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1. Nd8+, Ka8
[If 1 …, Ka6, then 2. a8=Q mate.]
2. b7+, K x a7
3. Nec6+, Ka6
4. b8=N mate.
Ravi
1. Nd8* – Ka8 2. b7*-Kxa7 3. Nec6*-Ka6 4. b8=N checkmate 🙂
yes, this was a real easy one since all moves are forced and we have the 4 moves limit.
btw Ravi, I thought that 1.. Ka6 a8=R X is nicer 🙂
nice 3-knight checkmate. but the problem is easy, the answer can be found by exclusion. exactly a case where the comp is not needed