No.1579 (MN & DP)

This is the kind of dual-avoidance complex that is a thrill to unravel. I see the logic this way:
We want to mate by Qg4# or Qd5#. So we need to put a bS on d7 (paralysed by a wS). We also need to take care of e3, either by guarding it or by blocking it with a bS – which we need to paralyse, so in any case one wS must move to control e3. Therefore bSd7 must come at B2, not B1, so as not to paralyse the wS that plays on W1.
For Qg4#, B1 must be S-e3 to stop both Qg1-d4 and Qd2-f4,g5 – but not Sg2-e3? because of Qxg3, so it must be Sc2-e3. W1 cannot be Sf6-d5? because of Qd8-g5,h4, so it must be played by Sb6. Therefore B2 cannot be Sc5-d7? because of Qa7-d4, so it must be Se5-d7. Therefore W1 cannot be Sc4? paralysing Se5, so it must be Sb6-d5.  
For Qd5#, B1 must stop Qg1-d4, but it cannot be Qg1-d1? because both Sc4 and Sg4 will be paralysed by Se5 leaving e3 unguarded. So it must be S-e3 – but not Sc2-e3 because of Qa2, so it must be Sg2-e3. B2 must be Sc5-d7 because Se5 will be paralysed by either Sc4 or Sg4. Therefore W1 cannot be Sc4 because of Qa7-c5,d4, so it must be Sg4.
As you can see, the logic is not 100% analogous, but it is 100% fascinating!
The fact that the position is ”illegal” (or rather impossible in a game) doesn’t bother me at all in fairy chess.
And I thought h#2 Madrasi was worked to death when it was stipulated in a WCCT many years ago…
(The first solution should read 1.- Sbd5 2.Sed7, and the second should read 2.Scd7.)