17th November 2025

Selected by Michael McDowell

Ado Kraemer

Deutsche Schachzeitung, 1952

4S3/2pp1p2/2p2B1p/s1rk2rR/1Pp2PBb/2p2P2/K3Ps2/8

Mate in 4

Kraemer shows an attractive idea in this problem, which he dedicated to his great friend and collaborator Eric Zepler.

The solution will appear here on Friday 21th November 2025.