I am disappointed with these comments. I thought somebody would have solved the difficulty I found and could not solve. I saw Rxe7+ and how mate follows 2… Kf8.
(By the way, Umesh, Prof Bhat, Venky, and Anonymous 6:16 — 2. … Kd8 is illegal. The white bishop at b5 controls that square.)
But if 2. … Kg8 then white does not have an immediate check on the next move, and after 3. Nx e6, black can interpose Bf7, and I don’t see a mate on the board for white. White has given up a rook for knight and pawn, and both White’s bishop and knight are en prise.
Dear mshroder, … Kd8 is not illegal. Only … Ke8 is illegal.As far as your difficulty is concerned after 1Rxe7+ Kxe7 2Rc7+, 2… Kg8 is impossible(illegal)as K would have to borrow power of a N. In my comments there was typographical error on second line.It should have been 1…. Kf8 or 1… Kg8 2R1c7. I am extremely sorry for the same.
Dear mshroder, If your problem was with 1… Kg8,then 1Rxe7+ Kg8 2R1c7 Bf5 to protect e6 If 2…. axb5 3Rxg7+ Kf8 4Nxe6+ Ke8 now it is not illegal 5R(any)e7# 3Rxg7+ Kf8 4R(c)f7# Now 4… Ke8 would be illegal.
Dear mshroder, After going through all comments I realized that your remarks referred to anonymous at 6:16:00 who wrote 1… Kg8 2Nxe6. So you are right to that extent.but 2R1c7 solves it.Now 2… Bf7 is punished by 3Rxf7.
1. Rxe7+ Kxe7
2. Rc7+ Kf8 (2… Kd8 3. Nxe6#)
3. Nxe6+ Kg8
4. Rxg7#
If 1… Kf8, 2. Rcc7, with Nxe6 and Rxg7 threatened.
Rxf7
1Rxe7+ Kxe7
1…. Rf8 or Rg8 2 R1c7
2Rc7+ and now
(a) 2…. Kd6 3Rd7#
(b) 2…. Kd8 3Nxe6#
(c) 2…. Kf8 3Nxe6+ Kg8 4Rxg7#
1. Re7+ Ke7
2. Rc7+ Kf8 (Kd6 3. Ne6#)
3. Ne6+ Kg8
4. Rg7#
mate in 4
…..
4. Rxg7#
RxN to break the last defence of king. Then Rc7+. Then everything is simple.Mustafiz
Hi Susan Polgar,
Well,nice but simple puzzle.
White wins the game in ease.[ Slight variations exist ]
Example One
===========
1.R*Ne7+ K*Re7
2.Rc7+ Kd8
3.N*e6++ Mate
Example Two
===========
1.R*Ne7+ Kf8
2.Rc7 a*Bb5
3.N*e6+ Kg8
4.R*g7++ Mate
Example Three
=============
1.R*Ne7+ K*Re7
2.Rc7+ Kd6
3.Rd7++ Mate
White wins the game in ease.
By
Venky [ India – Chennai ]
1. Rxe7+
1. … Kxe7
2. Rc7+
2. … Kd8
3. Nxe6#
2. … Kf8
3. Nxe6+ Kg8
4. Rxg7#
1. … Kf8
2. Rcc7, followed by Nxe6+, Rxg7#
1. … Kg8
2. Nxe6 any move
3. Rxg7#
Start with 1. Rxe7+. Then bring the other rook in, and with the black knight gone, the white rook, bishop and knight will combine to mate black.
I am disappointed with these comments. I thought somebody would have solved the difficulty I found and could not solve. I saw Rxe7+ and how mate follows 2… Kf8.
(By the way, Umesh, Prof Bhat, Venky, and Anonymous 6:16 — 2. … Kd8 is illegal. The white bishop at b5 controls that square.)
But if 2. … Kg8 then white does not have an immediate check on the next move, and after 3. Nx e6, black can interpose Bf7, and I don’t see a mate on the board for white. White has given up a rook for knight and pawn, and both White’s bishop and knight are en prise.
Mark
Dear mshroder,
… Kd8 is not illegal. Only … Ke8 is illegal.As far as your difficulty is concerned after 1Rxe7+ Kxe7 2Rc7+, 2… Kg8 is impossible(illegal)as K would have to borrow power of a N.
In my comments there was typographical error on second line.It should have been
1…. Kf8 or 1… Kg8 2R1c7. I am extremely sorry for the same.
Dear mshroder,
If your problem was with 1… Kg8,then
1Rxe7+ Kg8
2R1c7 Bf5 to protect e6
If 2…. axb5 3Rxg7+ Kf8 4Nxe6+ Ke8 now it is not illegal 5R(any)e7#
3Rxg7+ Kf8
4R(c)f7# Now 4… Ke8 would be illegal.
Dear mshroder,
After going through all comments I realized that your remarks referred to anonymous at 6:16:00 who wrote
1… Kg8 2Nxe6. So you are right to that extent.but 2R1c7 solves it.Now 2… Bf7 is punished by 3Rxf7.