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Offline Bill Spight
 Post subject: Yet another old problem
Post #1 Posted: Sat Feb 05, 2011 12:58 am 
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Maybe this one is easier, or maybe not. ;)

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O . . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X . O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


Edit: Sorry, no komi. Jigo is not a win. ;)

Enjoy! :)

OC, feedback welcome. :)

_________________
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 Sat Feb 05, 2011 7:38 am, edited 1 time in total.
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Offline Toge
 Post subject: Re: Yet another old problem
Post #2 Posted: Sat Feb 05, 2011 2:27 am 
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Bill Spight wrote:
Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 5 X X O X O O O O |
$$ | 7 X X 3 O O X X X |
$$ | 9 X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 6 8 0 . . . |
$$ -------------------[/go]



Reasons:
1. 1 point in sente with possibility to spoil 2 points in gote
2. Gote saving 2 points + also makes alley pushing not really sente
3. Gote which also makes the same 2 points
4. Next black of course runs the longer alley. Alleys have point values which approach 2 points, but never reach it. (reminds me of Zeno's paradox)
5. White runs alley which is now longer

White wins by komi (0,5 points).

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Offline HermanHiddema
 Post subject: Re: Yet another old problem
Post #3 Posted: Sat Feb 05, 2011 3:37 am 
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@Toge: there is no komi.

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O a b . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 5 X X 1 O O X X X |
$$ | 7 X O O O X . X . |
$$ | 9 X X X O X X X X |
$$ | 0 X O O O O O O O |
$$ | . X 2 4 6 8 c . . |
$$ -------------------[/go]


:w11:, :b12:, :w13: at a, b, c.

W1 keeps the option open of killing Black with a move at E9, if both corridors are destroyed. That means that the end of the black corridor is now sente, increasing its size to 2, and now the "longer corridor rule" no longer applies. If black does persist in playing the longer corridor, :b10: is mandatory, otherwise black dies. Then White can play the a-b exchange before taking tedomari at c.

White wins by 1 point.

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Offline Mnemonic
 Post subject: Re: Yet another old problem
Post #4 Posted: Sat Feb 05, 2011 8:47 am 
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Arg, this problem is too complex. Maybe I haven’t learned anything :sad: Every time I see a sequence and try starting to write it down I notice how black can counter :sad:

My conclusions so far:
Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 4 X X 5 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 6 7 . . . a a |
$$ -------------------[/go]

:w1: and :b2: are a must, I don't see any sequence that works for either player if they miss these plays. Now, if black blocks at 4 he loses, but if he takes the stone at 5, Whites only option is to start pushing down the black corridor. If black does likewise, white will get the last play and make the two marked points in the corner. It should be Jigo.
So 3 is a mistake and white should play it at 5?

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X 3 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 . . . . . . |
$$ -------------------[/go]

:b4: ruins whites chances. If he blocks he will reach a jigo. Trying to push down the corridors doesn't work because they are now the same length. Yes, white ends with sente, but there are no plays left and trying to change the sequence will just allow black to get some huge plays like 1.

This problem has me stumped. I really tried to read it all out, but I just don't see this working for white :sad:

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Offline Magicwand
 Post subject: Re: Yet another old problem
Post #5 Posted: Sat Feb 05, 2011 9:41 am 
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feedback:
this time i was more careful and spent close to 1 min to analize and check my initial thought.
my initial thought is correct and white win my 1.
very intresting problems.
you should write a book on endgame and publish it using such examples.
if it is too much problem publishing it, at least pdf it and sell it to the member in L19.

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The greater the unknown"
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Offline Bill Spight
 Post subject: Re: Yet another old problem
Post #6 Posted: Sat Feb 05, 2011 9:55 am 
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Mnemonic wrote:
Arg, this problem is too complex. Maybe I haven’t learned anything :sad: Every time I see a sequence and try starting to write it down I notice how black can counter :sad:

My conclusions so far:
Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 4 X X 5 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 6 7 . . . a a |
$$ -------------------[/go]

:w1: and :b2: are a must, I don't see any sequence that works for either player if they miss these plays. Now, if black blocks at 4 he loses, but if he takes the stone at 5, Whites only option is to start pushing down the black corridor. If black does likewise, white will get the last play and make the two marked points in the corner. It should be Jigo.
So 3 is a mistake and white should play it at 5?

