# [Retros] 288 Fen strings upon the given diagram! Yefim. 10/12/2004.

TregerYefim at aol.com TregerYefim at aol.com
Tue Oct 12 13:19:57 EDT 2004

Hello from Yefim.
While somebody is solving e.p. problem offered by Tom, let me tell you about
some funny observations concerning e.p. rule.
1. Somebody noticed that e.p. square in FEN symbolizes only a fact that a
last move was double pawn step. Yes, position after 1.e4 e5 looks as:
rnbqkbnr/pppp1ppp/8/4p3/4P3/8/PPPP1PPP/RNBQKBNR w KQkq e6 0 2
It has e6 square as possible e.p. move?!
2. But nothing can be taken because no pawns are near. So, that is Seeming
e.p. relation what may lead to another funny situation:
3. Position after 1.a4 a5 2.b4 b5 3.c4 c5 4.d4 d5 5.e4 e5 6.f4 f5 7.g4 g5
8.h4 h5:
has FEN:
rnbqkbnr/8/8/pppppppp/PPPPPPPP/8/8/RNBQKBNR w KQkq h6 0 9
and h6 e.p. square but all Black pawns are on 5 rank.
4. I noticed that a part of string above (and Position itself!) may be
obtained by different ways after 8 moves (how many?) and we will have 8 different
FENs depending which BP moved last.
5. That is also Seeming e.p. relations, like above only concerning many
pawns.
6. Let's consider the part of string (FEN):
rnbqkbnr/8/8/pppppppp/PPPPPPPP/8/8/RNBQKBNR
This part characterize Configuration of position (and may be given by
function, for example Algebraic notation).
Then a question is: how many Positions may exist upon the given
configuration?
That is a little bit difficult question than a question:
"What is maximum of FEN strings (without two last numbers) reflecting the
given configuration?"
That is 32+16*16=288,
where 32 reflects 2*16=32 combinations WITHOUT any e.p. square (2 reflects
"w" and "b"; 16 reflects castle combinations in "KQkq");
next 16*16 reflects "16" castle combinations WITH 16 e.p. given squares (if
a particular square is given it already defines turn of move, so we do not
count "w" and "b"...).
So, choose the right FEN and record your positions properly! :)
About Real E.P. relations I will talk next time.
Bye. Yefim. 10/12/2004.

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