Redstone problem 1

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topazg
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Re: Redstone problem 1

Post by topazg »

Bill, you're a real endgame guru, you scare me. Honestly, how easy is this stuff to you??

@Mark: Judging by how this sequence is progressing, I suspect there will be a strong correlation between Go strength and Redstone strength. :)

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O 9 2 0 |
$$ | O . X . O O O |
$$ | O O O O O 5 6 |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 7 |
$$ | X X X X X X X |
$$ | . X . X X O 8 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm11 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O 2 X 1 O O O |
$$ | O O O O O O X |
$$ | . O X X X X X |
$$ | O O O X X . C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


And now I'm back to something like this, which loses for White again, or this:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O 9 2 0 |
$$ | O . X . O O O |
$$ | O O O O O 6 8 |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 5 |
$$ | X X X X X X X |
$$ | . X . X X O 7 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm11 White to play
$$ ---------------
$$ | . O O O 1 . C |
$$ | O 3 X 2 O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X . C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Bm14 Black to play
$$ ---------------
$$ | . O O O O . C |
$$ | O C 2 1 O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X 3 C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


And now White wins. I'm amazing at playing to lose it seems :D

So...:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O . 2 . |
$$ | O . X . O O O |
$$ | O O O O O 7 8 |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 5 |
$$ | X X X X X X X |
$$ | . X . X X O 6 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | 5 O O O 1 X 2 |
$$ | O 4 X 3 O O O |
$$ | O O O O O O X |
$$ | 7 O X X X X X |
$$ | O O O X X 6 C |
$$ | X X X X X X X |
$$ | . X . X X 8 C |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm17 White to play
$$ ---------------
$$ | O O O O O 1 C |
$$ | O C 2 O O O O |
$$ | O O O O O O X |
$$ | O O X X X X X |
$$ | O O O X X X C |
$$ | X X X X X X X |
$$ | . X . X X X C |
$$ ---------------[/go]


See, I've done it again????

The key now seems to be to recognise there are an odd number of dame, to make sure you play one first instead of filling your own liberties. But, on the other hand, one of the key alternatives seems to be to fill your opponent's liberties, which always sounds like a better option. If you suicide two stones (by adding one to a single stone), you seem to buy yourself an extra move, unless your opponent responds by making the point unplayable. Unless....

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O 7 2 8 |
$$ | O . X . O O O |
$$ | O O O O O 5 6 |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O 4 X 2 O O O |
$$ | O O O O O O X |
$$ | . O X X X X X |
$$ | O O O X X O 1 |
$$ | X X X X X X X |
$$ | . X . X X O 3 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm13 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O C 1 . O O O |
$$ | O O O O O O X |
$$ | . O X X X X X |
$$ | O O O X X 2 C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


Playing in a two eye point to prevent your opponent getting a move in there too is winning, as it forces the opponent to play in their own space. Finally I get a variation where the person I'm trying to score a win for actually wins.

Is this even close to strongest play for both, or am I missing something fundamental yet again?
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Re: Redstone problem 1

Post by Bill Spight »

topazg, I hate to see you torture yourself. Your first move is wrong.

As for missing something fundamental, I suppose that if there were a book on Redstone, you might say that. But this is unfamiliar territory. I am more familiar, from working on no pass go and go infinitesimals. It took me a couple of hours to work out the eye values, to get the idea for the problem, and to work out the details. I could have done it a bit quicker, but I double checked myself at almost every step ;)

Edit: Taking a break, could be for a some hours. :)
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topazg
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Re: Redstone problem 1

Post by topazg »

No, the conclusion of learning is at its most satisfying when the journey is tortuous. It's got to the point where I'm not going to get beaten now ;)

Why the first move is wrong:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O . 2 . |
$$ | O . X . O O O |
$$ | O O O O O 5 . |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Of course, :w5: here is wrong, as White's job is presumably to make sure Black takes the first dame when it's even, leaving White the last one. I'm tempted to say D'oh at myself, but that statement may be wrong, so I won't.

In which case:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O 5 2 . |
$$ | O . X . O O O |
$$ | O O O O O . . |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Now there's an odd number of dame again, meaning Black has to take one, hoping that White takes another, giving himself the chance of taking the last one?

Click Here To Show Diagram Code
[go]$$Bm6 Black to play
$$ ---------------
$$ | . O O O O X . |
$$ | O . X 2 O O O |
$$ | O O O O O 1 3 |
$$ | . O X X X X X |
$$ | O O O X X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O X . |
$$ | O . X O O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Except now Black has his wish. As there are 4 dame and it's White to play, meaning he gets to take 2 and 4, leaving White the first person to fill. As such, Black wins.

