text:
Honinbo 1970, match 1: Why does white need to play toward the edge here?
People may remember this game from Strategtic Concepts of Go by Nagahara.
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . 2 . . . . . . . . . . . . . . .|
$$|. . . O X . . . . . . . . . . . . . .|
$$|. . O X . . . . . 3 . . . . . X . . .|
$$|. . O X . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . 1 . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . 4 . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

I want to know why white has to play toward the edge with 2. When black has gone and grabbed both star points, that just doesn't seem like a fair trade somehow. So why did white make that choice?

First of all, I want to insist that my answer is not 'dogmatic'. It is rather the way I like to think and rather a guideline instead of a deterministic answer.
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O X . . . . . . . . . . . . . .|
$$|. . O X . . . . . , . . . . . X . . .|
$$|. . O X . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . 1 . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

Let us analyze the different possible moves for White in this situation. Right now, Black is building up a large influence framework on the right. This strategies is supported by the fact that his stones on the upper left face toward the center.
However, to do so, he interrupted the large avalanche joseki, thus giving White the opportunity to take advantage of it locally.
Consequently, it seems attractive for White to play in this corner. Suppose White wants to pincer like in the following diagram.
$$W
$$---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O X . . . . . . . . . . . . . .|
$$|. . O X . . . . . 1 . . . . . X . . .|
$$|. . O X . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

In this case, Black can settle easily, for instance like in the next figure.
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . 1 . . . . . . . . . . . . . . .|
$$|. . 2 O X . . . . . . . . . . . . . .|
$$|. . O X . 3 . . . O . . . . . X . . .|
$$|. . O X . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

Of course, there are many many other possibilities for Black, but the important point is that White has no attack! Black has too much aji to make sabaki for White to hope to attack him. Hence, White's attack stone lost his meaning and is now floating inside Black's sphere of influence without suitable backup!
Hence, if White wants to play in this corner, it seems natural to kill all the possible Black's aji. The perfect move for this is White 1 in the next diagram.
$$W
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . 1 . . . . . . . . . . . . . . .|
$$|. . . O X . . . . . . . . . . . . . .|
$$|. . O X . . . . . , . . . . . X . . .|
$$|. . O X . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

This move has the several main positive points:

It takes a lot of points (the corner is worth about 15 points, probably more to come).
It completely kills Black's aji.
It threatens severly Black. Indeed, now a pincer would be very severe.

Consequently, Black answers at 2:
$$W
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . 1 . . . . . . . . . . . . . . .|
$$|. . . O X . . . . . . . . . . . . . .|
$$|. . O X . . . . . 2 . . . . . X . . .|
$$|. . O X . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

But still, the upper left part cannot be considered as Black's territory since Black is much too weak on the upper left corner. Hence the situation as to be considered as a yonrensei where White has now the initiative.
The Black stones in the upper left are in fact to be considered as light, forcing moves. Another way of seeing it, is to have a look at the following tewari. Suppose the following fuseki is taking place.
$$W
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . . . . . . . . . . .|
$$|. . . , . . . . . X . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . O . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

Wouldn't it be completely ajikeshi for Black to play like in the following diagram?
$$ ---------------------------------------
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . 6 . . . . . . . . . . . . . . .|
$$|. . . O 5 . . . . . . . . . . . . . .|
$$|. . 4 3 . . . . . X . . . . . X . . .|
$$|. . 2 1 . . . . . . . . . . . . . . .|
$$|. . O . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . , . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . O . . . . . , . . . . . X . . .|
$$|. . . . . . . . . . . . . O . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$|. . . . . . . . . . . . . . . . . . .|
$$ ---------------------------------------

Hope this helps ;-)

Not sure, but an additional thought might be to have black initiate large avalanche in case white tenukis. So black would get a locally even position too, which is too nice since he already took a spot elsewhere. By the sagari of W2, this is impossible, and of course it makes for good mid- and endgame too.

White is playing good moves that largely cancel Black's starpoint stones. Because it is closer to the edge than Black's stones, White's move 2 undercuts the zone that Black is trying to stake out with move 3.
Likewise, White's move 4 is an appropriate response to Black's star point move at 1. This prevents Black from expanding along the bottom from his san-ren-sei on the right. If he pincers White's stone (probably his best move), White makes a one point jump into the center that largely nullifies the influence of Black's two star point stones.
White could have taken the star point at the bottom, but White 4 is more aggressive. In general, star point stones are big moves at this stage of the game. But White had equally big moves with his descent to the edge (at 2), and his kakari at 4. As such, he doesn't "mind" having Black take the star points.


text:
Tsumego intermediate problem 11 - possible without winning ko two times?
This tsumego (intermediate problem 11) gave me hard time:

I actually only found a solution where black must win the ko two times, which is nasty. Is there a better solution?
It actually looks like "bent 5" :-)
My solution here:

Yes, that's the best you can do.

Can only agree :)

source: Stack Exchange

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sort: votes


text:
Tsumego elementary problem 633 - seki?
I'am trying to solve problem 633 of the elementary set (black's move):

It seems to me that after black takes the outside liberty, it's a seki. Or is there a better solution?

Actually, White is dead after you fill that liberty.  He can never attack the Black stones, but Black can capture the stone in the corner.  After that there is a ko that will kill White.  There's no way for White to win the ko - filling it in leaves the white stones in atari.  At the end of the game when White has no more ko threats, Black can take the ko, and then the white group.

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text:
Is there a site to play Go online against a computer, in 2018?
Apparently there's an old question Is there a site to play Go online against a computer? already but that's in 2014.
As we know AlphaGo/Master/Zero permanently changed the Go world, and many AI are competing the “2nd strongest” after AlphaZero retired. Is there a site to play Go online against one or (nowadays) more computers?

If it suffices to be able to play against a top bot offline, i.e. in your own machine, then there is e.g. the Lizzie interface for Leela Zero

You can play Leela Zero and Kata Go online at https://www.zbaduk.com The website also offers great tools to review your games.

The accepted answer in your linked question is still valid: ELF (the Facebook go bot, account name "ELF") or variants of LeelaZero play on the Kiseido Go Server (usually bots have "LeelaZero" or "LZ" in their names). You can find a list of these bots in the room "Computer Go" (Menu "Rooms" > Room List > Social > Computer Go)

Should vote to close this question as a duplicate of the other question since it was answered there or flag it if you don't have the vote to close privilege. .

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text:
How does White survive in this corner?
This is from a game between two high-ranking professionals I saw on TV.  Hopefully I copied it right.
There doesn't quite seem to be enough from after W1.  Let me show you [1]:
$$Wc
$$ . . . . . . . . . . |
$$ . . . X . . . . . . |
$$ . O X . X . . . 1 . |
$$ . . . . . . X . X . |
$$ . . . . . . . X X . |
$$ . X . . X X X O X O |
$$ . . . X O O O O O O |
$$ . . . . X . . . X . |
$$ --------------------|

If Black moves first, perhaps there's not enough from after B2 and W3 ?
$$Wc
$$ . . . . . . . . . . |
$$ . . . X . . . . . . |
$$ . O X . X . . . O . |
$$ . . . . . . X . X . |
$$ . . . . . . . X X . |
$$ . X . 3 X X X O X O |
$$ . . . X O O O O O O |
$$ . . . . X 2 . . X . |
$$ --------------------|

If White has just a little bit more time, we can put a white stone at A. Not even.
$$c
$$ . . . . . . . . . . |
$$ . . . X . . . . . . |
$$ . O X . X . . . O . |
$$ . . . . . . X . X . |
$$ . . . . . . . X X . |
$$ . X . . X X X O X O |
$$ . . . X O O O O O O |
$$ . . . . X 2 . . X . |
$$ --------------------|

So what are they thinking?

You say a White stone at A, but you have not marked A; perhaps you mean 2?

P.S. If you meant A15, perhaps you should add a c to your diagram headers to show the coördinates. It would also be interesting to know what happened next in the game – or was it just a snippet on a random news item? Anyway, like you, I have difficulty understanding what is going on. Maybe 1 makes a ko threat, but I do not know if that is relevant. The entire position (in the linked image) looks fairly complicated.

@PJTraill as far as I can tell... White is dead.  and maybe it's trying to put in some aji... would you like me to link you to the video?? [ here ]

It would indeed be worth linking to the video in your question. I see there that the B stone at Q5 was played after W 1. There is also a ko-fight going on at 12x{I,J}, so moves may be ko-threats; after Q5, White connects the ko, so he has ignored a Black threat. At 1:26:40 W apparently resigns during a countdown; this can also mean that previous moves were setting the scene for resignation: see my question What is looking for a place to resign in go? and https://senseis.xmp.net/?HowToResign.

