How evaluate double sente moves ?

[RobertJasiek]

Gerard, your case counting reminds me of one of my sins. As a mathematically interested pupil, I joined a week's excursion to university. One of our homeworks was a puzzle of dozens of tiles to be proven or disproven to fit into a square filling it. The puzzle was just small enough for my evening proof by enlisting each of some thousand cases:) The professor was not amused:) Of course, we were supposed to do an abstract proof:):) (I did better with my other homework.)

[Bill Spight]

A good example of where the belief in (intrinsic) double sente is detrimental. If they really exist, then the answer is plain. When your opponent plays the first double sente, then, instead of answering it, play one of the other double sente.

Bill: You can judge it much better than I can, so can I direct you to game 1703-01-11a in the GoGoD database. It seems to me both an example and a counter-example of what you say.

Thanks, John, I'll check the game out this weekend.
As for countre-examples (British spelling ) I said that was so if intrinsic double sente really exist; but they don't.

When Hayashi Genetsu played 128 he evidently expected Black to answer around there (as he did very soon after - Black 135), and equally clearly it a was reverse sente move in that he didn't want Black (Dochi) to around 128 himself.

But Dochi - only 13! - was alert and followed your advice. He played a different double sente: Black 129. But White had to answer that one: he couldn't sensibly go on a mutual damage rampage, even though a double sente in centre still existed. That is, he couldn't follow your advice! In short, 128 should first have been 129. But then Dochi would have had to answer in the corner - he couldn't divert to the double sente in the centre. Another counter-example?

Dochi was a genius, wasn't he?

But this was also a case of what you said in your other thread about pros being significantly better than even high-dan amateurs (or in this case a 4-dan pro) at the endgame. Both Sakata and Segoe picked up on this mistake. (For anyone else looking up the game, a White move at S4 is the killer threat.)

Well, some pros, anyway.

[Gérard TAILLE]

OK. Given a game, H = {d|e||f|g}, with d > e > f > g, we have a simple test for sente and gote. Let x = (d+e+f+g)/4. If x > e, then the game is a Black sente. If f > x, then the game is a White sente. If e > x > f, then the game is a gote. If d-g >> e-f, the game is unlikely to be gote. That would be threading the needle.

For the time being I tested several functions against couples of local endgames according to the kind of tree above.

OK. Let there also be another game, J = {p|q||r|s}, with p > q > r > s. And let there be an ideal environment with temperature t. Since H and J are considered to be double sente, let (d-e), (f-g), (p-q), and (r-s) all be greater than 2t. Thus, each will be played before a play in the environment, and the same goes for H and J.

It doesn't matter which side is to play, the lesson is basically the same. So let Black play first.

1) Let Black play in H first.

1a) Let White reply in H. Then the result after an even number of plays will be

e + q.

1b) Let White reply in J. Then we are left with {d|e} + {r|s}. Depending on which play is larger, our result will be

d + s, or
e + r.

2) Let Black play in J first.

2a) Let White reply in J. The result will be

e + q.

2b) Let White reply in H. Then the result will be either

p + g, or
q + f.

The environment does not play a role in this decision, as the reader may verify. Also, the result will be the same, whichever position Black plays first in, if White replies in it. So we treat the double sente as gote, to decide between them.

The question then becomes that of finding

max(max(d+s,e+r), max(p+g,q+f))

Black plays in H first if
(
d + s ≥ p + g, i.e., if
d - g ≥ p - s,

and

d + s ≥ q + f, i.e.,
d - f ≥ q - s
)
(Back around 50 years ago, when I solved this question, I called d - f the forward value for Black of playing in H. (I was still in the swing value camp back then.) And I called q - s the backward value for Black of playing in J.)

or if
(
e - g ≥ p - r

and

e - f ≥ p - q
)

OC, e - f and p - q were traditionally called the value of double sente for each play. Note that choosing which to play based upon that value alone is not correct. However, you could start with that comparison. Without loss of generality, let e - f ≥ p - q . Then if the backward value for H is greater than or equal to the forward value for J, play in H.

