Combination and Permutation Basics

Joseph Wood

11/30/2023

This document serves as an overview for attacking common combinatorial problems in R. One of the goals of RcppAlgos is to provide a comprehensive and accessible suite of functionality so that users can easily get to the heart of their problem. As a bonus, the functions in RcppAlgos are extremely efficient and are constantly being improved with every release.

It should be noted that this document only covers common problems. For more information on other combinatorial problems addressed by RcppAlgos, see the following vignettes:

For much of the output below, we will be using the following function obtained here combining head and tail methods in R (credit to user @flodel)

ht <- function(d, m = 5, n = m) {
  ## print the head and tail together
  cat("head -->\n")
  print(head(d, m))
  cat("--------\n")
  cat("tail -->\n")
  print(tail(d, n))
}

Introducing comboGeneral and permuteGeneral

Easily executed with a very simple interface. The output is in lexicographical order.

We first look at getting results without repetition. You can pass an integer n and it will be converted to the sequence 1:n, or you can pass any vector with an atomic type (i.e. logical, integer, numeric, complex, character, and raw).

library(RcppAlgos)
options(width = 90)

## combn output for reference
combn(4, 3)
#>      [,1] [,2] [,3] [,4]
#> [1,]    1    1    1    2
#> [2,]    2    2    3    3
#> [3,]    3    4    4    4

## This is the same as combn expect the output is transposed
comboGeneral(4, 3)
#>      [,1] [,2] [,3]
#> [1,]    1    2    3
#> [2,]    1    2    4
#> [3,]    1    3    4
#> [4,]    2    3    4

## Find all 3-tuple permutations without
## repetition of the numbers c(1, 2, 3, 4).
head(permuteGeneral(4, 3))
#>      [,1] [,2] [,3]
#> [1,]    1    2    3
#> [2,]    1    2    4
#> [3,]    1    3    2
#> [4,]    1    3    4
#> [5,]    1    4    2
#> [6,]    1    4    3

## If you don't specify m, the length of v (if v is a vector) or v (if v is a
## scalar (see the examples above)) will be used
v <- c(2, 3, 5, 7, 11, 13)
comboGeneral(v)
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]    2    3    5    7   11   13

head(permuteGeneral(v))
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]    2    3    5    7   11   13
#> [2,]    2    3    5    7   13   11
#> [3,]    2    3    5   11    7   13
#> [4,]    2    3    5   11   13    7
#> [5,]    2    3    5   13    7   11
#> [6,]    2    3    5   13   11    7

## They are very efficient...
system.time(comboGeneral(25, 12))
#>    user  system elapsed 
#>   0.054   0.013   0.067

comboCount(25, 12)
#> [1] 5200300

ht(comboGeneral(25, 12))
#> head -->
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#> [1,]    1    2    3    4    5    6    7    8    9    10    11    12
#> [2,]    1    2    3    4    5    6    7    8    9    10    11    13
#> [3,]    1    2    3    4    5    6    7    8    9    10    11    14
#> [4,]    1    2    3    4    5    6    7    8    9    10    11    15
#> [5,]    1    2    3    4    5    6    7    8    9    10    11    16
#> --------
#> tail -->
#>            [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#> [5200296,]   13   14   15   16   18   19   20   21   22    23    24    25
#> [5200297,]   13   14   15   17   18   19   20   21   22    23    24    25
#> [5200298,]   13   14   16   17   18   19   20   21   22    23    24    25
#> [5200299,]   13   15   16   17   18   19   20   21   22    23    24    25
#> [5200300,]   14   15   16   17   18   19   20   21   22    23    24    25

## And for permutations... over 8 million instantly
system.time(permuteGeneral(13, 7))
#>    user  system elapsed 
#>   0.023   0.014   0.037

permuteCount(13, 7)
#> [1] 8648640

ht(permuteGeneral(13, 7))
#> head -->
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#> [1,]    1    2    3    4    5    6    7
#> [2,]    1    2    3    4    5    6    8
#> [3,]    1    2    3    4    5    6    9
#> [4,]    1    2    3    4    5    6   10
#> [5,]    1    2    3    4    5    6   11
#> --------
#> tail -->
#>            [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#> [8648636,]   13   12   11   10    9    8    3
#> [8648637,]   13   12   11   10    9    8    4
#> [8648638,]   13   12   11   10    9    8    5
#> [8648639,]   13   12   11   10    9    8    6
#> [8648640,]   13   12   11   10    9    8    7

