Sente, gote and endgame plays

[RobertJasiek]

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

I work myself bottom-up:

Code: Select all

     B
    / \
D(1)   E
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.

Code: Select all

     B
    / \
D(1)   E(-8.5)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.

The tentative gote move value of E is 4.5. At the child F, the follow-up (gote) move value is 6.

The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.

Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.

Code: Select all

     B
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The tentative gote count of B is (1 + (-10)) / 2 = -4.5.

Code: Select all

     B(-4.5)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The tentative gote move value of B is 5.5. At the child E, the follow-up move value is what?

E is a local sente with the sente move value 3. So the follow-up move value is, I think, a sente move value and is 3.

However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?

Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Now, maybe this is a case of checking for traversal? We have B < F <=> -4.5 < -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following diagram.

I have been unsure about possibly checking traversal. Why would this be the right thing to do?

Code: Select all

     B
    / \
D(1)   I(-10)

This is what you suggested, except that you have replaced B by J. Why?

Now, this simplification tree looks like a simple gote. But is it? Or does it represent another type of local endgame that we should call a "traversal" rather than a "local gote"?

Now, what are the correct count and move value of the simplified game?

I will try the other example tree later.

[Bill Spight]

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

I work myself bottom-up:

Code: Select all

     B
    / \
D(1)   E
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.

Code: Select all

     B
    / \
D(1)   E(-8.5)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.

The tentative gote move value of E is 4.5. At the child F, the follow-up (gote) move value is 6.

The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.

Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.

Code: Select all

     B
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The tentative gote count of B is (1 + (-10)) / 2 = -4.5.

Code: Select all

     B(-4.5)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

The tentative gote move value of B is 5.5. At the child E, the follow-up move value is what?

That's not exactly the right question.

E is a local sente with the sente move value 3. So the follow-up move value is, I think, a sente move value and is 3.

In an actual go game, the play from B to E gains 5.5 pts., but then the local temperature drops to 3, as you point out. So Black may well stop there and not continue to F. The reason is that the play from E to F may become a ko threat. The slight risk is that White may be able to play the reverse sente to G. So normally Black will stop local play and play elsewhere.

But finding the mean value is a local matter, except for ko positions.

However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?

Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Yes, that is what we do.

Now, maybe this is a case of checking for traversal? We have B < F <=> -4.5 < -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following diagram.

I have been unsure about possibly checking traversal. Why would this be the right thing to do?

If you want to simplify, it is the right thing to do.

Is F >= B? That is the question.

Code: Select all

     B(-4.5)            +           -F
    / \                             / \
D(1)   E(-10)                     10   -2
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

If so, then B - F <= 0, which means that if Black plays first in the combination above, White can win or get jigo in an even number of plays.

First, suppose that Black plays in -F to 10, then White replies B -> E -> F -> -10, for a result of 0 in an even number of plays.

Second, suppose that Black plays in B to 1, then White replies in -F to -2, for a result of -1; White wins.

So F >= B and the play, B -> E reverses to I(-10).

Code: Select all

     B
    / \
D(1)   I(-10)

This is what you suggested, except that you have replaced B by J. Why?

Because B and J are different games. The question is whether they are equal. They are.

Now, this simplification tree looks like a simple gote. But is it? Or does it represent another type of local endgame that we should call a "traversal" rather than a "local gote"?

The simplified game, J, is a simple gote. B is not. As we have seen, in a real game Black is likely to leave the sente at E on the board for a while.

[RobertJasiek]

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E(-8.5)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.

The tentative gote move value of E is 4.5. At the child F, the follow-up gote move value is 6.

The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.

Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

The tentative gote count of B is (1 + (-10)) / 2 = -4.5.

Code: Select all

          A
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

The tentative gote move value of B is 5.5.

E is a local sente with the follow-up sente move value 3.

The tentative gote move value of B is larger than the follow-up sente move value at the child E, that is, 5.5 > 3. This condition identifies a local gote. (Besides, the follow-up gote move value at the child D is 0 for a pass. The tentative gote move value of B is larger than the follow-up gote move value at the child D, that is, 5.5 > 0. This condition also identifies a local gote.)