I agree 100% with you Bill. I used the same calculation taking (b,n,w) instead of (d,e,f,g) which is exactly equivalent because you can always choose arbitrary the value of one of your four d,e,f,g values.
And this calculation allowed me to know which amongs two sente moves I have to play first. Then I used this correct result to compare to the result got by the estimation function.

The point is the following: if you have only two double sente areas on the board you can read all possibilities as you did and you will find for sure the correct order. Fine. But if you have 3, 4 or more double sente moves on the board it becomes extremely difficult to read all possibilities and my view is that it could very useful to have a tool guessing the correct order with a quite high probablity to be the best one. After having viewed the result of this sequence you may, with your experience, look for a better sequence but you will gain a lot of time by beginning with this good (not best) guessing. BTW if you know that the tool is quite a good one and if you manage to find a better one your are almost sure it is not necessary to search another better sequence (unless you are a pro OC).

[Bill Spight]

The point is the following: if you have only two double sente areas on the board you can read all possibilities as you did and you will find for sure the correct order. Fine. But if you have 3, 4 or more double sente moves on the board it becomes extremely difficult to read all possibilities and my view is that it could very useful to have a tool guessing the correct order with a quite high probablity to be the best one.

In this and your Thermography thread you have suggested that having more than one global double sente on the board at the same time is common. Well, it is not uncommon to have a double sente arise on the board, but it is usually answered immediately, and if not, soon, after a Zwischenzug exchange. It is not common for it to be left unanswered for another one that is not the result of said Zwischenzug to arise. What does sometimes happen is a shared blindspot, so that a double sente is left unrecognized on the board for some time. That even happens to pros, as the bots now tell us. Then another on may arise and you have two. It is even rarer to have two shared blind spots. As a rule, the time to handle the kind of position you are suggesting is earlier in the play. (Patient: "Doctor, it hurts when I do this." Doctor: "Don't do that.") )

[Gérard TAILLE]

The point is the following: if you have only two double sente areas on the board you can read all possibilities as you did and you will find for sure the correct order. Fine. But if you have 3, 4 or more double sente moves on the board it becomes extremely difficult to read all possibilities and my view is that it could very useful to have a tool guessing the correct order with a quite high probablity to be the best one.

In this and your Thermography thread you have suggested that having more than one global double sente on the board at the same time is common. Well, it is not uncommon to have a double sente arise on the board, but it is usually answered immediately, and if not, soon, after a Zwischenzug exchange. It is not common for it to be left unanswered while another one that is not the result of said Zwischenzug to arise. What does sometimes happen is a shared blindspot, so that a double sente is left unrecognized on the board for some time. That even happens to pros, as the bots now tell us. Then another on may arise and you have two. It is even rarer to have two shared blind spots. As a rule, the time to handle the kind of position you are suggesting is earlier in the play. (Patient: "Doctor, it hurts when I do this." Doctor: "Don't do that.") )

OK Bill you mean that the position you mentionned by the link https://www.lifein19x19.com/viewtopic.p ... 35#p194535 is quite unusual?

[Bill Spight]

OK Bill you mean that the position you mentionned by the link https://www.lifein19x19.com/viewtopic.p ... 35#p194535 is quite unusual?

The link is broken, BTW. Copying without quoting is probably the reason. It would also work to jump to that page and then copy the URL. Here, again, is the link.
https://www.lifein19x19.com/viewtopic.p ... 35#p194535

The Nihon Kiin position I am absolutely sure was constructed to illustrate the importance of playing double sente. Unfortunately, it did not also illustrate the importance of sometimes not replying locally to double sente. Probably because if you're not supposed to answer it, how can it be double sente?

As for the position in the Nogami-Shimamura book, I strongly suspect that it was originally constructed for the same purpose as the Nihon Kiin position, but the third variation, where the double sente is not answered locally, was inserted before publication. Maybe that variation came from actual play, but I kind of doubt it. It would be interesting if it did. It would also be interesting to see what the bots say about it.

