Great work by Shankar
Finally, I've tried so many further (unconventional) approaches with the given stones, but all of them resulted in more than…
By removing stones (e.g. without bBh1) you might find other "valid" (shorter) solutions: 1.Qg8 2-5.Sf3-e1-c2-a1/a3 6.g1=S 7.Qf8 Q×d2# (via d1)
Without wPg6, we have an alternative solution: 1.Qg8 2.Sf3 3.Se5 4.Sg6 5.Sh8 6.Qf8 7.g1=R 8.Rg2 Q-d1xd2#
Without bBa5, we have a shorter solution: 1.Qg8 2.Sf3 3.g1=S 4.Se1 5.Sg2 Q-d1xd2#
Tested also by Jacobi with this input: cond marscirce stip ser-h#8 forsyth 2rkrq2/2P1PPQ1/6PK/b5P1/7p/7p/3pP1pP/6sb (in less than an hour)
Yes, you're right! The wQd2 can neither mate directly, she too requires the empty home square d1 for the mate…
I hardly dare to maintain my objection, but I still think this is NOT a unique solution! What about the…
Oh my god, how embarrassing! Yes your're absolutely right, it's a unique ser-h#8 then, how could I've lost sight of…
Honestly? A good problem is one that combines illegality (impossibility) of the diagram with disregard of standing principles (rules)? I…