Life In 19x19

Click Here To Show Diagram Code

[go]$$W White to play and win, no komi
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . e X O O O .
$$ . X X X X O O O X O .
$$ . X . O O r . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O b X . O .
$$ , . X X X X X O O O .
$$ . . X O O O a X X O .
$$ . . X X X X O O O O .
$$ . . X . O O c X . O .
$$ . . X X X X X O O O .
$$ . . . X O O d X . O .
$$ , . . X X X O O O O .
$$ . . . . . . . . . . .[/go]

Tryss got it right.

On the board we have:

r = {5 || -1 | -7}
a = {6 | -4}
b = {6 | -3}
c = {5 | -3}
d = {4 | -3}
e = {1 | 0}

The comparisons are these.

If
s > a
or
(s > a - d + e and t > a + e)
or
(s > a - b + c - d + e and t > a + c - d + e)

Then play in r

Else play in a.

Is 6 > 10 ? No.

Is 6 > 10 - 7 + 1 = 4 ? Yes.

Is 12 > 10 + 1 = 11 ? Yes.

The second condition is satisfied, so White should play in r.

Click Here To Show Diagram Code

[go]$$W White to play and win, no komi
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 6 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 1 . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O 3 X . O .
$$ , . X X X X X O O O .
$$ . . X O O O 2 X X O .
$$ . . X X X X O O O O .
$$ . . X . O O 4 X . O .
$$ . . X X X X X O O O .
$$ . . . X O O 5 X . O .
$$ , . . X X X O O O O .
$$ . . . . . . . . . . .[/go]

White wins by 1 pt., 13 to 12.

----

I decided to add two simple gote with score differences 9 and 8 (i.e., between 10 and 7) and to keep everything else the same, except for making the score for correct play come out -1. That kept the conditions for playing in r the same, because the comparisons that include the added gote never got triggered. The point being that White plays in r in each example in order to get the last play before the big temperature drop from 3½ to ½. This is really a last move problem.

OC, it need not be. I could have constructed an example with 5 gote so that the reasons for playing in r or a were not so clear. However, in these examples a play in a gains 5 pts. while a play in r gains only 4½ pts. Getting the last play before the temperature drop is the reason for taking the smaller initial gain.

Let's take a look at our little theory so far of how for White to play in r = {t | s || 0} or in a = {a | 0}, when the only other plays are in the simple gote, b = {b | 0}, c = {c | 0}, . . . , where the gote are labeled alphabetically in descending order of size.

For an odd number of gote we have this:

1) 1 gote

Play in r if s > a. (Else play in a understood.)

2) 3 gote

Play in r if (s > a)
or
((s > a - b + c) and (t > a + c)).

3) 5 gote

Play in r if (s > a)
or
((s > a - d + e) and (t > a + e))
or
((s > a - b + c - d + e) and (t > a + c - d + e)).

Now we can predict the conditions for 7 gote:

Play in r if (s > a)
or
((s > a - f + g) and (t > a + g))
or
((s > a - d + e - f + g) and (t > a + e - f + g))
or
((s > a - b + c - d + e - f + g) and (t > a + c - d + e - f + g)).

An so on for a larger odd number of gote.

For an even number of gote we have this:

1) 2 gote

Play in r if ((s > a - b) and (t > a)).

2) 4 gote

Play in r if ((s > a - d) and (t > a))
or
((s > a - b + c - d) and (t > a + c - d)).

We can predict the conditions for 6 gote.

Play in r if ((s > a - f) and (t > a))
or
((s > a - d + e - f) and (t > a + e - f))
or
((s > a - b + c - d + e - f) and (t > a + c - d + e - f)).

And so on.

-----

I have ordered the alternatives to start with the fewest operations. Note that each comparison with s gets easier to satisfy in each successive alternative, while the comparison with t gets harder to satisfy.

-----

Much later, edited for clarity.

So What?

OK, so we have this little theory about how to play with one position of the form, {t | s || 0}, and some number of simple positions of the form, {g | -g}. So what? It's a lot to remember for only one type of whole board position, and it's abstract. Even if it saves some reading, what does it have to do with playing go?

Well, remember I derived this little theory to show that it was possible to have a theory of endgame play that does not rely upon the concepts of gote, sente, and reverse sente. (I slipped and called some positions gote, My bad. ) I misremembered Tormanen's email and thought that in his new book he did without all three terms, but he uses sente, it's the other two he does not use. Let's introduce those concepts into the theory.