So B is a local gote and we keep its tentative gote move value as its gote move value.

We can check for traversal: We have B <= F <=> -4.5 <= -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following equal diagram.

Code: Select all

          A'
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   I(-10)

The line from B to I represents the sequence (move to E) - (move to F) - (move to I).

In the simplified game, the tentative gote count of A' is -8,75.

Code: Select all

          A'(-8.75)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   I(-10)

In the simplified game, the tentative gote move value of A' is 4.25. At the child B, the follow-up gote move value is 5.5.

The tentative gote move value of A' is smaller than the follow-up gote move value, that is 4.25 < 5.5. This condition identifies a local sente.

Therefore, we discard the tentative values of A' but determine the sente count of A' as the count of the sente follower I: -10.

Code: Select all

          A'(-10)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   I(-10)

The local endgame A' is a local sente with the already calculated sente count -10 and the sente move value 3.

Since A = A', we derive the values for original game A:

Code: Select all

          A(-10)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

The local endgame A has the sente count -10 and the sente move value 3.

***

Let us go back to the analysis of A before we used traversal to simplify the game:

Code: Select all

          A
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

We have already identified B as a local gote with gote count -4.5 and the gote move value 5.5.

Now, we proceed to analyse A.

The local endgame A has the tentative gote count -8.75.

Code: Select all

          A(-8.75)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

The tentative gote move value of A is 4.25. At the child B, the follow-up gote move value is 5.5.

The tentative gote move value of A is smaller than the follow-up gote move value, that is 4.25 < 5.5. This condition identifies a local sente.

Therefore, we discard the tentative values of A but determine the sente count of A as the count of the sente follower E: -10.

Code: Select all

          A(-10)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)

The local endgame A is a local sente with the sente count -10 and the sente move value 3.

Locally, without ko threat play, the sequence from A to I is traversed as Black's sente. The sente move value 3 at A and E and the follow-up move values 5.5 at B and 6 at F must be considered for the timing in the context of the global temperature.

***

For the local endgame A, traversal provides an optional technique with which type, count and move value of A can be determined. The type is a local sente. Traversal describes that, once Black has started the local sente sequence at a suitable global temperature, it may as well traverse to I.

Since the count of C equals the count of G, Black stopping at A incurs the same risk as stopping at E that White plays reverse sente. Since the follow-up move value of B is 5.5 and smaller than the follow-up move value 6 at F, if White should not stop at B allowing Black to achieve the count 1 at D, White should even less stop at F allowing Black to achieve the larger count 2 at H.

Ta-ta--ta---taaa!

[RobertJasiek]

(Continuing discussion from my previous post.)

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

B is easy. It equals J.

Code: Select all

     J
    / \
D(1)   I(-10)

The reason is that H(2) > D(1).

You sound as if this was sufficient reason for B equalling J. Please explain. Is such a condition general for all trees? If so, what is the general condition, that is, what counts to compare how to assess as sufficient reason that a local endgame simplified by traversal equals its simplified game?

We can do difference games to find the answer

To check that a local endgame and its traversal-simplification are equal? Why? This is a truth we (eh, should) know from CGT.

the count of A is indisputably 10

I'd even say it is indisputably -10. ;)

However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?
Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Yes, that is what we do.

I still wonder for what purposes exactly we a) compare the tentative gote move value of B to the follow-up sente move value at the child E and b) compare the tentative gote move value of B to the black gote move value from E to F.

Code: Select all

     B(-4.5)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

B is not [a simple gote.]

Let me see: B has the tentative gote move value 5.5 and the tentative sente move value D(1) - F(-4) = 5. The tentative gote move value 5.5 is larger than the tentative sente move value 5. This condition identifies a local sente. So B is White's local sente.

Furthermore, E is Black's local sente.

Oh!

The intermediate position E is involved in both White's local sente at B and Black's local sente at E. That's an alternative explanation why we may traverse the sequence from B to I if played at an appropriate temperature! The whole sequence is White's gote sequence but its parts are either player's sente sequences. Ugh. No suprise I found things to be so complicated. Have I understood it correctly now?

[Bill Spight]

(Continuing discussion from my previous post.)

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

B is easy. It equals J.

