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 Post subject: Re: This 'n' that
Post #561 Posted: Tue Apr 30, 2019 3:50 am 
Oza
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I understand everything you write, except the "suppose that "b" was {8 | -3 || -4}"

The CGT means Black plays two moves and gets 8, Black plays White replies -3, White plays -4. Where in the diagram are these moves?
Or is {8 | -3 || -4} some kind of reversal of { 4 | -3 || -8}, from White's perspective? Then shouldn't -3 be 3?

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 Post subject: Re: This 'n' that
Post #562 Posted: Tue Apr 30, 2019 6:44 am 
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Bill Spight wrote:
In his new book, Rational Endgame ( viewtopic.php?f=17&t=16567 ), Antti Tormanen does without the terms, sente and gote.

For better or worse, he does not do without the term sente; according to my PDF search the word is found on 40 of the book's 126 pages. I do not see gote, though.

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 Post subject: Re: This 'n' that
Post #563 Posted: Tue Apr 30, 2019 7:13 am 
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dfan wrote:
Bill Spight wrote:
In his new book, Rational Endgame ( viewtopic.php?f=17&t=16567 ), Antti Tormanen does without the terms, sente and gote.

For better or worse, he does not do without the term sente; according to my PDF search the word is found on 40 of the book's 126 pages. I do not see gote, though.


Thanks for the correction. Sorry, I misremembered his and my email exchange. It is reverse sente that he leaves out. If he uses sente, then he can simply regard non-sente plays and sequences as gain making without labeling them further. And if he labels some positions as sente, then the others are simply unlabeled non-sente positions.

Edit: I wonder how he defines sente, however. Even CGT, which uses (or used to use) the terms, equitable and excitable, regards equitable positions as more basic. Without gote positions and plays you do not have sente positions and plays.

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Last edited by Bill Spight on Tue Apr 30, 2019 7:45 am, edited 1 time in total.
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 Post subject: Re: This 'n' that
Post #564 Posted: Tue Apr 30, 2019 7:29 am 
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Knotwilg wrote:
I understand everything you write, except the "suppose that "b" was {8 | -3 || -4}"

The CGT means Black plays two moves and gets 8, Black plays White replies -3, White plays -4. Where in the diagram are these moves?
Or is {8 | -3 || -4} some kind of reversal of { 4 | -3 || -8}, from White's perspective? Then shouldn't -3 be 3?


Sorry for not being clear. :(

{8 | -3 || -4} is the same as {{8 | -3} | -4}. White to play moves to -4; Black to play moves to {8 | -3}, which, I trust, makes sense.

If you reverse the colors of the stones, {4 | -3 || -8} becomes { 8 || 3 | -4}. The ||s indicate non-scored positions with followers to the left for Black options and to the right for White options. { | -3} represents a position where White can move to a positions worth -3, while Black has no move. We don't really need to represent such positions. One possibility would be a completely surrounded White group with 4 one point eyes. White to play can fill one of the eyes and Black has no play.

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 Post subject: Re: This 'n' that
Post #565 Posted: Tue Apr 30, 2019 8:01 am 
Oza
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Bill Spight wrote:
{8 | -3 || -4} is the same as {{8 | -3} | -4}.


That I understand :)

Bill Spight wrote:
White to play moves to -4; Black to play moves to {8 | -3}, which, I trust, makes sense.


It's that bit, which, unfortunately, doesn't. Let me show the diagram for "b" again.

Click Here To Show Diagram Code
[go]$$B b diagram
$$ . . . X X X X O O O O O O . .
$$ . . . X O O b . X X X . O . .
$$ . . X X X X O O O O O O O . .
$$ . . X O O O . X . O . . . . .
$$ . . X . X X X O O O . . . . .
$$ . . X X X . . . . . . . . . .[/go]


A White play here gives White 8 points, so it "moves" to -8. A Black play moves to 4 | (4-7)=-3
But we had that already.

I just don't get what "suppose b is {8 | -3 || -4}" means in relation to diagram "b". I see no White move leading to 4 points, nor a Black move leading to a {8|-3} postion. I must be blind :(

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 Post subject: Re: This 'n' that
Post #566 Posted: Tue Apr 30, 2019 8:16 am 
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Knotwilg wrote:
Bill Spight wrote:
{8 | -3 || -4} is the same as {{8 | -3} | -4}.


