11-03


The problem was originally printed with a diagram error.
The correct version (above) was published as R386c in the October-December issue 2011.

Main plan: -1.b3-b4? (-1.Ba1-h8? -1…Kh8/g7-(xX)g8!) -1…h2-h1=S -2.Ba1-h8 O-O (legal retraction,
-2…K~xSg8 illegal) -3.Bf6xBa1 (White needs to provide Black with a last possible retraction, as the bR and bK cannot retract further moves after the uncastling) & 1.Rxd8#!
Notice that there are no duals in the main plan.

But -1…g2xSh1=S! refutes the main plan as now Black’s uncastling is illegal (in fact, any legal uncapture on h1 refutes the mainplan, but the stipulation is only defeated with -1…g2xSh1=S!).  Black has refuted the main plan by destroying his own uncastling right.

In the position arising after the unpromotion of the bS on h1, Black has captured all the missing white units with his pawns; thus both missing wPs had to promote. After -1.h2-h1=S White could have promoted his pawns on d8 and f8 without disrupting Black’s uncastling right (wPdxc>c6, bPb7xwPc6 was also possible), while still having a free uncapture available. However, after -1…g2xXh1=S, owing to the g/h-wP crossing, if White is to keep a free uncapture available, then both his promotions had to be on e8, thus disrupting Black’s uncastling right until the promotions on e8 are retracted.
               
Notice also that the sole reason why the main plan fails after -1…g2xSh1=S! is because Black cannot uncastle. White needs a foreplan that prevents this refutation while providing the right number of additional retractions for Black.


Solution: -1.Bg2-d5! h2-h1=S (the refutation has been prevented) -2.a5-a6 Ba6-a7 -3.b3-b4 Bb7-a6 -4.Ba1-h8 (-4.Ra6-a7? a7xRb6!) Ba6-b7 -5.Qb7-a8! (-5.Rb7-a7? but -5…a7xb6!;  -5.b7-b8 =S? but -5…O-O -6.Bf6xBa1? illegal uncapture) -5...O-O (legal retraction) -6.Bf6xBa1 & 1.Rxd8#!