Having added coördinates to your diagrams, I realise that I wrote A15 when I meant P1!

In the video, Q5 is not present when 1 is played.

It looks like 1 threatens Q4 Q5 T5 connecting out to 1.

To answer the original question: White's corner is unconditionally dead, it cannot be rescued even with White playing first. The points P1 and R1 are miai, when White takes one of them Black takes the other.
In the actual game White uses the aji left in the corner for some Ko threats, that's all White can get out of the local situation.


text:
Who has these intersections?
I am a beginner at go and we couldn't resolve this situation with my opponent. Neither of us want to play to these liberties because whoever plays loses their stones. Whose are these two intersections at the end of the game?
$$Bcm1 Whose are the points?
$$ |.....
$$ |XXX..
$$ |OOOX.
$$ |.OOX.
$$ |X.OX.
$$ +-----

The White's are dead if the surrounding Black's are alive. The White's are two moves away from being captured by Black. White can make moves to prolong its dying but not prevent it.

This is right, but the essential thing to explain to a raw beginner is that if Black plays (e.g.) A2 and White captures the two stones, then Black can play back at A1 and then A2. Inexperienced players often worry too much about individual stones being captured and have to learn to sacrifice them to gain territory or (as they improve) influence, and this is the stage before that: sacrificing in order to capture, in its most basic form. Read more in Sensei’s Library at Two eyes.

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text:
Go Surrounding Rules
I'm a bit confused by the rules of surrounding.
For example this board:
$$ +------------------+
$$ |.......OX.........|
$$ |.....XOOX.........|
$$ |....XOOX..........|
$$ |...,XOX..,....,...|
$$ |...XOOX...........|
$$ |...XOX............|
$$ |..XOOX............|
$$ |.XOO..............|
$$ |XOO.X.............|
$$ |XOOX.....,....,...|
$$ |OOX...............|
$$ |OXX...............|
$$ |XX................|
$$ |..................|
$$ |..................|
$$ |...,.....,....,...|
$$ |..................|
$$ |..................|
$$ |..................|
$$ +------------------+

What fields are now mine (assuming I'm white)?
In almost every example I found it was like this (X = captured) :
$$ +------------------+
$$ |???????OX.........|
$$ |?????ZOOX.........|
$$ |????ZOOX..........|
$$ |????ZOX..,....,...|
$$ |???ZOOX...........|
$$ |???ZOX............|
$$ |??ZOOX............|
$$ |?ZOO..............|
$$ |ZOO.X.............|
$$ |XOOX.....,....,...|
$$ |OOX...............|
$$ |OXX...............|
$$ |XX................|
$$ |..................|
$$ |..................|
$$ |...,.....,....,...|
$$ |..................|
$$ |..................|
$$ |..................|
$$ +------------------+

Why can't it be like this?
$$ +------------------+
$$ |.......OZ?????????|
$$ |.....XOOZ?????????|
$$ |....XOOZ??????????|
$$ |...,XOZ???????????|
$$ |...XOOZ???????????|
$$ |...XOZ????????????|
$$ |..XOOZ????????????|
$$ |.XOO??????????????|
$$ |XOO?Z?????????????|
$$ |XOOZ??????????????|
$$ |OOZ???????????????|
$$ |OZZ???????????????|
$$ |ZZ????????????????|
$$ |??????????????????|
$$ |??????????????????|
$$ |??????????????????|
$$ |??????????????????|
$$ |??????????????????|
$$ |??????????????????|
$$ +------------------+

Surrounding means disconnecting and connection works along the lines on the board. When you capture (connected clusters of) enemy stones, you disconnect them from vacant points; when you make territory, you disconnect it from your opponent’s stones. But this is not always played out: once both players agree what would happen, play stops. Your position has no captives (every stone is connected via its mates to a vacant point) and no territory (every vacant point is connected, possibly via other vacant points to stones of both colours). But White can easily capture several stones!

Note that when the second board image was converted to markup the black stone in A10 was not marked as "captured" (comment because I couldn't edit just that character).

The game is incomplete.
Literally nothing is "surrounded" here, and if the game ended in such a state nobody would get any points at all until you can mutually agree on which groups are "dead" and which ones are "alive", since right now, none of the groups are unconditionally alive.
Your first example presumes the inside black group is dead, the second example presumes the outside black group is dead. One could also make the case that both black groups are dead, meaning all the territory on the board is "surrounded" by white. Alternatively, one could argue that the white group is dead, meaning all the territory on the board is "surrounded" by black.
However, if everyone agrees that all the groups are alive (or, at least, can't agree that any of them are dead), then all territory is still disputed and nobody is actually surrounding anything.
If you can't tell which groups are alive and which ones are dead, keep playing until you can. Otherwise, any attempts at calculating the score would be guesswork at best.

how can you tell which groups are alive or dead? When they are fully surrounded?

@Miemels Generally when a group has (or can make) at least two eyes, but there's all sorts of caveats to that (e.g. false eyes or seki). Most life-and-death situations are pretty obvious with a bit of experience, but if you're not sure you can just keep playing until something actually dies.

If it's white's turn, you cannot argue that it's dead: White can either capture at 3-6, or at 5-2, or even both if it manages to finish the first capture with sente, and either capture is enough to live. There's also a gigantic weakness in black's center wall: White can built a solid connection up to the 8-8 point with a forced sequence, giving it a ton of potential in the center, or, alternatively, break out at the upper edge. I'd argue that white probably has the stronger position than black: All white stones are connected, and black is just a bunch of weaknesses...

The concept of "dead stones" is a short cut
It is possible to play any game of Go to the point where the board is divided into territories each of which is completely surrounded by a single color, none of which contain any 'dead' stones, and all mixed border territories have been filed in.
But playing the game all the way to this point is tedious if the players are experienced enough to know the outcome well in advance. So just as a grandmaster of chess will concede as soon as they see that checkmate is inevitable, even if the mate is several turns away, Go players won't fight out territory battles if they already know that they won't win them.
A "dead" stone is any stone inside an opponent's territory that you would lose if you were to fight out the battle to the end.  If you can't tell whether or not a stone is dead, then that means that you should fight the battle until the end, and you will find out. Your stones are only dead if you say they are.
One of the interesting features of Go is that trying to carve a space in an opponent's territory while your opponent defends it is point neutral (neither player gains any advantage) unless you win.  Therefore there is no* penalty for trying to save stones which are probably dead, but maybe aren't.
*The exception is that this is only neutral if both you and your opponent are playing in the territory. If your pieces are so dead that your opponent can play elsewhere while you try to save them before killing them anyway, that will cost you.

Of note, playing to save a dead group will only cost you if you territory scoring (i.e. japanese rules). there's no difference if you use area scoring (i.e. chinese rules).


text:
How to attack White’s corner in this beginner’s 9x9 game?

I'm playing as Black and wasn't able to prevent White from forming a solid area from here with the following sequence:
$$Bc
$$ ---------------------
$$ - . . 6 . . . . . . -
$$ - 8 0 . 2 5 . . . . -
$$ - . 4 O 3 X . O . . -
$$ - 9 7 1 . . . . . . -
$$ - . . . . X . . . . -
$$ - . . . . . . . . . -
$$ - . . X . . . O . . -
$$ - . . . . . . . . . -
$$ - . . . . . . . . . -
$$ ---------------------

Here's the full game.  As you can see, White was able to eventually create a two-eyed, safe area.  This happens a lot to me as a beginner player and I'm wondering if there was a way to prevent it.

This is a really broad question, and it depends on your level. The general idea is to reduce the opponent's eye space.  D8-C8-B6 would probably be a good way to start:
$$Bc
$$ ---------------------
$$ - . . . . . . . . . -
$$ - . . 2 1 . . . . . -
$$ - . . W . X . O . . -
$$ - . 3 . . . . . . . -
$$ - . . . . X . . . . -
$$ - . . . . . . . . . -
$$ - . . X . . . O . . -
$$ - . . . . . . . . . -
$$ - . . . . . . . . . -
$$ ---------------------

In the actual game, you can kill the group after A8 with B9-B8-E9. White can't play D9 because the three stones would be in atari, and that lets you take the second eye away:
$$Wcm8
$$ ---------------------
$$ - . 2 O . 4 . . . . -
$$ - 1 3 . O X . . . . -
$$ - . O O X X . O . . -
$$ - . X X . . . . . . -
$$ - . . . . X . . . . -
$$ - . . . . . . . . . -
$$ - . . X . . . O . . -
$$ - . . . . . . . . . -
$$ - . . . . . . . . . -
$$ ---------------------

Rather than worrying about specific sequences, the best thing to do to get better at handling invasions is lots of life and death problems.

