All of these values are averages, useful for heuristic purposes. Fortunately, they work the vast majority of the time in go.



Right. Since this ko threat costs one point to remove, it is a "-1 pt. sente". By itself it is worth nothing. However, in some ko fights it is possible that Black will play this threat and White will ignore it. That's unlikely, since the threat is so huge, but there are ko fights where it will still affect the analysis because it will effectively cancel one of the opponent's threats. There are also ko situations where it is right for White to eliminate the threat before the potential fight. We can now analyze all of these situations with thermography.



Formally we consider the temperature of an ideal environment. Reality may be different. When we combine games, such as kos and ko threats, we still make use of an ideal environment unless we can read things out, but the analysis can be rather complex.

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— Winona Adkins

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Let me give an example of thermography for a simple ko with your ko threat. It's a large threat, so let's look at a ko with large moves where that is the only threat. Suppose that White to play takes the ko. If Black does not play the threat and White wins the ko the local score in the ko is -41 pts. (for Black). Subtracting the 4 pts. in the unplayed threat the net score is -45 pts. Or if Black to play wins the ko for 4 pts., the net score is 0. Each play in the ko gains 15 pts. on average. That makes the ko a little hotter than the empty go board. Such kos do arise in actual games.

When I was learning go the textbooks said that the size of the ko threat was bigger than the ko, i.e., that 19 > 15 (although they doubled those numbers), so White should answer the ko threat. Thermography tells a somewhat different story. Suppose that White takes the ko, Black plays the threat, White wins the ko, and then Black kills the corner. The net score will be -7, which indeed is worse for White than the average value of the ko plus the White corner, which is -15. White loses 8 pts, on average; how can that be right?

Well, suppose that White does answer the ko threat and Black takes the ko back. At this point White plays in the environment and gains t, and then Black wins the ko. The net result at temperature t will be 0 - t = -t.

White's strategy should be to get min(-7,-t), for t up to 15 (above which the ko is not fought). Solving for t we get

-t = -7
t = 7

So when the temperature of the environment is less than 7 White should ignore the ko threat and win the ko.

OC, that is a heuristic. But if we know more about alternative plays, we can add them to the game and analyze that. Besides, the textbook advice to simply compare the size of the ko play with the size of the threat is a heuristic, as well. Just not as good a heuristic.

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

Visualize whirled peas.

Everything with love. Stay safe.

Ok for me. Seeing 19 > 15 it appears to me that white must answer to the ko threat.


Nothing wrong with that. White gains 2 * 15 but black gains 2 * 19. As a consequence, White loses 8 points in this sequence.





Here I do not understand your calculation.
Notation :
x = value of move in the ko (here x = 15)
y = value of a move in the ko threat (here y = 19)
t = temperature of the environment (t =14 on an empty board)

To simplying the reasonning let's take two gobans:
a first goban G1 on which it remains only two areas : the ko area itself (white can take the ko or black can connect the ko) and the ko threat area
a second goban G2 which is empty.

First of all, in an ideal environment, a player will play in the ko only if x >= t, and ideally when x = t.
If white takes the ko and ignores the ko threat then, on G1, black score is -2x + 2y and then white plays first on the G2 and black loses the game by t/2 points (7 points which is the supposed ideal komi). Finally black score in this first scenario is b1 = -2x + 2y - t/2
Suppose now white answers to the ko threat and black retakes the ko and then connects. On G1 black score is x points and on G2 white plays two moves before black plays its first move. Because with a pass move black loses t points, that means that, on G2, black loses the game by 3t/2 points. Finally black score in this second scenario is b2 = x - 3t/2.

Comparing b1 and b2, white has to ignore the black ko threat if
b1 < b2 => -2x + 2y - t/2 < x - 3t/2 => 2y < 3x - t
and taking into account x = t :
b1 < b2 => 2y < 2x => y < x
and I confirm the common sense : you do not answer the ko threat if the value of the ko threat (y = 19) is less if then the value of the ko (x = 15).

In an ideal environment (obviously we would avoid an evironment with only one big gote point or an environment with complete miai points) I do not see why you have to take into account the temperature in order to know if you have to answer a ko threat. Comparing x and y seems enough isn't it?

Where is the difference between our two approches : you expect to win t points when playing in the environment where I consider you will win only t/2 points. The point is to be well aware of the difference between the value of a move (temperature) and the score expected for the game (komi = t/2).

Maybe I am wrong but, for the time beeing, I like the common sense result of my calculation.

IIUC, you are estimating the final minimax result at the end of the game. That works, too. It's how I started out years ago.



OK. First suppose that White makes a mistake and plays in the environment, and then Black makes a mistake and wins the ko, and then play continues in the environment. Both plays in the environment gain approximately 14 pts., so we estimate the final score as

1) -15 - 14 + 15 - 14/2 = -21

Now suppose that White takes the ko and ignores Black's threat. We estimate the final score this way:

2) -15 - 15 + 19 - 15 + 19 - 14/2 = -14

Now suppose that White answers Black's threat and Black takes and wins the ko. We estimate the final score this way:

3) -15 - 15 + 19 - 19 + 15 - 14 + 15 - 14/2 = -21

Plainly White should answer the ko threat, given that we know nothing else about the environment.