Code: Select all

     J
    / \
D(1)   I(-10)

The reason is that H(2) > D(1).

You sound as if this was sufficient reason for B equalling J. Please explain. Is such a condition general for all trees? If so, what is the general condition, that is, what counts to compare how to assess as sufficient reason that a local endgame simplified by traversal equals its simplified game?

Given: Game B

Code: Select all

        B 
       / \
      D   E
         / \
        F   G
       / \
      H   x

where B, D, E, F, G, and H are games; B, E, and F are not numbers; x is a number; and H >= D.

To prove: B -> E -> F reverses to x. I.e., F >= B <=> B - F <= 0.

Code: Select all

        B         +        -F        <=  0
       / \                 / \
      D   E              -x  -H
         / \
        F   G
       / \
      H   x

Lines of play where Black plays first.
1) Black moves from B to D, then White moves from -F to -H. The result is D - H <= 0.
2) Black moves from -F to -x, then White moves from B to E, Black moves from E to F, and White moves from F to x. The result is x - x = 0.

When Black plays first White can always play to a final position that is less than or equal to 0.

QED.

Note that because x is a number, Black has no good move in 2) after White moves to E, except to move to F, by the number avoidance theorem.

We can do difference games to find the answer

To check that a local endgame and its traversal-simplification are equal? Why? This is a truth we (eh, should) know from CGT.

IMX, I have often found it easier to check that a game and its simplification are equal than to check that the conditions for reversal or dominance are met.

the count of A is indisputably 10

I'd even say it is indisputably -10.

Sans doute.

However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?
Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Yes, that is what we do.

I still wonder for what purposes exactly we a) compare the tentative gote move value of B to the follow-up sente move value at the child E and b) compare the tentative gote move value of B to the black gote move value from E to F.

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

F is not a number, so the fact that C = G is not enough to tell us that the play reverses through E. However, the play in B reverses through F.

No suprise I found things to be so complicated. Have I understood it correctly now?

Reverses are tricky. I think you have a good understanding.

[RobertJasiek]

Code: Select all

     B(-4.5)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

B has the tentative gote move value 5.5 and the tentative sente move value D(1) - F(-4) = 5. The tentative gote move value 5.5 is larger than the tentative sente move value 5. This condition identifies a local sente. So B is White's local sente.

This is still wrong, is it?

If indeed B is White's local sente, then B has the tentative sente count -4.

Code: Select all

     B(-4)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

Saying that B is White's local sente with the tentative sente move value 5 overlooks that, after the sente sequence from B to F, the follow-up gote move value at F is 6. The move value increases, that is, the tentative sente move value of B is smaller than the follow-up gote move value at F, that is, 5 < 6. Therefore, if the sente sequence from B to F is played, play does not stop but proceeds in alternation to I.

Since B is not a local sente but a traversal (that is, besides the types local gote, local sente, ambiguous and maybe local double sente, there is another type of local endgames: traversal), we discard the the tentative sente move value of B.

Code: Select all

     B
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

B, being a traversal, has as its count the average of the counts of its followers D and I, that is, (1 + (-10)) / 2 = -4.5.

Code: Select all

     B(-4.5)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

We had that number for the count earlier but it was wrong nevertheless because it was associated with the wrong meaning of tentatively being a local gote. Now the count -4.5 of B is correct because it is associated with the correct meaning of being a traversal. (We must not take as a count what deceiptively looks right because of accidentally having the same number.)

***

Since B is a traversal, we may simplify the game by pruning the traversed nodes to get B':

Code: Select all

     B'
    / \
D(1)   I(-10)

Note that B' does not have assigned its count yet. We determine it only now. B' is a simple gote so has the gote count -4.5.

Code: Select all

     B'(-4.5)
    / \
D(1)   I(-10)

CGT tells us that B = B' from a value perspective if there are / will be no kos. Therefore, if first we determine the gote count -4.5 of B', we may then also use this value as the count of B. However, in B, this same count has a different meaning! Whilst in B' the count -4.5 is a gote count, in B the count -4.5 is a traversal count!

