[Retros] It's that time again...

[Noam Elkies]

A few weeks ago I wrote:

< +-----------------+
< | . _ . _ . _ . _ | Noam D. Elkies, 22 Dec 2006
< | _ . _ . _ . _ . |
< | . _ . _ . _ . _ |
< | _ . _ . _ . _ . |
< | . _ P _ P _ . _ |
< | _ . N . R N _ . |
< | P P . P . P P P | 16 + 0
< | _ R B . _ B Q K | [stipulation should be easy to guess by now]
< |_________________|

<  (Rb1 Nc3 Bc1 Qg1 Kh1 Bf1 Nf3 Re3 Pa2 Pb2 Pc4 Pd2 Pe4 Pf2 Pg2 Ph2)

< Happy New Year,
< --NDE

Francois Labelle <flab at EECS.Berkeley.EDU> replied:


> Congratulations to Noam Elkies for his 2007-themed problem.

> In Popeye notation (C+):



> beginproblem

> forsyth 8/8/8/8/8/8/PPPPPPPP/RSBQKBSR

> stipulation ser-a=>b 12

> forsyth 8/8/8/8/2P1P3/2S1RS2/PP1P1PPP/1RB2BQK

> option NoBk

> endproblem



> Here's one with our beloved theme: Pawn to-the-(1st)-rank mate.

> How many solutions?



> beginproblem

> forsyth rsbqkbsr/ppp1pp1p/3p4/8/7P/6P1/PPPPPP2/RSBQK2q

> stipulation dia 6.0

> endproblem



> Happy New Year!


Thank you.  I surmise that the count is the same, and that you found
the position by your usual technique of exhaustive computer search
(trimmed to exclude positions where Black is too far from promotion
by move 6) and thus the answer should not be expected to have a clean
mathematical derivation.

Meanwhile my colleague Mark Watkins (now at Bristol) suggested another
starting point for constructing a proof game with 2007 solutions,
and I eventually succeeded in getting it to work:

  +-----------------+
  | . r . q k b n r |  NDE 1.i.2007, version 3.i.2007
  | p b p n p Q p p |
  | . p . _ . _ . _ |
  | _ . _ . _ . _ . |
  | . _ . P p B . _ |
  | _ B _ . _ . _ . |
  | P P P _ . P P P |  15 + 15
  | R N _ . K . N R |  helpgame in 6.5: how many solutions?
  |_________________|
  White Ra1 Nb1 Bf4 Qf7 Ke1 Bb3 Ng1 Rh1 Pa2 Pb2 Pc2 Pd4 Pf2 Pg2 Ph2
  Black Rb8 Nd7 Bb7 Qd8 Ke8 Bf8 Ng8 Rh8 Pa7 Pb6 Pc7 Pe4 Pe7 Pg7 Ph7

(The only change between the two versions is the added final half-move.)
Unlike the other two 2007-solution problems, this one also satisfies
Richard Stanley's criterion that the move set be unique, with only the
acceptable permutations to be enumerated.

When I sent my earlier 2007-solution problem to the students in my
Freshman Seminar on chess and mathematics, two of them, Yi Sun and
Nick Wage, came up with their own enumerative puzzle, "a bit ahead
of schedule" as they put it:

   +-----------------+
   | . _ . _ . _ . _ |  Yi Sun & Nick Wage, January 2007
   | _ . _ . _ . _ P |
   | . _ . _ . _ . B |
   | _ Q _ P _ . _ . |
   | . _ . _ . _ . _ |
   | _ . _ . _ . _ . |
   | P P P . P P P _ |  16 + 0
   | R N _ K _ B N R |  How many minimal sequences?
   |_________________|

(Ra1 Nb1 Kd1 Bf1 Ng1 Rh1 Pa2 Pb2 Pc2 Pe2 Pf2 Pg2 Qb5 Pd5 Bh6 Ph7)

Enjoy,
--NDE