Let's start with positions of the form, {g | -g}. These are gote, and each has an average territorial value of 0. Other positions of the form, {g+m | -g+m}, are also gote, with an average territorial value of m. We may think of them as the sum, {g | -g} + m. The form, {g | -g}, is the prototype of simple gote, and all simple gote may be reduced to the sum, {g | -g} + m. BTW, m may be any number, while g ≥ 0. A play by either player in such a simple gote gains, on average, g.

For instance, {6 | -3} = {4.5 | -4.5} + 1.5. It has an average territorial value of 1.5 and an average gain for each player of 4.5. One basic characteristic of all gote, not just this prototype, is that the average gain for each player is the same.

BTW, is it better pedagogically, to start with the algebra, or to start with the arithmetic? My tendency has been to start with actual go positions, then apply the arithmetic to them, then generalize with the algebra. But that is for a general audience. OC, if you are reading this note, you are familiar with the concept of gote and you are not afraid of algebra.

Let's leave aside more complex positions for now and suppose that the whole board consists of a number of simple gote with average gains, a > b > c > .... We know that correct play is to take the gote with the largest average gain. If Black plays first the total average gain, TB, is this.

1) TB = a - b + c - d + . . .

OTOH, if White plays first, we have

2) TW = -a + b - c + d . . .

The difference in the total average gains between the two scenarios is this.

3) TB - TW = 2a - 2b + 2c - 2d + . . .

Don't tell anyone, but 2a, 2b, etc., may be considered as deiri values. But in the theory we simply consider them as differences between local scores. Now, when writing out the theory I represented the local score differences as a, b, c, etc., so for instance we have the comparison, s > a - b + c - d + e. But the idea is the same. s is greater than the score difference with 5 simple gote when Black plays first versus when White plays first.

Taking a break. Reverse sente comes next.

"We know that correct play is to take the gote with the largest average gain."

What is your take on a proof a) without environment, b) with a non-ko environment?

RobertJasiek wrote:"We know that correct play is to take the gote with the largest average gain."

What is your take on a proof a) without environment, b) with a non-ko environment?

The same proof using a difference game can apply to both. As far as kos are concerned a miai pair of simple gote with equal gains can act as a ko threat. Without kos or potential kos we can ignore them.

One more note about simple gote

For values of sequences of simple gote I use the following notation.

a' = a - b + c - . . .

This notation allows us to condense the theory as follows.
For an odd number of gote we have this:

1) 1 gote

Play in r if s > a. (Else play in a understood.)

2) 3 gote

Play in r if (s > a) or ((s > a') and (t > a + c)).

3) 5 gote

Play in r if (s > a) or ((s > a - d') and (t > a + e)) or ((s > a') and (t > a + c')).

4) 7 gote:

Play in r if (s > a) or ((s > a - f') and (t > a + g)) or ((s > a - d') and (t > a + e')) or ((s > a') and (t > a + c')).

An so on for a larger odd number of gote.

For an even number of gote we have this:

1) 2 gote

Play in r if ((s > a') and (t > a)).

2) 4 gote

Play in r if ((s > a - d) and (t > a)) or ((s > a') and (t > a + c')).

3) 6 gote.

Play in r if ((s > a - f) and (t > a)) or ((s > a - d') and (t > a + e')) or ((s > a') and (t > a + c')).

And so on.

Note that with this notation some of the conditions remain the same with different numbers of simple gote on the board, even numbers with different parity.

What about reverse sente?

We can write a simple potential reverse sente, r, for White as

r = {z | 0 || -s}, z > 0, s ≥ 0

If z > 2s, r is a reverse sente for White. It has a count of 0. We can add or subtract a constant to all results and change only the count. So

{z+m | m || m-s} = {z | 0 || -s} + m

The average gain for White for playing the reverse sente is s.

If 2s > z, then r is a gote. It has a count of (z-2s)/4. The average gain for either player is (z+2s)/4.

Such a value does not appear in our little theory, for a very simple reason. The theory is concerned with exact results, not averages. (I think that making this clear to people would nip in the bud the idea that it is always right to take the largest play. Taking the largest play is simply playing the averages. It does not guarantee correct play. )

Sente gains nothing

The saying goes that sente gains nothing. That is because, if a player makes a play with sente, he gains nothing more than he ends up with. And, because the play is sente, that is what he is entitled to when it is his turn to play.