To recollect, suppose we are here when analysing B:

Code: Select all

     B
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

From the count -4.5 of B', we know that this is also the count of B:

Code: Select all

     B(-4.5)
    / \
D(1)   E(-10)
      / \
 F(-4)   G(-13)
    / \
H(2)   I(-10)

From the earlier analysis of the game B (yes, B, not B'), we also know that B is a traversal so the assigned count of B is a traversal count derived from the counts of D and I.

Despite numerical equality, it would be wrong to say that the count of B could simply be calculated as the average of the counts of D and E. This would miss the meaning of the count, which is NOT a gote count but is a traversal count.

Did I say traversal was difficult? It is! :)

[Bill Spight]

From the earlier analysis of the game B (yes, B, not B'), we also know that B is a traversal so the assigned count of B is a traversal count derived from the counts of D and I.

Despite numerical equality, it would be wrong to say that the count of B could simply be calculated as the average of the counts of D and E. This would miss the meaning of the count, which is NOT a gote count but is a traversal count.

Right you are.

Did I say traversal was difficult? It is!

Right again.

[Bill Spight]

Reading https://senseis.xmp.net/?Count I thought I would have understood traversal (what confusingly that page and CGT calls reversal). However, when trying to apply it to the following examples, I notice that I have understood nothing.

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

The second example is one move earlier than the first example. The second example can be represented as what looks like a hane-and-connect "sente" sequence.

How to distinguish and identify local gote, simple local sente and traversal from each other? What is the exact general procedure? Which tentative or final - gote or sente - counts and move values to calculate for which nodes? How and procedurally when? Which conditions determine the initial positions' types? What distinguishes a long sente sequence (more than 2 moves) from a traversal sequence?

I understand the conditions for simple local gote and simple local sente. However, when I try to apply them to the examples, I am confused.

Earlier, I said that you should always check for reverses. IMX with calculating positions and plays from the scores, reverses can lead to errors. However, that is not so with thermography, where everything comes out in the wash, at least for non-ko positions.

If we think in terms of thermography, we may be able to reduce our efforts.

Let's look at the first example.

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

For non ko positions the maximum lines of the thermograph are either of the form v = x - t or v = x, and similarly, the minimum lines are of the form v = x + t or v = x. That means that we can tell the maximum and minimum possibilities for the count from following alternating lines of play to the end. For B the maximum value is 1 and the minimum value is -10.

For the left wall of the thermograph of B, the line is v = 1 - t, up to the local temperature, above which it is v = b, for some value of b. For the right wall there are also three lines which may be relevant, v = -10 + t, v = 2 - t, and v = -13 + 2t. The coefficient of t for the line for G is 2 because it takes two plays by White to get to a score of -13. The line for G will not form part of the thermograph for B, but it may determine a relevant line where it intersects with the line for H or I. The line for H will not form part of the thermograph for B, because it is to the left of the line for D, which indicates the maximum possible values.

The intersection of v = 1 - t and v = -10 + t occurs at v = -4.5 and t = 5.5. That gives us possible values for the count and miai value of B. The intersection of v = 2 - t and v = -10 + t, which gives us the count and miai value of F, will occur with greater values for v and t and is thus irrelevant.

The intersection of v = -10 + t and v = -13 + 2t occurs at v = -10 and t = 3. We know that the miai value of F is greater than 5.5, which is greater than 3, so E is a Black sente with a count of -10.

All of this confirms that the count of B is -4.5 and its miai value is 5.5.

Note that we can, as mentioned above, ignore the H score because it is greater than the D score, which means that we can treat E as sente, even if it is not.

Now let's take a look at A.

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

The count of A lies between -13 and -10. The basic thermographic lines are v = -13 + t, v = -10, v = 1 - 2t, and v = 2 - 2t. Note that G and C generate the same thermographic line. Note also that the maximum count of both G and C is -10. Since the intersection of v = 1 - 2t and v = -10 occurs at t = 5.5, which is greater than 3, which is the temperature at which v = -10 and v = -13 + t intersect, we conclude that C is a Black sente. And since H(2) is greater than D(1), E is also a Black sente with a count of -10, which is then the count of A.

Edit: Note that if either D or H had a value of -8 while the other stayed the same, then the count of A would be -11.

