On an empty chessboard
let the total number of moves a King can make = K
let the total number of moves a Queen can make = Q
let the total number of moves a Rook can make = R
let the total number of moves a Bishop can make = B
let the total number of moves a Knight can make = N
let the total number of moves a Pawn can make = P

eg. R = (64)(14) = 896
because, wherever you place the Rook, it always attacks 14 squares.

1a) What is the relationship between R, B, and N? Is this a coincidence?

1b) Are there other simple relationships between any of the six pieces?

On an nxn chessboard
let the total number of moves a King can make = K(n) for n > 0
let the total number of moves a Queen can make = Q(n) for n > 0
let the total number of moves a Rook can make = R(n) for n > 0
let the total number of moves a Bishop can make = B(n) for n > 0
let the total number of moves a Knight can make = N(n) for n > 1
let the total number of moves a Pawn can make = P(n) for n > 3

2) Find K(n), Q(n), R(n), B(n), N(n), P(n)

On an nxm chessboard
let the total number of moves a King can make = K(n,m) for n,m > 0
let the total number of moves a Queen can make = Q(n,m) for n > m
let the total number of moves a Rook can make = R(n,m) for n,m > 0
let the total number of moves a Bishop can make = B(n,m) for n > m
let the total number of moves a Knight can make = N(n,m) for n,m > 1

3) Find K(n,m), Q(n,m), R(n,m), B(n,m), N(n,m)

Let the total number of moves a Pawn can make on an rxf board = P(r,f) for r > 3, f > 0

4) Find P(r,f)

5) Prove that there is only a finite number of boards such that R(i,j) = B(i,j) + N(i,j)


6) For what size boards are there such simple relationships between the pieces?

Define a Jumper as a Knight which can jump pxq (p =< q) rather than only 1x2.
Let the total number of moves a Jumper can make = J(n,m,p,q) for n,m >= q)

7) Find J(n,m,p,q) when (i) p = 0 (ii) p = q (iii) 0 < p < q

.

These are good questions...I am trying to chip them away slowly.

For 1A:

By my calculations I get B = 280 and N = 336, so together they satisfy

2B + N = R.

Not a coincidence.

Nope, your Bishop formula must be wrong.

By the way, P(n,m) includes captures and en passant.

Originally posted by THUDandBLUNDER
Nope, your Bishop formula must be wrong.

By the way, P(n,m) includes captures and en passant.

Originally posted by davegage
If that's not right, then I think I am missing something...

Originally posted by THUDandBLUNDER

Why do you think it is not a coincidence?