Global sente

Now, the above test for whether r = {z | 0 || -s} is a sente for Black is that z > 2s, It makes no reference to other plays and it defines a local or intrinsic characteristic of r. It determines the local count and helps us to play the averages. But our little theory is not concerned with playing the averages, it wants exact results. r is global sente when, if Black plays to {z | 0}, White has nothing better, in terms of the score, than to reply to 0.

On this board with r and some number of simple gote of the form {a | 0}, {b | 0}, . . . , r is global sente for Black when z ≥ a. Our question is whether White should play in r or in a. If White plays in a, that leaves a board with r and zero or more simple gote. With no gote left, r is of course a global sente for Black. In our example, r = {5 | -1 || -7} is local gote with a count of -2.5. But if r is a global sente we may regard it as having a count of -1, and White's reverse sente gains 6 pts. instead of the average gain of 4.5.

Suppose that z ≥ b, which makes r a global Black sente at that point. Then when White takes a, Black will play sente and gain nothing, and then the rest of the board will be played out. The total gain for White will be this:

1) T1 = a'/2 ; Remember that in {g | 0} each player gains g/2

OTOH, if White plays in r to -s, the total gain for White will be

2) T2 = s - a'/2

White should play in r if

2) T2 > T1 <=> s - a'/2 > a'/2 <=> s > a'

This test appears when we have two or more gote. (Note: It applies in all of them if we let a' apply to a single gote. OC, with only one gote we may regard r as a global sente if White takes a. )

For 2 gote the rule is this:

Play in r if ((s > a') and (t > a)).

Let's do some rewrites.

Play in r if ((s > a - b) and (s + z > a)).

Play in r if ((s > a - b) and (s > a - z)).

If z ≥ b then a - b ≥ a - z and the rule reduces to

Play in r if ((s > a').

And that rule applies to all cases with two or more gote.

Well, if b > z then r is a global gote. However, we can still apply the idea of comparing z with the values of local gote. Doing so gives the theory is this form:

Play in r if

((z ≥ b) and (s > a')) OR

((z ≥ c) and (s + z > a + c')) OR

((z ≥ d) and (s > a - d')) OR

* * * *

This form is simpler and easier to comprehend. In fact, it was the first form of the theory that I came up with in the 1970s. But back then I was not constrained to avoid the concepts of gote and reverse sente.

N.B. When considering whether to play r as a reverse global sente, we do not worry about whether it is so now, but whether it will be on Black's next turn.

----

Estimation

Suppose that r is not a global reverse sente and there are so many gote that we do not know how many there are and z is so small that we do not know which comparison to make. In that case we can estimate our comparison as this:

s >? a - z'

That is some help, but we still have to estimate z'. It has a maximum value of z and a minimum value of 0. The estimate that minimizes its maximum error is z/2. That gives us this comparison.

s + z/2 >? a

I.e, we are comparing the swing value of r as a gote with the swing value of the largest simple gote. Back to playing the averages.

What about possible sente?

Let the positions on the board be of the form, r = {z | 0 || -s}, and some number of simple gote, ordered as before, with Black to play. Then our little theory is this.

Black should play in r if

((z ≥ b) and (s + z > a + b')) OR
((z ≥ c) and (s > a - c')) OR
((z ≥ d) and (s + z > a + d')) OR
((z ≥ e) and (s > a - e')) OR
* * * *

This is slightly different from the little theory I derived back in the 1970s, in that there is no test for global sente, i.e., (z ≥ a). The idea was that Black should play in r if

((z ≥ a) and (s > b'))

That condition is not lost because if it is satisfied then ((s + z > a + b') and (z ≥ b)), so the first condition in the present theory is satisfied.

I found the (s > b') condition interesting for the following reason. Our estimate of b' is b/2, which is the average gain for {b | 0}. s is the average gain for the reverse sente. Now, both traditional go theory and combinatorial game theory identify the urgency of a sente play by the size of the reverse sente. A so-called 3 pt. sente is one where the reverse sente gains 3 pts. on average. OC, the sente may be played earlier, but in theory it is OK to wait until s > a/2. Well, that's the standard theory, but the (s > b') condition says don't wait quite that long. (OC, with no ko or potential ko nothing is lost by playing the sente early, as long as doing so is not aji keshi. But that's another story.) The 1970s theory says the proper comparison is with b, not a. The modern theory says that, z, the score difference of Black's threat, also matters: (s + z > a + b') will often be satisfied when (b' ≥ s), when z is so much larger than a. The condition may be considered a compromise between (z ≥ a), i.e., play sente early, and (s > b'), i.e., play sente late.