So, presumably White has to play to force Black to take the first dame when there's an even number remaining, so he can get the last one:

Click Here To Show Diagram Code
[go]$$Bm6 Black to play
$$ ---------------
$$ | . O O O O X 2 |
$$ | O . X . O O O |
$$ | O O O O O 1 3 |
$$ | . O X X X X X |
$$ | O O O X X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O 4 X 2 O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X O 1 |
$$ | X X X X X X X |
$$ | . X . X X O 3 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm13 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O C 1 . O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X . C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


Now White wins, but as :w9: and :w11: are miai, Black can resist, by forcing White to take the last dame, creating one with a suicide:

Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O 4 X 3 O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X O 1 |
$$ | X X X X X X X |
$$ | . X . X X O 2 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm13 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O C . O O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X . C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


Again White loses. Which means I'm struggling to see how White can force Black to take the last dame.


Why, ironically, my original suspicion may have been right:

Back to the original position:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O . 1 . |
$$ | O 5 X 4 O O O |
$$ | O O O O O a a |
$$ | a O X X X X X |
$$ | 2 O O a X 3 a |
$$ | X X X X X X X |
$$ | . X . X X O a |
$$ ---------------[/go]


All the points marked "a" are miai, so White has to play at :w5:

Click Here To Show Diagram Code
[go]$$Bm6 Black to play
$$ ---------------
$$ | . O O O . O . |
$$ | O C 1 2 O O O |
$$ | O O O O O . . |
$$ | . O X X X X X |
$$ | X O O . X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Now there are an even number of dame, and it's Black's turn. There are no non-dame, so White wins. Black has no resistance.

Tell me I'm close?? ;)
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Re: Redstone problem 1

Post by Bill Spight »

topazg wrote:No, the conclusion of learning is at its most satisfying when the journey is tortuous. It's got to the point where I'm not going to get beaten now ;)

Why the first move is wrong:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O . 2 . |
$$ | O . X . O O O |
$$ | O O O O O 5 . |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Of course, :w5: here is wrong, as White's job is presumably to make sure Black takes the first dame when it's even, leaving White the last one. I'm tempted to say D'oh at myself, but that statement may be wrong, so I won't.

In which case:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O 5 2 . |
$$ | O . X . O O O |
$$ | O O O O O . . |
$$ | . O X X X X X |
$$ | 1 O O 4 X 3 . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Now there's an odd number of dame again, meaning Black has to take one, hoping that White takes another, giving himself the chance of taking the last one?

Click Here To Show Diagram Code
[go]$$Bm6 Black to play
$$ ---------------
$$ | . O O O O X . |
$$ | O . X 2 O O O |
$$ | O O O O O 1 3 |
$$ | . O X X X X X |
$$ | O O O X X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O X . |
$$ | O . X O O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Except now Black has his wish. As there are 4 dame and it's White to play, meaning he gets to take 2 and 4, leaving White the first person to fill. As such, Black wins.

So, presumably White has to play to force Black to take the first dame when there's an even number remaining, so he can get the last one:

Click Here To Show Diagram Code
[go]$$Bm6 Black to play
$$ ---------------
$$ | . O O O O X 2 |
$$ | O . X . O O O |
$$ | O O O O O 1 3 |
$$ | . O X X X X X |
$$ | O O O X X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O 4 X 2 O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X O 1 |
$$ | X X X X X X X |
$$ | . X . X X O 3 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm13 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O C 1 . O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X . C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


Now White wins, but as :w9: and :w11: are miai, Black can resist, by forcing White to take the last dame, creating one with a suicide:

Click Here To Show Diagram Code
[go]$$Wm9 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O 4 X 3 O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X O 1 |
$$ | X X X X X X X |
$$ | . X . X X O 2 |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$Wm13 White to play
$$ ---------------
$$ | . O O O O . C |
$$ | O C . O O O O |
$$ | O O O O O X X |
$$ | . O X X X X X |
$$ | O O O X X . C |
$$ | X X X X X X X |
$$ | . X . X X . C |
$$ ---------------[/go]


Again White loses. Which means I'm struggling to see how White can force Black to take the last dame.


