Originally posted by Bowmann
Very approximately as:

321^6300 = 10^15,790

So what's this equal to in laymans terms then Bow?

Originally posted by jimslyp69
So what's this equal to in laymans terms then Bow?

Well let's say that, on average, a player has a choice of around 30 possible moves each turn.

Originally posted by Siebren
On a chessboard we've got a total of 16 pawns good for 7 moves each.
We've got a total of 30 pieces wich can be captured (capture the pawns after they've been promoted.) good for 1 move each.
Knowing this you should say the longest ...[text shortened]... ngest possible game by your means is 7.100 - 350 = 6.750 moves.

I agree, except that:

1) Pawns are good only for 6 moves, including promotion.

2) I believe 8 captures are necessary in order to create 16 passed pawns.

Thus we get [(16 x 6) + 30 - 8] x 50 = 5900

3) Your method assumes that only Black is doing the capturing and pawn-pushing.

Originally posted by jimslyp69
Ooops. Slight revision
Kings can't be taken so that's - 49 * 2
Also we need minimum piece for check mate. ie one rook so that - 49
so its 5773 moves folks.

No way. Its massively bigger than that. Think about it. Players can make 49 knight hops around the centre of the board in between each pawn move/capture.

I worked it out in some thread, but I can't remember for the life of me what thread it was posted in.

As a bit of an aside, there are more possible games of chess than there are grains of sand on every beach in the world.

D

Originally posted by THUDandBLUNDER
I agree, except that:

1) Pawns are good only for 6 moves, including promotion.

2) I believe 8 captures are necessary in order to create 16 passed pawns.

Thus we get [(16 x 6) + 30 - 8] x 50 = 5900

3) Your method assumes that only Black is doing the capturing and pawn-pushing.

I agree with your analysis, except that since the draw can be claimed after 100 consecutive half-moves, shouldn't it be:

[16*6 + 30 - 8]*49.5 (instead of 50) = 5841?

I think it must be 49.5 instead of 50 because only 99 half-moves can be made before one of the [16*6 +30 - 8] pawn/capture moves are made.

Originally posted by davegage
I think it must be 49.5 instead of 50 because only 99 half-moves can be made before one of the [16*6 +30 - 8] pawn/capture moves are made.

But if every capture/pawn move is made on the 100th turn then each cycle is exactly 50 moves long, right?

It is only 49.5 when the capturing side changes from Black to White, or vice versa.
I think it can be proved that this must happen at least four times during the game, shaving at least 2 moves off the maximum length.

Originally posted by THUDandBLUNDER
But if every capture/pawn move is made on the 100th turn then each cycle is exactly 50 moves long, right?

It is only 49.5 when the capturing side changes from Black to White, or vice versa.
I think it can be proved that this must happen at least four times during the game, shaving at least 2 moves off the maximum length.

ah...yes, you're right -- 100 moves per cycle if they don't switch, at most 99 if they do.

the problem is a bit trickier than i thought. the capturing/pawn-moving side must switch at some points -- that is obvious because half of the pawns/pieces-to-be-captured are black, half white.

as you say, the question is what is the fewest number of times the switch must occur. i'll have to give that some more thought.

Originally posted by THUDandBLUNDER

I think it can be proved that this must happen at least four times during the game, shaving at least 2 moves off the maximum length

Yes...I agree that the capturing/pawn-moving sides must switch a minimum of 4 times. I think there are many ways to do it, but in the longest game, the capturing/pawn-moving moves would go something along these lines:

Side A would position its pawns such that most (but not all) of Side B's pawns could get past them (it's possible to do all, but not in only 8 capturing moves by Side B pawns -- in this case, it's vastly better to do another switch than another simultaneous pawn-move/capture move). Side A could also use it's pieces to capture all of Side B's pieces except for Side B's pawns and King.

Switch 1 occurs: then Side B moves most (but not all) of his pawns past Side B's pawns with capturing moves (one capturing move per pawn). He goes on to promote these pawns and uses them to capture all of Side A's pieces except for Side A's pawns and king.

Switch 2 occurs: Side A promotes all of his pawns, only some of which need to use capturing moves to get past Side B's remaining pawns (total capturing moves for both sides to get their passed pawns will be 8). He uses the promoted pawns to capture Side B's earlier promoted pawns (only Side B's king is left, along with a few pawns).

Switch 3 occurs: Side B promotes his remaining pawns and uses them to capture Side A's remaining pieces. At this point, only Side A's king remains for side A.

Switch 4 occurs: Side A uses his king to capture the remaining Side B pieces. Once the last capture is done, only kings are left and the game is drawn on insufficient mating material.

I think there are several possible initial pawn configurations that could be used, but I don't see any way to get it down to fewer than 4 switches.

So, i think the total number of moves in the longest game would be:

[(6*16) + 30 - 8]*50 - 4(0.5) = 5898.

Originally posted by Bowmann
The Web likes the number 5949.

The Web likes many numbers.

I believe the above one comes from playing on with only two kings.