Function as.function.permutation() in the permutations package: group actions

Robin K. S. Hankin

as.function.permutation

## function (x, ...) 
## {
##     a <- NULL
##     x <- as.matrix(as.word(x))
##     if (nrow(x) == 1) {
##         return(as.function(alist(a = , x[, a])))
##     }
##     else {
##         return(as.function(alist(a = , x[cbind(seq_len(nrow(x)), 
##             a)])))
##     }
## }
## <bytecode: 0x64414f1c07b0>
## <environment: namespace:permutations>

To cite the permutations package in publications, please use Hankin (2020). The permutations package was intended to manipulate and combine permutations, but often one wants to consider the effect of a permutation on the underlying set, taken to be \(\left[n\right]=\left\lbrace 1,2,\ldots,n\right\rbrace\). In other words, we wish to consider a permutation as a function. In package idiom, coercing a permutation to a function is straightforward:

g <- as.cycle("(45)(127)")
as.function(g)(4)

## [1] 5

Above we see that permutation \((45)(127)\) maps 4 to 5. We can see from the function body, at the top of the page, that permutations are coerced to word form. Function as.function.permutation() uses as.matrix() to stop “x[a,]” dispatching to [.word() and use matrix extraction instead. It might be argued that unclass() would be better coding.

Coercion is vectorized:

as.function(g)(1:7)

## [1] 2 7 3 5 4 6 1

as.function(allperms(4))(3)

##  [1] 3 4 2 4 2 3 3 4 1 4 1 3 2 4 1 4 1 2 2 3 1 3 1 2

as.function(rperm(7,8))(1:7)

## [1] 6 2 2 7 3 2 2

The second and third forms use the alist(a = , x[cbind(seq_len(nrow(x)),a)]) construction. We now discuss the extent to which the underlying permutation group is represented in package idiom. Consider the following construction:

(p <- cyc_len(2))

## [1] (12)

as.function(p)(3)

## Error in x[, a]: subscript out of bounds

On the one hand, object p is a permutation on the set \([2]=\left\lbrace 1,2\right\rbrace\). The action of this permutation on 3 is not defined, and the package returns an error. Above we effectively see

t(1:2)[,3]

## Error in t(1:2)[, 3]: subscript out of bounds

which is the origin of the error. On the other hand, one might reasonably hold that the action of \((12)\) on 3 should be 3, on the grounds that \((12)\) transposes elements 1 and 2 and leaves all other elements unchanged. To realise this interpretation we need to ensure that p has underlying set including 3, in this case \(\left\lbrace 1,2,3\right\rbrace\). This is straightforward with as.word():

as.function(as.word(p,n=3))(3)

## [1] 3

as.function(id)(4)

## Error in x[, a]: subscript out of bounds

as.function(as.word(id,n=4))(4)

## [1] 4