Valid bishop chess moves

Done in collaboration with Ahaan Rungta.

Consider the diagonals in the $\nearrow \swarrow$ direction. We seek the number of total pairs in this direction and then we use symmetry. In a diagonal with $k$ cells, we have $2 \cdot \dbinom {k}{2}$ orderings. We multiply by $2$ to account for ordering $x, y$ .

We have to count the long diagonal once and the other diagonals twice, and then multiply the sum by $2$ . So the answer is $2 \cdot \left( 2 \cdot \dbinom {8}{2} + 4 \cdot \displaystyle\sum_{k=2}^{7} \dbinom {k}{2} \right ) = \boxed {560}$

Done in collaboration with Sreejato Bhattacharya.

Consider the diagonals in the $\nearrow \swarrow$ direction. We seek the number of total pairs in this direction and then we use symmetry. In a diagonal with $k$ cells, we have $2 \cdot \dbinom {k}{2}$ orderings. We multiply by $2$ to account for ordering $x, y$ .

We have to count the long diagonal once and the other diagonals twice, and then multiply the sum by $2$ . So the answer is $2 \cdot \left( 2 \cdot \dbinom {8}{2} + 4 \cdot \displaystyle\sum_{k=2}^{7} \dbinom {k}{2} \right ) = \boxed {560}$

I try to draw for $8 \times 8$ in here

For yellow $(2 \times 2)$ square, There are $4$ ordered pairs (x,y) up diagonal and down diagonal. We can choice $7^2$ posible ordered pairs.

For pink $(3 \times 3)$ square ,There are $4$ ordered pairs (x,y) up diagonal and down diagonal. We don't counting $2 \times 2$ square because, we was do it in $2 \times 2$ . We can choice $6^2$ posible ordered pairs.

For green $(4 \times 4)$ square , There are $4$ ordered pairs (x,y) up diagonal and down diagonal. We don't counting $2 \times 2$ square and $3 \times 3$ square. because, we was do it in $2 \times 2$ . We can choice $5^2$ posible ordered pairs.

If we continue this step $4 \times 4$ square until $8 \times 8$ square we will will get posible ordered pairs, $4^2,3^2,2^2,1^2$ and the number every square are $4$

So, We get posible ordered number are $(1^2+2^2+...+7^2) \times 4=140 \times 4=\fbox{560}$

Just like we draw diagonal lines in Aufbau principle(if you remember), we may draw the same for the chess board. So, there will be 15 lines on the board going down to up (like this- / ). For every diagonal covering 'n' squares on the board, there will be n values for x and (n-1) values for y.Therefore, the no of ordered pairs (x,y) would be = [2 { ∑ (n x (n-1)) from n=0 to 7} + 8 x 7]

But, since the bishop can move along diagonals going up to down (like this- \ ), we have to multiply this by 2.

On solving, we would get the result as 560.

chess

this is the link to the image "http://pbrd.co/1iypkmC"

just add the numbers

this digram can be easily made withing 2 or 3 minutes as a quick pattern emerges

Casework by displacement. If the bishop moves $(\pm k, \pm k)$ then there are a total of $(8-k)^2$ such ways. So the answer is $4(7^2+6^2+\dots +1^2) = 4\cdot 7 \cdot 8 \cdot 15 / 3 = \boxed{560}$ , since there are 4 possibilities for signs.