That I understand :)

Bill Spight wrote:
White to play moves to -4; Black to play moves to {8 | -3}, which, I trust, makes sense.


It's that bit, which, unfortunately, doesn't. Let me show the diagram for "b" again.

Click Here To Show Diagram Code
[go]$$B b diagram
$$ . . . X X X X O O O O O O . .
$$ . . . X O O b . X X X . O . .
$$ . . X X X X O O O O O O O . .
$$ . . X O O O . X . O . . . . .
$$ . . X . X X X O O O . . . . .
$$ . . X X X . . . . . . . . . .[/go]


A White play here gives White 8 points, so it "moves" to -8. A Black play moves to 4 | (4-7)=-3
But we had that already.

I just don't get what "suppose b is {8 | -3 || -4}" means in relation to diagram "b". I see no White move leading to 4 points, nor a Black move leading to a {8|-3} postion. I must be blind :(


Sorry. I meant, suppose that b were a different position with those values, not the one in the diagram. The comparison would still be 12 pts. vs. 10 pts.

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 Post subject: Re: This 'n' that
Post #567 Posted: Thu May 02, 2019 2:09 am 
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Here is another small problem with a position we do not call reverse sente. ;)

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . . X O O O .
$$ . X X X X O O O X O .
$$ . X . O O . . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O . . O . .
$$ , . X X X X X O O O .
$$ . . X O O O . X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


No komi. All play and scoring is only in corridors. The framing stones are alive.

Enjoy. :)

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Last edited by Bill Spight on Thu May 02, 2019 11:35 am, edited 1 time in total.
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 Post subject: Re: This 'n' that
Post #568 Posted: Thu May 02, 2019 4:27 am 
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I know how to play and win, but not based on CGT
1|0 = 1
5|-1||-7 = (5+1)/2 + 7 = 10
6|-1 = 7
6|-4 = 10

I can't distinguish between b or d here, while I can see I need b to win.
b = -7
d = 6
c = -1
a = 1


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 Post subject: Re: This 'n' that
Post #569 Posted: Thu May 02, 2019 8:50 am 
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Knotwilg wrote:
I know how to play and win, but not based on CGT
1|0 = 1
5|-1||-7 = (5+1)/2 + 7 = 10
6|-1 = 7
6|-4 = 10

I can't distinguish between b or d here, while I can see I need b to win.
b = -7
d = 6
c = -1
a = 1


The little theory I am illustrating could be considered part of CGT. These corridors are combinatorial games, and it is a theory. ;) I suppose that your final figures are the result of reading out the board with alternating play.

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 Post subject: Re: This 'n' that
Post #570 Posted: Thu May 02, 2019 10:58 am 
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New to this, but trying out the formula you gave:

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . a X O O O .
$$ . X X X X O O O X O .
$$ . X . O O b . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O c . O . .
$$ , . X X X X X O O O .
$$ . . X O O O d X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


a: {1 | 0} => 1 (1-0)
b: {5 | -1 || -7} => 12 (5 - (-7))
c: {6 | -1} => 7 (6 - (-1))
d: {6 | -4} => 10

So the order to play is in order of those values: b, d, c, a.

That'd give:
Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 4 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O 2 X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


So black gets 7 more points from the outside, white gets 8.

But doesn't white still lose? White has 4 points on the outside already, and black has 8 points already, so white needed to get 5 more points. From the whole board, it's black +3?

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , C C X X X X C C C .
$$ . C C X . 4 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 1 . X X O .
$$ . X X X X X O O O O .
$$ . C X O O O 3 . O C .
$$ , C X X X X X O O O .
$$ . C X O O O 2 X X O .
$$ . C X X X X O O O O .
$$ . . . . . . . . . . .[/go]



Or are we only counting points on the inside of the shape?

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 Post subject: Re: This 'n' that
Post #571 Posted: Thu May 02, 2019 11:27 am 
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Standard calculations for thems as is innarested. :)

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . . X O O O .
$$ . X X X X O O O X O .
$$ . X . O O . . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O . . O . .
$$ , . X X X X X O O O .
$$ . . X O O O . X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


Labeling the positions, as Knotwilg does, from top to bottom, we have

a = {1 | 0} Count = ½. Gain = ½.

b = {5 | -1 || -7} Count = -2½. Gain = 4½.
Black follower = {5 | -1} Count = 2. Gain = 3.

c = {6 | -1} Count = 2½. Gain = 3½.

d = {6 | -4} Count = 1. Gain = 5.