Hi Bill,

How to score ?

Click Here To Show Diagram Code

[go]$$B 5.5 komi, zero prisoners
$$ ---------------------
$$ | . X O O . X X X . |
$$ | X . X O O O X X X |
$$ | X X X X X O X X X |
$$ | O O O O X O O O O |
$$ | . O . O X O O O O |
$$ | X X X O X O X X . |
$$ ---------------------[/go]

EdLee wrote:Hi Bill,

How to score ?

$$B 5.5 komi, zero prisoners
$$ ---------------------
$$ | . X O O . X X X . |
$$ | X . X O O O X X X |
$$ | X X X X X O X X X |
$$ | O O O O X O O O O |
$$ | . O . O X O O O O |
$$ | X X X O X O X X . |
$$ ---------------------

Click Here To Show Diagram Code

[go]$$B 5.5 komi, zero prisoners
$$ ---------------------
$$ | . X O O . X X X . |
$$ | X . X O O O X X X |
$$ | X X X X X O X X X |
$$ | O O O O X O O O O |
$$ | . O . O X O O O O |
$$ | X X X O X O X X . |
$$ ---------------------[/go]

seems B+0.5 by area scoring (B 28, W 27.5).



prior 9x9 (trying to fit certain size constraints)

$$B 0.5 komi, zero prisoners
$$ ---------------------
$$ | X X X X X X X . O |
$$ | X X X X X X X O O |
$$ | X X X X X X X O O |
$$ | O O O X X . X O O |
$$ | . O . O X O X O . |
$$ | O O O O X O X O O |
$$ | X X X X O X X X X |
$$ | . X . X O O O O O |
$$ | O O O X O . O . O |
$$ ---------------------

Click Here To Show Diagram Code

[go]$$B 0.5 komi, zero prisoners
$$ ---------------------
$$ | X X X X X X X . O |
$$ | X X X X X X X O O |
$$ | X X X X X X X O O |
$$ | O O O X X . X O O |
$$ | . O . O X O X O . |
$$ | O O O O X O X O O |
$$ | X X X X O X X X X |
$$ | . X . X O O O O O |
$$ | O O O X O . O . O |
$$ ---------------------[/go]

seems B+0.5 by area scoring (B 39, W 38.5).

By traditional territory scoring with a group tax, I get 0 for both boards (sans komi).

Hi Bill,

Thanks. I'm not familiar with the group tax or its history ( or when it was eliminated in Japan... by Sai's era? ). Without it, how does territory scoring handle the LL corner ? my counts:

9x6: B: 2 (or 3?); W 0, before komi. same winner as area scoring.
9x9: W: 4 (or 5?); B 0, before komi. different winner v. area scoring.

trying 'traditional territory with group tax':
9x6: B = (2-2)+(-2)+(-2)+(-1) = -5 before komi.
W = (-2)+(-2) = -4 before komi.
I don't know how to score or tax the 15-stone group here.
Maybe I completely misunderstood sensei's page on group tax.

EdLee wrote:Hi Bill,

Thanks. I'm not familiar with the group tax or its history ( or when it was eliminated in Japan... by Sai's era? ).

I'm not familiar with its history, either. But apparently the oldest game records in China for which we have scores are consistent with territory scoring with a group tax. Before I was aware of that, I worked out that the form of scoring that is closest to modern combinatorial game theory, is territory scoring with a group tax.

If you think of a point of territory as a point on which you can play a stone without danger (as in the capture game) then you cannot afford to fill one of the two eyes necessary for the life of a group, so from the total number of empty points which you control you have to subtract the number that you cannot fill. That's like a tax on the territory you count for the group, hence the term, group tax. There is a similar argument for stone scoring.

Without it, how does territory scoring handle the LL corner ?

Despite the removable ko threats, Japanese rules still regard those Bent Four positions as dead. I think Korean rules do, too.