For on-line life and death problems, see links on the go tag-info page, e.g. GoProblems.com. Useful books are Life and Death and Tesuji, both by James Davies.

TimK's answer is right on point on how to kill white in the corner, but your question belies a more general concern: Thinking "I need to kill all the things" is a very common beginner's mistake.
Take a look at your final outcome; yeah, white lives. He lives small, blocked into the corner. You've got pretty much the entire lower left quadrant on lock, and you're poised to drive a wedge deep into white's right-side territory around G5 or diving into the top right corner and living there like he lived in yours.
Meanwhile, white is alive with three points.
As trade-offs go, this result really doesn't seem all that bad for black. If anything, I'd say black came out ahead here; territory-wise you're still about equal but black has a huge advantage in any future fighting since all of his stones are still ready to rumble while most of white's are locked up holding those three points in the corner. And black has sente, so how to bring this overwhelming force into play is entirely up to you. I really don't envy white here.
In other words, there's absolutely nothing wrong with letting white live in your corner, just make sure you're getting something at least as good if not better for the trouble.
(of course, if you can also kill it so you still have all your points and you also take all his points that's cool too)

Another plus for Black: E9 is sente (forces a response) against the White group, which may help you attacking the top right, As this answer says, you do not have to kill, you just have to make more territory.


text:
In Go (board game) under Chinese rulings how does the scoring of empty territory work?
I am reading up on the Chinese rules of Go and have trouble understanding the ‘area scoring’ rules they use.
I get that each player gets a point for each stone that is alive, for each empty field in their territory and for each empty field in neither territory, but how exactly is any of this figured out?
As far as I understand it, after a game both players agree on which stones are dead and which empty positions count towards which player's territory. This doesn't seem very satisfying.
If I were to write a computer program that is supposed to score a given board state at the end of the game, how would such a machine make these decisions? It can't, can it?
I can see how this would work in case of eyes, as there is no way of placing an opponent's stone there, but for everything bigger then that it seems obscure to me how that decision (black, white, neutral territory) is made.

Where did you get the impression that each empty field in neither territory gives a point? Since they would cancel out anyway it would make no difference, but it is not in the area scoring rules they use.

@PJTraill here in Section 7 determining the winner

I see – but it is Section 9, and says “Vacant points situated between both sides' living stones are shared equally”, meaning they get half each. But what it actually says is that you only count the score for one player, and compare that with 180½, so the reason for counting them although they cancel is to be able to compare with 180½. But your question is of course actually about how one recognises dead stones and surrounded territory, and has been adequately answered: if one cannot agree, play it out; see also Chapter 3. Section 21. The end of the game (with some unclear points!).

Note also that the rules you linked to do not only define the score but specify a specific method of counting it; Sensei’s Library, to which I linked, gives a definition, and describes the Chinese method in a separate article, Chinese Counting; see also Counting methods.

It's already clear for you that a group with two eyes is alive.
By extension, a group that has enough room for two eyes is alive, even if the eyes are not formed.
At the end of the game, the human players would indicate - by a click for example - which stones are dead. The computer can then do the scoring.
If you expect the computer to figure out which stones are dead, then it needs to know how to play. There is now open source software that can do it, such as Leela Zero.

But what if there is room for two eyes and a potential invasion of the opponent that can at least create a seki (I think that's the name)? If both play sub-optimal enough this situation might arise. Also how would one deal with scoring weird "edge cases", e.g. two players make one move and then both decide the game is over?

If one player thinks an invasion may succeed, they need to try that invasion before passing at the end of the game.  If both players agree the game is over after one move, then the whole board is a seki - there are no points surrounded by white or black stones.

Same as any game really, if there's a winning sequence and the player doesn't see it then it's moo. That's why the standard sequence on online go sites is to ask the players to mark dead stones (try OGS for example). In chess, you could have a case where the software predicts checkmate in 5 moves but the player never sees the sequence.

Just to give you an example, a heated discussion on the status of a corner territory: https://forums.online-go.com/t/question-on-3-3-invasion/15523 . Go was chosen as experimental field for AI for a reason!

@TimK do I understand that right? If the game ends and both parties disagree on who's territory it is then that region is seki? (Surrounding is relative, there is this joke with the mathematician catching a lion...)

@FirefoxMetzger No, not if they disagree. In go rulesets there are complicated ways of resolving disagreements after the end of the game. I meant if they both agree that the game is over after one move each :).

The rules of Go: http://www.usgo.org/rules-go

@FirefoxMetzger In Chinese rules and in most occidental rules, it's the other way around. A group that is still on the board when the game ends is alive, unless both players agree that it is dead. If the players "disagree", then the game actually isn't over, and play resumes until all dead groups have actually been removed from the board during the game. There is no" judgement": if there is at least one disagreement, then the game resumes, and when the game ends again there won't need to be have agreements at all. All stones still on the board will be considered alive.

@FirefoxMetzger "But what if there is room for two eyes and a potential invasion" It doesn't matter whether there is "potential" or not. Once the game is over, if the opponent didn't invade your territory, then it's your territory, regardless of whether there was room for invasion or not. During the game, if the opponent invades, then it's up to you to succeed or fail to kill the invading stones.

First remove all dead groups. Remember, groups are only dead by player agreement, therefore a computer cannot possibly perform this step independent of the players.
Once the dead groups are removed the algorithm for scoring empty intersections is as follows:
For each empty intersection. If all possible paths along the lines run into white stones, the intersection is scored for white, if they all run into black stones then it is scored for black. Otherwise it has no score.
QED
For actual computer programmers: Use Dijkstra's Algorithm where each intersection is a node and each line is an edge. If a search for a white stone fails from that intersection, then it is scored for black. If a search for a black stone fails, it is scored for white. If both succeed, then it is not scored. (Note that the only way the search could fail for both is if the board is empty.)

This makes sense. Arguably the opponent could have continued playing if they thought that an invasion in an area might succeed. Algorithm wise flood-fill on "empty intersection" is the better option I think. That way you get the entire connected region and can decide who's territory it is based on the total number (and type) of colors encountered along the border.

In Wikipedia it says Dijkstra’s algorithm is for determining the shortest path between two nodes: is that what you meant? I suppose the idea in common is recursively checking all neighbours until you have your answer (though it may be implemented with a queue rather than recursively).

It can also be used to determine if there is a path between two nodes. Basically, the algorithm checks the 4 adjacent points, then the points adjacent to those points, etc. If a white stone, for e.g., can't be reached then that intersection can't possibly count toward white's score.

It's more "flood-fill" than "Dijkstra's", because (1) we're not interested in actually finding a shortest path; (2) if an intersection belongs to a player, then all intersections in the same territory also belong to the same player, so we don't need to look for a path from those intersections.


text:
Newbie go question: why can I do one move but not the other one
I am playing with SmartGo and I don’t understand one situation.
I am playing as Black.
$$Bcm82
$$ +---------+
$$ |O.O.OX.X.|
$$ |OOOOOXXXX|
$$ |.OXXOXXO.|
$$ |OOOXXXXXX|
$$ |COOOOOXXX|
$$ |2XO.OOOX.|
$$ |XXXOOXOXO|
$$ |nXOOOXXXO|
$$ |1XXO.XyXO|
$$ +---------+

Why am I allowed to play at G1 (marked y), but not at A2 (n)? I am surrounded by my own stones in both cases.

It is not what you were asking about, but note that both G1 and A2 (if your rules allow suicide) lose you points: it is normally only good to play inside your territory if it saves some of your stones from capture, usually by connecting them to the rest or my making (two eyes)[https://senseis.xmp.net/?TwoEyes]. Otherwise you are passing up on a chance to expand your area, and may even help your opponent capture your stones.

I am glad if it helps – I have just noticed I got the markup for a link wrong, but it seems clear enough anyway.

yes, I already knew that page. There is a tremendous amount of info on the sensei's library site!

It's not the stone that you play that matters, but the resulting string of solidly connected stones.

In other words: for the same reason that, for example, the already present Black stone at g6 doesn't get removed from the board (or at b3, etc.).

Playing in the space above 82 would result in that black group having no liberties while the white groups surrounding it do still have liberties, so the group would immediately be captured by white. According to Wikipedia, most rulesets prohibit moves like that. If you play in the bottom right hole, on the other hand, that black group still has multiple liberties.