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X 3 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 . . . . . . |
$$ -------------------[/go]

:b4: ruins whites chances. If he blocks he will reach a jigo. Trying to push down the corridors doesn't work because they are now the same length. Yes, white ends with sente, but there are no plays left and trying to change the sequence will just allow black to get some huge plays like 1.

This problem has me stumped. I really tried to read it all out, but I just don't see this working for white :sad:


Now you know what does not work. :)

Hint:

Question your assumptions.

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

Visualize whirled peas.

Everything with love. Stay safe.
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Offline Bill Spight
 Post subject: Re: Yet another old problem
Post #7 Posted: Sat Feb 05, 2011 10:11 am 
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Magicwand wrote:
feedback:
this time i was more careful and spent close to 1 min to analize and check my initial thought.
my initial thought is correct and white win my 1.
very intresting problems.
you should write a book on endgame and publish it using such examples.
if it is too much problem publishing it, at least pdf it and sell it to the member in L19.


Thank you, Magicwand. :)

I am glad you find them interesting. I had thought that the lack of tesuji made them relatively uninteresting, but I am rethinking that. And after all, if I don't write such a book, who will? ;)

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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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Offline Mnemonic
 Post subject: Re: Yet another old problem
Post #8 Posted: Sat Feb 05, 2011 10:25 am 
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@ Bill Spight
Sorry, I've already looked at Herman Hiddema's solution. (I never thought about killing the black group because it had _so_ much eye space. So I looked at D5 as first move and concluded that black could just take the 3 points on top and that would ruin anything I could do :( )
Other question: Would you allow me to copy your problem and show it to my club? (I don't want to try and sell it or profit from it or anything :))

If you do decide to write that book I will definitely get me a copy :tmbup:

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"Cutting? Killing? Poking out eyes? What the hell are you playing?"
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Offline lorill
 Post subject: Re: Yet another old problem
Post #9 Posted: Sat Feb 05, 2011 10:47 am 
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A6 & C1 are miai, so the problem is to decide whether D9 or D5 first.

if we play D5, black D9, result is jigo. If we play D9 first, either black take or protect, white does the other one, and wins by 1.

Edit: after reading the other posts, i see why i'm wrong.

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Offline Bill Spight
 Post subject: Re: Yet another old problem
Post #10 Posted: Sat Feb 05, 2011 11:42 am 
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@ Mnemonic
Mnemonic wrote:

Other question: Would you allow me to copy your problem and show it to my club? (I don't want to try and sell it or profit from it or anything :))


Sure, please do. :)

I once set it up on a spare board for some pros and they liked it. :)

Quote:
If you do decide to write that book I will definitely get me a copy :tmbup:


Many thanks. :) I do plan to write it.

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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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Offline jts
 Post subject: Re: Yet another old problem
Post #11 Posted: Sat Feb 05, 2011 12:40 pm 
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Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O a . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | c X X O X O O O O |
$$ | . X X b O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X d . . . . . . |
$$ -------------------[/go]


A is 2pt with a 2 pt follow-up, B is 2pts, the corridors starting with C and D both sum to slightly less than 2 (D being worth more). White wants to start with A; that makes B and the followup miai, so then W gets the last 2 pt move. After that, B gets (in effect) 1 pt; after his move at D the corridors are the same length and W and B will have to get equal points from them.

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 6 X X O X O O O O |
$$ | . X X 3 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 5 . . . . . |
$$ -------------------[/go]


No, that's jigo. I think B and W have to get the same number of points on the bottom, but I'll try it this way...

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 5 X X O X O O O O |
$$ | 7 X X 3 O O X X X |
$$ | 9 X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 6 8 0 . . . |
$$ -------------------[/go]


Oh I see, there's some sort of L&D issue, isn't there? If we change up the order of play, then B will have to blink first on the corridor.

