Six (6) months ago, I played a game with Carcer (whose blog is now linked to on the right) with six Chinese handicaps to show that manego (or mirror-go) cannot be beaten. A lot of things were tried in that game, including sacrificing a whole ton of stones to play under them. Nothing worked. A few hours after that game, I found out I was wrong. It's been a while, but I recently recalled the solution and here it is:
Looks intimidating, doesn't it?
Here's the proposed 'cannot be beaten' manego.
Still looks intimidating, doesn't it?
What we'll do is take advantage of Black's existing stones. Anything that turns out differently for both sides playing mirrored moves is a step towards breaking mirror go. In this case, we force Black to capture a stone with just three (3) moves.
Giving a second stone? Really?
Same thing here. We force Black to capture another stone with just three moves. Where are we going with this?
Aha! It's broken! ...
Yes! We have what we want! Black cannot play another move to keep a mirrored board! ... or so we think...
Ah zut! (darn in French)
Black has yet to play the traditional tengen which starts the manego, so he can play it now. What to do?
Try and try again
If it worked before, it'll work again. This time we let Black capture two (2) stones with four (4) moves, while White captures one (1) stone with the same moves. We'll repeat this process again in the next diagram:
Finally broken!
Yes, this time it is finally broken. Other than a pass, which I consider illegal in a real unfinished game of go, Black has no move to keep the board mirrored. Black will have to think for himself.
Aftermath
Analysis? It's not as pretty for White as breaking less than six stones manego. He doesn't get to capture tengen. So I'd say Black's mirror go certainly paid off. White must do this sequence early in the game because it does take this many moves to break it. (As far as I can tell.) Considering this is a six handicap game, White might be able to come back, because this isn't a lopsided board. These moves are forced, and that's nearly a sixth (1/6) of the game already played, so Black should feel good; he's managed to keep ahead. In most 6hc games, Black will lose a lot of his advantage in the opening. Black ends here in sente, covering the center of two thirds (2/3) the board. Where would I play from here? One of the star points on the sides, not the corner. This helps get into White's side and marks out a huge portion of territory for himself on the sides. The corner is really not that big now with that mass in the middle.
White's battle is tough, but he can do it if the difference in strengths really is six stones.
Note that this solution works for eight (8) Chinese handicaps of similar configuration. In that case, Black would instantly lose two stones because he'd end up with the same result as these six handicaps.
btw: five (5) and seven (7).
Update: I just had to look... sensei's has a
better solution. Look all the way at the bottom for the 6hcp example. In the solution, you kill the six handicap stones too!
