Generating function for a chess knight and king

Generating function for a chess knight and king on an arbitrarily-dimensioned chessboard

Usage

knight(d=2)
king(d=2)

Arguments

d

Dimensionality of the board, defaulting to 2

Value

Returns the generating function of the piece in question.

Author

Robin K. S. Hankin

Note

The pieces are forced to move; if they have the option of not moving, add 1 to the returned spray. The vignette contains a short discussion.

Examples

knight()  # default 2D chess board
#>            val
#>  -1  2  =    1
#>   2 -1  =    1
#>  -2  1  =    1
#>  -1 -2  =    1
#>  -2 -1  =    1
#>   1  2  =    1
#>   1 -2  =    1
#>   2  1  =    1
king()    # ditto
#>            val
#>   1  1  =    1
#>   0  1  =    1
#>  -1  1  =    1
#>   1  0  =    1
#>  -1  0  =    1
#>   1 -1  =    1
#>   0 -1  =    1
#>  -1 -1  =    1

knight()^2  # generating function for two knight's moves
#>            val
#>   3  3  =    2
#>  -1 -3  =    2
#>  -1  1  =    2
#>  -2 -4  =    1
#>   3 -3  =    2
#>  -1 -1  =    2
#>  -3 -3  =    2
#>   2 -4  =    1
#>   1 -1  =    2
#>  -1  3  =    2
#>   1 -3  =    2
#>  -3  3  =    2
#>   4 -2  =    1
#>   3  1  =    2
#>  -2  0  =    2
#>   2  4  =    1
#>   3 -1  =    2
#>   0 -4  =    2
#>   0  0  =    8
#>   4  0  =    2
#>  -3  1  =    2
#>   1  3  =    2
#>  -4 -2  =    1
#>   0  2  =    2
#>   4  2  =    1
#>  -3 -1  =    2
#>   1  1  =    2
#>   0  4  =    2
#>  -2  4  =    1
#>   2  0  =    2
#>  -4  0  =    2
#>   0 -2  =    2
#>  -4  2  =    1

## How many ways can a knight return to its starting square in 6 moves?
constant(knight()^6)
#>           val
#>  0 0  =  5840

## How many in 6 or fewer?
constant((1+knight())^6)
#>           val
#>  0 0  =  8481

## Where does a randomly-moving knight end up?
d <- xyz(2)
kt <- (1+knight())*d^2/9
persp(1:25,1:25,as.array(d*kt^6))



## what is the probability that a 4D king is a knight's move from
##   (0,0,0,0) after 6 moves?

sum(coeffs(((king(4)/80)^4)[knight(4)]))
#> [1] 0.06367969