[RobertJasiek]

Before entering study of linear algebra called thermography with 1000 questions, I have more fundamental questions relevant for validity of count / miai value calculations both without or with linear algebra.

We suspect non-existence of local double sente but what is a proof for this? Maybe it is straightforward, but I have not tried it yet. I have, however, recognised a necessity of clarification. If there is ambiguity between local gote and local double sente (as there is for certain shapes, such as mutual sacrifice of 3 stones and recapture), we need not speak of an optional strategy of local double sente because we can still treat it as local gote. In general, when distinguishing local gote, ambiguous, local sente and maybe local double sente, we use the comparison Mgote ? Msente for these tentative move values. More specifically, we must be considering a particular player's Msente. If it is Black's tentative local sente, we have the comparison Mgote ? M_B_sente. If it is White's tentative local sente, we have the comparison Mgote ? M_W_sente. A non-ambiguous local sente has Mgote > Msente. Therefore, Black's tentative local sente has Mgote > M_B_sente and White's tentative local sente has Mgote > M_W_sente. To possibly prove the non-existence of local double sente, we have to show that Mgote > M_B_sente AND Mgote > M_W_sente <=> (B - W) / 2 > Sb - W AND (B - W) / 2 > B - Sw is FALSE for all values. Do we have B >= Sb >= Sw >= W to help us proving?

My second problem is from which counts along "long" alternating sequences started by Black or White to calulate the count of the initial position. Sometimes, for either particular alternating sequence, we need the count of the direct follower. Sometimes we need the count of the "last" interesting follower. When to use which and why? To emphasise again, this choice occurs for both alternating sequences.

My third problem: Earlier I was naive to assume that we could distinguish local gote and sente by one condition Mgote ? Msente. Now, I think that (unless for the trivial cases of short alternating sequences before encores), we always need to check the two conditions Mgote ? M_B_sente and Mgote ? M_W_sente. Am I right?

[Bill Spight]

Before entering study of linear algebra called thermography with 1000 questions, I have more fundamental questions relevant for validity of count / miai value calculations both without or with linear algebra.

We suspect non-existence of local double sente but what is a proof for this?

Every finite combinatorial game has a mean value; if the game is not a number, there is a non-negative temperature at which the result of play will be equal to the mean value, no matter who plays first. For these statements let me refer you to On Numbers and Games.

Every sente sequence produces a vertical thermographic line, v = x, where x is the numerical result of the sequence. Suppose that there is a local double sente, such that when Black plays first the result is b and when White plays first the result is w, and b > w. In that case there is no non-negative temperature at which the results of play will be equal to the mean value, no matter who plays first. (You will have two vertical thermographic lines which do not intersect.) Therefore there can be no such double sente. There may be local double sente where b <= w. And I do not think that anyone has proven that there may not be local double sente with kos or superkos. In 1998 I showed how to find the thermographs of multiple kos and superkos, but I assumed that they had a mast value. Kos and superkos are basically non-finite games for which the rules may or may not produce a value. If the rules do not do so, then they cannot be scored, and they could be local double sente. In the original draft of my 1998 paper I pointed out that my method would not work for molasses ko unless the rules gave it a value. AGA rules do, other rules may not do so.

My second problem is from which counts along "long" alternating sequences started by Black or White to calulate the count of the initial position.

You stop when you reach a number.

My third problem: Earlier I was naive to assume that we could distinguish local gote and sente by one condition Mgote ? Msente. Now, I think that (unless for the trivial cases of short alternating sequences before encores), we always need to check the two conditions Mgote ? M_B_sente and Mgote ? M_W_sente. Am I right?

Since there is no local double sente that gains any points, you can start out by assuming that the position is gote and then possibly disprove that.

Edit: Or you can use thermography. You may start out with two vertical lines for best play, but if the position is not a number, at some temperature another line will intersect one of those lines and you then draw a gote mast line for higher temperatures.

[RobertJasiek]

In Black's tentative local double sente where b <= w, Black prefers to a) pass instead of playing his "sente sequence" or b) play his longer gote sequence to achieve some b' >= w so in practice, there is no local double sente.

I need think through your other explanations.

