Return all possible 5 x 4 matrices with four ones and the remaining zeroes

December 31, 2021

Consider the following simplified example, with all possible 2 x 2 matrices with one 1 and the remaining 0s.

library(arrangements)

generate_matrices <- function(nrow, ncol, ones_count) {
  
  vectors <- permutations(c(
    rep(1, ones_count),
    rep(0, nrow * ncol - ones_count)
  ))
  
  # remove redundancies
  vectors <- vectors[!duplicated(vectors),]
  
  # list of matrices
  out <- list()
  
  for (i in 1:ncol(vectors)) {
    out[[i]] <- matrix(
      data = vectors[,i],
      nrow = nrow,
      ncol = ncol,
      byrow = TRUE
    )
  }
  return(out)
}

generate_matrices(nrow = 2, ncol = 2, ones_count = 1)

[[1]]
     [,1] [,2]
[1,]    1    0
[2,]    0    0

[[2]]
     [,1] [,2]
[1,]    0    1
[2,]    0    0

[[3]]
     [,1] [,2]
[1,]    0    0
[2,]    1    0

[[4]]
     [,1] [,2]
[1,]    0    0
[2,]    0    1

Now let’s extend this to the case given in the question title:

> generate_matrices(nrow = 5, ncol = 4, ones_count = 4)
Error in permutations(c(rep(1, ones_count), rep(0, nrow * ncol - ones_count))) : 
  too many results

My guess is that the lines

vectors <- permutations(c(
    rep(1, ones_count),
    rep(0, nrow * ncol - ones_count)
  ))

takes too long to run and/or there is not enough memory on my laptop to store these. Is there a more efficient way to implement this?

EDIT: It is worth noting that I would like to eventually extend this to the 6 x 5 case with 4 ones, and 8 x 5 case with 8 ones.

>Solution :

You could use the function developed here to get all possible matrices of dim 5×4, and then filter by the number of 1s using sum.

f = function(nrow, ncol) lapply(asplit(do.call(expand.grid, rep(list(0:1), nrow * ncol)), 1), matrix, nrow, ncol)
list = f(5,4)
list[lapply(list, sum) == 4]