# Parity of number of pieces on the black squares of a chessboard [closed]

On a chess board there is an odd number of pieces in each row and in each column. What can we say about the total number of pieces on a black square on the board?

1. The number must be even.
2. The number must be odd.
3. The number can be either both even or odd.
• $2)$ can be ruled out with the constellation $a1-b2-c3\cdots h8$ – Peter Mar 29 '18 at 16:46
• I guess a parity-argument shows $1)$, but I have no clue yet how. – Peter Mar 29 '18 at 16:49

Orient the board so that a white square is in the top left hand corner. Label each square with a piece on it with $1$, and $0$ otherwise. Let $R_1,R_2,\ldots, R_8$ be the rows, and $C_1,C_2,\ldots, C_8$ be the columns.

Add together $R_1+R_3+R_5+R_7+C_1+C_3+C_5+C_7$, which is even. Each white square in these squares gets accounted for $2$ times, so they have an even contribution to the sum. So this leaves the black squares, and an even minus an even is even, and all the black squares are accounted for exactly once, so there are an even number of pieces on the black squares.

The answer is $1$).

• Ï thought this worked, but then I drew the diagram and I couldn't follow your reasoning. If the square at $\langle R_1, C_1\rangle$ is black, then your $R_1+R_3+\ldots+C_1+\ldots$ decomposition counts each white square once, not twice. And if $\langle R_1, C_1\rangle$ is white, then some white squares get counted twice and some not at all. Image here – MJD Mar 29 '18 at 17:20
• hi, thank you for the answer. i think you meant: R2+R4+...+R8+C1+...+C7, right? – Omer Mar 29 '18 at 17:22
• @omer that's the same, if you just put a different color square in the upper-left corner. – MJD Mar 29 '18 at 17:23
• @MJD you must orient the board so that awhitesquare is in the top left hand corner, ill update. It doesnt matter if some whites dont get counted, it only matters that the one that are counted have an even contribution to the sum – ultrainstinct Mar 29 '18 at 17:27
• @Omer I updated my answer, that is correct if the top left hand corner is black. – ultrainstinct Mar 29 '18 at 17:30

It depends. If diagonals are all black, $B$ should be always odd. Else if diagonals are all white, $B$ should be even. Assume the former case. Because there are odd number of pieces, the chess board is of size $(2n+1)×(2n+1)$. Let the position of $i$-th piece be $(a_i , b_i)$ $(1\le i \le 2n+1)$. Then it is on the black square if $a_i +b_i$ is even and on the white square otherwise. Then $S:=\sum_{i=1}^{2n+1} {(a_i +b_i )}$ is odd if $B$ is even and $S$ is even if $B$ is odd. But as pieces are on each row and each column, $S=\sum_{i=1}^{2n+1}{(i+i)}$ is even. Hence $B$ must be odd.

• not true. $8$ black pieces on black diagonal is even number, but each row/column has $1$ piece – Vasya Mar 29 '18 at 17:10
• Ramansa is considering a more general case. The first sentence says “If diagonals are all black”. This occurs on a square board with odd side length. – MJD Mar 29 '18 at 17:11
• As there are odd number of pieces, diagonal must have odd number of squares, not 8. – Ramanasa Mar 29 '18 at 17:13
• no, there is an odd number of pieces on each row and column. at least, that's how I understand the question. – Vasya Mar 29 '18 at 17:15