This 'n' that

Talk about improving your game, resources you like, games you played, etc.
Post Reply
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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.
Last edited by Bill Spight on Thu May 02, 2019 11:40 am, edited 2 times in total.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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

Visualize whirled peas.

Everything with love. Stay safe.
Top
Kirby
Honinbo
Posts: 9553
Joined: Wed Feb 24, 2010 6:04 pm
GD Posts: 0
KGS: Kirby
Tygem: 커비라고해
Has thanked: 1583 times
Been thanked: 1707 times

Re: This 'n' that

Post by Kirby »

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?
be immersed
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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

Visualize whirled peas.

Everything with love. Stay safe.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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

Visualize whirled peas.

Everything with love. Stay safe.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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! :)
Last edited by Bill Spight on Fri May 03, 2019 8:55 am, edited 1 time in total.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
Top
User avatar
Knotwilg
Oza
Posts: 2432
Joined: Fri Jan 14, 2011 6:53 am
Rank: KGS 2d OGS 1d Fox 4d
GD Posts: 0
KGS: Artevelde
OGS: Knotwilg
Online playing schedule: UTC 18:00 - 22:00
Location: Ghent, Belgium
Has thanked: 360 times
Been thanked: 1021 times
Contact:
Contact Knotwilg

Re: This 'n' that

Post by Knotwilg »

Are you sure that top white chain is connected? For the problem to have a 2 step endgame, that point should be empty.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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.
Top
Tryss
Lives in gote
Posts: 502
Joined: Tue May 24, 2011 1:07 pm
Rank: KGS 2k
GD Posts: 100
KGS: Tryss
Has thanked: 1 time
Been thanked: 153 times

Re: This 'n' that

Post by 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
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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

Visualize whirled peas.

Everything with love. Stay safe.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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. :)
Last edited by Bill Spight on Sun Jan 10, 2021 8:36 am, edited 1 time in total.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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. :oops: ) 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. :D

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

Visualize whirled peas.

Everything with love. Stay safe.
Top
RobertJasiek
Judan
Posts: 6273
Joined: Tue Apr 27, 2010 8:54 pm
GD Posts: 0
Been thanked: 797 times
Contact:
Contact RobertJasiek

Re: This 'n' that

Post by RobertJasiek »

"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?
Top
Bill Spight
Honinbo
Posts: 10905
Joined: Wed Apr 21, 2010 1:24 pm
Has thanked: 3651 times
Been thanked: 3373 times

Re: This 'n' that

Post by Bill Spight »

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

Visualize whirled peas.

Everything with love. Stay safe.
Top
Post Reply

Return to “Amateurs”