Mid-summer endgame problem

Bill Spight · Post by **Bill Spight** » Sun Jul 28, 2013 11:22 am

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

No komi. White to play and win.

Enjoy!

Edited to add coordinates. I hope I didn't mess up your comment, jts.

jts · Post by **jts** » Sun Jul 28, 2013 3:52 pm

Kirby · Post by **Kirby** » Sun Jul 28, 2013 8:50 pm

This is not a cute problem!

As I see it, there are something like 8 areas to play here. In a real game, I could try evaluating the score by iterating through each of the permutations of moves to the end of the game, but that's not feasible.

After everyone's answered, instead of getting a solution, could it be possible to explain the process of deducing the precise optimal solution from this example?

lightvector · Post by **lightvector** » Sun Jul 28, 2013 9:45 pm

Ooooh, I think I got it. I think there's a good chance this is the answer, though I might have overlooked a detail.

Knowing Bill, this is going to be a tedomari problem. So let's first make an observation:

Click Here To Show Diagram Code: [go]$$Wc Two different kinds of 1-point moves. $$ --------------------------- $$ | . . O X O . . . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X a O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X . . . X X X O . | $$ | W O X . O O O O O O O O . | $$ | X b X O X X . X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X . O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

In the above position, with the marked white stone added, consider "a" and "b". They're both worth 1 point, but they're different. Whereas "a" is a simple 1 point gote, "b" is asymmetric. If black plays first, it's finished, but if white plays first, it leaves behind another 1 point gote.

Now, if you're trying to get tedomari and snag the last 1 point play, who is this asymmetry good for?

I think it can only be good for Black. For example, if "a" and "b" are the only two endgame moves, black is guaranteed the last 1 point play regardless of who goes first. If black plays first, he takes "a", then white takes "b", leaving one last move for black. If white goes first, then regardless of what he takes, black will take the other. Positions like "b" hinder white from getting the last play because you can only get the last play in cases where your move actually finishes the position, but if white tries to play at "b" it only generates another endgame move for black rather than finishing it.

So in general, if there are asymmetric endgame moves like "b", then due to possible tedomari, they are potentially better for the side that can finish the play there.

Now, keeping this in mind, let's calculate the miai values of all the moves.

Click Here To Show Diagram Code: [go]$$Wc Wow there are a lot of choices! $$ --------------------------- $$ | . h O X O . d . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X e O . | $$ | O O O O X . . j . X O O . | $$ | O X X X X g . . X X X O . | $$ | b O X a O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X c O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

a: 2 points. If white plays, he leaves a 1 point gote.
b: 2 points. If white plays, he leaves behind a 1 point asymmetric move favoring black (because black's play afterwards finishes it while white's play afterwards leaves yet another move).
c: 1 point. This is an asymmetric move favoring white. Black must play 3 times in a row (each worth 1 point) to finish the position, whereas at any of those stages a single play by white finishes the position.
d: 1 point. This is an asymmetric move favoring black. Black playing 'd' finishes things up immediately after white ataries and black captures, whereas white playing 'd' leaves an extra move there.
e: 1 point gote.
f: 7/8 point. It's one of those corridors you learn about on sensei's library.
g: 7/8 point. I think "g" is the correct move for both white and black, based finding this position on sensei's library and by working out and verifying the variations as follows:
* If white plays there, black can make 1 point in gote afterwards, whereas white could play "j" to reduce it to 0. So if white plays first, the position is 1/2 point for black.
* If black plays first, then an additional play by black makes 3 points in gote, whereas a play by white reduces it to 1.5 points in gote. So if black plays first, the position is 2.25 = 9/4 points for black.
* So then we take (9/4 - 1/2) / 2 and get that a move is worth 7/8 for the original position.
h: 1/2 point gote.

Okay, now knowing the above, we can deduce that best play should go something like this:

Click Here To Show Diagram Code: [go]$$Wc First grab the 2-point moves. $$ --------------------------- $$ | . h O X O . d . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X e O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X g . . X X X O . | $$ | 2 O X 1 O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X c O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

White grabs a big 2-point move and black takes the other. White carefully makes sure to take this one with :w1:

rather than the one on the left edge, because it's better to leave a plain gote move rather than an asymmetric move that favors black.

