# Longest possible chess game

David113
Posers and Puzzles 27 Aug '06 16:08
1. 27 Aug '06 16:08
Hi,
I have another question. It is a little bit long...

I once read that if after 50 moves without capture or pawn move the game ends, then the longest possible game is 5948.5 moves long. This is calculated as follows:
Maximal number of captures - 30
Maximal number of pawn moves - 96
At least 8 pawn moves must be captures, so we may think the maximal length is 50+50(96+30-8)=5950; but this answer assumes the first capture/pawn move is at half-move (= single move, ply) number 100, the second capture/pawn move is at half-move number 200, etc., which is impossible, since it assumes only black captures. Each time we swap from a sequence of captures/pawn moves every 50th move by one player to such a sequence by the other player we lose one tempo. WE MUST HAVE AT LEAST 3 SUCH SWAPS, so the correct answer is 5950-1.5=5948.5.
Now, finally, my question...
I understood all the above argument, exept the capitalized words. I don't understand why we must have 3 swaps from a sequence of captures/pawn moves every 50th move by one player to such a sequence by the other player. Why isn't one swap sufficient? and can somebody show how 3 swaps are sufficient? Thanks...
2. 27 Aug '06 19:271 edit
From http://www.chesscafe.com/text/bruce23.pdf

“It's a bit trickier than that because there must be several changes in whose turn it is to make the pawn move or capture through the game. Assuming Black makes the first capture, we need a switch to White making the captures so that White can get pieces out and give them up on squares which double White's pawns on files to leave gaps for Black's pawns to pass through. Then we need another change back to Black making these captures. At this stage both sides have unpromoted pawns so we need another switch for White to promote those pawns and take Black's pieces, and a final switch for Black to take White's surviving pieces. Each switch costs half a move, so on Black's 5898th move, a king capture of White's remaining piece, the game is drawn as only two kings are left and FIDE Law 1.3 applies immediately. (Does the USCF have this law too?)
3. 27 Aug '06 20:021 edit
Is this correct?

Do we really need 4 switches, and 3 are not sufficient?

I don't know...

Is the link you gave OK? It doesn't work.
4. 27 Aug '06 20:17
Originally posted by David113
Is this correct?
Do we really need 4 switches, and 3 are not sufficient?
I don't know...
Is the link you gave OK? It doesn't work.
Which of the four switches do you think is unnecessary?