Why is this tsumego solution wrong?

Question

I'm trying to understand why my solution to this tsumego is incorrect. It's "Basic problem #26" from Tsumego Pro and looks like this:

$$ ----------------------------
$$ . . . . . . . . X . O X . . |
$$ . . . . . . . . X O . O O O |
$$ . . . . . . . X . X O O X . |
$$ . . . . , . . . . X X X X . |
$$ . . . . . . . . . . . . . . |
$$ . . . . . . . . . . . . . . |
$$ . . . . . . . . . . . . . . |

The built-in correct solution is this:

$$ ----------------
$$ . . X 3 O # 1 2 |
$$ . . X O . O O O |
$$ . X . X O O X . |
$$ . . . X X X X . |

(alternatively, after (2), with black playing at the marked position again and white taking once more by playing at (1), and only then black playing at (3)). I understand why this solution is correct, just not why mine isn't.

The following is my solution, which the program marks as incorrect.

$$ ----------------
$$ . . X 1 O X 2 . |
$$ . . X O . O O O |
$$ . X . X O O X . |
$$ . . . X X X X . |
$$ . . . . . . . . |

(white's response at (2) is built into the program, the rest of the white moves aren't).

$$ ----------------
$$ . . X X O . O . |
$$ . . X O a O O O |
$$ . X . X O O X b |
$$ . . . X X X X . |
$$ . . . . . . . . |

If black now takes the ko at (a), the only thing white can do (that I can see) is play at (b) in order to try escaping on the right side. It can't play anywhere on its inside because that would essentially be suicide, and because for black winning the ko means connecting its two groups, I don't see any possible white threat that black would have to respond to.

$$ ----------------
$$ . . X X O . O . |
$$ . . X O 3 O O O |
$$ . X . X O O X 4 |
$$ . . . X X X X . |
$$ . . . . . . . . |

$$ ----------------
$$ . . X X O . O . |
$$ . . X 5 X O O O |
$$ . X . X O O X O |
$$ . . . X X X X 6 |
$$ . . . . . . 7 8 |
$$ . . . . . . 9 0 |

This could now continue all the way to the bottom of the board (not shown), at which point the white group has no liberties left except for the eye-and-a-half at the top. If white decides to stop running away, I still don't see anything forcing it could play to stop black from closing (e.g. at (10)).

In the end black can take:

$$m37
$$ ----------------
$$ . . X X O 1 O 3 |
$$ . . X X X O O O |
$$ . X . X O O X O |
$$ . . . X X X X O |
$$ . . . . . . X O |
$$ . . . . . . X O |

with no useful response (38) for white.

What is it that I'm missing? Where could white have lived?

Answer 1

In tsumegos where you should kill a group, the best (and thus correct) answer is always the solution where you kill unconditionally (if possible). Only if there is no unconditional kill the ko would be the best (correct) solution.

The basic idea is that a tsumego is a local fighting position which you should solve without knowledge of the rest of the board. In your example, the rest of the board could either be that white has more ko threats or that black has more ko threats. Thus, your solution would depend on the unknown rest of the board, whereas the unconditional kill works always.

In general, if there is a position with equal outcome you should always prefer the way without ko. Even if you have more ko threats now, the position might change over time and therefore it is better to avoid the ko from the beginning.

Answer 2

The Black move 1 reduces the "three in a row" from two eyes to one. White cannot protect his "second" eye at 3 until he kills the marked stone (in red) with 2, because it would be self atari. Black then quashes the potential second eye with 3, and White is dead (for sure).

If Black plays 1 at 3, White will play in the middle of "three in a row" to create a sure eye in the corner, and "half" an eye on the third space from the edge. To prevent the "half" eye from becoming a second eye, Black has to fight a ko at A. He's likely to win it, and kill the White group, but White will be able to make a ko threat elsewhere on the board to get several points of compensation. Having to fight a ko to kill the White group (and losing "several" points) is not nearly as good as killing it for sure, as above, without additional loss, which is why the second answer is incorrect.