Click Here To Show Diagram Code: [go]$$Wc Get rid of black's advantageous asymmetric move as quick as possible. $$ --------------------------- $$ | . h O X O . 3 . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X e O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X g . . X X X O . | $$ | X O X O O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X 4 O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

Now we're on 1-point moves. As soon as possible, :w3:

gets rid of the asymmetric black-favoring position at the top, transforming it into a 1 point gote. Black tries to do the same to white with :b4:

.

Click Here To Show Diagram Code: [go]$$Wc White gets the last 1 point move $$ --------------------------- $$ | . h O X O 5 O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X 9 O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X g . . X X X O . | $$ | 2 O X 1 O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X 7 O X . X . 8 6 O . | $$ | . X X X X X . X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

But white's is better than black's, because black needs to play there yet again with :b6:

to convert it to an ordinary 1 point gote. So when the remaining 1 point gote moves trade off, white gets the last one with :w9:

and wins tedomari.

Click Here To Show Diagram Code: [go]$$Wcm9 Now the corridor and that weird center area $$ --------------------------- $$ | . h O X O O O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O . O . | $$ | . O . O X X X X X X O O . | $$ | O O O O X . . 6 . X O O . | $$ | O X X X X 2 4 5 X X X O . | $$ | X O X O O O O O O O O O . | $$ | X . X O X X 3 X X X O . . | $$ | . X X O O X 7 X . X X O . | $$ | . X X X X X . X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

Now, I think it doesn't matter which of the two 7/8-point moves black takes with :b10:

. I tried a few variations and they all seem to come out the same. Either way, white takes the other one, and we end up with the same result.

Click Here To Show Diagram Code: [go]$$Wcm15 White wins $$ --------------------------- $$ | . 2 O X O O O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O . O . | $$ | . O . O X X X X X X O O . | $$ | O O O O X . . X . X O O . | $$ | O X X X X X X O X X X O . | $$ | X O X O O O O O O O O O . | $$ | X . X O X X O X X X O . . | $$ | . X X O O X O X . X X O . | $$ | . X X X X X 3 X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

And we are left with a 1/2 point gote that pairs off perfectly with the 1/2 point gote in the upper left. White wins by 1 point.

Bill Spight · Post by **Bill Spight** » Mon Jul 29, 2013 6:28 am

@ lightvector

lightvector wrote:Knowing Bill, this is going to be a tedomari problem. So let's first make an observation:

$$Wc Two different kinds of 1-point moves.
$$ ---------------------------
$$ | . . O X O . . . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O X O . |
$$ | . O . O X X X X X X a O . |
$$ | O O O O X . . . . X O O . |
$$ | O X X X X . . . X X X O . |
$$ | W O X . O O O O O O O O . |
$$ | X b X O X X . X X X O . . |
$$ | . X X c O X . X O O . O . |
$$ | . X X X X X . X X . O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wc Two different kinds of 1-point moves. $$ --------------------------- $$ | . . O X O . . . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X a O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X . . . X X X O . | $$ | W O X . O O O O O O O O . | $$ | X b X O X X . X X X O . . | $$ | . X X c O X . X O O . O . | $$ | . X X X X X . X X . O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

In the above position, with the marked white stone added, consider "a" and "b". They're both worth 1 point, but they're different. Whereas "a" is a simple 1 point gote, "b" is asymmetric. If black plays first, it's finished, but if white plays first, it leaves behind another 1 point gote.

Now, if you're trying to get tedomari and snag the last 1 point play, who is this asymmetry good for?

I think it can only be good for Black. For example, if "a" and "b" are the only two endgame moves, black is guaranteed the last 1 point play regardless of who goes first. If black plays first, he takes "a", then white takes "b", leaving one last move for black. If white goes first, then regardless of what he takes, black will take the other. Positions like "b" hinder white from getting the last play because you can only get the last play in cases where your move actually finishes the position, but if white tries to play at "b" it only generates another endgame move for black rather than finishing it.