The total count is 1½. So White needs to pick up 2½ pts. to win.

The hottest position is d, closely followed by b.


CGT hint:
There is a big temperature drop between either c and a (3 pts.) or the Black follower of b and a (2½ pts.). That suggests treating this as a last move problem, to get the last move before the temperature drop. In that case we may represent c and d as STARs (*) and b as a DOWN (v). The sum of two *s is 0, so * + * + v = v. So to get that last play before a White would play in b.

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Last edited by Bill Spight on Thu May 02, 2019 11:40 am, edited 2 times in total.
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 Post subject: Re: This 'n' that
Post #572 Posted: Thu May 02, 2019 11:34 am 
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Kirby wrote:
New to this, but trying out the formula you gave:

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . a X O O O .
$$ . X X X X O O O X O .
$$ . X . O O b . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O c . O . .
$$ , . X X X X X O O O .
$$ . . X O O O d X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


a: {1 | 0} => 1 (1-0)
b: {5 | -1 || -7} => 12 (5 - (-7))
c: {6 | -1} => 7 (6 - (-1))
d: {6 | -4} => 10

So the order to play is in order of those values: b, d, c, a.

That'd give:
Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 4 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O 2 X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


So black gets 7 more points from the outside, white gets 8.

But doesn't white still lose? White has 4 points on the outside already, and black has 8 points already, so white needed to get 5 more points. From the whole board, it's black +3?

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , C C X X X X C C C .
$$ . C C X . 4 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 1 . X X O .
$$ . X X X X X O O O O .
$$ . C X O O O 3 . O C .
$$ , C X X X X X O O O .
$$ . C X O O O 2 X X O .
$$ . C X X X X O O O O .
$$ . . . . . . . . . . .[/go]



Or are we only counting points on the inside of the shape?


Sorry. Yes, we are counting only points inside the shape.

As for the formula I gave, it was for Black to play against only one simple gote. This theory gives exact results, but it requires many formulas to do so. Still, the theory requires fewer comparisons than reading alone.

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— Winona Adkins

Visualize whirled peas.

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 Post subject: Re: This 'n' that
Post #573 Posted: Thu May 02, 2019 12:15 pm 
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Bill Spight wrote:
Kirby wrote:
New to this, but trying out the formula you gave:

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . a X O O O .
$$ . X X X X O O O X O .
$$ . X . O O b . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O c . O . .
$$ , . X X X X X O O O .
$$ . . X O O O d X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


a: {1 | 0} => 1 (1-0)
b: {5 | -1 || -7} => 12 (5 - (-7))
c: {6 | -1} => 7 (6 - (-1))
d: {6 | -4} => 10

So the order to play is in order of those values: b, d, c, a.

That'd give:
Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 4 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O 2 X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


So black gets 7 more points from the outside, white gets 8.

But doesn't white still lose? White has 4 points on the outside already, and black has 8 points already, so white needed to get 5 more points. From the whole board, it's black +3?

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , C C X X X X C C C .
$$ . C C X . 4 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 1 . X X O .
$$ . X X X X X O O O O .
$$ . C X O O O 3 . O C .
$$ , C X X X X X O O O .
$$ . C X O O O 2 X X O .
$$ . C X X X X O O O O .
$$ . . . . . . . . . . .[/go]



Or are we only counting points on the inside of the shape?


Sorry. Yes, we are counting only points inside the shape.

As for the formula I gave, it was for Black to play against only one simple gote. This theory gives exact results, but it requires many formulas to do so. Still, the theory requires fewer comparisons than reading alone.


OK. Do you mean that the theory doesn't apply to this problem?