[Bill Spight]

Forward values and backward values

Actually, I think I was on to something all those years ago. I was still in the swing value camp, where I took the value of a play in this simple gote, {a|b}, given a > b, as a - b. I also was aware that the best play was not always the largest play as traditionally calculated. I also became aware of the concept of the environment, which, as we know, I modeled as a set of simple gote. I took the value of the largest such gote as my standard for evaluation.

From that perspective, how to evaluate {d|e||f|g}, with d > e > f > g? Well we compare it with a simple gote, {a|b}, given a > b, in an environment of temperature, t, to modernize my thinking slightly. It does not affect the argument. OC, a-b > 2t, but also d-e > 2t and f-g > 2t. Otherwise it is not an environment. And it doesn't hurt to assign a number to a couple of variables. So let's evaluate D = {d|e||0|-g}, given g > 0, with A = {a|0}, with Black to play.

1) Black starts in A. The result is

a + 0 - t/2 = a - t/2.

2) Black starts in D, with sente. The result is

e + a - t/2, which is better for Black than 1), OC. But White does not have to answer in D. Let White play in A.

3) Black starts in D, White answers in A. The result is

d + 0 - t/2 = d - t/2.

So we compare a with d. a is the swing value of A, our standard of comparison. And d is the forward value of D.

OC, if d - e > a, then d > a, so the case of sente is covered. And for the case of gote, the forward value of D is easier to calculate than {d + e + f + g}/4. In fact, we do not have to worry about how to classify D.

In hindsight, we can see that I was actually anticipating the calculation of the sides of the thermograph at different temperatures in terms of the play in an ideal environment. Which is how I redefined thermography in 1998.

----

Whether to answer a double sente, given two of them, of the form {j|k||m|n}.

Let A = {a|0||-b|-c} and D = {d|0||-e|-f}, and let Black play in A to {a|0}. Should White answer?

Well, we already know the answer. We compare a with f, the swing value of Black's threat with the backward value of D (from Black's point of view).

----

OC, I haven't worked anything out in detail, at least not recently, but let's explore Gérard's hypothetical with several such plays where the temperature of the environment is low enough.

First, let's compare threats and find the largest one. Next, compare its swing value with the backward values of each of the others. If it is greater than all of them, it's a good bet that Black should make that threat.

And, as things get more complicated, let's not forget that traditional evaluation arose for a reason. The more double sente there are, the more likely it is that they act as a rich environment for each other, and the traditional evaluation becomes more relevant. And by traditional I do not mean the traditional double sente value, since with even two of them we cannot assume that one of them will be sente against the other, much less double sente.

Edit: Difference games.

Also, as things get more complicated, since difference games allow us to draw conclusions about play in every non-ko environment, they can be useful. OC, working them out on the table can be daunting, but I have done some of the work for us.

Let A = {a|0||-b|-c} and D = {d|0||-e|-f}, and let Black play in A to {a|0}. Should White answer?

Well, when there are only two of them with a sufficiently low ambient temperature, White should answer if a is at least as big as the backward value of D (for Black, that is). But what about non-ko environments? When is answering in A at least as good as playing in D, in every non-ko environment? When a is greater than or equal to both the backward value of D and the forward value of D. Vive la difference game!

[Bill Spight]

With this sample of 67 998 area couples I tested various estimation function in order to mesure the pourcentage of good guess reached by this function.
Here are my first results:

f(b, n, w) = n => pourcentageOK = 81,47%
f(b, n, w) = n + b => pourcentageOK = 81,94%
f(b, n, w) = n + b/2 => pourcentageOK = 87,17%
f(b, n, w) = n + b + w => pourcentageOK = 87,93%
f(b, n, w) = n + 0.5b + 0.4w => pourcentageOK = 94,54%

Any idea to improve the model? the sample of area couples? the function itself?
I can easily test other configurations if you are interested in of course.

It is interesting that n + 0.5b + 0.4w worked best, as that is almost the same as treating the double sente as gote and comparing swing values. The swing value of the gote being n + b/2 + w/2.