I read the whole section on Wikipedia, it makes sense now. Thanks!


text:
Best move for this pattern, in Go
I am the black side. It’s my turn to play.
$$Bcm4
$$ +---------+
$$ |.........|
$$ |.........|
$$ |..X...X..|
$$ |.........|
$$ |....,.2..|
$$ |....1Xa..|
$$ |..XbOOX..|
$$ |.........|
$$ |.........|
$$ +---------+

I see this pattern developing very often in early games.
Am I better to play a to  make a chain; or, is it better to be pro-active on the left side, and try to block White from going left with b?
If I go left, the program sacrifices these two stones, and I don’t know why.
$$Bcm6
$$ +---------+
$$ |.........|
$$ |.........|
$$ |..X...X..|
$$ |..4..2...|
$$ |....,.O..|
$$ |....XX...|
$$ |..X1OOX..|
$$ |....3....|
$$ |.........|
$$ +---------+

Whereas, if I play on the right side:
$$Bcm6
$$ +---------+
$$ |.........|
$$ |.........|
$$ |..X...X..|
$$ |.........|
$$ |...3,.O..|
$$ |...2XX1..|
$$ |..X4OOX..|
$$ |.........|
$$ |.........|
$$ +---------+

It keeps fighting for the area.
Can someone tell me what move is best and why?

Four stones is an enormous handicap on a 9x9 board. With perfect play by Black, White probably can't make a living group. If you're taking four stones, you presumably aren't going to play perfectly. The main thing to do here is stay out of trouble. Connecting below 5 is a nice solid move. Playing to capture the two stones is fine, too. As long as you keep one of the upper corners alive after that, there's no way for White to win. White has to play very unreasonable moves in a situation like this, so thinking too much about them won't teach you much.

I’m a very beginner (played my first game 3 weeks ago and had a baby jn the meantime :)) so 4 stones handicap worked quite well. I’m learning to recognize patterns (a few common with “go game” on iOS, much less with “smart go” iOS). I am starting to realize that when you have the upper hand in an area it can be valuable to leave it as it and focus somewhere else as you can use it for pressure later. But the balance doesn’t seem easy.

@Thomas As for finding the right balance: that's indeed very difficult, but if you start to progress in that area, it seems like you are learning a lot. Good luck :D

I cannot tell you, what move is best, but here is how I (around 7k) think about it:
$$Bcm4
$$ +---------+
$$ |.........|
$$ |.........|
$$ |..X...X..|
$$ |.........|
$$ |....,.2..|
$$ |...c1Xa..|
$$ |..X3OOX..|
$$ |.....de..|
$$ |.........|
$$ +---------+

The white move at G5 is obviously a threat to cut at G4. I guess, that won't surprise you.
The threat can be countered, either by connecting solidly at G4, or by the move at D3. Again, that's what you understood yourself.
Capturing the two white stones removes any possibility for a cut.
The key to understanding D3 is that it actually makes the cut impossible: There are two theoretical cutting points in blacks structure, one at a, and the other at c. White cannot play at c because you would catch it in a ladder, so your two 2-stone groups are solidly connected. White can play at a, but then you just catch the two whites stones in another ladder at d. If white really wanted to give you trouble at a, they would need to play directly at e. This would lead to a sequence like this:
$$Wcm7
$$ +---------+
$$ |.........|
$$ |.........|
$$ |..X...X..|
$$ |.........|
$$ |....,.O..|
$$ |...cXX248|
$$ |..XXOOX37|
$$ |.....d196|
$$ |.......5.|
$$ +---------+

White could succeed in living in the corner, but it would not be able to cut your stones apart. And it would basically force you to build a solid black wall on top of its mediocre little life, which creates a giant mojo for you at the top. Not good for white.
Playing at a instead of b directly gives you a physical connection. It also puts immediate pressure on the white stone at G5. However, it does allow white to attack the other possible cut at c:
$$Bcm4
$$ +---------+
$$ |.........|
$$ |.........|
$$ |..X...X..|
$$ |.........|
$$ |....,.O..|
$$ |...3XX1..|
$$ |.5X2OOX..|
$$ |..4...67.|
$$ |.....8...|
$$ +---------+

I'm not sure whether that really looks better for white, it seems easier to make a life, though. And, somewhat ironically, the white stone at G5 remains a bit stronger than if you had connected at b.

Bottom line, either move can be made to work well, either one has a possibility for white to live at the lower border, but the move at b puts much more pressure on the white stones. This is why your computer does give up those two stones in the following sequence: It knows that your stones are connected, and trying to live beneath your black wall does hurt its position on the rest of the board. Better to leave the lost battle behind and try to gain some ground elsewhere. Which is precisely the story that your second diagram tells.

thanks a lot for analyzing this scenario. However, I am a bit confused at one point since you introduce the point 'b' in your text, but it is not on the diagram

The point b is the one that you marked yourself as b, which is marked as move 6 in my diagram.

Disclaimer I am about 7 kyu, and seldom play  9×9, so I defer to those with higher ranks or more experience.
How would KataGo play?
I asked KataGo, running as an engine for Sabaki, and this is what it thinks:

The first image shows that it rates the connection at G4 as best, expecting you to win by ~60 points if you play there; the second shows how it would play both sides after that. As you can see, White ends up separated into four chains, none of which is going to survive, and so the score is much higher. This could be because KataGo was perhaps not trained on 9×9 boards and/or not trained against such large handicaps.
You might suppose that Black should have won by 81 to 0, but Sabaki, when asked, shows the score as 71½, because it gives White 4 points to compensate the handicap stones and 5½ “komi” (to compensate for Black starting). KataGo evidently thinks slightly differently, but it all makes little difference.
I also tried playing 6 at G4 and then KataGo’s suggestions, and again Black (who is about to capture at B2) wiped the board with White, while after G3, White did manage to get one live group:

How should you play?
What can you take away from this? There are many moves that probably lead to a crushing victory for Black, if only they play right. As Tim K’s answer rightly says, White may well try weird or desperate tricks to confuse Black. Your best bet is to strike a balance between keeping things simple and keeping the pressure or White. The connection at G4 keeps your stones connected and makes it harder for White to connect his, so it does both. The fact that KataGo estimates it to be best is probably less important, as that could depend on some really complicated reading, and you can afford to drop a point or two and still win.
As your handicap reduces the balance will shift and you will not be able to simplify quite as much because you will not be able to afford to drop so many points. I quite liked D4 (rather than D3), as it prepares to grab a chunk of the top left while still making life difficult for White. I am glad to see this at least rated second best, even if the first estimate was about 6 points worse than G4.
How good were your moves?
I was not so happy with your move 8 in your first variation, as I expect you to capture those two stones anyway; by going on there, you let White play 7 and 9 in the top half, which might be enough to cancel out the advantage of your handicap stones: he has 3 stones to your 2, but it is your turn. Your 8 keeps it simple, but the cost may prove too high. As you say in a comment, once you have an advantage you can play somewhere else; actually it is more that an area gets played out, and for either side to play the next move in that area is only a small amount better than letting the opponent do so. Later in the game the overall “temperature” drops and one can return to those areas, or their temperature may increase as they are destabilised by nearby developments.
In your second variation the position looks better balanced: you can strengthen your top position while keeping some pressure on his stones at the bottom, but you do have some cutting points you might need to defend. If you can play C5, stop White living in the top right and make sure you either defend the cutting point at E5 or live in the bottom right, I think you should win.
Finally
As mafu said in a comment, you seem to be picking up some important principles; I hope you continue to improve and, most important, to enjoy playing. You might want to join one of the Go servers if you have not already done so – and when coronavirus calms down, your nearest Go club.


text:
What activity is there in Computer Go tournaments?
At the moment, the 17th TCEC superfinal is running, Stockfish vs. Leela. It is considered the unofficial world championship of computer chess, and games are frequently publicly commentated by professional chess players. TCEC runs tournaments continuously, 2-4 of them every year.
Given the advance of Go programs, I wonder whether there are well-organized computer Go tournaments. I found the Computer Go UEC Cup, but Wikipedia says it took place annually from 2007-2017 and has stopped; also, its website has not been updated for several years. There is slightly confusing further information about a follow-up tournament in 2019, which points to Japanese websites.
Hence my questions:

1) Where can one get information in English about the UEC Cup? Do most state-of-the-art programs participate?
2) Are there other computer Go tournaments?

I am surprised that this has been voted both down and to close (2/5) with no comments whatsoever. While I have the impression the questioner could have found out more themself, I do not know why it should be so poor or inappropriate.