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 Post subject: Re: This 'n' that
Post #574 Posted: Thu May 02, 2019 12:31 pm 
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Kirby wrote:
Bill Spight wrote:
Kirby wrote:
New to this, but trying out the formula you gave:

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . a X O O O .
$$ . X X X X O O O X O .
$$ . X . O O b . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O c . O . .
$$ , . X X X X X O O O .
$$ . . X O O O d X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


a: {1 | 0} => 1 (1-0)
b: {5 | -1 || -7} => 12 (5 - (-7))
c: {6 | -1} => 7 (6 - (-1))
d: {6 | -4} => 10

So the order to play is in order of those values: b, d, c, a.

That'd give:
Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 4 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O 2 X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


So black gets 7 more points from the outside, white gets 8.

But doesn't white still lose? White has 4 points on the outside already, and black has 8 points already, so white needed to get 5 more points. From the whole board, it's black +3?

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , C C X X X X C C C .
$$ . C C X . 4 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 1 . X X O .
$$ . X X X X X O O O O .
$$ . C X O O O 3 . O C .
$$ , C X X X X X O O O .
$$ . C X O O O 2 X X O .
$$ . C X X X X O O O O .
$$ . . . . . . . . . . .[/go]



Or are we only counting points on the inside of the shape?


Sorry. Yes, we are counting only points inside the shape.

As for the formula I gave, it was for Black to play against only one simple gote. This theory gives exact results, but it requires many formulas to do so. Still, the theory requires fewer comparisons than reading alone.


OK. Do you mean that the theory doesn't apply to this problem?


Oh, the theory applies, but each type of board uses a different formula, or formulas. Which to use may be derived from the theory. If I had started off with the theory instead of simple examples, people's eyes would have glazed over.(If they didn't anyway. ;))

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 Post subject: Re: This 'n' that
Post #575 Posted: Fri May 03, 2019 12:31 am 
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OK, here is the theory. :)

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . c 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O a X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


I have labeled the play with the Black follower as r, for reverse sente. It is not actually a reverse sente, but it has a shape which could be. The other plays are simple gote, which I have labeled according to their score differences, from largest to smallest.

The largest is a = {6 | -4}, with a score difference of 10.
Next is b = {6 | -1}, with a score difference of 7.
Next is c = {1 | 0}, with a score difference of 1.

r = {{5 | -1} | -7} has a Black follower, {5 | -1}, which is a simple gote. There are two score differences of interest, t = 5 - (-7) = 12, when Black plays first in the follower and there are 3 net plays between the two scores, what is called the tally, and s = -1 - (-7) = 6, when White plays first in the follower and the tally is 1. I have labeled the two score differences for convenience.

How many simple gote there are matters. The theory asks up to the same number of questions as there are simple gote. And, OC, who plays first matters. With White to play and an odd number of simple gote, the first question is always the same. (You can ask the questions in any order, but there is one question which may decide which play to make.) That question is this:

s >? a

Where a stands for the score difference for the position a, the simple gote with the largest score difference. Filling in the blanks,

6 >? 10

Is 6 greater than 10? If so, then White should play in r. OC, 6 is less than 10, and usually as a practical matter a score difference with a tally of 1 that has not yet been played will be smaller than the largest score difference with a tally of 2. But if it is greater, we are done.

Next there are two comparisons with the two score differences for r, which must favor r for White to play there. The first one is this:

s >? (a - b + c)

In the right side the score differences for all three simple gote appear in descending order.

Filling in the blanks,

6 >? (10 - 7 + 1) = 5

Since 6 > 5 we ask the third question,

t >? (a + c)

Note that we skip b for this comparison. Filling in the blanks,

12 >? (10 + 1) = 11

Both comparisons favor r, so that is where White plays.

Here is the diagram with optimal play.

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 4 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O 2 X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


White wins by 1 pt., 8 to 7.

Here is the failure diagram.

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X a 5 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 2 4 X X O .
$$ . X X X X X O O O O .
$$ . . X O O O 3 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O 1 X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


This is a 5 to 5 jigo. With Chinese, Ing, or AGA scoring Black can fill the dame at "a" and win.

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

Visualize whirled peas.

Everything with love. Stay safe.

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 Post subject: Re: This 'n' that
Post #576 Posted: Fri May 03, 2019 1:30 am 
Honinbo

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Where are we with the theory so far?