So in general, if there are asymmetric endgame moves like "b", then due to possible tedomari, they are potentially better for the side that can finish the play there.

I think that usually you are right that it is preferable for White to take the play that leaves the STAR (*) at "c" instead of the UP (^) at "b". However, UP and STAR are confused, which means that it is possible to construct a problem in which it is right to leave the UP. Suppose, for instance, that the only play left on the rest of the board was the STAR at "a". As in the next diagram.

Click Here To Show Diagram Code: [go]$$Wc A different problem $$ --------------------------- $$ | . . O X . X O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X . O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X . . . X X X O . | $$ | 1 O X 2 O O O O O O O O . | $$ | X 3 X O X X . X X X O . . | $$ | . X X . O X . X O O O O . | $$ | . X X X X X . X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . O . O O X . . . . X . | $$ ---------------------------[/go]

Now

gets the last play, and a play at :w2:

does not.

Now, keeping this in mind, let's calculate the miai values of all the moves.

$$Wc Wow there are a lot of choices!
$$ ---------------------------
$$ | . h O X O . d . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O X O . |
$$ | . O . O X X X X X X e O . |
$$ | O O O O X . . j . X O O . |
$$ | O X X X X g . . X X X O . |
$$ | b O X a O O O O O O O O . |
$$ | X . X O X X f X X X O . . |
$$ | . X X . O X . X O O . O . |
$$ | . X X X X X . X X c O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wc Wow there are a lot of choices! $$ --------------------------- $$ | . h O X O . d . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X e O . | $$ | O O O O X . . j . X O O . | $$ | O X X X X g . . X X X O . | $$ | b O X a O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X c O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

a: 2 points. If white plays, he leaves a 1 point gote.
b: 2 points. If white plays, he leaves behind a 1 point asymmetric move favoring black (because black's play afterwards finishes it while white's play afterwards leaves yet another move).
c: 1 point. This is an asymmetric move favoring white. Black must play 3 times in a row (each worth 1 point) to finish the position, whereas at any of those stages a single play by white finishes the position.
d: 1 point. This is an asymmetric move favoring black. Black playing 'd' finishes things up immediately after white ataries and black captures, whereas white playing 'd' leaves an extra move there.
e: 1 point gote.
f: 7/8 point. It's one of those corridors you learn about on sensei's library.
g: 7/8 point. I think "g" is the correct move for both white and black, based finding this position on sensei's library and by working out and verifying the variations as follows:
* If white plays there, black can make 1 point in gote afterwards, whereas white could play "j" to reduce it to 0. So if white plays first, the position is 1/2 point for black.
* If black plays first, then an additional play by black makes 3 points in gote, whereas a play by white reduces it to 1.5 points in gote. So if black plays first, the position is 2.25 = 9/4 points for black.
* So then we take (9/4 - 1/2) / 2 and get that a move is worth 7/8 for the original position.
h: 1/2 point gote.

Well done!

As for "g", that'll teach me to put my discoveries on SL.

Okay, now knowing the above, we can deduce that best play should go something like this:

$$Wc First grab the 2-point moves.
$$ ---------------------------
$$ | . h O X O . d . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O X O . |
$$ | . O . O X X X X X X e O . |
$$ | O O O O X . . . . X O O . |
$$ | O X X X X g . . X X X O . |
$$ | 2 O X 1 O O O O O O O O . |
$$ | X . X O X X f X X X O . . |
$$ | . X X . O X . X O O . O . |
$$ | . X X X X X . X X c O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wc First grab the 2-point moves. $$ --------------------------- $$ | . h O X O . d . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X e O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X g . . X X X O . | $$ | 2 O X 1 O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X c O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

White grabs a big 2-point move and black takes the other. White carefully makes sure to take this one with rather than the one on the left edge, because it's better to leave a plain gote move rather than an asymmetric move that favors black.