@PJTraill To be fair a what happens a lot on stack exchange in general is when someone posts what they think wrong with a post people tend to take it personal, assume things and attack the person who posted

The  Computer Go Server1 is currently a very active test bed for bots. The Computer Go page at Sensei’s Library refers, under Competitions, to various competitions including the Computer Go Server, with its own page at Sensei’s Library, which gives the URL1. On that site, you will find daily updated tables of bots for 9×9, 13×13 and 19×19, These refer to the Bayes ELO ratings pages (on the same site) where for example the 19×19 page as I write (2020-04-19 22:17) shows 1064 programmes ranging in strength from DunceBot at 538 to KG30b11_406_2x2060 at 3387. The non-Bayes page shows the date+time of the most recent game by each bot, showing that a game was played some two hours ago.
The page http://www.computer-go.info/events/index.html, though was described at Sensei’s Library as showing “all competitions” in fact covers the dates 1920–20172.
I know no more than you of what has become of the UEC Cup. Surprisingly, none of the AlphaGo variants appear although one convincingly defeated a top professional, Lee Sedol, in 2016; more recent bots such as LeelaZero (released 2017-10-25) and KataGo (released 2019) appeared after the UEC Cup apparently stopped.
1  The current URL is http://www.yss-aya.com/cgos/ . Sensei’s also gave the URL http://cgos.boardspace.net/, but this appears not to work, which I have noted in Sensei’s.
2 I have updated Sensei’s accordingly.

I'd guess none of the commercial vendors would like to compete against another. There is little to gain for either party, but much to lose. The only bots that were compared are those from OSS and (to a degree) academia.


text:
What is the easiest way to play DarkForest, probably the strongest publicly available go AI?
The strongest publicly available go AI is probably this: https://github.com/facebookresearch/darkforestGo, since there is currently no way to play AlphaGo.  Is there any way to play this without building it on a local machine?  Is it available on a server somewhere?  Looks like the name on KGS is darkfmcts3 .

Do you want source?  Stand alone application?  Website?  Do you have a preferred order among those three?  Not to clear what you are wanting.

I would like easy to run source with good docs on installation and not too many dependencies, standalone, or access through webapp to a server.  Not too picky on how to interact with it just wasn't able to build darkforest from github instructions.

I looked at the installation docs.  They are very standard for this kind of researchy code.  There are some messages about gcc versioning issues on the boards that you may wish to dig into.  Building from source takes a lot of skill (it shouldn't.  But it does in my experience.)   If you haven't done much programming (and it looks like you haven't from your stack overflow profile) you're going to have problems dealing with source.

I got it working, I just had to figure out how luarocks worked.  It wasn't that difficult.

That is most heartening to hear!  Congratulations!

Define publicly, CrazyStone is publicly available. It need to be bought 80$ but available on computers et mobiles.

@JonBarker Would you consider self-answering your question with the process that you used? This is likely to be of interest to others.

github page of DeepForest claims it's KGS 5d. Leela (https://sjeng.org/leela.html), while not open-source, is available for free and seems to be KGS 8d. Am I missing something?

Hactar go https://play.google.com/store/apps/details?id=net.gowrite.hactarLite (android application) offers darkforest as opponent (darkforest running on server side). Not free, but 2€/3 month subscription.

(Edit: my answer is incorrect since these bots seem to have been taken offline).
There are a number of bots on kgs that run DarKForest that you can play against.
These are its accounts:
darkfmcts3
darkfmcts2
darkfmcts1
darkfores2
darkfores1
darkforest
Playing against this bot on KGS is probably the easiest way for most people to play against it.

Are these always online?  I have not seen them online the few times I have logged on.

Hmm, it seems that they have all been taken offline.  Perhaps the creator will put out a new version of the bot.

Anyone could in theory rebuild the bot and deploy a bot to the KGS servers though correct?


text:
What's the longest possible forced repetition in Go?
Assume no superko rules so the game would end with no result. And both players are willing to play the shortest possible cycle as long as they don't give their opponent a chance to win.
There are 4 moves in the example of eternal life, 6 moves in triple ko. And I think I have seen a 8 move one. Is there a known longer one? Is that proven to be the longest possible?

Molasses ko: http://denisfeldmann.fr/bestiary3.htm#mol
"The semeai in figure 11 is probably the worst known case of repetition. Known as "molasses ko", ..."
Denis Feldmann's Go Bestiary is a compendium of various things (interresting semeai, life and death, edge cases, incredible problems, rules edge cases, weird sekis, etc).


text:
End game deciding on dead stones
In a game of Go, we were in a strange position. If my opponent moved at one point to put my pieces in atari, his pieces would get taken. If I moved at that same point my pieces would’ve been taken.
So if neither of us move at that point, are our stones dead?
How do you count this kind of position in the counting phase?

Welcome to B&CG.SE! Thanks for your question. Is the game you are asking about called Dead Stones? It's a little hard to decipher what your specific question is.

I think this is asking about Go

Without zero punctuation, it is hard to determine the question. Rather unclear.

With the reasonable edit of Toon Krijthe the question is a valid and good question for beginners of go. The situation is described well now, though a board position would be nice. The situation is called seki. The territory is neutral.

@havogt it may not be a seki !

@Kii If you include board positions for both cases in your answer it would be really helpful for beginners. I would upvote.

@havogt : it is done :)

As TimK pointed out, the situation could be but may not be a Seki but without a diagram to show to us, it's not easy for us to guess what happened.
Seki : no one die, everyone live
$$cm1
$$ +---------------------------------------+
$$ | X . O X . . . . . . . . . . . . . . . |
$$ | X . O X . . . . . . . . . . . . . . . |
$$ | X X O X . . . X . . . . . . . . . . . |
$$ | O O X X . . . . . . . . . . . . . . . |
$$ | . . O O X X . . . . . . . . . . . . . |
$$ | . O . . O . . . . . . . . . . . . . . |
$$ | . . . O . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |

Some patterns may look like a Seki but can be a simple Bulky Five for example. In this case, even if both opponents can attari it is not a seki.
Bulky-five : Black dies after capturing because he cannot make two eyes.
See this variation on eidgogo to explain why the bulky five is dead.
$$cm1
$$ +---------------------------------------+
$$ | . . O X . O . X O . . . . . . . . . . |
$$ | . . O X . . X X O . . . . . . . . . . |
$$ | . . O X X X X O O . . . . . . . . . . |
$$ | . . O O O O O O . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |

There's also a Bent-three : Black dies
$$cm1
$$ +---------------------------------------+
$$ | . . O X X X X O . . . . . . . . . . . |
$$ | . . O X O . X O . . . . . . . . . . . |
$$ | . . O X . X X O . . . . . . . . . . . |
$$ | . . O X X X O O . . . . . . . . . . . |
$$ | . . O O O O O . . . . . . . . . . . . |

@kii under what ruleset is your first example not a seki?  As I see it neither player will play in the corner so neither can claim the two spaces as territory.  Note also that here: http://senseis.xmp.net/?Seki the exact same board shape is described as a "simple seki (no eyes)".

@BrettPontarelli the first example IS a seki. It is written just above. I added an link to see why the bulky five is dead.

See also the tag-info on the new tag [tag:seki].

This situation is called Seki.  Scoring depends on the ruleset you're using.

Can you expand on this, and at least give a brief summary of the contents of the page you linked to? Right now, this answer is just a link to an external resource and a non-informational statement about scoring.

It may not be a Seki ! See my answer.

The situation you are describing is called a seki or possibly just a point of contention on the board.  There are two ways to look at the situation.  The first addresses the your questions and the second is a little more theoretical.
First

It is the end of the game and both players have passed.
If you both agree that that area is unplayable and a seki then those spaces are not counted for either player.  For example, in Japanese counting you might just fill those spaces with stones from your bowls; move areas around for easier counting; and then determine the score.
If either player contests the area then play resumes and that player will attempt to play in and capture the area.  Play then ends again when both players have passed.

As you can see the area may or may not be in an absolute sense a seki, but that doesn't matter for the purposes of playing.  Players can only play at their own level and see how to play through positions at that level.  So, it is possible to incorrectly agree that an area is a seki or dead and belonging to either white or black, or any number of other errors.  But, for the purposes of playing and learning, as long as both players agree that is how that area will be counted.  You will especially see this kind of thinking during tournaments were there is no official observer for each game.  It is then the losing players job, having agreed to the loss, to report the outcome of the game.
Second
Often times during post game analysis with a stronger player (s)he might explain that the area was not in fact a seki (or dead in the case of a life and death situation that was not fully played out) and that one player could have proceeded along some sequence of moves to capture the other.  In these cases it doesn't change the outcome of the game it only changes the players' understanding of the game and how it "could" have changed.


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Dominant strategy in chess
Is there any proof ensuring there can be no dominant strategy for chess? By dominant strategy, I mean a way of playing that always works regardless what the opponent do.

Quite the opposite... there IS a dominant strategy, and it can be proven that there is one.

One of the following is true:

There is a dominant strategy for White.
There is a dominant strategy for Black.
There are strategies for both players that guarantee they don't lose, i.e. perfect play results in a draw (e.g. as in Tic-Tac-Toe).

