Where to poke your nose?

Bill Spight · #1

A tricky endgame problem.

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

A fun reading problem.

Or, if you know the theory, the first move is obvious.

Enjoy!

Edit: Solution, found by ez4u.

emeraldemon · #2

hyperpape · #3

Theory? I don't know no theory. But I think I heard a rule of thumb once.

Codexus · #4

I think I got it...

jts · #5

Bill Spight · #6

Well, no cigar yet. Here's a hint.

synopsis · #7

If all plays are one point, why does the order make a difference?

Cassandra · #8

synopsis wrote:

If all plays are one point, why does the order make a difference?

E.g. the number of subsequent threats to (e.g.) destroy another point must be considered.

Bill Spight · #9

synopsis wrote:

If all plays are one point, why does the order make a difference?

Well, if all plays gain one point, order can matter. See http://senseis.xmp.net/?PlayingInfinitesimals

But in this case all plays gain less than one point, and the largest play is correct.

Bill Spight · **#10**

Well, the long open corridor seems to be attractive. But White can do better.

Here are some variations to show how Black gets jigo if White starts in that corridor.

Edit: Can't we show an 11x11 board?

ez4u · **#11**

Bill Spight wrote:

Edit: Can't we show an 11x11 board?

No, it's a limitation in eidogo. We checked it in another thread but I forgot where.

Meanwhile.

I have never been able to understand the jargon and mathematics of infinitesimals but if I have the basic idea right...

I think White wants to start here. The problem seems to come down to the fact that Black sooner or later can atari the White stones and force White to connect at 9 in the first diagram below. As a result, the plays by and against the White center stones are worth slightly more than plays elsewhere. So here :w1:

is worth more than :w5:

although they both push into a space of the same length. The center corridor looks like it should be worth more, but note that after :b2:

and

, it is another 4-space corridor that is worth the same as the one on the left edge. Therefore at best for Black it is a corridor worth something less than the one with :w1:

. Below is one example continuation where Black goes ahead and pushes to create the atari

Click Here To Show Diagram Code: [go]$$Wc Black goes for the atari, White wins \n regardless of what follows. $$ ----------------------- $$ | . . 6 O O O O 9 . . . | $$ | X X X O X O X O O O O | $$ | . X O O X X X X X X O | $$ | . X X O 2 . . . 7 3 O | $$ | . X O O X X X X X . O | $$ | 5 X X O O 8 . . O O O | $$ | O O X 1 X X X X X X O | $$ | . O X 4 . . X O O O O | $$ | . O X X X X X O . . . | $$ | O . O X . O X O O O . | $$ | . . O X . . X X X O . | $$ -----------------------[/go]

Another example

Click Here To Show Diagram Code: [go]$$Wc Black goes for the biggest points and \n never gets the atari $$ ----------------------- $$ | . 9 7 O O O O . . . . | $$ | X X X O X O X O O O O | $$ | . X O O X X X X X X O | $$ | . X X O 2 . . . 4 3 O | $$ | 8 X O O X X X X X . O | $$ | 5 X X O O . . . O O O | $$ | O O X 1 X X X X X X O | $$ | . O X 6 . . X O O O O | $$ | . O X X X X X O . . . | $$ | O . O X . O X O O O . | $$ | . . O X . . X X X O . | $$ -----------------------[/go]

The importance of the atari

Click Here To Show Diagram Code: [go]$$Wc Note that 7 and 8 above are not interchangable. \n Jigo after "a" and "b". $$ ----------------------- $$ | . . 8 O O O O a . . . | $$ | X X X O X O X O O O O | $$ | . X O O X X X X X X O | $$ | 9 X X O 2 . . . 4 3 O | $$ | 7 X O O X X X X X . O | $$ | 5 X X O O 0 . b O O O | $$ | O O X 1 X X X X X X O | $$ | . O X 6 . . X O O O O | $$ | . O X X X X X O . . . | $$ | O . O X . O X O O O . | $$ | . . O X . . X X X O . | $$ -----------------------[/go]

There are a lot of continuations that I have not confirmed, but this is my best guess.