First, what is the theory we are talking about? We are given a position where the player with the move globally can play to a locally scorable position, while the opponent can play to a simple gote. When White has the move we can write that position as {t | s || r}, where t ≥ s ≥ r, from Black's point of view. For convenience let r = 0, so that we have two score differences of interest, s and t. Optimal play will be unaffected. (The fact that we do not have to worry about t - s is a virtue of the theory. :)) We label this position as r. We also have some number of simple gote, {a | 0}, {b | 0}, {c | 0}, ... , with score differences a > b > c > . . . . We leave out equal score differences, as we can treat them as strict miai. We label the gote with their respective score differences. The question before us is whether White, the player with the move, should play in r or a.

We have already seen an example with only one simple gote. The question for that case is this:

s >? a

If so, White plays in r.

I skipped over the case with two gote The questions for that case are these:

s >? a - b

t >? a

Both comparisons must favor r for White to play there.

With three gote the questions are these:

s >? a

If so, White plays in r. If not we ask these questions:

s >? a - b + c
t >? a + c

If both comparisons favor r, White plays there.

We do not have enough information to say how to decide cases with more gote. I'll put up the next problem soon. :)

Meanwhile, we can simplify the latest example by chilling it. We do so by penalizing each move by 1 pt. Correct play in chilled go is also correct in regular go, as long as there are no kos.

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . c 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 . O . .
$$ , . X X X X X O O O .
$$ . . X O O O a X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]


After chilling we have the following:

r = {3 | -1 || -6}
a = {5 | -3}
b = {5 | 0}
c = ½

Neither player will play in c, so it is simply scored as ½ pt. for Black. That leaves us with two gote.

The two questions are these:

5 >? (8 - 5) = 3

9 >? 8

The answer in both cases is yes, so White plays in r. :)

Chilling also reduces this board to only two gote. :)

Click Here To Show Diagram Code
[go]$$W White to play and win
$$ . . . . . . . . . . .
$$ . . . . . . . O O O .
$$ , . X X X X X . . O .
$$ . . X . . . X O O O .
$$ . 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 .
$$ . . X O O O b . O . .
$$ , . X X X X X O O O .
$$ . . X O O O a X X O .
$$ . . X X X X O O O O .
$$ . . . . . . . . . . .[/go]

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

Visualize whirled peas.

Everything with love. Stay safe.

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 Post subject: Re: This 'n' that
Post #577 Posted: Fri May 03, 2019 8:42 am 
Honinbo

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Another easy one. :)

Click Here To Show Diagram Code
[go]$$W White to play and win, no komi
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . . X O O O .
$$ . X X X X O O O X O .
$$ . X . O O . . X X O .
$$ . X X X X X O O O O .
$$ . . X O O O . X . O .
$$ , . X X X X X O O O .
$$ . . X O O O . X X O .
$$ . . X X X X O O O O .
$$ . . X . O O . X . O .
$$ . . X X X X X O O O .
$$ . . . X O O . X . O .
$$ , . . X X X O O O O .
$$ . . . . . . . . . . .[/go]


Plays and scores are inside the corridors only.

Enjoy! :)

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

Visualize whirled peas.

Everything with love. Stay safe.


Last edited by Bill Spight on Fri May 03, 2019 8:55 am, edited 1 time in total.
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 Post subject: Re: This 'n' that
Post #578 Posted: Fri May 03, 2019 8:46 am 
Oza
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Are you sure that top white chain is connected? For the problem to have a 2 step endgame, that point should be empty.


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 Post subject: Re: This 'n' that
Post #579 Posted: Fri May 03, 2019 9:00 am 
Honinbo

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Knotwilg wrote:
Are you sure that top white chain is connected? For the problem to have a 2 step endgame, that point should be empty.


Corrected, thanks. :)

I used SL to generate the diagram. Setting it to show 0 moves made it place that stone there. :scratch:

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

Visualize whirled peas.

Everything with love. Stay safe.

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 Post subject: Re: This 'n' that
Post #580 Posted: Fri May 03, 2019 11:59 am 
Lives in gote

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KGS: Tryss
Click Here To Show Diagram Code
[go]$$W
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . 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]


Result : W+1


Failure :

Click Here To Show Diagram Code
[go]$$W
$$ . . . . . . . . . . .
$$ , . . X X X X . . . .
$$ . . . X . 7 X O O O .
$$ . X X X X O O O X O .
$$ . X . O O 2 6 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 1 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]


Jigo


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