No one knows which is true. Most experts guess that perfect play leads to a draw, and a few believe White can always win. It is unlikely Black has a dominant strategy, but no one has been able to rule it out either. Theoretically, a computer working for long enough could just try everything and tell us the answer, but with today's best technology and algorithms that would take prohibitively long.
Here are some important features of chess that together guarantee that one of the 3 cases above is true:

Chess is a game of complete information, i.e. both players always know everything that is going on.
Players take turns (as opposed to playing simultaneously, e.g. as in Rock Paper Scissors)
The game always ends eventually (at least under tournament rules as of 2014, after enough moves are made that aren't irreversibly advancing the game, it's a draw).
There is no randomness.
There are only 2 players.

Intuitively, these facts let you completely reason out the consequences of everything that could possibly happen (if I do A, my opponent could do B or C, in which case I could do D or E...). For a more formal explanation of why this implies there are either dominant or drawing strategies, see Zermelo's Theorem.
[Parts of this answer are synthesized from earlier answers/comments.]

Very nicely broken down.  (Am I correct in assuming you have some CGT background?)  I've researched this issue a little, and recall that White seems to win by a narrow statistical margin, but, as you note, the game remains unsolved.  I found a wiki on first move advantage the questioner may find useful: https://en.wikipedia.org/wiki/First-move_advantage_in_chess

essentially, it seems that in a more complex game, it is more likely that there will be a draw. Connect-Four is the most complex game we have solved so far, and although the first player wins, that is due to the mathematical nature of the game (i.e., it is impossible to go back to a previous position like for one side of chess). There is no proof that either side wins in a perfect game of chess, however, a draw is very likely since most expert games are beginning to end that way. We don't have to explore every single chess game possible--just the strongest moves. (NH3 is stupid, guys. It doesn't help us at all)

If you've never opened Ammonia as White, you've never truly lived. +1

The term "dominant strategy" is generally used for games with incomplete information (or simultaneous moves, which can be analyzed as a game with incomplete information). In a deterministic game where players take turns and there is no secret information, the term "perfect play" is generally used, rather than "dominant strategy". The term "dominant strategy" refers to a strategy that is better than every other strategy, regardless of what the other player does. However, in chess, you always know what your opponent did on their last turn, so the "regardless of what your opponent does" part is a bit redundant. The term "strategy" is often used to refer to the entire decision tree, including what you would have done if your opponent had done something different, but even there, the "regardless of what your opponent does" part doesn't add much; two strategies can't possibly have different results unless they differ with regard to moves actually taken.
And technically speaking, there is a weakly dominant strategy in chess (that is, a strategy that does as well as or better than every other strategy), but there isn't a strictly dominant strategy in chess (that is, a strategy that does better than every other strategy).

Since chess is open-information deterministic game, it is bound to have some kind of optimal strategy for both players, which would result in same outcome every time, provided both players play perfectly. IMO this outcome would probably be a draw, but could be victory for either side.

So, what about today's best computer players? I mean, did we find the best strategy for chess?

No, we didn't. Chess is too complex to fully compute, even for today's computational powers. Computers just exceeded the ability of top players to find better options.

This seems basically correct, though it's proven true, so I would remove the "bound to". Calling it an "optimal strategy" in the general case is a little misleading though. If the outcome is "white wins", then if white plays perfectly, any strategy for black results a loss, so it doesn't make sense to call one "optimal".

Thanks, those are valid points, should I edit? Although I'm not sure about changing "optimal". If expected outcome of perfect play is a draw, both players indeed have to play optimally to achieve it. I don't see better wording to cover all the cases, without going into too much details.

If it was my answer, I would reorder it. The crux is that if one player can force a win, only that player has an optimal strategy, and if neither player can force a win, both need an optimal strategy to force a draw (assuming win/loss/tie are the only outcomes we care about). So, you could talk about the possible outcomes, and then about who needs an optimal strategy to achieve those outcomes. Also, the theorem I linked requires that the game has alternating turns, so I would suggest mentioning that characteristic of chess too.


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Go game set with balls or spheres
Once in a documentary I saw an exquisite Go Set.
The stones were not double or single convex, they were actually balls (round spherical stones).
The board had carved channels or grooves, so as you put a ball to an intersection it stayed there.
I tried to find such a board on the internet and could not even find a picture.
Does anybody know of such boards?
(I am not interested kurokigoishi boards, they might be exquisite, but they are not balls).

GO ARK , a 9x9 Go set designed by Japanese designer Junji Masamitsu, fits your description.  The set is also sold on amazon.jp, shown as "out of stock" at the moment.

Good for a start. However what I the board was from wood and the marbles where not translucent.


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Overplay in orthodox fuseki - Leela AI fails miserably?
I'm not sure how much go is discussed on StackExchange, but currently I prefer trying it here than an actual go forum. This question concerns two topics:
Recently the free go AI Leela has been released, which is thought to be fairly strong (up to 9d KGS with proper hardware). I appear to have tricked it with a fairly simple position though, which puzzles me deeply. The starting position is:

This is a fairly common situation resulting from the orthodox fuseki. My concern is with black resisting this in a vulgar way: Black plays 1. bQ13 2. Q14 3. P13 4. O13 5. P12. The first question now is: How does white punish this overplay? I think there is a way to net black, which I've seen before somewhere but I can't remember nor find it. I was hoping the engine would do this for me but that was quite wrong:
The AI Leela (version 0.9.0 on fairly crappy hardware) gives this as a >60% winning percentage for white and suggests the continuation 6. R13 7. Q12 8. S12 wheruppon it suggests for black to let white connect under. However in case black plays 10. S11 its suggestion (with still some 60% white win) is to cut and "loose ladder", resulting in the following position ("1" in the second diagramm = move 6):
Black now pushes at P11 (= move 1 in diagramm 3) and white blocks. Leela says that white has a big advantage, but only considers black capturing T12. If black pushes out, the engine tries to "ladder" the group, each time suggesting black to take the S12 stone, and white's winning percentage dropping rapidly every time black (instead of taking S12) just keeps pushing out the futile "ladder" attempt by white. I conclude that the engine doesnt't consider the "pushing out" move by black and incorrectly asseses the entire variation (which is somewhat surprising).
What are your thoughts on this, especially as to correct play in the various positions? (In particular after diagramm 1 move 5)? I suspect O12 is a big mistake but can't be sure (my rank is 1k on KGS).

W1 should be N11, then black dies, I'd claim from memory.
W1 at R13 is probably not a complete disaster in itself, but O12 should then be O11, or maybe P10 should be O10. That would force black to capture and white would at least build some influence -- not thickness, as she can't protect all cuts.
Even before that, white should probably descend at T13 and later throw in to induce the infamous "hammer" shape, which would probably give white one or two additional sente moves.
That Leela is not seeing your sequence does seem like a clear case of a bug to me. Maybe Leela figured something like "black is losing the ladder, so black must capture. And because playing out ladders that do not work is bad, black will capture there immediately!" (pure speculation on my part).


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How to count non fully walled territory?
I just started to learn Go. I still have some problems to properly understand how to count territory when the game has finished (by both players passing). For example, if the game ends in the following situation, what part of the upper area would count towards black's territory?
$$c13
$$ -------------
$$|...........X.
$$|...........XX
$$|..XXXXX....X.
$$|..X.X.XXXXWXX
$$|XXXXXXX...O..
$$|..........OOO
$$|..........OO.
$$|OOOOOOOOOOO.O
$$|..........OOO

I know, usually black would just place a stone at L11 befor finally passing but what happens if black forgets to do so? (and white also misses the opportunity to exploit this)
The resources I read about counting did not answer this question for me.

simply put, if you reach this situation then counting doesn't really matter. Learning the rules would come first!

Simple answer: There is no such thing as "non-fully walled territory".
So far as the upper left is concerned, black has exactly two points: D10 and F10.  All the rest is not territory, because it's not enclosed.  Sure, it's all potential territory, but it only becomes actual territory when black takes that critical L11 point to actually enclose it.
So if your game actually ends in such a state, then black just made a huge 29-point blunder, whether white did anything to exploit it or not.
(of course if it's a casual game and you're nice you'll let him fix it because hey everyone makes mistakes but then again you're allowed to be a rule-abiding jerk too because winning is everything it's up to you)

This isn't really a finished game.  If both players passed, then all of the points around E13 would count as dame just like the ones around H8, and wouldn't be territory for Black.  This wouldn't happen in a serious game, so in a beginners' game, you would notice the problem during the scoring and probably just fix it.


