# [Retros] Happy New Year!

Noam Elkies elkies at math.harvard.edu
Fri Jan 13 12:20:29 EST 2006

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