## Factors are preserved
permuteGeneral(factor(c("low", "med", "high"),
               levels = c("low", "med", "high"),
               ordered = TRUE))
#>      [,1] [,2] [,3]
#> [1,] low  med  high
#> [2,] low  high med 
#> [3,] med  low  high
#> [4,] med  high low 
#> [5,] high low  med 
#> [6,] high med  low 
#> Levels: low < med < high

Combinations/Permutations with Repetition

There are many problems in combinatorics which require finding combinations/permutations with repetition. This is easily achieved by setting repetition to TRUE.

fourDays <- weekdays(as.Date("2019-10-09") + 0:3, TRUE)
ht(comboGeneral(fourDays, repetition = TRUE))
#> head -->
#>      [,1]  [,2]  [,3]  [,4] 
#> [1,] "Wed" "Wed" "Wed" "Wed"
#> [2,] "Wed" "Wed" "Wed" "Thu"
#> [3,] "Wed" "Wed" "Wed" "Fri"
#> [4,] "Wed" "Wed" "Wed" "Sat"
#> [5,] "Wed" "Wed" "Thu" "Thu"
#> --------
#> tail -->
#>       [,1]  [,2]  [,3]  [,4] 
#> [31,] "Fri" "Fri" "Fri" "Fri"
#> [32,] "Fri" "Fri" "Fri" "Sat"
#> [33,] "Fri" "Fri" "Sat" "Sat"
#> [34,] "Fri" "Sat" "Sat" "Sat"
#> [35,] "Sat" "Sat" "Sat" "Sat"

## When repetition = TRUE, m can exceed length(v)
ht(comboGeneral(fourDays, 8, TRUE))
#> head -->
#>      [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8] 
#> [1,] "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Wed"
#> [2,] "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Thu"
#> [3,] "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Fri"
#> [4,] "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Sat"
#> [5,] "Wed" "Wed" "Wed" "Wed" "Wed" "Wed" "Thu" "Thu"
#> --------
#> tail -->
#>        [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8] 
#> [161,] "Fri" "Fri" "Fri" "Fri" "Sat" "Sat" "Sat" "Sat"
#> [162,] "Fri" "Fri" "Fri" "Sat" "Sat" "Sat" "Sat" "Sat"
#> [163,] "Fri" "Fri" "Sat" "Sat" "Sat" "Sat" "Sat" "Sat"
#> [164,] "Fri" "Sat" "Sat" "Sat" "Sat" "Sat" "Sat" "Sat"
#> [165,] "Sat" "Sat" "Sat" "Sat" "Sat" "Sat" "Sat" "Sat"

fibonacci <- c(1L, 2L, 3L, 5L, 8L, 13L, 21L, 34L)
permsFib <- permuteGeneral(fibonacci, 5, TRUE)

ht(permsFib)
#> head -->
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]    1    1    1    1    1
#> [2,]    1    1    1    1    2
#> [3,]    1    1    1    1    3
#> [4,]    1    1    1    1    5
#> [5,]    1    1    1    1    8
#> --------
#> tail -->
#>          [,1] [,2] [,3] [,4] [,5]
#> [32764,]   34   34   34   34    5
#> [32765,]   34   34   34   34    8
#> [32766,]   34   34   34   34   13
#> [32767,]   34   34   34   34   21
#> [32768,]   34   34   34   34   34

## N.B. class is preserved
class(fibonacci)
#> [1] "integer"

class(permsFib[1, ])
#> [1] "integer"

## Binary representation of all numbers from 0 to 1023
ht(permuteGeneral(0:1, 10, T))
#> head -->
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,]    0    0    0    0    0    0    0    0    0     0
#> [2,]    0    0    0    0    0    0    0    0    0     1
#> [3,]    0    0    0    0    0    0    0    0    1     0
#> [4,]    0    0    0    0    0    0    0    0    1     1
#> [5,]    0    0    0    0    0    0    0    1    0     0
#> --------
#> tail -->
#>         [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1020,]    1    1    1    1    1    1    1    0    1     1
#> [1021,]    1    1    1    1    1    1    1    1    0     0
#> [1022,]    1    1    1    1    1    1    1    1    0     1
#> [1023,]    1    1    1    1    1    1    1    1    1     0
#> [1024,]    1    1    1    1    1    1    1    1    1     1

