Yeah, 26. A proof should not be that hard to sketch up...

The other one is not true. What if B=0? Maybe you mean

W =< M - B, which is trivially true (add B to both sides)

Yeah, 26. A proof should not be that hard to sketch up...

The other one is not true. What if B=0? Maybe you mean

W =< M - B, which is trivially true (add B to both sides)

Tapani írta:

Yeah, 26. A proof should not be that hard to sketch up...

The other one is not true. What if B=0?

You are right that no side can promote more than 8 pawns, therefore no side can have more than 15 pieces. So the correct formula would be:

max ( W ) = min ( M - B , 15 )

By this formula being always true I really meant that is it always legal for white to have min ( M - B , 15 ) pieces?

[ post edited : ]

Let's assume that there are no pawns on the board!

I think it's fewer than that, I dont see how you can have more than 8 total.

for every promoted pawn, there should be at least one capture? Else, how will it ever get past its counterpart on the same file?

[edit] of course I mean 8 promoted pieces, total 22

rooperi wrote:

I think it's fewer than that, I dont see how you can have more than 8 total.

for every promoted pawn, there should be at least one capture? Else, how will it ever get past its counterpart on the same file?

[edit] of course I mean 8 promoted pieces, total 22

My initial thought was 22, a total of 8 promotions, assuming none of the inital pieces are captured (which wouldn't make a difference as far as I can see).

Are we assuming neither side has no pawns at all? Then yes, I think it is correct. Even if I would write it as:

W <= min { M - B, 15 }

The maximum material difference with the maximum material on board would be with:

White - 1 King, 9 Queens, 2 Rooks, 2 Bishops, 2 Knights

Black - 1 King, 1 Queens, 2 Rooks, 4 Bishops, 4 Knights

and min with:

White - 1 King, 7 Queens, 2 Rooks, 2 Bishops, 2 Knights

Black - 1 King, 7 Queens, 2 Rooks, 2 Bishops, 2 Knights

Well, let's see.

Each side has 4 knights. 4 Bishops. 2 Kings. 4 Rooks. 16 Pawns. 2 Queens.

We have to note that whenever we promote, we give AWAY one piece, and get BACK one piece. So, this would be the highest total number:

32

Robert0905 написал:

Well, let's see.

Each side has 4 knights. 4 Bishops. 2 Kings. 4 Rooks. 16 Pawns. 2 Queens.

We have to note that whenever we promote, we give AWAY one piece, and get BACK one piece. So, this would be the highest total number:

32

If you read the original post again, you'll notice that he specified pieces OTHER THAN kings and pawns.

rooperi написал:

I think it's fewer than that, I dont see how you can have more than 8 total.

for every promoted pawn, there should be at least one capture? Else, how will it ever get past its counterpart on the same file?

[edit] of course I mean 8 promoted pieces, total 22

White's a, c, e, g pawns capture black's b, d, f, h pawns. White can now promote all 8 pawns, black can still promote the remaining 4, therefore 12 total.

If nobody captures anthing then the most pieces on the board between black and white will always be 32, to include promotions

Juergenwerner, first of all, kings don't count.

second of all, to promote a pawn it must get past it's counterpart on the other side, only way is to capture. A capture will either result (assuming the pawn later promotes) in an equal outcome - as in no net promotion/loss of pieces, or a plus outcome (again, assuming we play smartly).

@rooperi, not every en passant/take can only produce another piece, what if you get double pawns on an open file?

i'm not that good at chess files and stuff, i did the thing on lichess- i didn't save and lost the move list but trust me on the final position

s23bog wrote:

Er, 26. Kings aren't really considered "pieces".

Kings are pieces. You start the game with 8 pieces and 8 pawns!

I wonder what is the maximum number of pieces on board in a legal position. By 'piece' I mean any chessman apart from pawn and king ( ie. knight, bishop, rook, queen ). I'm not interested in the maximum number of promotions but the maximum net gain that can be achieved by promotions in the total number of pieces on board. My gut feeling is that the maximum net gain is 12 so you can have a maximum of 26 pieces simultaneously on board in a legal position but I await the final word by theorists.

I'm also interested in the distribution of pieces between white and black. Supposing that the maximum number of pieces is 'M', white has 'W' number of pieces and black has 'B' number of pieces then is the below formula always true:

max ( W ) = M - B

I'm asking these questions because I got involved to some extent in puzzle creation is some thread and I felt insecure about how many pieces I can have on board by both sides without endangering the legality of the positon.