2 moves or NOT?

i don't think it's possible for white to mate in less than 10 moves.. i might have overlooked something though..

edit: i think it's possible for white to mate in about 5 moves.. but more likely 8..

1. hxg5 hxg5 2. Ke3+ (discovered check from the bishop) Kh6 3. Rb7 g4 (forced move) 4. Bxg4 g5 5. Re6# (but if 4... Kg5, it takes much longer i think.. 5. Kf3 Kh6 6. Kf4 g5+ 7. Kf3 Kg6 8. Re6#

1. Rf4 and Black is in zugzwang and must move the Bishop.

If 1... Bf2 then 2. Kxf2#

If 1... Be3 then 2. Kf2#

If 1... Bishop anywhere else (including Bxh4) then 2. Ke3#

 hmm you're right.. as i said.. i would have overlooked something.. silly me, kept wanting to take the queen.. @.@

 i think there's something wrong with your moves.. the bishop's on e1, how to go Be3?? and in the 2nd line, if Bd2, it's not possible to go Ke3.. it has to be Ke4.. in fact, i think the 2nd line is better like this:

If 1... Bishop anywhere else (including Bxh4), then 2. Ke4#

After Qxb5 the queen will block on e2 so it is not mate in 2, but 3.

What about 1. Rxe1? That basically leaves black with 2 options: 1...Qxb5 or 1...Kxh4.  If Kxh4, then 2. Rh1 is mate. If Qxb5, then 2. Kf4+, Qe2 3. Bxe2#.

Unless I'm missing something, it seems that mate is forced in 3 moves, but could happen in 2 if black moves his king. Can anyone find a line that refutes this?

Oops, that should read:

If 1... Bd3 then 2. Kf2#

not

If 1... Be3 then 2. Kf2#

Note that if the Bishop is on d3 then Ke4# isn't available, which is why Kf2# is played instead in this case.

no.. mate can be forced in 2.. if we go 1. Rf4, black HAS to move his bishop or use the queen to take on b5.. then moving the king would simply force mate (by discovering a check from the bishop)..

Got it! Rf4 is not correct. How about Rbe5... I think that solves. If Qf5+ Ke3# and if Qxe5 Ke3#! if the Bf2 Kxf2#  if Bg3 Kg2#  if Bxh4.... uh wait I thought... oh noes!!!

fail.

Maybe the answer is, not.

how Rf4 fails:

1.Rf4 Qxb5 2.Kg2+ (or Ke4+) Qf2+ and there is no mate in two in that line.

Ahh, you are correct sir.  Back to the drawing board.

 hmm?? Bd3 is also not possible.. bishops can only move diagonally, so since it's on e1, the only moves are Bf2, Bg3, Bxh4, Bd2, Bc3, Bb4 or Ba5.. and in any case, Ke4# would be the correct one.. look carefully ;)

you mean 2... Qe2+ i assume?? but you're right.. it delays mate for another move..

and i've yet found mate in 2.. maybe there isn't any??

its quite easy really white will move his king to the only square it can go (f3-e2) black cant do anything about it so the only thing black thinks is right is to get the only pawn white has out of the way with his bishop (d1xh4) so all blacks peices is now around the queen now all white has to do is move his king and it will be mate 

                                               or

white will move his king to the only spot open  (f3-e2) black will take his rook check (g5xb5+)  white will of course move out of the check anywere possible because the bishop on square d1 is going to mate no matter were is king goes as long as its not f3 this way hes blocking the discover check which is mate

mwb21089, your solution has the same problem as mine: in your second variation black has the delaying 2... Qe2+

Well, seeing as the title says '2 moves or NOT?', then the answer is NOT. And, as we have seen, there are several ways to mate in 3. So there we go. 

i think the logical move for black would be the second variation.. nobody holding black would ever take a pawn (and trap his own king) instead of taking a rook.. and as TheGrobe mentioned, black can delay mate by 1 move by 2...Qe2+

btw, the more i look at it, the more i think it's impossible to mate in 2.. the best is mate in 3..

Wait a minute, what about 1. Rc4?

Any Bishop move would follow my original line [edit: with Ke4 in place of Ke3], and any Queen move can be met with either Kg2# or Ke3# depending on the circumstances.  Most importantly, if 1. ...Qxb5 2. Ke3# is checmate becasue the rook on c4 now blocks the delaying interposition.

Can anyone see a flaw in this?

Important to note that Bg3 or Bxh4 is met with Ke4 (which has now been vacated by the rook) in the above solution.

Great job, Grobe! Well done. I love it when the fail-trial near-successes slowly lead to insight into the position and ultimately the solution.

I also love that no one blew-up our fun with a computer generated answer, probably takes Rybka on a decent machine less than .5 seconds to find this.