Working with Multisets

Sometimes, the standard combination/permutation functions don’t quite get us to our desired goal. For example, one may need all permutations of a vector with some of the elements repeated a specific number of times (i.e. a multiset). Consider the following vector a <- c(1,1,1,1,2,2,2,7,7,7,7,7) and one would like to find permutations of a of length 6. Using traditional methods, we would need to generate all permutations, then eliminate duplicate values. Even considering that permuteGeneral is very efficient, this approach is clunky and not as fast as it could be. Observe:

getPermsWithSpecificRepetition <- function(z, n) {
    b <- permuteGeneral(z, n)
    myDupes <- duplicated(b)
    b[!myDupes, ]
}

a <- as.integer(c(1, 1, 1, 1, 2, 2, 2, 7, 7, 7, 7, 7))

system.time(test <- getPermsWithSpecificRepetition(a, 6))
#>    user  system elapsed 
#>   1.261   0.022   1.283

Enter freqs

Situations like this call for the use of the freqs argument. Simply, enter the number of times each unique element is repeated and Voila!

## Using the S3 method for class 'table'
system.time(test2 <- permuteGeneral(table(a), 6))
#>    user  system elapsed 
#>   0.000   0.000   0.001

identical(test, test2)
#> [1] TRUE

Here are some more general examples with multisets:

## Generate all permutations of a vector with specific
## length of repetition for each element (i.e. multiset)
ht(permuteGeneral(3, freqs = c(1,2,2)))
#> head -->
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]    1    2    2    3    3
#> [2,]    1    2    3    2    3
#> [3,]    1    2    3    3    2
#> [4,]    1    3    2    2    3
#> [5,]    1    3    2    3    2
#> --------
#> tail -->
#>       [,1] [,2] [,3] [,4] [,5]
#> [26,]    3    2    3    1    2
#> [27,]    3    2    3    2    1
#> [28,]    3    3    1    2    2
#> [29,]    3    3    2    1    2
#> [30,]    3    3    2    2    1

## or combinations of a certain length
comboGeneral(3, 2, freqs = c(1,2,2))
#>      [,1] [,2]
#> [1,]    1    2
#> [2,]    1    3
#> [3,]    2    2
#> [4,]    2    3
#> [5,]    3    3

Parallel Computing

Using the parameter Parallel or nThreads, we can generate combinations/permutations with greater efficiency.

library(microbenchmark)

## RcppAlgos uses the "number of threads available minus one" when Parallel is TRUE
RcppAlgos::stdThreadMax()
#> [1] 8

comboCount(26, 13)
#> [1] 10400600

## Compared to combn using 4 threads
microbenchmark(combn = combn(26, 13),
               serAlgos = comboGeneral(26, 13),
               parAlgos = comboGeneral(26, 13, nThreads = 4),
               times = 10,
               unit = "relative")
#> Warning in microbenchmark(combn = combn(26, 13), serAlgos = comboGeneral(26, : less
#> accurate nanosecond times to avoid potential integer overflows
#> Unit: relative
#>      expr        min         lq       mean    median        uq       max neval cld
#>     combn 134.471317 133.255350 100.725417 131.10378 73.933281 66.475301    10 a  
#>  serAlgos   2.968905   2.973074   2.462118   2.94302  2.117913  1.987334    10  b 
#>  parAlgos   1.000000   1.000000   1.000000   1.00000  1.000000  1.000000    10   c

## Using 7 cores w/ Parallel = TRUE
microbenchmark(
    serial = comboGeneral(20, 10, freqs = rep(1:4, 5)),
    parallel = comboGeneral(20, 10, freqs = rep(1:4, 5), Parallel = TRUE),
    unit = "relative"
)
#> Unit: relative
#>      expr      min       lq     mean   median      uq      max neval cld
#>    serial 3.236333 2.853772 2.763475 2.823088 2.77055 1.491217   100  a 
#>  parallel 1.000000 1.000000 1.000000 1.000000 1.00000 1.000000   100   b

