Proof Game - Is it possible?

Black to Move.

Originally posted by timmydoza
[fen]rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR b KQkq - 0 1[/fen]

Black to Move.

No, it is not possible. White and Black have obviously both made an even number of moves [if they have moved at all]. Parity only matches AFTER Black moves.

EDIT: Nevermind... I was wrong...

Originally posted by Gastel
I think it is possible, if the knights reversed positions. (I did this in my head, and I think I may have injured myself.)

That still leads to an even number of total moves - both N's have made an odd number of moves if they switched!

Originally posted by SwissGambit
That still leads to an even number of total moves - both N's have made an odd number of moves if they switched!

Damn you SG for your speedy replies... I was editting to remove my blunder and there you go.

I was doing doing the whole move, not stopping at a half move.

Edit: swiss gambit posted while i was typing

no.

in order to reach the position either black has made x moves and white x+1; so some body has made an odd number of moves.

The rooks have only one square they can move to (after a knight moves out) so the rook moves away and moves back w times wich is a=2x an even number of times.

if the knights are are on their original sqares: it requires a knight an even number of moves to move from one square to a square of the same color; which gives b, even

so if the knights are on original squares total moves are a1+a2+b1+b2
which is a sum of even numbers which is even.

if the knights have switched: a kight requires an odd number of moves to move from a square to a square of opposite color; which gives c, odd. but the other knigt must also have moved to a square of opposite color. total knight moves are c1+c2=d which is a sum of 2 odd numbers, so d is even.

so if the knights have swithed; total moves are a1+a2+d which is a sum of even numbers which gives an even number.

so neither player could have made an odd number of movers.