[Bill Spight]

In Black's tentative local double sente where b <= w, Black prefers to a) pass instead of playing his "sente sequence" or b) play his longer gote sequence to achieve some b' >= w so in practice, there is no local double sente.

I need think through your other explanations.

Code: Select all

            A
           / \
          /   \
         B     C
        / \   / \
      13   5 7  -3


            D
           / \
          /   \
         E     F
        / \   / \
       7   3 3  -5 


            G
           / \
          /   \
         B     C
        / \   / \
       5  -3 2  -8

A, D, and G are all numbers. Numbers of type D occur frequently in go. Numbers of type G occur occasionally. Number of type A are rare.

[RobertJasiek]

Code: Select all

            A
           / \
          /   \
         B     C
        / \   / \
      13   5 7  -3


            D
           / \
          /   \
         E     F
        / \   / \
       7   3 3  -5 


            G
           / \
          /   \
         H     I
        / \   / \
       5  -3 2  -8

A, D, and G are all numbers. Numbers of type D occur frequently in go. Numbers of type G occur occasionally. Number of type A are rare.

[Edited your letters in the citation.]

Let me see:

Code: Select all

            A
           / \
          /   \
         B     C
        / \   / \
      13   5 7  -3

Code: Select all

            A
           / \
          /   \
         B(9)  C(2)
        / \   / \
      13   5 7  -3

Mgote = 5.5. M_B_sente = 3. M_W_sente = 2.

We have Mgote > M_B_sente, M_W_sente.

Therefore, from either player's sente perspective, it is his local sente. The starting Black achieves 5. The starting White achieves 7. White's start is more favourable for Black than Black's start therefore Black wants to pass and let White start. White's start results in a positive count favouring Black and White prefers to pass, let Black start and let Black achieve the smaller positive count. So White does not help Black. If White starts, he passes and lets Black start his sente sequence to 5. Therefore, this is not a local double sente but is Black's local sente.

Since A = {5|7} and not 5 >= 7, A is a number. But what number? I forgot.

Code: Select all

            D
           / \
          /   \
         E     F
        / \   / \
       7   3 3  -5 

Code: Select all

            D
           / \
          /   \
         E(5)  F(-1)
        / \   / \
       7   3 3  -5 

Mgote = 3. M_B_sente = 4. M_W_sente = 1.

We have Mgote < M_B_sente so it is not Black's local sente. We have Mgote > M_W_sente so it is White's local sente. The sente count 3 of D is inherited from the leaf 3 of F.

Since D = 3, D is a number.

Code: Select all

            G
           / \
          /   \
         H     I
        / \   / \
       5  -3 2  -8

Code: Select all

            G
           / \
          /   \
         H(1)  I(-3)
        / \   / \
       5  -3 2  -8

Mgote = 2. M_B_sente = 0. M_W_sente = 1 - 2 = -1 is correctly calculated as Black's minus White's value and is negative so White's start is his loss.

We have Mgote > M_B_sente, M_W_sente.

Therefore, from either player's sente perspective, it is his local sente.

However, since White's sente move value M_W_sente = -1 is negative, he prefers to pass. He does not use his loss-making sente sequence. Therefore, we do not have a local double sente.

Black may, but need not, start his sente sequence because the move value is 0. G is Black's local sente so the sente count of G is -3. Black's sente sequence transforms the count -3 of G into the count -3 of the black sente follower.

What are sample go positions for these games?

How many, and what, mistakes have I made?

[Bill Spight]

Code: Select all

            A
           / \
          /   \
         B     C
        / \   / \
      13   5 7  -3


            D
           / \
          /   \
         E     F
        / \   / \
       7   3 3  -5 


            G
           / \
          /   \
         B     C
        / \   / \
       5  -3 2  -8

A, D, and G are all numbers. Numbers of type D occur frequently in go. Numbers of type G occur occasionally. Number of type A are rare.

A = 5

D = 3

G = 0

Numbers of type D are frequent. They are miai, and may be double ko threats.

G is a seki.

Here is an example of type A.



When Black plays first, we may assume the best result for Black, i.e., that White cannot win the potential ko. It is still not as good for Black as when White plays first.

