[Retros] Happy New Year

[Noam Elkies]

"Juha Saukkola" <juha_saukkola at hotmail.com> writes:


> g3, c4->Nc3->Rb1, Nf3->Ne5->f3->Kf2->Ke3->Kd4->Kc5->Kb5



> 12 x 11!/3!8! = 12 x 165 = 1980



> g3, Nf3->Ne5->f3->Kf2->Ke3->Kd3/d4->Kc4->Kb5->c4->Nc3->Rb1



> 12 x 2 = 24



> 1980 + 24 = 2004


Yes, that's what I meant.


> It's nice!


Thanks!


> What's the position (in shortest number of moves) with 2004 sols?

> Or with n?

> What's the lowest number n without that kind of position?


I suspect that in general

1) This function of n is quite unpredictable,
2) The minimal position(s) are too "accidental" to be constructed
 directly from n (as I constructed the above case with n=2004),
3) The minimal unreachable n is quite large,
4) The methods used by Francois Labelle
 <www.cs.berkeley.edu/~flab/chess/problems-moves.html>
 can answer this question for impressively large values of n.
 (Thanks to Richard Stanley for alerting me to this site a few days ago.)

As corroborating evidence for (1), (2), and (3), consider just the
positions in which only one White Knight moves (if I did this right,
each n from 1 through 10 arises in at most 4 moves, followed by
a random-looking selection of n's for 5-move games).  Then try
the same thing with both White Knights, but nothing else, moving.
Then imagine how many complicated modifications can be made by inserting
just one or two pawn moves.

Much the same can be said for two-sides orthodox PG's.  For instance,
consider the following position (unfortunately composed too late for
the Bonsdorff contest):

+-----------------+

| . n . k . . . r |  NDE 11/03

| . . p p . p p p |

| . . . b b n . . |

| . p . . p r q . |

| p P . . P . B . |

| . . . . B P K . |

| P . P P . R P P |  SPG-13.5

| R N . Q N . . . |  how many solutions?

|_________________|


Enjoy!
--NDE