Edit: We can accommodate traditional theory this way, can't we?

if w ≥ 2n + b then f = 2n + b ; Black reverse sente or ambiguous
else if b ≥ 2n + w then f = b ; Black sente or ambiguous
else f = n + b/2 + w/2 ; gote

[Gérard TAILLE]

With this sample of 67 998 area couples I tested various estimation function in order to mesure the pourcentage of good guess reached by this function.
Here are my first results:

f(b, n, w) = n => pourcentageOK = 81,47%
f(b, n, w) = n + b => pourcentageOK = 81,94%
f(b, n, w) = n + b/2 => pourcentageOK = 87,17%
f(b, n, w) = n + b + w => pourcentageOK = 87,93%
f(b, n, w) = n + 0.5b + 0.4w => pourcentageOK = 94,54%

Any idea to improve the model? the sample of area couples? the function itself?
I can easily test other configurations if you are interested in of course.

It is interesting that n + 0.5b + 0.4w worked best, as that is almost the same as treating the double sente as gote and comparing swing values. The swing value of the gote being n + b/2 + w/2.

Edit: And since, to justify the double sente nature of the plays, we assume that n is quite small by comparison with b and w, and therefore that it is likely that the double sente is local sente for one side or the other, we might consider trying traditional sente values. I.e.,

2n + max(b,w)

My suspicion is that that won't work too well, statistically, since it tends to leave sente on the board which it would be fine to play.

Your idea 2n + max(b,w) is interesting but not as good on average than the swing value of the gote:
f(b, n, w) = 2n + max(b,w) => pourcentageOK = 89,68%
f(b, n, w) = n + 0.5b + 0.5w => pourcentageOK = 93,24%

[Bill Spight]

With this sample of 67 998 area couples I tested various estimation function in order to mesure the pourcentage of good guess reached by this function.
Here are my first results:

f(b, n, w) = n => pourcentageOK = 81,47%
f(b, n, w) = n + b => pourcentageOK = 81,94%
f(b, n, w) = n + b/2 => pourcentageOK = 87,17%
f(b, n, w) = n + b + w => pourcentageOK = 87,93%
f(b, n, w) = n + 0.5b + 0.4w => pourcentageOK = 94,54%

Any idea to improve the model? the sample of area couples? the function itself?
I can easily test other configurations if you are interested in of course.

It is interesting that n + 0.5b + 0.4w worked best, as that is almost the same as treating the double sente as gote and comparing swing values. The swing value of the gote being n + b/2 + w/2.

Edit: And since, to justify the double sente nature of the plays, we assume that n is quite small by comparison with b and w, and therefore that it is likely that the double sente is local sente for one side or the other, we might consider trying traditional sente values. I.e.,

2n + max(b,w)

My suspicion is that that won't work too well, statistically, since it tends to leave sente on the board which it would be fine to play.

Your idea 2n + max(b,w) is interesting but not as good on average than the swing value of the gote:
f(b, n, w) = 2n + max(b,w) => pourcentageOK = 89,68%
f(b, n, w) = n + 0.5b + 0.5w => pourcentageOK = 93,24%

How about traditional theory?

if w ≥ 2n + b then f = 2n + b ; Black reverse sente or ambiguous
else if b ≥ 2n + w then f = b ; Black sente or ambiguous
else f = n + b/2 + w/2 ; gote

[Gérard TAILLE]

I begin to understand why you claim double sente doesn't exist.
Let me take an example (it is certainly not the best one example but it is enough to explain the point)

$$B
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . . . . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |

Click Here To Show Diagram Code

[go]$$B
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . . . . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |[/go]

Let's look at point "a" with different hypothesis.

$$B
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . . 1 . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |

Click Here To Show Diagram Code

[go]$$B
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . . 1 . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |[/go]

After black the white territory is open in the right and as a consequence another black move at "a", though big, would not threaten great damage on white territory and it does not look double sente.

$$B
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . 2 1 . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |

Click Here To Show Diagram Code

[go]$$B
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . 2 1 . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |[/go]