Click Here To Show Diagram Code
[go]$$Wc White to play and win - F1 is 11
$$ -------------------
$$ | O . O 2 . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 5 X X 1 O O X X X |
$$ | 7 X O O O X . X . |
$$ | 9 X X X O X X X X |
$$ | 0 X O O O O O O O |
$$ | . X 4 6 8 1 . . . |
$$ -------------------[/go]


-1 :D

Click Here To Show Diagram Code
[go]$$Wc White to play and win
$$ -------------------
$$ | O . O 2 . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 5 X X 1 O O X X X |
$$ | 7 X O O O X . X . |
$$ | 9 X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 6 8 0 . . . |
$$ -------------------[/go]

Click Here To Show Diagram Code
[go]$$W11c White to play and win
$$ -------------------
$$ | O . O X 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | O X X O X O O O O |
$$ | O X X O O O X X X |
$$ | O X O O O X . X . |
$$ | O X X X O X X X X |
$$ | 1 X O O O O O O O |
$$ | . X X X X X 3 . . |
$$ -------------------[/go]


-2


This problem was a lot of fun! Probably great for a 10k.

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Offline mw42
 Post subject: Re: Yet another old problem
Post #12 Posted: Sat Feb 05, 2011 1:08 pm 
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I thought this problem was harder. My first attempt I got jigo, so I sat down and thought about it. (I know I also got your first problem wrong, but that was only because of a silly mistake! :oops: )

Please, let me know if I made any errors in my solution for this problem.

[sgf-full](;GM[1]FF[4]CA[UTF-8]AP[CGoban:3]ST[2]
RU[Japanese]SZ[9]KM[0.00]
AW[aa][ca][ha][bb][cb][hb][ib][ac][bc][hc][dd][fd][gd][hd][id][ee][fe][cf][df][ef][eg][ch][dh][eh][fh][gh][hh][ih]AB[ga][db][fb][gb][cc][dc][ec][fc][gc][bd][cd][ed][be][ce][ge][he][ie][bf][ff][hf][bg][cg][dg][fg][gg][hg][ig][bh][bi]LB[ci:A]C[Preface - Corridors

If we want to calculate how much a move at A is worth, let us first consider a simpler case.

After reviewing the lower branches, we see that the value of a move like A increases from 0,5 as the corridor increases, but is there an upper-bound? If black plays at A, before miai consideriations, he has destroyed one potential point of white territory. Therefore, the value of this play cannot exceed one.

Therefore, the value of a corridor play like A is bound by one point and a half-point, increasing in value for a longer corridor.

/waves hand]
(;B[]LB[da:A][ea:E][ad:C][de:B][ci:D]C[Looking at the size of the moves, we see that A is worth 2p, B and E are worth 1p, and C and D are worth between 0,5p and 1p with D being slightly larger.

Therefore, if white plays first the natural sequence would be A B E D C...

So, we see that the "natural" sequence does not work. Then how should white play?

If white plays B giving black the larger move at A will his sacrifice be enough to win?]
(;W[da]C[This is a mistake by white.]
(;B[de]C[This is a mistake by black.]
;W[ea]
;B[ci]
;W[ad]LB[ae:B][di:A]C[Now, they trade corridor plays. Remember, it is black's advantage to play A instead of B since it is slightly larger.]
(;B[di]
;W[ae]
;B[ei]
;W[af]
;B[fi]
;W[ag]C[Black must respond at A2 or he will die.]
;B[ah]
;W[gi]TW[ba][ia][ab][ic][hi][ii]TB[dd][gf][if][ai]C[So, in this scenario white wins by a point.

