Seeing (in "The most difficult yose problem" topic) how easy it is to build a long sequence by using dame and pass moves I am wondering what is the longest problem (and maybe the most easiest problem ?) we can build in japanese rule (where all dame has to be filled to avoid a seki). The only constraint is to prove that each move can be considered as the best move (I mean no move loses points).

Assume a komi = 0.5 points.

Without giving you my best record I managed to get more than 300 moves. What is your record?

This restriction alone is NOT sufficient!!!

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The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

What do you mean? Be aware that with other restrictions the resulting problem could be a completly different one.

Just an example on a small board.

Click Here To Show Diagram Code

[go]$$ Black to play, komi 0.5
$$ +-----------+
$$ | . . O . . |
$$ | O O O X . |
$$ | . O . X . |
$$ | . O . X O |
$$ | . O . X . |
$$ +-----------+[/go]

Click Here To Show Diagram Code

[go]$$B pass
$$ +-----------+
$$ | . . O 1 . |
$$ | O O O X . |
$$ | . O . X . |
$$ | . O . X O |
$$ | . O . X . |
$$ +-----------+[/go]

After it is clear that black will win by 0.5 points but black cannot pass because the japanese rule will declare a seki and white will win according to the komi.

Black must continue the game :

Click Here To Show Diagram Code

[go]$$B pass, pass, pass
$$ +-----------+
$$ | . . O 1 . |
$$ | O O O X 9 |
$$ | . O 3 X 8 |
$$ | . O 5 X O |
$$ | . O 7 X . |
$$ +-----------+[/go]

and black wins by 0.5 points after 9 moves.

Note the constraint I mentionned : neither player made a mistake.

Something like
https://senseis.xmp.net/?MolassesKo

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

That's a good point Cassandra. Molasses ko is a very special position which changes completly the logic of the go. In that sense it is a good idea to treat molasses ko independantly. I will try to find another sequence with a molasses ko.

BTW I forget to mentionned an (obvious?) point : the potential winner of the game tries always to gain in the minimum number of moves.

Have you considered trying constructions like the following? I have not attempted to maximize this construction or prevent having value-gaining moves on the rest of the board, this is just an attempted proof of concept. Every one step in the ladder serves as a threat for the entire ko chain, so the longest resistance path results in *multiplying* the two together.



I seem to remember seeing a construction sort of like this that achieved tens of thousands of moves somewhere on the internet, but it was many years ago and I don't know how to find it again.

If I understand correctly your example there is some misunderstanding: my problem does not allow non optimal moves (I mean a move which loses some points). In your example if black wins the ladder I presume that playing the ladder with white cannot be the best moves for white because white will probably lose points won't she?

In general, when a ladder takes place, one of the players has to avoid playing this ladder => in general a ladder does not allow to play long optimal sequences but only allows long sequence before the capture of a group of stones which is quite different.

You can make the ladder entirely inside an area that would be guaranteed territory for black except for white being able to try this resistance. Then there is no loss, it is score-optimal play. But under white's longest resistance, black is only guaranteed to own this territory if they do continue to follow along every time white runs the ladder or plays the ko snake, otherwise of course white gains a lot compared to optimal play and white wins instead.

Then this means you can easily construct 10000+ long move sequence where if black wants to win the game, black must follow the entire sequence and can never pass until the finish, and if white wants to resist as long as possible, white can indeed force the entire 10000 moves, and in the entire sequence both players never lost points.

You can also pick any other method you wish besides a ladder of giving one player a very large but finite number of ko threats, and where all the ko threats are non-point-losing ko threats. A ladder inside someone's territory is simply one way of doing so that can be made particularly dense and numerous.

The point is that every ko threat now adds the entire length of the double-ko-death-like ko snake in additional moves to the longest variation. So this ko-snake construction allows you to multiplicatively expand the number of forced moves that a final sequence of the game takes, easily to many thousands of moves.

I have this one: https://online-go.com/game/41383743

Ooh thanks! Ah ok shorter than I remembered, not 10000+ moves at all. Still, 3600 moves isn't bad. May need some more modification or a redesign since it doesn't meet Gerard's criteria, but certainly this is suggestive that 1000+ moves should be possible even with Gerard's criteria with the right effort and construction.

Put four 8-tuple kos on the board each with two ko mouths open for the same player. Under superko-like rules and according to a theorem by Spight, Rickard, Davies, a correct sequence has some 19,580,000 moves.

OC this is interesting.
Actually in my first post I stated we are under japanese rules which is different.

Play in Berlin tournaments using Japanese rules and positional superko.

EDIT: Alternatively, of course, Japanese no-result rules permit the players to play such a long cycle to create a no-result and possibly win a tournament if, e.g., that game is treated like a jigo. It may require stamina though.

I begin to look at sequences using molasses ko but I am not sure how to count the number of moves with this special configuration.
Let's try on a small example. How many moves white can impose to black? What is your understanding of what is allowed by the japanese rule?

Very good point Lightvector. I understand.
Before trying this idea what about solving a far more easier problem. What is the longest winning sequence without any capture?

I think if you are using Japanese rules, a long cycle would mean that the game goes on forever, so the sequence is infinite. I believe that the meaning of »no result« is that the players don't finish it.

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A good system naturally covers all corner cases without further effort.

Yes Harleqin I know. That is the exact reason why my question was "What is the longest WINNING sequence?".
If one player can force a "no result" then the sequence cannot be a winning one, can it?

It is not quite clear for me what is the result of a molasses ko under japonese rule.
It looks to me like a "no result" game. Is it true?