Bill Spight · **#12**

ez4u wrote:

Bill Spight wrote:

Edit: Can't we show an 11x11 board?

No, it's a limitation in eidogo. We checked it in another thread but I forgot where.

Meanwhile.

I have never been able to understand the jargon and mathematics of infinitesimals but if I have the basic idea right...

I think White wants to start here. The problem seems to come down to the fact that Black sooner or later can atari the White stones and force White to connect at 9 in the first diagram below. As a result, the plays by and against the White center stones are worth slightly more than plays elsewhere. So here :w1:

is worth more than :w5:

although they both push into a space of the same length. The center corridor looks like it should be worth more, but note that after :b2:

and

, it is another 4-space corridor that is worth the same as the one on the left edge. Therefore at best for Black it is a corridor worth something less than the one with :w1:

. Below is one example continuation where Black goes ahead and pushes to create the atari

Click Here To Show Diagram Code: [go]$$Wc Black goes for the atari, White wins \n regardless of what follows. $$ ----------------------- $$ | . . 6 O O O O 9 . . . | $$ | X X X O X O X O O O O | $$ | . X O O X X X X X X O | $$ | . X X O 2 . . . 7 3 O | $$ | . X O O X X X X X . O | $$ | 5 X X O O 8 . . O O O | $$ | O O X 1 X X X X X X O | $$ | . O X 4 . . X O O O O | $$ | . O X X X X X O . . . | $$ | O . O X . O X O O O . | $$ | . . O X . . X X X O . | $$ -----------------------[/go]

Another example

Click Here To Show Diagram Code: [go]$$Wc Black goes for the biggest points and \n never gets the atari $$ ----------------------- $$ | . 9 7 O O O O . . . . | $$ | X X X O X O X O O O O | $$ | . X O O X X X X X X O | $$ | . X X O 2 . . . 4 3 O | $$ | 8 X O O X X X X X . O | $$ | 5 X X O O . . . O O O | $$ | O O X 1 X X X X X X O | $$ | . O X 6 . . X O O O O | $$ | . O X X X X X O . . . | $$ | O . O X . O X O O O . | $$ | . . O X . . X X X O . | $$ -----------------------[/go]

The importance of the atari

Click Here To Show Diagram Code: [go]$$Wc Note that 7 and 8 above are not interchangable. \n Jigo after "a" and "b". $$ ----------------------- $$ | . . 8 O O O O a . . . | $$ | X X X O X O X O O O O | $$ | . X O O X X X X X X O | $$ | 9 X X O 2 . . . 4 3 O | $$ | 7 X O O X X X X X . O | $$ | 5 X X O O 0 . b O O O | $$ | O O X 1 X X X X X X O | $$ | . O X 6 . . X O O O O | $$ | . O X X X X X O . . . | $$ | O . O X . O X O O O . | $$ | . . O X . . X X X O . | $$ -----------------------[/go]

There are a lot of continuations that I have not confirmed, but this is my best guess.

Bravo, Dave! :clap:

of the solution gains 15/16 of a point instead of 7/8, like the other plausible moves.

The theory of corridors with the intruder vulnerable to a possible atari was developed by David Wolfe. See Mathematical Go, by Berlekamp and Wolfe. (We are talking about corridors without stones in them.) When you have one closed corridor of length 1, atari is not a threat and everything acts as usual. The same thing is true with two closed corridors of length 2, which are miai to produce a closed corridor of length 1. And the same thing is true with four closed corridors of length 3, etc.

Otherwise, when you have both open and closed corridors, one pair of corridors acts as a unit: the longest closed corridor and the shortest open corridor. They act like a closed corridor with a length equal to the sum of their lengths minus two.

In the problem the longest closed corridor has length 4 and the shortest open corridor has length 3, together they act like an open corridor of length 5. The move in it gains 15/16 of a point. But we could have had the longest closed corridor with length 3 and the shortest open corridor with length 4, and the same would be true.

This example illustrates that.

Where to poke your nose?

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