Lasker-Maas rules and Spight rules allow the ko fight to be played. I assume that traditional territory scoring with a group tax in China did, as well, since there is no evidence of any Japanese-style scoring of seki in China, to the best of my knowledge. Anyway, allowing the ko fight to be played means that, with best play, it will not be played.

Hi Bill, Thanks.

Despite the removable ko threats, Japanese rules still regard those Bent Four positions as dead.

Thanks. Removable and unremovable ko threats are also unclear for me (need to study sensei's).
But here, the LL is seki -- I don't know how the 'Japanese rules' square this circle.

( still trying to squeeze to a certain size. )

Click Here To Show Diagram Code

[go]$$B 5.5 komi, zero prisoners
$$ -------------------
$$ | . X O O . X X . |
$$ | X . X O O O X X |
$$ | X X X X X O X X |
$$ | O O O O X O O O |
$$ | . O . O X O X O |
$$ | X X X O X O X . |
$$ -------------------[/go]

Bill Spight wrote:

EdLee wrote:Hi Bill,

How to score ?

$$B 5.5 komi, zero prisoners
$$ ---------------------
$$ | . X O O . X X X . |
$$ | X . X O O O X X X |
$$ | X X X X X O X X X |
$$ | O O O O X O O O O |
$$ | . O . O X O O O O |
$$ | X X X O X O X X . |
$$ ---------------------

Click Here To Show Diagram Code

[go]$$B 5.5 komi, zero prisoners
$$ ---------------------
$$ | . X O O . X X X . |
$$ | X . X O O O X X X |
$$ | X X X X X O X X X |
$$ | O O O O X O O O O |
$$ | . O . O X O O O O |
$$ | X X X O X O X X . |
$$ ---------------------[/go]

If the ko fight for the Bent Four is allowed, neither player will make a play with correct play. Result: 0.

This depends upon how the Bent Four position was treated, something that we do not really know.

If the Bent Four is dead, White can capture the two Black stones in the bottom right, since he does not need a ko threat. That means that White has 5 pts. of territory in the seki to 1 pt. for Black, net 4 pts. for White. (The way that modern stone scoring in the 19th and 20th centuries applied the group tax was different, but they applied it to both groups in seki, so the net result was the same.)

Click Here To Show Diagram Code

[go]$$W Bent Four dead
$$ ---------------------
$$ | . X O O . X X X . |
$$ | X . X O O O X X X |
$$ | X X X X X O X X X |
$$ | W W W W X O O O O |
$$ | . W . W X O O O O |
$$ | X X X W X O B B C |
$$ ---------------------[/go]

Here is the latest board for no pass go with prisoner return.

Click Here To Show Diagram Code

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

Click Here To Show Diagram Code

[go]$$B Black first. ii
$$ -------------------
$$ | . X O O 8 X X . |
$$ | X . X O O O X X |
$$ | X X X X X O X X |
$$ | O O O O X O O O |
$$ | 5 O O O X O X O |
$$ | 4 3 7 O X O X 6 |
$$ -------------------[/go]

Result after an even number of plays: Black has captured 9 White stones and has 9 pts. of territory in which she can play. Black score: 18.

White has captured 6 Black stones and has 13 additional pts., 6 dead stones and 7 pts. in which to play. White score: 19.

Net result for Black: -1

Click Here To Show Diagram Code

[go]$$W White first
$$ -------------------
$$ | . X O O . X X . |
$$ | X . X O O O X X |
$$ | X X X X X O X X |
$$ | O O O O X O O O |
$$ | 4 O 5 O X O X O |
$$ | X X X O X O X 1 |
$$ -------------------[/go]

throws in, captures the throw-in stone

But then starts the Bent Four ko.

Click Here To Show Diagram Code

[go]$$W White first, ii
$$ -------------------
$$ | . X O O . X X . |
$$ | X . X O O O X X |
$$ | X X X X X O X X |
$$ | O O O O X O O O |
$$ | 8 O O O X O O O |
$$ | 7 6 0 O X O . O |
$$ -------------------[/go]

returns one prisoner

After an even number of moves Black has 18 pts., as above. White has 6, for the six remaining prisoners. Net result: +12

Whoever plays first loses, so this is a jigo.

----

With komi, instead of playing on the board White can reduce the komi by 1 pt. and win the game.

Edit: Correction. With 5½ komi White can reduce the komi to 5 pts. and win the game. Black can increase the komi to 6 pts., after which White can reduce it to 5 pts. and win.