$$Wc Get rid of black's advantageous asymmetric move as quick as possible.
$$ ---------------------------
$$ | . h O X O . 3 . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O X O . |
$$ | . O . O X X X X X X e O . |
$$ | O O O O X . . . . X O O . |
$$ | O X X X X g . . X X X O . |
$$ | X O X O O O O O O O O O . |
$$ | X . X O X X f X X X O . . |
$$ | . X X . O X . X O O . O . |
$$ | . X X X X X . X X 4 O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wc Get rid of black's advantageous asymmetric move as quick as possible. $$ --------------------------- $$ | . h O X O . 3 . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X e O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X g . . X X X O . | $$ | X O X O O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X . O X . X O O . O . | $$ | . X X X X X . X X 4 O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

Now we're on 1-point moves. As soon as possible, gets rid of the asymmetric black-favoring position at the top, transforming it into a 1 point gote. Black tries to do the same to white with .

$$Wc White gets the last 1 point move
$$ ---------------------------
$$ | . h O X O 5 O . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O X O . |
$$ | . O . O X X X X X X 9 O . |
$$ | O O O O X . . . . X O O . |
$$ | O X X X X g . . X X X O . |
$$ | 2 O X 1 O O O O O O O O . |
$$ | X . X O X X f X X X O . . |
$$ | . X X 7 O X . X . 8 6 O . |
$$ | . X X X X X . X X X O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wc White gets the last 1 point move $$ --------------------------- $$ | . h O X O 5 O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O X O . | $$ | . O . O X X X X X X 9 O . | $$ | O O O O X . . . . X O O . | $$ | O X X X X g . . X X X O . | $$ | 2 O X 1 O O O O O O O O . | $$ | X . X O X X f X X X O . . | $$ | . X X 7 O X . X . 8 6 O . | $$ | . X X X X X . X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

But white's is better than black's, because black needs to play there yet again with to convert it to an ordinary 1 point gote. So when the remaining 1 point gote moves trade off, white gets the last one with and wins tedomari.

$$Wcm9 Now the corridor and that weird center area
$$ ---------------------------
$$ | . h O X O O O . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O . O . |
$$ | . O . O X X X X X X O O . |
$$ | O O O O X . . 6 . X O O . |
$$ | O X X X X 2 4 5 X X X O . |
$$ | X O X O O O O O O O O O . |
$$ | X . X O X X 3 X X X O . . |
$$ | . X X O O X 7 X . X X O . |
$$ | . X X X X X . X X X O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wcm9 Now the corridor and that weird center area $$ --------------------------- $$ | . h O X O O O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O . O . | $$ | . O . O X X X X X X O O . | $$ | O O O O X . . 6 . X O O . | $$ | O X X X X 2 4 5 X X X O . | $$ | X O X O O O O O O O O O . | $$ | X . X O X X 3 X X X O . . | $$ | . X X O O X 7 X . X X O . | $$ | . X X X X X . X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

Now, I think it doesn't matter which of the two 7/8-point moves black takes with . I tried a few variations and they all seem to come out the same. Either way, white takes the other one, and we end up with the same result.

After

, the last 1 pt. move, the unplayed regions of the board add up to an integer (5 pts. for Black), with no kos. That means that correct play will yield 5 pts. for Black, no matter who plays first. (With plays worth 1 pt. or more, that would not necessarily be the case.

)

$$Wcm15 White wins
$$ ---------------------------
$$ | . 2 O X O O O . . . . . . |
$$ | O X X X X X O O O O O O . |
$$ | O O O O O X X O X O . O . |
$$ | . O . O X X X X X X O O . |
$$ | O O O O X . . X . X O O . |
$$ | O X X X X X X O X X X O . |
$$ | X O X O O O O O O O O O . |
$$ | X . X O X X O X X X O . . |
$$ | . X X O O X O X . X X O . |
$$ | . X X X X X 3 X X X O O O |
$$ | X X O O O X . X . X X X O |
$$ | O O . . O X X . . . . . X |
$$ | . . . . O O X . . . . X . |
$$ ---------------------------