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How to apply the ko rule (Go)
I’m having trouble understanding how to apply the ko rule. For instance, in the description in Wikipedia, there is an example where the ko rule can potentially apply. But suppose there were two, or three, or four such scenarios on the board. Does one really have to keep track of all the past states of the board? Would you just keep going round-and-round, and whoever created the last original board configuration wins those points?

If such a situation arises in your game it would be great if you could put in a photo of it. We could try and analyze it, and I think it might be easier than you expected. It feels (not sure) like you have wrongly interpreted how to treat that situation. Maybe not though, ko tends to be good for surprises :)

Could I ask you to check out my new (very belated!) answer, which I feel covers more ground than the one you have currently accepted.

The theoretical answer depends whose rules you are using, because different countries use different rules, and have different ways of dealing with the multi-ko1 situation you describe.
In practice this almost never arises, and keeping track is usually easy, as something else happens to change the whole-board state, such as a move on a previously unoccupied point – note that the rule applies to the state of the entire board. It therefore also usually makes no difference which rules you use. Also, though it is common to end a game with many half-point kos2, there is no advantage to fighting more than one of them: correct play is to leave them all till there are no other points to be made, then connect if you can, take if you can’t and only fight the very last one.
Rules currently in use
All bodies forbid immediate recapture. The main approaches to complex repetitive positions are:

Superko: You are not allowed to recreate a whole-board position that has already occurred in the game. Different bodies forbid this in different circumstances:

Positional Superko (China, in theory): always.
Chinese Superko (China, in practice): as a result of immediate recapture or “Sending two, returning one” (and I do not know what they do otherwise).
Situational Superko (US, NZ): if it leaves the same player to move.
Natural Situational Superko (UK): as the result of a move (not a pass) by the same player.

No result (Japan, Korea): If a position is repeated, the game has “no result” and must usually be replayed.
Long cycle rule (Computer Olympiad): If a position is repeated, the game stops and the result is purely determined by the number of captures by both sides since it first occurred.

Other answers

The currently accepted answer (by GendoIkari) is a misleading, because it only deals with the “basic ko rule” (forbidding immediate recapture), on which all bodies agree, and its tactical consequences.
Brian Tyler’s answer explains how computer programmes detect repetitions with Zobrist hashing.

Further reading
(As of 2020-04-08) Wikipedia gives more details in the article Rules of Go than at the link you gave, but for a comprehensive discussion you are better off with the article Superko in Sensei’s Library. For the correct order of play in half-point kos at the end of the game, see the article. Half-point Ko (also Sensei’s Library).
P.S.
It is a shame you could not share with us the game with your son where this arose, which would have been interesting.
Notes
1 A “ko” refers here to a single position where capture-recapture would be possible if not forbidden.
2 A half-point ko is one where all that is at stake is the capture itself. This excludes kos for the life of a group or just to decide whether a point of territory has to be filled.
3 “Sending two, returning one” is a situation (usually part of a seki) in which one side can sacrifice two stones, and then recapture, recreating the previous position. This could be used to prolong the game indefinitely, though each cycle costs one point if territory scoring is used.

Thank you for this tour de force of an answer -- saves me the trouble of making something half as good.

Ko only prevents a player from making a move that would have the effect of returning the board to the position that it was in just before his opponent's last move only. So the only thing you ever need to keep track of is what the board was like just before your opponent made his last move; you cannot move such that it would recreate that position.
All the further description in the wiki is explaining what a player will do if the move they want to make it prevented by ko. They make a move somewhere else on the board that forces black to respond to it; thus freeing them to make the play that they couldn't have played the previous turn due to ko.

This is the simple ko rule, which is not the most common. The super ko rule variants prohibit any previous board position. Link: http://en.wikipedia.org/wiki/Rules_of_Go#Repetition

Thanks... so, for the more common and complex rule, one would really have to keep track of all previous configurations? I don't know whether it's because I'm a beginner, but my son and I end a game with four or five such "reversible" configurations.

@adam.baker Yes, you are right. However, such a situation is really rare. If you have a situation with multiple ko at the same time, usually one player will simply close one of them and ignore the other (whatever gives him the best result globally). It is extremely unusual that all ko are connected and one player absolutely cannot give in. In such a situation, superko would apply, and you would have to keep track of the previous situations, or whatever your used rule book states (it's not something everyone agrees on...). Hope that clears it up a bit.

@adam.baker: you will seldom need to keep track of many previous positions, as most of the time there are more stones (of more stones of some colour) on the board than at any previous point in the game. Only when you start playing out multiple kos does it become tricky.

Super Ko can be detected efficiently using Zobrist hashing.
https://en.wikipedia.org/wiki/Zobrist_hashing
https://www.gnu.org/software/gnugo/gnugo_11.html#SEC144
The idea is to pick a random 64-bit integer to represent each of the 2x19x19 = 722 positions on the board. The empty board has a hash value of 0. When you add a stone to the board you XOR the current hash of the board by the hash value that represents the stone at that position. You do the same for removal (XOR being self inverse). Now you have compressed the game state into a single 64-bit integer and you simply check to see if you have seen this integer before.
The assumption is that there are no hash collisions. This is clearly not true because there are 3^361 board states (although many will be illegal) and your hash is < 64^2. However, the assumption is that in your game of a few hundred moves, the chances of getting a collision is so small that it isn't worth worrying about. If you get a collision, you could check it if you store the full state, but it is so unlikely that you would find anything but a perfect match, that you may as well not bother.

You need something like three kos to do this.
But let's say that there are two players, X and Y. Player X takes the first ko. Player Y takes a second ko as a "ko threat." Player X takes a third ko in response. Player Y re-takes the first ko. Etc.
The ko rule serves to prevent repetition between one or two kos. But if there are enough kos so that you can take them in a "round robin," then the game could go on endlessly.

You have ignored the growing prevalence of stronger ko rules designed specifically to prevent such situations. link


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What's the name of the theory that the player with the most options has the advantage?
A long time ago, I remember reading about a theory that sounds like this:

Players A and B have access to options X and Y. Player A also has access to option Z. Player A is, at worst, equal to and, at best, at an advantage to player B.

Ok, so that was poorly worded. Let me put it another way:

Having more options available to you in a two player game can't possibly hurt you.**

Example
Player A can attack from both sides. Player B must attack from the left. If attacking to the left is the best choice, then both players are equal. If attacking from the right is the best choice, then Player A has a distinct advantage. Player A is either tied or winning. Player B is either tied or losing.
Can anyone tell me the name of the theory I described?
** "... can't possibly hurt you" assumes optimal play from both players. It's not always the case the player with the most options will use them wisely. Intuition tells us that sometimes having too many options can make it difficult for the player to find the best one. Just assume that both players are making the best choices when this theory applies.

Indeed, if we don't assume optimal play, then the result could easily be the exact opposite! Player B only has access to option X, and Player A has option X and another, worse option. Player B is in the better position.

Are you sure this is actually part of formal game theory, or just a theory about games?

To readers: this question has spurred me to open the following meta question: http://meta.boardgames.stackexchange.com/questions/1162/the-game-theory-tag-is-consistently-misused-how-do-we-resolve-this

@JonathanHobbs I'm not well-versed in formal game theory. When I wrote the question, I felt like a theory that relates "options available" to "advantage" should fall into that realm. Since it was the [tag:identify-this-game] tag that led me to even ask this question here, I thought it would be appropriate to retag with a brand new tag called [tag:identify-this-theory].

I suspect this "theory" isn't much of a thing. It's pretty trivial: if my opponent doesn't have any options I don't have, then they don't have a better option than my best.

@murgatroid99 Others have voiced similar suspicions. You may be correct, but there isn't a tag called [tag:identify-this-theory-if-it-exists], and so we work with what we've got.

That comment would mean more if not for the fact that you also invented the tag that you actually used. And I'm not saying that the question is mis-tagged. I'm arguing that a question asking for the name of a trivial set theory result doesn't belong here at all, even if you use the words "player" and "option".

This question appears to be off-topic because it is about a trivial set theory result with the word "player" substituted in. It is not related to any particular board or card game.

@murgatroid99 I don't see how triviality plays a role here. I asked for the proper terminology so that I can use it (if it exists). This is the second time someone on this site has misinterpreted "I don't think a good answer exists." as "This question is off-topic." If all you saw in my question was "I have two sets, and one of them is a subset of the other.", then I think you missed the point of the question.

OK then, let's ignore the triviality aspect. Your question is still of the class "Here is a description of . Is there a word for it?" I don't think that belongs here.

@murgatroid99 I can respect the view that the question is only loosely related to games. I hope others don't feel the same way.