However, what if black isn't forced to take the "lesser" point at a2, what will the result be?])
(;B[ae]
;W[di]TW[ba][ia][ab][ic][ei][fi][gi][hi][ii]TB[dd][af][gf][if][ag][ah][ai]C[White will win by a point.]))
(;B[ea]C[This move is also worth a point and now black is alive.]
;W[de]
;B[ci]
;W[ad]
(;B[di]
;W[ae]
;B[ei]
;W[af]
;B[fi]
;W[ag]
;B[gi]
;W[ah]
;B[hi]TW[ba][ia][ab][ic]TB[fa][eb][gf][if]C[This results in jigo! So, if white plays A first then black can make jigo.])
(;B[ae]
;W[di]TW[ba][ia][ab][ic][ei][fi][gi][hi][ii]TB[fa][eb][af][gf][if][ag][ah][ai]C[White will win by a point.])))
(;W[de]
;B[da]
(;W[ci]
;B[ad]TW[ba][ia][ab][ic][di][ei][fi][gi][hi][ii]TB[ea][fa][eb][ae][af][gf][if][ag][ah][ai]C[Trading D for C is not enough for white to win. This is jigo.])
(;W[ad]
(;B[ae]
;W[ci]TW[ba][ia][ab][ic][di][ei][fi][gi][hi][ii]TB[ea][fa][eb][af][gf][if][ag][ah][ai]C[Black cannot play this way, white wins by 1.])
(;B[ci]
;W[ae]
;B[di]
;W[af]
;B[ei]
;W[ag]C[Here is the crucial moment that white's play at A was look toward. Must black play a2? ]
(;B[fi]
;W[ah]
;B[ea]
;W[gi]TW[ba][ia][ab][ic][hi][ii]TB[fa][eb][gf][if]C[If he does not, he loses by 2, clearly the worst result for black thus far.])
(;B[ah]
;W[fi]TW[ba][ia][ab][ic][gi][hi][ii]TB[ea][fa][eb][gf][if][ai]C[W+1 and black could not resist. So this is the best sequence for white.])))))
(;B[ci]
(;AB[di][ei][fi][gi][hi]LB[ii:A]C[In this case, a move at A is dame and is, therefore, worth 0p.])
(;W[]
;B[di]
(;AB[ei][fi][gi]LB[hi:A]C[In this case, a move at A is worth 1p gote or 0,5p.])
(;W[]
;B[ei]
;AB[fi]LB[gi:A]C[Here, the move at A is worth 1,5p gote or 0,75p.]))))[/sgf-full]

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Offline Bill Spight
 Post subject: Re: Yet another old problem
Post #13 Posted: Sat Feb 05, 2011 1:36 pm 
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@ mw42

You solved the problem, but the calculations are a bit tricky. I'll post them later. :)

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— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.
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Offline mw42
 Post subject: Re: Yet another old problem
Post #14 Posted: Sat Feb 05, 2011 1:40 pm 
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Bill Spight wrote:
@ mw42

You solved the problem, but the calculations are a bit tricky. I'll post them later. :)


:grumpy: I look forward to the post. I've never done endgame problems before so these are certainly a learning experience.
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Offline daal
 Post subject: Re: Yet another old problem
Post #15 Posted: Sun Feb 06, 2011 3:54 am 
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Bill Spight wrote:
Maybe this one is easier, or maybe not. ;)

Click Here To Show Diagram Code
[go]$$Wc
$$ -------------------
$$ | O . O 1 d . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | b X X O X O O O O |
$$ | . X X c O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X a . . . . . . |
$$ -------------------[/go]


I think 1 is the biggest, because if black doesn't answer it, a further move at d takes away two points. Since nothing else is bigger, 1 is sente. Of the remaining points, I will boldly claim that c is worth one point. A and b are...tunnels and the longer the tunnel the more it's worth due to some cumulative factor. How much? I forget, but I think more than one. Thus:

Click Here To Show Diagram Code
[go]$$Wc
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 5 X X . X O O O O |
$$ | 6 X X 4 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 3 . . . . . . |
$$ -------------------[/go]


Looks like a done deal.


ARRGh. I forgot to count the prisoner. Next try:
.
.
.
Three tries later, white always loses or jigo. Conclusion 1 is wrong. The long tunnel must be the biggest. First I will strain my brain to figure out what a is really worth:

"a" is worth an average of the end result of black or white playing it. If black plays it, he gets 3 points. If white plays it, we still don't know the end result, which depends on whether black or white plays on d. If black plays on d, black gets 2 points. If white plays on d, black gets 0 points. This averages to 1 point. How to leverage this? 'a' is worth 3 or 1 from black's perspective. Thus it is worth 2 points. :-? (I suspect this is wrong - please mark my errors in red).

[hide]
Click Here To Show Diagram Code
[go]$$Wc
$$ -------------------
$$ | O . O a d . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | b X X O X O O O O |
$$ | . X X c O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 1 . . . . . . |
$$ -------------------[/go]


Following the logic (?) above, 1 is worth an average of 6 and whatever, which makes it worth more than a. :-? :roll: :-? :-? This is not accurate, but appears to lead to a win.

