# [Retros] Happy New Year!

Andrew Buchanan andrew at anselan.com
Fri Jan 16 23:16:37 EST 2004

Francois,

Excellent stuff with the counting.

(1) The fact that you have reached ply 8 is fortuitous, because it may allow
you to confirm the answer to a very simple question.
What is the number of "at home" SPGs of minimum length?
Explanation:
- an "at home" SPG is an SPG where all the surviving pieces are apparently
on their start squares. (aka "deletion", "chez soi".)
- a number of deletion SPGs are known at ply 8. (See
http://www.geocities.com/anselan/CHEB.html)
- it is strongly suspected that there are no shorter deletion SPGs (except
the trivial one at ply 0).
Can you confirm this, and give the list of all deletion SPGs at ply 8? That
would be impressive.

(2) With the 2004 stuff.
It's usual practice in SPGs not to have any text e.g. "both sides can still
castle". Moreover, since all play is retrograde, the castling convention
won't come into play. So you can't *assume* that both sides can castle. So
what's the quickest *diagram* which can be generated by exactly 2004 proof
games? (i.e. no restriction on castling (or e.p.!)) Is there one in ply =<
8?

Cheers for now,
Andrew.

PS: Please give my best regards to Elwyn Berlekamp, if you are in touch with
him.

----- Original Message -----
From: "Francois Labelle" <flab at EECS.Berkeley.EDU>
To: <retros at janko.at>
Sent: Tuesday, January 13, 2004 2:36 AM
Subject: [Retros] Happy New Year!

Noam Elkies is right, I can search for positions with n solutions for up
to ply 8.

For n=2004, there is a proof game in 7 plies:

r n b q k b n r
p p p p . p p p
. . . . p . . .
. . . . . . . .
. . . . . . . .
. . P . P . . .
P P . P . P P P
R N B Q K B N R

(with the requirement that castling is still possible for both sides)

Unfortunately, with examples found by computer there is no guarantee that
counting the number of solutions is easy for a human, but in this case
we're probably lucky that it's not painfully hard.

Here's my reasoning: the position can obviously be obtained in 3 plies, so
White and Black must each waste 2 plies by moving an officer forward and
backward. For each side the officer can be a knight, the f-bishop, or the
queen, but not the king because of the requirement that castling is still
possible.

We can count White's moves and Black's moves independently, and then take
into account contention for the squares a3 and a6.

Skipping the details, I get 92*22 - 10*1 - 2*3 - 2*2 = 2004.

In fact there are 4 such proof games for ply 7: the problem also works if
one or both e-pawns did a double jump.

There are 2232 proof games with 2004 solutions for ply 8.

Francois Labelle

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