|b||a), c>b>a

GOTE VALUES:

C(c|b) = (c+b)/2
M(c|b) = (c-b)/2
C(c|b||a) = ((c+b)/2 + a) / 2
M(c|b||a) = ((c+b)/2 - a) / 2

CASE STUDY:

Gote if ((c+b)/2 - a) / 2 > (c-b)/2 <=>
c/4 + b/4 - a/2 > c/2 - b/2 <=>
c/2 + b/2 - a > c - b <=>
-a > c/2 - 3b/2 <=>
a < 3b/2 - c/2.

Sente if ((c+b)/2 - a) / 2 < (c-b)/2 <=> a > 3b/2 - c/2.

BLACK SENTE VALUES (a > 3b/2 - c/2):

C(c|b||a) = b (inherited from the leaf b)
M(c|b||a) = (c-b)/2 (inherited from (c|b))

********************************************************************************************

GAME (a|b||b|c), a>b>c

GOTE VALUES:

C(a|b) = (a+b)/2
M(a|b) = (a-b)/2
C(b|c) = (b+c)/2
M(b|c) = (b-c)/2
C(a|b||b|c) = ((a+b)/2 + (b+c)/2) / 2 = (a+2b+c)/4
M(a|b||b|c) = ((a+b)/2 - (b+c)/2) / 2 = (a-c)/4

CASE STUDY:

Gote if (a-c)/4 > (a-b)/2 AND (a-c)/4 > (b-c)/2.

(a-c)/4 > (a-b)/2 <=>
a/4 - c/4 > a/2 - b/2 <=>
a/2 - c/2 > a - b <=>
-a/2 > -b + c/2 <=>
a/2 < b - c/2 <=>
a < 2b - c

AND

(a-c)/4 > (b-c)/2 <=>
a/4 - c/4 > b/2 - c/2 <=>
a/2 - c/2 > b - c <=>
a/2 > b - c/2 <=>
a > 2b - c.

a < 2b - c AND a > 2b - c is FALSE.

Therefore, the game is never gote but always local sente.

BLACK SENTE VALUES:

C(a|b||b|c) = b (inherited from the left b leaf)
M(a|b||b|c) = (a-b)/2 (inherited from (a|b))

WHITE SENTE VALUES:

C(a|b||b|c) = b (inherited from the right b leaf)
M(a|b||b|c) = (b-c)/2 (inherited from (b|c))

GENERAL CASE:

C(a|b||b|c) = b

However, the move value depends on the moving player. Therefore, if |a| <> |c|, the players can start choosing to play in the game at different ambient temperatures. Playing a reverse sente is playing a sente.

So far we have been studying simple gote and simple sente but, even if we restrict choices per position to one best black and one best white move, there are also complicated gote and complicated sente, namely such with sequences of more than two successive plays.

According to http://senseis.xmp.net/?Count complicated gote have been called [CGT] reversal. I dislike this word because go players use reversal with a different meaning: creation of the same position by another sequence. Besides, reversal of complicated gote is similar to complicated sente: the uninterrupted walking along successive plays. I suggest to call such 'traversal'. There are gote traversal and sente traversal.

To evaluate all non-ko go game CGT trees, one must know all existing value types and have an algorithm to identify the types and calculate the counts and move values. Are simple gote, simple sente, gote traversal and sente traversal all types (since reverse sente is implied from sente) or which other types can exist? Is there an algorithm or do we still need to develop it?

For the trees, we have the leaves and their counts if we know them. Otherwise, we can consider stable followers as pseudo-leaves and replace and approximate counts by the territory counts of positional judgement. Hence, construction of the trees and counts for the leaves are straighforward algorithmic steps. The real fun starts afterwards.

If we have a traversal path from position A to position Z, if Y is the parent of Z and B is the simple gote child of A, then I hope that everything is simple as follows:

C := count, M := move value.


WHITE-SIDE GOTE TRAVERSAL:



C(A) = (C(B) + C(Z)) / 2

M(A) = (C(B) - C(Z)) / 2 (gote move value)


WHITE SENTE TRAVERSAL:



C(A) = C(Z)

M(A) = M(Y) (sente move value)

Black-side gote traversal and Black sente traversal are similar.

Is it this simple and are the calculations correct? How about, what naively might be called, "double sente" traversal?

Correct.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

I am unaware that go player have adopted the term, reversal, in either sense. IIUC, the creation of the same position by another sequence is called a transposition in English.



Minimax play in an ideal environment with a certain temperature is enough to detect reversals. Even experienced human calculators can overlook reversals. It can help to work bottom up, but you cannot throw away necessary information.

Example: Given: the game, G = {a ||| b | c || d}, a, b > c > d

Let's work bottom up. First, suppose that {b | c || d} is sente. Then its mean value is c and the mean value of G is (a+c)/2 and its temperature is (a-c)/2.

Now suppose that {b | c || d} is gote, with a mean value of (b+c+2d)/4. That yields this equation for the local temperature of G:

a - t = max(min((b+c)/2, c+t), (b+c+2d)/4 + t)

Edit: Corrected parentheses in the equation.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Reversal (reversion, revert) is a go term in my stated meaning, is used only occasionally and therefore not known by everybody, and is used e.g. in tewari or joseki contexts.

I guess you are fluent with minmax termperature equations. For me, they are more or less new. I needed ca. 2 to 4 hours per maths message I have written in this thread. So I will have to think a lot about your current message and think more when trying to answer most of my own questions while wondering what I might be reinventing or inventing.

I am surprised about maths discussion on L19. Off-topic fun maths attracts lots of participants in discussions. Contrarily, the more go-related and more generally applicable and relevant for one's go playing maths becomes the fewer participants remain in discussions. Although it has been like this for 20 years now, I hope this changes because our understanding could be accelerated. In particular, there is still a lot to be revealed about sente.