Now let play continue in the environment until play ends as expected with a score there of -7, and there are no additional ko threats. t = 0. At this point let White take the ko and Black play the threat.

Suppose that White answers the threat and Black takes and wins the ko. Then the final score will be this.

4) (-7 - 15) - 15 + 19 - 19 + 15 - 0 + 15 = -7

That's the best result for Black so far. Suppose instead that White does not answer the threat but wins the ko. The final score will be this.

5) (-7 - 15) -15 + 19 - 15 + 19 = -14

This is better for White by 7 pts. So when t = 0 White's best play is different from best play when t = 14, as a rule. White should as a rule answer the threat at a high ambient temperature but ignore it at a low temperature. Where is the crossover point?

Suppose now that play continues in the environment until t = 7, with no additional ko threats. We estimate that White has gained (14 - 7)/2 = 3½ pts.

First, suppose that White takes the ko and answers Black's ko threat and Black takes and wins the ko. This is our estimate of the final score.

6) (-3½ - 15) - 15 + 19 - 19 + 15 - 7 + 15 - 7/2 = -14

Next, suppose that White takes the ko and ignores Black's ko threat. This is our estimate of the final score.

7) (-3½ - 15) - 15 + 19 - 15 + 19 - 7/2 = -14

The result is the same, so we have found our crossover point. Below an ambient temperature of 7 White should ignore Black's ko threat, as a rule. QED.

Note also that Black gains when White does not play the ko fight, so White should play the ko fight before the ambient temperature drops. As a rule, OC.

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

Visualize whirled peas.

Everything with love. Stay safe.

Good news Bill, I think we are completly in line!

With your last post I now understand where is our previous misunderstanding

When introducing the ko threat problem you wrote:



When reading the last sentence I understood you started the ko as soon as the value of a move in the ko (here 15) was just a little hotter than the temperature of the environmment (here 14).In this hypothesis I explained (remember I supposed x = t) you always has to answer a ko threat if this ko threat is greater than the ko (19 > 15).
I understand now that your hypothesis was different. You do not analyse a ko with the value of the ko just a little hooter than the temperature but also in the case where the value of the ko move (15) is far highter than the temperature (7 in your analysis).

Taking my calculation with your hypothesis it remains
b1 < b2 => -2x + 2y - t/2 < x - 3t/2 => 2y < 3x - t

but now I cannot put x = t.
I that case:
b1 < b2 => t < 3x - 2y
if x = 15 and y = 19 then b1 < b2 => t < 7 which is exactly your result.

When you look at the formula
t < 3x - 2y
you see that you have to compare the temperature to the global value of the ko (3x) minus the global value of the ko threat (2y) and that does not hurt my common sense though it seems not a known rule.

Do you agree with that?

Yes. Thermography is not just a tool to discover the average value of a position or combination of positions, and how much a play gains, but also to discover different strategies at different ambient temperatures.



I discovered that rule and others in the early 1970s, but as a mere amateur shodan I had no thought of publishing anything. This and other rules are derived in the 1996 Mueller, Berlekamp, Spight paper: http://citeseerx.ist.psu.edu/viewdoc/su ... .1.34.6699

_________________
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 test if you agree with my calculation in order to know when starting a ko in an ideal environment.

Let's suppose that the environment is made of pure gote areas with values equal to g1, g2, g3, g4, ...
Because I suppose an ideal environmment I expect
g1 - g2 + g3 - g4 ... = g1 / 2
g2 - g3 + g4 - g5 ... = g2 / 2
etc.

if x is the value of a ko move then
1) In absence of ko threat you start the ko when 3x >= 2g1 + g2
2) If your opponent has one ko threat you start the ko a little earlier when 3x >= 2g1 + g4
3) If your opponent has two ko threats you start the ko still a little earlier when 3x >= 2g1 + g6
4) If your opponent has a lot of ko threats you can even start the ko as soon as 3x >= 2g1

Note : if you are not in the endgame phase and if the value of the ko is equal to the temperature then, you may consider g1 = g2 = g3 = g4 ...,
the number of ko threat is irrelevant and there are no stress at all for this ko. It's equivalent to play in the ko or to play in the environment. In this case keep all your threats for another occasion and play where you want, in the ko or the environment.

Bravo!

This was basically my first result, many moons ago. Note that each ko threat must be so large that it must be answered.

Technically, this result does not fit the model of thermography, because each gote and each threat that you take into account actually adds to the game with the ko. They are not considered part of the environment. (Edit: In thermography the right side is always 2g1 + t.) However, this result can be made more abstract to produce what Professor Berlekamp dubbed pseudo-thermography. A different form of pseudo-thermography was developed by Kim Yonghoan.

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

Visualize whirled peas.

Everything with love. Stay safe.