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Problem ID: 0021252
it's a double solution in the 2nd move, for there are no difference between 2... Qh5 and 2... Qh6. However, if u put white Q at c1, it's a difference!
I couldn't spot my mistake either after Qh5...
There are two solutions as 2. ... Qh6 comes to exactly the same thing.
Adding a black pawn to h6 seems a good and natural-looking way to save the problem.
LydiaBlonde and DarkPhobos are right!
New game3r2k1/1b5p/pB3pq1/1p6/8/1P1p2P1/P2NrP2/1Q1R1RK1 b - - 0 1Analysis by Rybka 3 32-bit-4:1. -+ (-#3): 2...Qh6 2. -+ (-#3): 2...Qh5 (, 03.01.2010)
The problem is changed now. The comments above are not valid any more.
Disappointed I missed this. Maybe I outta get some rest.
Doesn't 1... Rxd2 2. Rxd2 Bf3 provide a crushing advantage?
Probably not. You figure that if you try to move the queen to h3, you drop d3, and then f3 is attacked, and then f3 is possible as a resource. That simply doesn't work.
OK, maybe d3 drops, but not until after Black collects the house with 1... Rxd2 2. Rxd2 Bf3 3. a3 (what else?) Qf5 4. a4 (what else?) Qh3 5. Rxd3 Qg2#.
In hindsight Black may need to interpose 2... Bb4 to discourage 3. Rfd1.
Well first your rook on d8 drops, if anything else, or Rxd3 straight away (Q and R on), then your bishop has to move on f3...? I was thinking more of Rxd2, Rxd2, Qh6, but that didnt work :S
I mean i think it does work, just not as good as solution
darn, I tried Rxd2 as well!
Misclicked and played Rxh2 lol
so easy cause i know windmill
lol, i went 2.qg3
if hxB Q gives mate on h1?
Qh5 still works no?
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