Click Here To Show Diagram Code
[go]$$Wcm15 White wins $$ --------------------------- $$ | . 2 O X O O O . . . . . . | $$ | O X X X X X O O O O O O . | $$ | O O O O O X X O X O . O . | $$ | . O . O X X X X X X O O . | $$ | O O O O X . . X . X O O . | $$ | O X X X X X X O X X X O . | $$ | X O X O O O O O O O O O . | $$ | X . X O X X O X X X O . . | $$ | . X X O O X O X . X X O . | $$ | . X X X X X 3 X X X O O O | $$ | X X O O O X . X . X X X O | $$ | O O . . O X X . . . . . X | $$ | . . . . O O X . . . . X . | $$ ---------------------------[/go]

And we are left with a 1/2 point gote that pairs off perfectly with the 1/2 point gote in the upper left. White wins by 1 point.

daal · Post by **daal** » Mon Jul 29, 2013 7:16 am

Kirby wrote:This is not a cute problem!

As I see it, there are something like 8 areas to play here. In a real game, I could try evaluating the score by iterating through each of the permutations of moves to the end of the game, but that's not feasible.

After everyone's answered, instead of getting a solution, could it be possible to explain the process of deducing the precise optimal solution from this example?

Yes, and please show us the easy way to do it. :mrgreen:

Kirby · Post by **Kirby** » Mon Jul 29, 2013 2:14 pm

daal wrote:
Kirby wrote:This is not a cute problem!

As I see it, there are something like 8 areas to play here. In a real game, I could try evaluating the score by iterating through each of the permutations of moves to the end of the game, but that's not feasible.

After everyone's answered, instead of getting a solution, could it be possible to explain the process of deducing the precise optimal solution from this example?

Yes, and please show us the easy way to do it.

Well, easy or hard, I want to know the correct way to do it. I see people talking about some move being worth X points or this stuff being sente or gote, but it's confusing to me whether these methods give you the precise answer all of the time, or if it's really just an approximation.

I can understand reading out permutations, but I don't understand how this fits up with assigning point values, because as far as I know, this can lead to error... maybe?

Maybe this example is too hard, but as a general rule for these type of problems, I don't really know how to do them unless I just try to estimate the point values in each location, and maybe pick the biggest from each spot.

With the issue of sente/gote, I just try to pick sente moves whenever possible, defining sente as saying that the point value that I gain by playing again is greater than what he'd get if he took a different spot.

It all just seems imprecise, I feel, so I wonder if it's correct.

Based on Bill's responses to lightvector's answer, is lightvector's approach the correct way to do this kind of a problem? Is it exact?

At what point do you start reading out variations instead of calculating some sort of points for various areas? All in all, I'm confused.

Kirby · Post by **Kirby** » Mon Jul 29, 2013 2:41 pm

I'll give a simpler example to explain what I mean. Linked on the SL page for tedomari, I see the following problem:

Click Here To Show Diagram Code: [go]$$B Tedomari problem 1 $$ ------------------- $$ | . . X X a X O . . | $$ | X X . X O . O . . | $$ | X O X X O O O O O | $$ | X O O O . O X X b | $$ | O O . O O O O O X | $$ | . O O X O X O X X | $$ | . O X X X X X . X | $$ | O O c O X . . X . | $$ | . X . X X . . . . | $$ -------------------[/go]

The problem is small enough that *maybe* I can read out each combination of play to see who will win in the end.

But the solution shows this approach:
1.) Calculate miai values of each location - seems easy enough. Maybe this is simple to do following the way of calculating miai values.

2.) Get overall count - kind of makes sense how this is done, but I don't see what the point is in terms of finding optimal play.

3.) Get tedomari and win. The solution shows how using the greedy approach (the one I would usually use unless I can read it out exhaustively) fails, and how getting the last play gives you a score that wins the game.

----

OK, easy example, and I can see what is said is true, for the most part.

