# The most difficult 2-mover problem I've ever seen!

TasmanianTiger wrote:

Nf4 + Nxf4 2. Nxe7#?

There's a bishop waiting to take a knight on e7...

Move one:

Move two:

Remove your King from  the board while sneezing.

White wins irregardless of move(s)

Seriously though, nice position to post.

Asorski wrote:
Cnacnel wrote:

Nb6 Rd4 (rook from e4, dunno how to note it)?

Then Black king can escape at either c6 or e6. Please try again.

He can't escape c6 but yeah, e6 is legit ^^

Would be nice if it was on a board on which u can move pieces.

Ugh, I keep seeing mate in 3's.

Scottrf wrote:

There's a bishop waiting to take a knight on e7...

That is a fine defense Scootrf.

Keep trying.

PortlandPatzer wrote:

Move one:

Move two:

Remove your King from  the board while sneezing.

White wins irregardless of move(s)

Seriously though, nice position to post.

This is not a joke. I hope somebody hit the solution ASAP.

Keep trying dude! Keep trying.

I think I see it! Is it f=Q!, (if NxQ then Re5) if not Qd8#!

Or am I missing something? :D

Geticus wrote:

La problemele de sah ,,ortodoxe" cu sah in 2;3;4;.......n mutari se cere

1 Cheie (prima mutare) UNICA !    Cele cu 2,3 .....solutii se considera DEFECTE! Se admit mai multe soluti la problemele cu ,,mat ajutor" si la cele ,,ortodoxe"  cu  mai mult de  4x dar numai daca autorul indica ASTA,existenta de mai multe solutii si trebuie sa mentioneze cate!

2 POZITIE LEGALA (Rezultata  dintr-o partida chiar imaginara) ;ori la problema dvs. aceasta regula nu se respecta! albul are pe tabla peste 16 piese !!!iar pozitia pionilor negrii putea rezulta dupa inca cateva capturi, am calculat eu 4 deci 20 piese existente +4 capturate rezulta 24 !translate pls ROUMAIN LANGUAGE  FOLOSING! Merci mister ASORSKI  [Sper ca era o problema ,,ortodoxa"  nu heterodoxa sau ,,ferica" ?]

Geticus, I don't understand your words.

Anyone please translate in english language what Geticus is saying?

Ne7+  Nxe7 2. Nf4#

DM_CaptainObvious wrote:

I think I see it! Is it f=Q!, (if NxQ then Re5) if not Qd8#!

Or am I missing something? :D

1. f8=Q?? Black simply plays ...bxc4!!

1.Rc7 black can only move the knight at g6, the bishop at a3, pawn at b5 and Kxe4. If he moves knight 2.Nxf4#, if he moves pawn Rc5#. If he moves bishop to other than b4 or d6 then 2.Rc5# If Bb4 2.Nxb4# If Bd6 2.Nb6#. If Kxe4 2.Nb4#

DalaiLuke wrote:

Ne7+  Nxe7 2. Nf4#

1. Nxe7+ Bxe7!!

DM_CaptainObvious wrote:

I think I see it! Is it f=Q!, (if NxQ then Re5) if not Qd8#! Or am I missing something? :D

1...bxc4

Martin0 wrote:

1.Rc7 black can only move the knight at g6, the bishop at a3, pawn at b5 and Kxe4. If he moves knight 2.Nxf4#, if he moves pawn Rc5#. If he moves bishop to other than b4 or d6 then 2.Rc5# If Bb4 2.Nxb4# If Bd6 2.Nb6#. If Kxe4 2.Nb4#

If bxc4, Rc5 is illegal.

[COMMENT DELETED]

He can't play bxc4 if the rook isn't there.

Martin0 wrote:

1.Rc7 black can only move the knight at g6, the bishop at a3, pawn at b5 and Kxe4. If he moves knight 2.Nxf4#, if he moves pawn Rc5#. If he moves bishop to other than b4 or d6 then 2.Rc5# If Bb4 2.Nxb4# If Bd6 2.Nb6#. If Kxe4 2.Nb4#

If bxc4, Rc5 is illegal.

1.Rc7 b4 (bxc5 in illegal) 2.Rc5#

1.Rc7 Nxh4 2.Nxf4#

Martin0 wrote:

1.Rc7 black can only move the knight at g6, the bishop at a3, pawn at b5 and Kxe4. If he moves knight 2.Nxf4#, if he moves pawn Rc5#. If he moves bishop to other than b4 or d6 then 2.Rc5# If Bb4 2.Nxb4# If Bd6 2.Nb6#. If Kxe4 2.Nb4#

Very nice try Martino! Excellent! But: 1. Rc7! Bc5!!