Using arguments lower and upper

There are arguments lower and upper that can be utilized to generate chunks of combinations/permutations without having to generate all of them followed by subsetting. As the output is in lexicographical order, these arguments specify where to start and stop generating. For example, comboGeneral(5, 3) outputs 10 combinations of the vector 1:5 chosen 3 at a time. We can set lower to 5 in order to start generation from the 5th lexicographical combination. Similarly, we can set upper to 4 in order to only generate the first 4 combinations. We can also use them together to produce only a certain chunk of combinations. For example, setting lower to 4 and upper to 6 only produces the 4th, 5th, and 6th lexicographical combinations. Observe:

comboGeneral(5, 3, lower = 4, upper = 6)
#>      [,1] [,2] [,3]
#> [1,]    1    3    4
#> [2,]    1    3    5
#> [3,]    1    4    5

## is equivalent to the following:
comboGeneral(5, 3)[4:6, ]
#>      [,1] [,2] [,3]
#> [1,]    1    3    4
#> [2,]    1    3    5
#> [3,]    1    4    5

Generating Results Beyond .Machine$integer.max

In addition to being useful by avoiding the unnecessary overhead of generating all combination/permutations followed by subsetting just to see a few specific results, lower and upper can be utilized to generate large number of combinations/permutations in parallel (see this stackoverflow post for a real use case). Observe:

## Over 3 billion results
comboCount(35, 15)
#> [1] 3247943160

## 10086780 evenly divides 3247943160, otherwise you need to ensure that
## upper does not exceed the total number of results (E.g. see below, we
## would have "if ((x + foo) > 3247943160) {myUpper = 3247943160}" where
## foo is the size of the increment you choose to use in seq()).

system.time(lapply(seq(1, 3247943160, 10086780), function(x) {
     temp <- comboGeneral(35, 15, lower = x, upper = x + 10086779)
     ## do something
     x
}))
#>    user  system elapsed 
#>  26.438  11.890  38.333

## Enter parallel
library(parallel)
system.time(mclapply(seq(1, 3247943160, 10086780), function(x) {
     temp <- comboGeneral(35, 15, lower = x, upper = x + 10086779)
     ## do something
     x
}, mc.cores = 6))
#>    user  system elapsed 
#>  30.185  15.874   9.810

GMP Support

The arguments lower and upper are also useful when one needs to explore combinations/permutations where the number of results is large:

set.seed(222)
myVec <- rnorm(1000)

## HUGE number of combinations
comboCount(myVec, 50, repetition = TRUE)
#> Big Integer ('bigz') :
#> [1] 109740941767310814894854141592555528130828577427079559745647393417766593803205094888320

## Let's look at one hundred thousand combinations in the range (1e15 + 1, 1e15 + 1e5)
system.time(b <- comboGeneral(myVec, 50, TRUE,
                              lower = 1e15 + 1,
                              upper = 1e15 + 1e5))
#>    user  system elapsed 
#>   0.003   0.002   0.004

b[1:5, 45:50]
#>           [,1]      [,2]      [,3]     [,4]      [,5]       [,6]
#> [1,] 0.5454861 0.4787456 0.7797122 2.004614 -1.257629 -0.7740501
#> [2,] 0.5454861 0.4787456 0.7797122 2.004614 -1.257629  0.1224679
#> [3,] 0.5454861 0.4787456 0.7797122 2.004614 -1.257629 -0.2033493
#> [4,] 0.5454861 0.4787456 0.7797122 2.004614 -1.257629  1.5511027
#> [5,] 0.5454861 0.4787456 0.7797122 2.004614 -1.257629  1.0792094

User Defined Functions

You can also pass user defined functions by utilizing the argument FUN. This feature’s main purpose is for convenience, however it is somewhat more efficient than generating all combinations/permutations and then using a function from the apply family (N.B. the argument Parallel has no effect when FUN is employed).

