Day #37 December 11th 5 yrs ago no solutions to the advanced problem were posted. Maybe because the issues with the previous series. So solve it this time on your own risk as in real life.
This ko is better for black. If white takes the ko and black takes the two stones, white still has to connect. In the other ko, white connected, then black took 2 stones.
Prodigious wrote:Here's an unconditional solution .........
Wow. you make me happy. OC I tried to refute your solution but the next variant for fails.
$$Bc a&b are miai for B after this sequence $$ +--------------------- $$ | . X . . . . . . . . $$ | O O O O O X . X . . $$ | O X X O X X . . . . $$ | X . X X O O . X X , $$ | . X X O B . O O . . $$ | . 2 3 O . . . . . . $$ | . O 1 O b O . . . . $$ | . a 4 5 . . . . . . $$ | . . . . . . . . . .
[go]$$Bc a&b are miai for B after this sequence $$ +--------------------- $$ | . X . . . . . . . . $$ | O O O O O X . X . . $$ | O X X O X X . . . . $$ | X . X X O O . X X , $$ | . X X O B . O O . . $$ | . 2 3 O . . . . . . $$ | . O 1 O b O . . . . $$ | . a 4 5 . . . . . . $$ | . . . . . . . . . .[/go]