so the critical trick is that 1. c7 Rxc7! 2. bxc7 c3 3. c8=Q c2!

The pawn is untouchable because of stalemate, and white can’t approach the black king with his knight or king without allowing black to queen. (The same trick works if white first plays b7)

White needs to get his knight down to blockade black’s pawn before it queens, but since blacks pawn is too fast, white’s goal must really be to reach the forking square on b3 to gain tempo.

1. c7 Rxc7 2. Nxc7! c3 3. Na6! c2 4. Nc5!

now if black tries to move his king to cover b3…

4… Kb2? c8=Q!

Black still covers c1 with another fork, and if… 4… Kb1? Nd3!

The knight covers the queening square, and can’t be chased away quickly enough.

Underpromoting with 3.c8=R gets around the stalemate problem. White can still get his knight back in time to deal with Black’s pawn.

1. c7 Rxc7 2. Nxc7! c3 3. Na6! c2 4. Nc5!

In this line 3.Nb5 c2 4.Nd4 c1=Q 5.Nb3+ works just as well.

So for the moment, I vote “easy,” because White seems to have a choice between winning moves at his second turn. This can’t be all there is to it, though — easy compositions with dual solutions don’t tend to survive long.

so the critical trick is that

1. c7 Rxc7!

2. bxc7 c3

3. c8=Q c2!

The pawn is untouchable because of stalemate, and white can’t approach the black king with his knight or king without allowing black to queen. (The same trick works if white first plays b7)

White needs to get his knight down to blockade black’s pawn before it queens, but since blacks pawn is too fast, white’s goal must really be to reach the forking square on b3 to gain tempo.

1. c7 Rxc7

2. Nxc7! c3

3. Na6! c2

4. Nc5!

now if black tries to move his king to cover b3…

4… Kb2? c8=Q!

Black still covers c1 with another fork, and if…

4… Kb1? Nd3!

The knight covers the queening square, and can’t be chased away quickly enough.

err… 4… c8=Q! should probably just be b7, as the pawn isn’t on c7. =P

1. c7 Rxc7!2. bxc7 c3

3. c8=Q c2!

Underpromoting with 3.c8=R gets around the stalemate problem. White can still get his knight back in time to deal with Black’s pawn.

1. c7 Rxc72. Nxc7! c3

3. Na6! c2

4. Nc5!

In this line 3.Nb5 c2 4.Nd4 c1=Q 5.Nb3+ works just as well.

So for the moment, I vote “easy,” because White seems to have a choice between winning moves at his second turn. This can’t be all there is to it, though — easy compositions with dual solutions don’t tend to survive long.

After 1.c7 does 1…Rh6+ bail Black out?

Excellent analysis in the previous posts.

If,

1.b7 Rxb7

2.cxb7 c2 and is draw no matter in what white promotes his Pawn.

However, what if,

1.c7 Rh8!

What happens now? White wins but not so straight forward.

Pharaoh

A typo above, i mean

1.b7 Rxb7

2.cxb7 c3

Pharaoh

If,1.b7 Rxb7

2.cxb7 c3 and is draw no matter in what white promotes his Pawn.

I beg to differ. 3.c8=Q is drawn, but 3.c8=R wins for White.

Grrr. Apologies, Pharaoh. Maybe I should try reading before posting. In my haste I thought you’d started with 1.c7.

Yes, 1.b7 Rxb7 2.cxb7 draws. The line involving the underpromotion is:

1.c7 Rxc7

2.bxc7 c3

3.c8=R!

Now Black is helpless. For example:

3…Kb2

4.Nb6 c2

5.Nc4+

… and the pawn soon falls.

There is still the case with,

1.c7 Rh8!

you may check that out.

Pharaoh

1.c7 Rh8

2.b7 c3

3.c8=Q Rxc8

4.bxc8=R

For the same reason as in the 1…Rxc7 line. 4.bxc8=Q would only draw.

4…Kb2

5.Nb6 c2

6.Nc4+, etc.

Yes, actually the case with,

1.c7 Rh8!

2.b7 c3 somewhat transposes to the other lines above.

Pharaoh

What about simply 1.Nc7 c3 2.Nb5 c2 3.Nd4 c1Q 4.Nb3+

(1..Rc7: 2.bc c3 3.c8Q c2 4.Qa6+)

What about simply 1.Nc7 c3 2.Nb5 c2 3.Nd4 c1Q 4.Nb3+(1..Rc7: 2.bc c3 3.c8Q c2 4.Qa6+)

You’re certainly right that:

1.Nc7 Rxc7?

2.bxc7 c3

3.c8=Q c2

4.Qa6+

… is an easy win for White. But in your main line:

1.Nc7 c3

2.Nb5 c2

3.Nd4

… Black unleashes his rook:

3…Rh6+!

4.Kd7 Rxc6!

… and draws.