#> Given an array, I'd like to add elements whose location vectors are
#> permutations of one another.  For example, in a 3-dimensional array, I'd
#> add the elements in positions (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2)
#> and (3,2,1); those in positions (1,2,4), (1,4,2), (2,1,4), (2,4,1),
#> (4,1,2), and (4,2,1); and so on.  Elements with repeated positions
#> -- (1,1,2) or (1,4,4), for example -- can be ignored.
#> 
#> In two dimensions, what I want can be done with (x + t(x))[lower.tri(x)]. 
#
#One idea is to use a matrix index, of course.  The following is
#not optimal by any means, but if you don't push it too far it
#should work well enough.  First a couple of helper functions
#whose purpose is self-evident (and the first now pretty
#well-known):
#
subsets <- function(r, n, v = 1:n)
  if(r <= 0) NULL else
if(r >= n) v[1:n] else
rbind(cbind(v[1], Recall(r - 1, n - 1, v[-1])),
      Recall(r, n - 1, v[-1]))

permutations <- function(n, v = 1:n) {
  if(n == 1)
    return(v[1])
  X <- NULL
  for(i in 1:n)
    X <- rbind(X,
               cbind(v[i], permutations(n - 1, v[-i])))
  X
}

#The following example uses a 3-dimensional array but the
#generalization to any number of dimensions is no more
#complicated.  It does assume that you are dealing with an array
#with all dimensions having the same ranges.  If this is not the
#case you will need to do some additional pruning, but that too
#can be vectorized.
#
> A <- array(1:(4^3), dim = c(4,4,4))  # let's say

First generate the non-repeated index vectors as the rows of a
matrix:

> ind <- subsets(3, 4)
> ind
     [,1] [,2] [,3] 
[1,]    1    2    3
[2,]    1    2    4
[3,]    1    3    4
[4,]    2    3    4

Now generate the sums (this next step is the nub):

> sums <- apply(ind, 1, function(x) sum(A[permutations(3, x)]))

Finally give the vector names that identify which sum it is:

> names(sums) <- do.call("paste", c(data.frame(ind), sep = ","))

and here is what we have:

> sums
 1,2,3 1,2,4 1,3,4 2,3,4 
   132   174   216   258

Disclaimer: this was actually done in S, but you should never know!

> I'll summarize to the list.

Yes, please do, but I thought I would post this one anyway as my
puzzle of the month...

Bill Venables.