nice check of plainperms()

[RobinHankin]

Knuth points out that the plain permutation algorithm produces alternating odd and even permutations and verifying this has nice R idiom:

> apply(plainperms(5),2,function(x){is.even(as.cycle(as.word(x)))})
  [1]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [13]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [25]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [37]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [49]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [61]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [73]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [85]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
 [97]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE
[109]  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE

[RobinHankin]

but including this in the tests would require the partitions package to be added to the Depends list

[RobinHankin]

Observing here that we are using partitions::plainperms(); also there is no need to coerce to cycle form:

apply(plainperms(5),2,function(x){is.even(as.word(x))})

[RobinHankin]

A nice way to test the above would be to use standard recycling, or compare each element with the next:

library("partitions")
suppressMessages(library("permutations"))
x <- apply(plainperms(5),2,function(x){is.even(as.word(x))})
all(x == c(TRUE,FALSE))
#> [1] TRUE
all(x[-1] != x[-length(x)])
#> [1] TRUE

Above, I think the first test is "cuter" but is defective because it relies on: (1), the vector starting with TRUE, and (2), the vector is of even length. Neither of these assumptions is necessarily true. The second test just uses the fact that each element is the negation of the previous. Actually, maybe the second test is cuter. I am clearly overthinking this.