But this still leaves me confused:
1.) Why do you need overall count to play optimally?
2.) In this example, trying to "get" tedomari seemed to work to give you the right answer. But:
a. Is tedomari always optimal?

b. In general, is this the approach: calculate miai values, try to get tedomari? Let's say there are 10 spots on the board. So I calculate the miai value of each of those 10 spots and then I try to play to get the last play?

Basically, I wonder if this is imprecise, and when it fails, condition the method fails, and so on.

My current understanding:
* You can use miai values as a heuristic to estimate the order of play. Naïve way to go about this is to then take greedy approach, taking largest miai value as you get it.

* Tedomari is another heuristic that (usually?) works that lets you say, "OK. I'm not going to take the biggest moves in order, but I'm going to give up a big move so I can get the last play, and probably have more points..."

If my current understanding is correct, it would seem that my endgame approach should be as follows:
1.) If possible, read out entire game tree with all combinations to see best sequence (many search possibilities, but straightforward).

2.) If #1 is not feasible, calculate miai values for each spot. Then iterate all combinations of miai value plays until I see which gives greatest result. Is this approach optimal?

3.) If #1 and #2 are not feasible, try to figure out ordering that gives me tedomari. This approach is not always optimal...?

4.) If #1, #2, and #3 are all not feasible, order miai values largest to smallest, and play in that order.

5.) If #1, #2, #3, and #4 are not feasible... Try to keep sente.

6.) If #1, #2, #3, #4, and #5 are not feasible... Try to play big spots.

7.) If #1, #2, #3, #4, #5, and #6 are not feasible... Resign...?

---

Is this the best approach?

Basically, to me it sounds like we have a bunch of heuristics and stuff, and I want to define precisely when an approach is optimal, and if we don't have time for the optimal approach, the next best approach to getting the solution.

Am I making any sense, or am I just typing?

daal · Post by **daal** » Mon Jul 29, 2013 2:50 pm

Kirby wrote: All in all, I'm confused.

There's at least a bookfull of things to understand, and each of the 8 areas in the problem could probably fill a chapter.

edit: link to the quite useful example Kirby quoted above: http://senseis.xmp.net/?EndgameProblem24

Kirby · Post by **Kirby** » Mon Jul 29, 2013 3:01 pm

daal wrote:There's at least a bookfull of things to understand, and each of the 8 areas in the problem could probably fill a chapter.

...

I suppose that makes sense. Given that, are each of these techniques (tedomari, lightvector's mention of asymmetry, etc.) just estimates that help us refine our endgame skills to be more widely applicable to more and more problems, or are some techniques superior to others?

For example, if tedomari and this asymmetry stuff are both heuristics (are they?), then can I get a different candidate move by trying to get tedomari and by trying to avoid some sort of asymmetrical position? If so, is one usually superior?

I haven't really gotten the knack for when these techniques are precise and when they are guesses, so it's hard for me to tell the utility of using one method over another.

That's probably why I usually just try to play the moves that have the biggest miai value (if I can calculate it), in order.

xed_over · Post by **xed_over** » Tue Jul 30, 2013 1:14 am

Kirby wrote:That's probably why I usually just try to play the moves that have the biggest miai value (if I can calculate it), in order.

As I read lightvector's post, he's doing just that.. calculating the largest play(s) and playing those in order.

Hanmanchu · Post by **Hanmanchu** » Sun Aug 11, 2013 9:25 am

g: 7/8 point. I think "g" is the correct move for both white and black, based finding this position on sensei's library and by working out and verifying the variations as follows:

@ Lightvector, Bill Spight:

Where do you find the information on SL on how to calculate the value of "g"? I am asking because I got it wrong

Bill Spight · Post by **Bill Spight** » Sun Aug 11, 2013 2:27 pm

Hanmanchu wrote:
g: 7/8 point. I think "g" is the correct move for both white and black, based finding this position on sensei's library and by working out and verifying the variations as follows:

@ Lightvector, Bill Spight:

Where do you find the information on SL on how to calculate the value of "g"? I am asking because I got it wrong

http://senseis.xmp.net/?OneMoreNumber

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