[hide]
Click Here To Show Diagram Code
[go]$$Wc
$$ -------------------
$$ | O . O 3 4 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 2 X X O X O O O O |
$$ | . X X 5 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 1 . . . . . . |
$$ -------------------[/go]

Quote:
feedback welcome. :)

Even if I got it wrong, having fewer options than the last problem made me willing to tackle it.

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Patience, grasshopper.

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Offline mw42
 Post subject: Re: Yet another old problem
Post #16 Posted: Sun Feb 06, 2011 5:45 am 
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@daal

:b2: @ :w3: is better for black.
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Offline Bill Spight
 Post subject: Re: Yet another old problem
Post #17 Posted: Sun Feb 06, 2011 10:34 am 
Honinbo

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A deceptive problem. :)

Solution:

Click Here To Show Diagram Code
[go]$$Wc Tempting
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X . O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


The sente of :w1: is very tempting. If all of the plays were independent, it would be the dominant play. :) But the plays are not independent.

The problem is that :b2: makes two eyes, rendering the Black group invulnerable. That reduces the size of the plays in the center and on the left side. :w1: is aji keshi.

Click Here To Show Diagram Code
[go]$$Wc Aji Keshi
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X 3 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 4 . . . . . . |
$$ -------------------[/go]


After :b2:, :w3: is best. But then :b4: gets jigo. The rest is miai.

Click Here To Show Diagram Code
[go]$$Wc Solution
$$ -------------------
$$ | O . O . . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


:w1: takes away Black's potential eye in the center. Now Black will have to make an eye on the left side. OC, Black will have no trouble doing so, but that makes plays on the left side slightly bigger than in the first diagram. That slight difference is enough for the win. :)

Click Here To Show Diagram Code
[go]$$Wc Canonical play
$$ -------------------
$$ | O . O 2 . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 4 X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 5 . . . . . . |
$$ -------------------[/go]


This diagram shows technically correct play. White gets :w5: to win by 1 point.

Click Here To Show Diagram Code
[go]$$Wc Stubborn resistance
$$ -------------------
$$ | O . O . . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | . X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 2 . . . . . . |
$$ -------------------[/go]


:b2: may be technically inferior, but it puts up a stubborn resistance. White still must avoid the aji keshi on the top side.


Click Here To Show Diagram Code
[go]$$Wc Stubborn resistance, continued
$$ -------------------
$$ | O . O . . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 5 X X 1 O O X X X |
$$ | 7 X O O O X . X . |
$$ | 9 X X X O X X X X |
$$ | 0 X O O O O O O O |
$$ | . X 2 4 6 8 . . . |
$$ -------------------[/go]


Black continues to resist, but :w9: is sente.

Click Here To Show Diagram Code
[go]$$Bcm10 Death
$$ -------------------
$$ | O . O . 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | O X X O X O O O O |
$$ | O X X O O O X X X |
$$ | O X O O O X . X . |
$$ | O X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X X X X X 1 . . |
$$ -------------------[/go]


If :b10: pushes again, :w11: kills.

Click Here To Show Diagram Code
[go]$$Bcm10 Stubborn resistance, finis
$$ -------------------
$$ | O . O 2 3 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | O X X O X O O O O |
$$ | O X X O O O X X X |
$$ | O X O O O X . X . |
$$ | O X X X O X X X X |
$$ | 1 X O O O O O O O |
$$ | . X X X X X 4 . . |
$$ -------------------[/go]


So :b10: makes the eye, :w11: plays sente on the top side, and then :w13: wins by 1 point. :)

_________________
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 Sun Feb 06, 2011 4:31 pm, edited 1 time in total.
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Offline Bill Spight
 Post subject: Re: Yet another old problem
Post #18 Posted: Sun Feb 06, 2011 12:24 pm 
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Now to count. :)

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


To count this, let's first review closed corridors.

