[Retros] Happy New Year!

[Noam Elkies]

Joost writes:


> There are probably easier solutions, but here are mine:


These look right, of course; The only improvement I can offer
is that in the first problem:

   +-----------------+
   | . n b q . r k _ |   Noam D. Elkies, 12/2005
   | _ p p p n p p p |
   | . _ . _ r _ . _ |
   | p . b . p . _ . |
   | . _ . _ . P . _ |
   | _ . _ P _ N _ . |
   | P P P B P _ P P |   16+16
   | R N K Q . B R . |   SPG-7.5: How many solutions?
   |_________________|

The count of 34 Black sequences can be obtained more easily
by starting from  Binomial(7,3) = 35  -- the total number
of ways to mathematically combine the three-move sequence
a5, Ra6, Re6 and the four-move sequence e5, Bc5, Se7, O-O
-- and substracting the one impossible combination
a5, Ra6, Re6, e5, Bc5, Se7, O-O.

In the second problem:

   +-----------------+
   | _ . _ . _ . _ . |   Noam D. Elkies, 12/2005
   | . _ . _ . _ . _ |
   | _ . _ . _ . _ . |
   | . K . _ . _ . _ |
   | _ . P . _ R _ P |
   | . _ . _ . P . _ |
   | P P Q P P . P . |   16+0
   | R _ B N . B N _ |   OSPG-14: How many solutions?
   |_________________|

the factor of 2 due to the interchangeability of Qc2/Sc3
applies throughout, so one could also count the total as
2*[Binomial(14,4)+2].  Note that the four-move sequence
had to keep the White King from reaching b5 in five moves
via d1-c2.  It so happens that the Kb5/Pc4 device also
appeared in my 2004-solution problem, where it was used
to obtain the factor 167 as  Binomial(11,3) + 2.

NDE