Matti wrote:
Gérard TAILLE wrote:
Matti wrote:
In addition the martian problem is supposed to start with a position ignoring all the previous moves. As a consequence you cannot make the assomption that a move is unplayable due to a ko.
You didn't state this in the problem.
- Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X . X O |
$$ | X X O c . O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O . O |
$$ | O O O . O X W . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X b O X . X |
$$ | X . . . X X O O X . |
$$ | X X . X X a X O X X |
$$ -----------------------[/go]
Yes Matti, of course I realize that I did not mentionned this point, sorry for that.
Anyway, taking your point into account, I looked for this version of the martian problem and I concluded that this new problem is far easier that the original one, though it is quite interesting. Instead of just excluding a move due to a simple ko you can also chose to exclude a move due to a superko.
With this in mind I built the following solution:
White to play
- Click Here To Show Diagram Code
[go]$$W
$$ ------------
$$ | . . O O O |
$$ | O O O O O |
$$ | O O X X X |
$$ | . O X X X |
$$ | O X X X . |
$$ -------------[/go]
starting from this position white captures all black stones and continue until position:
- Click Here To Show Diagram Code
[go]$$W
$$ ------------
$$ | O O O O O |
$$ | O O O O O |
$$ | O O O O O |
$$ | . O O O O |
$$ | O O O O O |
$$ -------------[/go]
here black captures all white stones and now you can build the following very simple martian position
Black to play
- Click Here To Show Diagram Code
[go]$$W
$$ -------------
$$ | . . O O O |
$$ | O O O O O |
$$ | O O X X X |
$$ | X O X X X |
$$ | a X b X . |
$$ -------------[/go]
japonese rule : black must pass to avoid giving an additional prisoner
AGA rule : black must play "a" to save its stones and reach soon a seki
chinese rule : black must play "b" to reach an immediat seki because white cannot answer "a" due to superko
Note : for the original martian problem I searched for a solution with a molasses ko but I had to give up. The two solutions I found do not use this molasses ko scheme.