Since this is a number, we could in theory just score it as 3 pts. for White, but the threat to score it as 0 forces White to play first, except perhaps in extraordinary circumstances.

[RobertJasiek]

In your three-points-without-capturing shape, scoring rules can define whatever score to the initial position... So if the scoring rules are CGT territory scoring defined for terminal positions, the score and therefore counts of the black and white terminal followers are -5 and -3, respectively. So in CGT the initial position's count is -3.

You suggest that other scoring rules could score the initial position as 0, but why 0? Why not, e.g., 1 or -1 or -4? How do you derive the 0? By rules-style scoring definitions of the kind "territory only if surrounded by one player's independently alive stones and scoring dead stones only as prisoners or in territory"? This would alter tree analysis. We would not any longer derive counts of non-terminal nodes from counts of terminal nodes but all nodes would get their counts as the scores of the current nodes' positions!

E.g., with area scoring, the initial local score is 4. If Black starts, the terminal score is -1; this is 5 worse for Black. If White starts, the terminal score is 1; this is 3 better for White or 3 worse for Black. For Black, the optimal strategy is to let White achieve "3 worse for Black".

***

I am still trying to understand whether a local double sente with b <= w exists. Your example trees are not local double sentes. For simple tentative local double sentes with simple gote follow-ups, your example trees represent all cases 1) b, w > 0 (or by symmetry b, w < 0), 2) b = w, 3) b < 0 and w > 0 (or by symmetry b > 0 and w < 0). My reasoning when analysing your example trees applies regardless of the values (provided b <= w and the black-black follower being >= b and the white-white follower being <= w). Therefore, I think that also for b <= w there is no local double sente.

However, you write "There may be local double sente where b <= w.". Why? Which? With which example position? Why is my reasoning (kos excluded) insufficient for excluding the possibility of local double sente in which either starting player gains? Or have you just meant to say that, for b <= w, there may be local double sentes in which at least one player does not gain (but loses or keeps constant) from starting?

Ah, then you also write: "Since there is no local double sente that gains any points". So has my reasoning been a) right and b) what you have wanted to imply?

Even so, how about long-sequence local double sentes? Might they exist even with simple local double sentes not existing?

***

"Every finite combinatorial game has a mean value;"

Ok.

"if the game is not a number, there is a non-negative temperature at which the result of play will be equal to the mean value, no matter who plays first."

This description itself is easy to understand, thank you for the translation! I am still struggling though with relating the theorems in ONAG to your translation because I am not used to understanding the ONAG annotation. For now, I need to believe it.

"Every sente sequence produces a vertical thermographic line, v = x, where x is the numerical result of the sequence. Suppose that there is a local double sente, such that when Black plays first the result is b and when White plays first the result is w, and b > w. In that case there is no non-negative temperature at which the results of play will be equal to the mean value, no matter who plays first. (You will have two vertical thermographic lines which do not intersect.) Therefore there can be no such double sente."

With "sente sequence", I think you do not just mean an alternating 2-play sequence but additionally presume a requirement of it belonging to a simple sente.

I do not exactly understand the annotation v = x yet. Is this an abbreviation for v(t) = x, with t >= 0 being the temperatures? I.e., v is a mapping (abbildung)?

Your proof is by contradiction, ok.

"In that case there is no non-negative temperature at which the results of play will be equal to the mean value, no matter who plays first."

This summarises two cases: Black plays first; White plays first. For each case, we have to prove "there is no non-negative temperature at which the results of play will be equal to the mean value".

"In that case there is no non-negative temperature at which the results of play will be equal to the mean value"

So this is just a sketch of what must be worked out as a proof in detail.

Part I: We have to calculate the mean value. Since it is defined for some multiple of the local endgame, we have to study multiples of it until we find a suitable number of multiples of it to get the mean value.

Part II: Suppose a temperature t >= 0. We have the ensemble of the local game tree and an environment with the temperature t, that is, playing first in the environment gains t/2 for the player playing first in it. So we have to study the local lines of play with the addition of accounting the gain of the first play in the environment. For each such sequence of "global" play, we have to calculate the result and find that it is unequal to the mean value.

Ugh. Much work to be done to transform the sketch of a proof into a proof!