Click Here To Show Diagram Code
[go]$$ Closed corridors
$$ ----------------------
$$ . O X X X . . . . . .
$$ . O u . X . . . . . .
$$ . O X X X X . . . . .
$$ . O v . . X . . . . .
$$ . O X X X X X . . . .
$$ . O w . . . X . . . .
$$ . O X X X X X . . . .
$$ . O . . . . . . . . .[/go]


Corridor "u" is worth 1/2 point for Black, corridor "v" is worth 1 1/4 points for Black,
corridor "w" is worth 2 1/8 points for Black, etc.

A corridor of length, L, is worth L - 2 + 2^(1 - L). A play in the corridor gains 1 - 2^(1 - L).

In the problem diagram, corridor "d" has a length of 7, so we count it as 5 1/64 points for White, and a play there gains 63/64.

Corridor "b" has a length of 6, but it is not independent.

Click Here To Show Diagram Code
[go]$$Bc Black first
$$ -------------------
$$ | O . O 2 3 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X W X O O O O |
$$ | . X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


First, suppose that Black takes :wc:. Now Black has an eye in the center and one on the top side, and does not need one on the left side. That makes the left side independent, and we can count it as 4 1/32 for Black. :w2: is sente, so we can count the top side as 2. With the center, that gives Black a local count of 8 1/32.

Click Here To Show Diagram Code
[go]$$Wc White first
$$ -------------------
$$ | O . O 5 6 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 4 X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


Next, suppose that White plays first. After :w1: Black must make an eye on the left side, and plays there gain 1 point. :w3: and :b4: each gain 1 point, so the count remains the same. The same goes for :w5: and :b6:. Black has a local count of 6.

Putting it all together, the Black group in the original diagram has a local count of 7 1/64. A move there gains 1 1/64.

We may count the left side corridor as the average of 4 1/32 and 4, or 4 1/64. :)

Click Here To Show Diagram Code
[go]$$Wc Solution
$$ -------------------
$$ | O . O . . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


In the original position Black has an overall count of 9 1/64 and White also has an overall count of 9 1/64. The board is even.

:w1: plays to a count of -1 1/64. At this point the best Black can do is to "round up" to -1, which is a win for White.

Click Here To Show Diagram Code
[go]$$Wc Solution
$$ -------------------
$$ | O . O 2 . . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | 3 X X O X O O O O |
$$ | 4 X X 1 O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X 5 . . . . . . |
$$ -------------------[/go]


Then :b2:, :w3:, and :b4: each gain 1 point, yielding a count of -1/64. :w5: "rounds down" to a net score of -1. :)

Click Here To Show Diagram Code
[go]$$Wc Aji keshi
$$ -------------------
$$ | O . O 1 2 . X O . |
$$ | . O O X . X X O O |
$$ | O O X X X X X O . |
$$ | . X X O X O O O O |
$$ | . X X . O O X X X |
$$ | . X O O O X . X . |
$$ | . X X X O X X X X |
$$ | . X O O O O O O O |
$$ | . X . . . . . . . |
$$ -------------------[/go]


After :w1: and :b2: the center and left side are independent, with local counts of 1 and 4 1/32, respectively. The overall count for Black is 9 1/32. That makes the overall count of the board 1/64. ( :w1: is a losing sente, losing 1/64.) Now the best White can do is to "round down" to 0.

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

Visualize whirled peas.

Everything with love. Stay safe.

This post by Bill Spight was liked by: topazg
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Offline mw42
 Post subject: Re: Yet another old problem
Post #19 Posted: Sun Feb 06, 2011 3:56 pm 
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I think you are a wizard. :-P Thanks for your great post. After reading it, I feel insecure about my simplistic way of thinking. :-)
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Offline Numsgil
 Post subject: Re: Yet another old problem
Post #20 Posted: Wed Feb 09, 2011 12:46 pm 
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Bill Spight wrote:
Quote:
If you do decide to write that book I will definitely get me a copy :tmbup:


Many thanks. :) I do plan to write it.[/hide]


I'd also get one, for the record. The "Elementary Series of Go" book on the endgame is all there is at present, and I feel like there's wide room for improvement in presentation. I'd also say that you shouldn't feel like the problems are boring if they don't have any special tesujis; even handling straightforward whole-board endgame problems is more than most SDKs can handle! No need to complicate things.

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