Solution
The solution is as follows:
- 1.b4!
- Threatening 2.Rf5 any 3.Rf1# or 2.Rd5 any 3.Rd1# (with possible prolonging of both by 2...Rc5 3.bxc5 any 4.R mates). White cannot begin with 1. Rf5 because Black's 1.... Rc5 would pin the rook.
- Now there are multiple possible half-defences defending only one of threats and one secondary non-thematical defence:
- 1...Rxc2 2.Nxc2! a2 3.Rd5 (or Rf5) a1Q 4.Nxa1 any 5.R mates.
- 1...Rc5+ 2.bxc5!
- Threatening 3.Rb1#.
- 2...a2 3.c6!
- Again with the same threats as on move one, i.e. 4.Rf5 any 5.Rf1# or 4.Rd5 any 5.Rd1#.
- 3...Bc7
- Because both Rd5 and Rf5 are threatened; the alternative moves 3.... Bf6 and 3.... Bg5 would only defend against one or the other, i.e. they would half-defend. The given move does defend against Rd5 in the sense that 4.Rd5 Bxg3 5.Rd1+ Be1 6. Rdxe1# takes more than the required five moves, and similarly for 4.Rf5 Bf4.
- 4.cxb7 any 5.bxa8=Q/B#.
- The mate is delivered with the pawn which starts on b2.
Any problem which features a pawn moving from its starting square to promotion in the course of the solution is now said to demonstrate the Excelsior theme. Nowadays it is most usually shown in helpmates and seriesmovers.
Read more about this topic: Excelsior (chess Problem)
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