Why, ironically, my original suspicion may have been right:

Back to the original position:

Click Here To Show Diagram Code
[go]$$Wm1 White to play
$$ ---------------
$$ | . O O O . 1 . |
$$ | O 5 X 4 O O O |
$$ | O O O O O a a |
$$ | a O X X X X X |
$$ | 2 O O a X 3 a |
$$ | X X X X X X X |
$$ | . X . X X O a |
$$ ---------------[/go]


All the points marked "a" are miai, so White has to play at :w5:

Click Here To Show Diagram Code
[go]$$Bm6 Black to play
$$ ---------------
$$ | . O O O . O . |
$$ | O C 1 2 O O O |
$$ | O O O O O . . |
$$ | . O X X X X X |
$$ | X O O . X O . |
$$ | X X X X X X X |
$$ | . X . X X O . |
$$ ---------------[/go]


Now there are an even number of dame, and it's Black's turn. There are no non-dame, so White wins. Black has no resistance.

Tell me I'm close?? ;)


Bingo! :)

Moves 1 and 2 each gain 1/2 point, but differ by a dame. ;)
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Visualize whirled peas.

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Re: Redstone problem 1

Post by MarkSteere »

:)
topazg wrote:@Mark: Judging by how this sequence is progressing, I suspect there will be a strong correlation between Go strength and Redstone strength. :)

That would be the best possible outcome :)
finito ludos regula
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Re: Redstone problem 1

Post by Bill Spight »

Solution:

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


:w1: makes two points in the corner. :b2: prevents a White point on the left side. :w3: does not make a point difference, but guarantees the last move. Next the dame are played. As captures, :b8: and :w9: are played with red stones.

Click Here To Show Diagram Code
[go]$$B Black to play
$$ ---------------
$$ | . O O O . O . |
$$ | O . X . O O O |
$$ | O O O O O X O |
$$ | X O X X X X X |
$$ | X O O O X . + |
$$ | X X X X X X X |
$$ | . X . X X . + |
$$ ---------------[/go]


We have now reached a scorable board. Black has 4 points, minus 2 for the group tax, and White also has 4 points, minus 2 for the group tax, for a net score of 0. Zero is a second player win, so Black to play loses.

----

Why is the three point White eye with the Black stone in the center worth 1 point (move) for White?

First, suppose that White plays first in the eye.

Click Here To Show Diagram Code
[go]$$W White to play
$$ ---------------
$$ | . O O O . O . |
$$ | O 1 X 2 O O O |
$$ | O O O O O X O |
$$ | X O X X X X X |
$$ | X O O O X . + |
$$ | X X X X X X X |
$$ | . X . X X . + |
$$ ---------------[/go]


:b2: is a red stone, leaving this position.

Click Here To Show Diagram Code
[go]$$ One point eye
$$ ---------------
$$ | . O O O . O . |
$$ | O O . + O O O |
$$ | O O O O O X O |
$$ | X O X X X X X |
$$ | X O O O X . + |
$$ | X X X X X X X |
$$ | . X . X X . + |
$$ ---------------[/go]


Plainly White has one point now in the eye, after zero net plays.

Now suppose that Black plays first.

Click Here To Show Diagram Code
[go]$$B Black to play
$$ ---------------
$$ | . O O O . O . |
$$ | O 1 X 2 O O O |
$$ | O O O O O X O |
$$ | X O X X X X X |
$$ | X O O O X . + |
$$ | X X X X X X X |
$$ | . X . X X . + |
$$ ---------------[/go]


Click Here To Show Diagram Code
[go]$$B Black to play
$$ ---------------
$$ | . O O O . O . |
$$ | O 3 4 + O O O |
$$ | O O O O O X O |
$$ | X O X X X X X |
$$ | X O O O X . + |
$$ | X X X X X X X |
$$ | . X . X X . + |
$$ ---------------[/go]


leaving

Click Here To Show Diagram Code
[go]$$B One point eye
$$ ---------------
$$ | . O O O . O . |
$$ | O . + + O O O |
$$ | O O O O O X O |
$$ | X O X X X X X |
$$ | X O O O X . + |
$$ | X X X X X X X |
$$ | . X . X X . + |
$$ ---------------[/go]


Again we have a 1 point eye after zero net plays. :)

Now, what kind of play is :w3:?

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


:w3: leaves a Black point plus a dame in the Black eye.

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


If Black plays there instead, Black also gets one point in the eye. So :w3: does not change the number of points in the eye. What it does in this case is to guarantee that White gets the last move before the players start reducing their own points. Like a dame, :w3: is an infinitesimal play. See http://senseis.xmp.net/?GoInfinitesimals for more about infinitesimals in regular go.

Gotta run.

{More later.} :)
Last edited by Bill Spight on Mon Feb 20, 2012 9:42 pm, edited 3 times in total.
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— Winona Adkins

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Re: Redstone problem 1

Post by MarkSteere »

Nice :) Thanks Bill.
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