No. 1676 Bojan Bašić
original - 22.12.2021
(after D. Baibikov & W. Keym)
a) Add wK, bK, 10 wSs and 8 bSs for an illegal cluster.
b) Add wK, bK, 10 wSs and 7 bSs for an illegal cluster.
Solution: (click to show/hide)
In comparison to the Baibikov & Keym’s problem (feenschach III–IV/2019, Nr. 12009; also WJP 2019, 2.-3. Place ex æquo), here we have twice as many pieces in the initial position! 😀 But this is compensated by the following refinements:
· The twinning is more subtle: the requirement to add only one knight less makes all the diference.
· The number of pieces that are to be added is greater. In fact, in the twin a) the maximal possible total number of 20 knights is reached, and 19 knights in the twin b).
· The two solutions are more diverse. Not only that in the twin a) Black gives check and in the twin b) it is White who does it, but also the twin a) is built around legalizing the retraction of the knight’s move Sg6(x)h8+, while the twin b) is built around legalizing the retraction of the underpromotion h7-h8=S+!