After and the context is quite different. The white territory is now closed on the right and a black move at "a" would be a greater threat on white territory and we may call a black move at "a" a double sente move. Here is the point : instead of saying that "a" here is double sente move it may be better to say that white is simply sente, implying black answer at "a" and with this wording a double sente move doesn't exist.
More generally the so called double sente may be only the result of the previous move we can thus consider sente!

$$W
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . . 1 . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |

Click Here To Show Diagram Code

[go]$$W
$$ -----------------------------------------
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . a . . . . . . 1 . . . . . . |
$$ | . . . . X . O . . . . O . X . . . . . |
$$ | X X X X X X O . . . . O X X X X X X X |
$$ | . . . . . . O O O O O O . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |[/go]

We can see the same phenomenon here
When white plays closing the right side of her territory instead of saying that the point a is double sente we can just say that is sente and black should answers immediately at "a".

In that sense double sente doesn't exist. It is only a part of the flow of the game as an answer to a previous sente move.
I am sure you will able to find more relevant example showing this fact.

Is it your feeling concerning the non existence of double sente or do you have something else in mind?

[Bill Spight]

Is it your feeling concerning the non existence of double sente or do you have something else in mind?

Practical, accidental double sente certainly exist. and occur quite frequently. They may be illustrated thermographically when both sides of the thermograph are vertical at the same temperature.

However, theoretical, essential double sente do not exist, at least on a finite board. The point is well illustrated by the Nihon Kiin example and the Nogami-Shimamura book. If it is wrong to answer a double sente, how is it a double sente? Your statistical result on estimating the value of double sente moves says that evaluating them as gote moves works very well. (Edit: Rather better than the double sente value.) If so, why consider them to be double sente?

If a play is double sente, it is so only with respect to the rest of the board. It is not intrinsically double sente.

[Gérard TAILLE]

Is it your feeling concerning the non existence of double sente or do you have something else in mind?

Practical, accidental double sente certainly exist. and occur quite frequently. They may be illustrated thermographically when both sides of the thermograph are vertical at the same temperature.

However, theoretical, essential double sente do not exist, at least on a finite board. The point is well illustrated by the Nihon Kiin example and the Nogami-Shimamura book. If it is wrong to answer a double sente, how is it a double sente? Your statistical result on estimating the value of double sente moves says that evaluating them as gote moves works very well. If so, why consider them to be double sente?

If a play is double sente, it is so only with respect to the rest of the board. It is not intrinsically double sente.

Yes Bill I agree with you but how to adapt thermography to this fact?

$$B
$$ -----------------
$$ | . . . . . . O |
$$ | X X . . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------

Click Here To Show Diagram Code

[go]$$B
$$ -----------------
$$ | . . . . . . O |
$$ | X X . . . . O |
$$ | . X O O O O O |
$$ | . X X X O . . |
$$ | . . . X O O O |
$$ | . . . X X X X |
$$ | . . . . . . . |
$$ -----------------[/go]

at temperature t < 1 the two lines are vertical and the local area is seen double sente ... but local double sente does not exist!

I perfectly understand why thermography says it is double sente at temperature t < 1 but that is not the point.
If we are saying double sente does not exist and they may even be estimate as gote point (value n + b/2 + w/2) we have two problems with thermography:
1) the wording "double sente" in presence of two vertical lines must be clarified
2) two vertical lines does not really exist because eventually any so called double sente have an estimate value of n + b/2 + w/2 which may be far above t but not INFINITY.

OC nobody (including me) wants to change this beautiful thermography theory but how can we be consistant?

[Gérard TAILLE]

Your statistical result on estimating the value of double sente moves says that evaluating them as gote moves works very well. (Edit: Rather better than the double sente value.) If so, why consider them to be double sente?

Good news Bill.
By changing my model to a far bigger one, the best formula converges towards b + 0.5n + 0.5w
That means that the formula b + 0.5n + 0.4w was only the consequence of a too small model.
Isn't it an interesting result?

[RobertJasiek]

No local double sente is not a feeling but a proved theorem. See https://www.lifein19x19.com/viewtopic.p ... 33#p260633 for an example, for which you should calculate move value and follow-up move values. You can hardly get any closer to an alleged double sente, except in doubly ambiguous shapes.