@murgatroid99 Now that my question is on the verge of being closed due to your campaign on meta that has drawn larger than average attention to my question, I feel entitled to further explanation. It sounds like you view my question as "Set A is larger than set B and also contains set B. What is that called?" Now consider this: If talking about how many nails are in your leg, larger is worse. If talking about options in a game with optimal play, larger is better. Therefore, the context is necessary. Your oversimplification of my question is therefore inaccurate!

I'm sorry, my campaign on meta? You can see from my user profile that I have had no activity on meta for months. "Better" and "worse" don't matter, either way the question is about the same mathematical fact. Regardless, that's not why this question should be closed. It should be closed because it's not about specific board or card games and is therefore not on-topic on this site.

@murgatroid99 Your campaign on meta was both good and helpful. I was merely pointing to the fact that it drew extra attention to the question, and therefore a higher number of close votes if you assume a constant rate of close votes per viewer. Are you saying that every question on this site must be about a specific board or card game, and that questions about games in general are not welcome here? If so, then I sorely misinterpreted the topic of this stack exchange.

I seriously have no idea what meta campaign you are talking about. In any case, the on-topic help page says "For a question to be on topic, it must relate to a game that is on topic."

@murgatroid99 You collected like 7 questions into one post on meta and then called for them to all be retagged. I call that a campaign, but you want to argue linguistics so we'll call it a "thing". Moved discussion to meta: http://meta.boardgames.stackexchange.com/q/1167/6692

@Rusher You've made a mistake - it was me who opened that meta topic. And, point of order: it's not a campaign, just a tag cleanup, which is a pretty common thing to do on SE sites. That and your question has been sitting on the front page for days; I doubt the meta topic was a big contributor to the close votes.

I'm not sure this theory, as you initially state it, is meaningful for most kinds of turn based game. Consider a chess board where white has a rook and a king, and black has a queen and a king. White goes first, and uses his rook to capture the queen. In the following turns, he will force a checkmate. Even though black had more options in the starting state, he loses.

This is in fact a well-established theorem in Conbinatorial Game Theory; it's a piece of proving the so-called Simplest Form Theorem.  If I remember when I get home I can chase down proper references, but the canonical source is the book Winning Ways For Your Mathematical Plays.

@Kevin In your hypothetical situation, the black player does not actually have all of the options available to the white player. The white player will change the board state before the black player even gets to move, and so the black player's future options are unknown. An alternative way to think about it is that the black player only has a single immediate option: Do nothing until the turn ends. In other words, your assumption that black had the option to move his Queen, or would ever have that option in the future, is false.

Yes, I agree. I had similar thoughts while writing that comment. So, generally, for turn based games where the active player can make a choice that limits the other player's options, it's very hard to characterize both players as "having all the same options, except one option that only one player always has", since both player's options fluctuate up and down depending on whose turn it is.

I happened to come across this Question a few days ago, and then entirely unrelatedly came across this paragraph in the wikipedia article on the GameTheory of the game Chicken: One tactic in the game is for one party to signal their intentions convincingly before the game begins. For example, if one party were to ostentatiously disable their steering wheel just before the match, the other party would be compelled to swerve. This shows that, in some circumstances, reducing one's own options can be a good strategy. Link: https://en.wikipedia.org/wiki/Chicken_(game)#Pre-commitment

This sounds a lot like a strategy-stealing argument. It's often used to prove that one side of a game has an advantage, even if optimal strategy is not known. As a broad example, consider a game with the following properties:

It is a turn-based game
The starting positions are identical except Player A goes first
Player A has the option of skipping the first turn

Assuming optimal play, the first player cannot be at a disadvantage. The proof is a strategy-stealing argument: Any hypothetical strategy that Player B could use to get an advantage could be countered by Player A just skipping the first turn and then using that exact strategy.

Is this principle even true? Consider the following game:
A, B, and C, are dividing $60. All players left standing at the end of the game receive an equal split.
Each player may shoot one other player. If a player gets shot twice, that player can no longer participate in the split.
Alternatively, C has the option to shoot another player, and then (after the first shots resolve) shoot both other players twice.
Obviously, the optimal strategy for C is to take the second option, but then the optimal strategy for A and B is to both shoot C (if they don't, they will get nothing). This means that if all players act rationally, A and B will get $30 each and C will get $0.
In other words, C having more options leads to a worse result for C. So the theory isn't even true, which I think is decent evidence that it doesn't have a name, especially given that nobody else who's come here so far has heard of it. :-P

First, this isn't even an answer. Second, your counterexample involves more than two players. The theory still holds for two players. I edited the question to clarify. I know that it invalidates half of your answer, but it should have been a comment to begin with.

SE sites don't usually let you comment on questions with only 1 reputation. Is this one different? I know this isn't quite what you were looking for, but I thought it was worth sharing.

I think the game must also have no hidden information and no randomness. I can elaborate if you're interested.

I didn't mean to imply that this site was any different. If the choice was post in the wrong place, or don't post at all, I would choose the latter. But since you already posted, if you still want to share then I am interested.

This place doesn't work like a forum - check our [tour] for more information. Answers are expected to answer the question (in this case, what's the theory called?), whereas this just seems to just be discussing the theory itself. You're welcome to discuss that as part of an answer that otherwise answers the question, though.

This seems like a valid answer to me. If the theory really didn't have a name, what would the correct answer look like? "I read every game theory paper in the world, and none of them had this theory in them"? The next best thing you could do is a probabilistic argument: "theories that aren't true are usually not named; this theory isn't true; therefore, it probably doesn't have a name". (whether his disproof of the theory is correct or not is another matter; even if it wasn't, the post would still be an answer, just a wrong answer.)

If this player gets shot once, he's going straight to the hospital. I probably shouldn't play games where I can get shot.

Moderator: This answer passes the bar for being an answer to the question.  It can be downvoted as not useful, but it shouldn't be deleted.

This doesn't pass the bar. At what point is this actually answering the question of what the theory is called? Even if it's an inaccurate or incorrect theory, it still may have a name. If the theory doesn't have a name, there either is no answer, or the answer is "it doesn't have one, and I can say that because {insert authoritative citation here}" - so, yes, the answer @Kevin describes is roughly what a correct answer would look like. Not having a name doesn't turn this into an opportunity for this to be a discussion forum.

I'm not deleting this. Attempting to prove the theory invalid as an argument for why it doesn't exist is a reasonable answer. Whether the proof is any good is a different question. Feel free to vote appropriately.


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What is the optimal strategy to solve Hanoi Solitaire?
I recently discovered for the first time a variation of Solitaire called Hanoi (sometimes spelled Hanoy or Hanoii). My Solitaire app says it's rated easy, but I have never solved it because I only get so far before I get stuck. Is there a simple strategy I can employ to finally start winning?

Can you point is to your solitaire app?  Is this the card solitaire version or the disk version?  What is the goal?  I suspect that I can provide the answer but need more detail.

It's the card version. 9 cards of one suit are arranged randomly in 3 stacks, and the object is to sort them into one stack by moving only one card at a time and never placing a higher ranked card on top of a lower one.

Does the initial setup possibly have cards that are not in a legal position at the start of the game?  Example: 5 on top of 2.

Yes, as a matter of fact it does.

The biggest piece of strategy is knowing how to move a pile of cards from one stack to the next.
The basic algorithm is a recursive algorithm.
To move n cards from stack a to stack b, you need to:

move the n-1 cards from stack a to stack c
move the nth card to stack b
move the n-1 cards from stack c to stack b.

To move a stack of three from a to b you need to move the two cards from a to c.   To move the two cards from a to c you need to move the one card from a to b.  Move the second card from a to c.  Move the first card from b to c.  Move the third card from a to b.  To move the two cards from c to b, we need to move the first card from c to a.  Move the second card from c to b.  Move the first card from a to b.
The key strategy is to keep one ordered stack.  When a new card is unearthed, put it in to the stack in the order that it belongs by moving your stack that belongs on top of the new card to it and then moving it to the top of your ordered stack.
This strategy will not likely lead you always to the shortest solution but will lead you to a correct solution.
There are more basic strategies for the Towers of Hanoi peg and disc game on the Wikipedia article http://en.m.wikipedia.org/wiki/Tower_of_Hanoi but I'm not sure how explicitly applicable they are to this game.

Actually, I've had a look at it the past few days and have come up with some strategies of my own. If I focus on the little parts, which fits well with the idea of recursion but less technical for the layperson, and if I have different moves for the top card in an odd stack versus an even stack, I've found I can finish the game quite easily now. For higher cards such as 9 which are somewhere in the middle, it helps to move one of those cards to an empty space as soon as I can.

make a table called input-output and rule and ask your kids to grab a deck of cards and only pick out the ace,the 2,3,4 the rule is to double each number, for example, double 12 = 24.start with the bottom