I know the answer and the method of arriving at that answer, but... do you want me to post it so soon, or let others have some fun puzzling over it?

# The longest possible chess game

I didn't solve it myself... so if anyone can figure it out you did better than me!

So lets start with something basic... would you agree with me that this reasoning is good?

After 50 moves without a capture or pawn move the game ends in a draw. There are 30 pieces that can be captured, and there are 96 possible pawn moves (each pawn can take 6 moves to queen, and there are 16 pawns).

So we can start with the idea that a chess game definitely can't be longer than 30x50 + 96x50 = 6300 moves.

But you cannot make all 6 moves with one pawn without ANY captures - in such case your pawns will be stopped by your opponent's pawns. So you need to capture at least one time with pawns. As I understand - you do not need to make such captures with each pawn - half of it will be enough. And it means that the number of pawn moves will be reduced by 8, Am I right?

So let's look at the basic mechanism just so everyone is clear on the idea.

So from this, we know that black has to make the first pawn move or capture, and that each pawn move or capture represents the 99 ply that came before it. In other words 50 moves. So we can start thinking like this.

So now we try to compose a game where we can keep this sort of thing going as long as possible... which was too hard for me to do (and come to the right answer) but maybe someone else can!

But you cannot make all 6 moves with one pawn without ANY captures - in such case your pawns will be stopped by your opponent's pawns. So you need to capture at least one time with pawns. As I understand - you do not need to make such captures with each pawn - half of it will be enough. And it means that the number of pawn moves will be reduced by 8, Am I right?

Yeah, at least one move will be both a pawn move and a capture (otherwise the pawns can never get past each other)

But how many times must a pawn make a capture? 8? Less than 8? And why? I'm not given any hints hehe.

I do not know how to make less than 8 captures with pawns and free way for others. Here is my idea how to do it.

PS - between each caprute there will be 50 moves with other figures)))

Yeah, I think it has to be 8. I wrote down the logic and solution so to be sure I'd have to look it up in my notes. But that's no fun, so let me try to think...

Logically a pawn capture will always change 2 files (no more and no less).

So it can (ideally)

1) put itself on an open file and

2) open the file it was on previously for another pawn to use

Also we know every file it opens will only aid 1 pawn. Any extra pawn who use the same file must make a capture to get there.

So a pawn capture, at maximum, can only aid 2 pawns... so we need at least 8 pawn captures. I think that's correct.

So, as a result we have:

- 6 moves for each of 8 pawns without captures = 48 moves
- 5 moves for each of another 8 pawns who capture = 40 moves
- pawns should not capture each other. they need to capture only the figures
- moves with capture of all other figures (except kings) = 30 moves
- between each of the moves above should be 50 "simple" moves

40+48+30=118*50=5.900. Is it correct? Or maybe there are some moments I missed)))

There's still one more idea, and it's one that makes getting the right answer pretty tricky. You can discover it yourself when you try to compose a game. For example...

And putting all the pawns on the 4th rank is just an example. I don't mean you have to do that

But when you start to make a game like this you'll notice something...

Ka3 ba - and we have an open B file to move white pawn from b2 to b8 and receive queen. This queen can be sacrifised on c3 by capturing dc. And again. This way we can open files B, D, F and H. Then it is time for black to sarcifice their figures on fields b3 (opens file A for black pawns), then d3, f3 and h3.

Now let's add more fun into our calculations. Found an interesting chess rule in Wikipedia.

"The seventy-five-move rule is related to the fifty-move rule in as such it looks at a series of consecutive moves without capture or pawn move. For the fifty-move rule, a **player can request a draw** after a series of at least 50 such moves each. The seventy-five-move rule states that after such a series of 75 moves each, this, as a checkmate, immediately ends the game and the game is drawn. So contrary to the fifty-move rule, there is no draw request, the draw is valid automatically. The arbiter must enforce the draw in such a case."

So, do you mean that in case when on the next move it is time to move white pawn instead the black one the number of full moves between them decreases from 50 to 49.5? Then (logically) when we move black pawn after white it should be 50.5.

So, do you mean that in case when on the next move it is time to move white pawn instead the black one the number of full moves between them decreases from 50 to 49.5?

Yes, exactly

Then (logically) when we move black pawn after white it should be 50.5.

But we're not allowed to go over 50. If the 50th move isn't a pawn move or capture the game immediately ends, and it wont be the longest game possible.

T say it another way... once white makes a pawn move or capture, as long as white continues to keep making pawn moves or captures each of white's moves will represent 50.

But when it switches back to black it will be 49.5 again.

So to make the longest game possible, you need to find the minimum number of times this sort of switching happens, because every time it happens you lose 0.5 moves.

Who can calculate - if we follow all standard chess rules what will be the maximum number of moves can be done during one chess game? I mean the rules of 50 moves without captures and pawn move, same position repeats 3 times, etc.