funCustomComb = function(n, r) {
    combs = comboGeneral(n, r)
    lapply(1:nrow(combs), function(x) cumprod(combs[x,]))
}

identical(funCustomComb(15, 8), comboGeneral(15, 8, FUN = cumprod))
#> [1] TRUE

microbenchmark(f1 = funCustomComb(15, 8),
               f2 = comboGeneral(15, 8, FUN = cumprod), unit = "relative")
#> Unit: relative
#>  expr     min      lq     mean   median       uq      max neval cld
#>    f1 5.21106 4.97984 4.616931 4.989338 5.036799 2.015578   100  a 
#>    f2 1.00000 1.00000 1.000000 1.000000 1.000000 1.000000   100   b

comboGeneral(15, 8, FUN = cumprod, upper = 3)
#> [[1]]
#> [1]     1     2     6    24   120   720  5040 40320
#> 
#> [[2]]
#> [1]     1     2     6    24   120   720  5040 45360
#> 
#> [[3]]
#> [1]     1     2     6    24   120   720  5040 50400

## An example involving the powerset... Note, we could
## have used the FUN.VALUE parameter here instead of
## calling unlist. See the next section.
unlist(comboGeneral(c("", letters[1:3]), 3,
                    freqs = c(2, rep(1, 3)),
                    FUN = function(x) paste(x, collapse = "")))
#> [1] "a"   "b"   "c"   "ab"  "ac"  "bc"  "abc"

Using FUN.VALUE

As of version 2.5.0, we can make use of FUN.VALUE which serves as a template for the return value from FUN. The behavior is nearly identical to vapply:

## Example from earlier involving the power set
comboGeneral(c("", letters[1:3]), 3, freqs = c(2, rep(1, 3)),
             FUN = function(x) paste(x, collapse = ""), FUN.VALUE = "a")
#> [1] "a"   "b"   "c"   "ab"  "ac"  "bc"  "abc"

comboGeneral(15, 8, FUN = cumprod, upper = 3, FUN.VALUE = as.numeric(1:8))
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7]  [,8]
#> [1,]    1    2    6   24  120  720 5040 40320
#> [2,]    1    2    6   24  120  720 5040 45360
#> [3,]    1    2    6   24  120  720 5040 50400

## Fun example with binary representations... consider the following:
permuteGeneral(0:1, 3, TRUE)
#>      [,1] [,2] [,3]
#> [1,]    0    0    0
#> [2,]    0    0    1
#> [3,]    0    1    0
#> [4,]    0    1    1
#> [5,]    1    0    0
#> [6,]    1    0    1
#> [7,]    1    1    0
#> [8,]    1    1    1

permuteGeneral(c(FALSE, TRUE), 3, TRUE, FUN.VALUE = 1,
               FUN = function(x) sum(2^(which(rev(x)) - 1)))
#> [1] 0 1 2 3 4 5 6 7

Passing additional arguments with ...

As of version 2.8.3, we have added the ability to pass further arguments to FUN via ....

## Again, same example with the power set only this time we
## conveniently pass the additional arguments to paste via '...'
comboGeneral(c("", letters[1:3]), 3, freqs = c(2, rep(1, 3)),
             FUN = paste, collapse = "", FUN.VALUE = "a")
#> [1] "a"   "b"   "c"   "ab"  "ac"  "bc"  "abc"

This concludes our discussion around user defined functions. There are several nice features that allow the user to more easily get the desired output with fewer function calls as well as fewer keystrokes. This was most clearly seen in our example above with the power set.

We started with wrapping our call to comboGeneral with unlist, which was alleviated by the parameter FUN.VALUE. We then further simplified our usage of FUN by allowing additional arguments to be passed via ....

S3 methods

As of version 2.8.3, we have added several S3 methods for convenience.

Take our earlier example where we were talking about multisets.

a <- as.integer(c(1, 1, 1, 1, 2, 2, 2, 7, 7, 7, 7, 7))

## Explicitly utilizing the freqs argument and determining the unique
## values for v... Still works, but clunky
t1 <- permuteGeneral(rle(a)$values, 6, freqs = rle(a)$lengths)

## Now using the table method... much cleaner
t2 <- permuteGeneral(table(a), 6)

identical(t1, t2)
#> [1] TRUE

There are other S3 methods defined that simplify the interface. Take for example the case when we want to pass a character vector. We know underneath the hood, character vectors are not thread safe so the Parallel and nThreads argument are ignored. We also know that the constraints parameters are only applicable to numeric vectors. For these reason, our default method’s interface is greatly simplified:

We see only the necessary options. With numeric types, the options are more numerous:

There is also a list method that allows one to find combinations or permutations of lists:

comboGeneral(
    list(
        numbers   = rnorm(4),
        states    = state.abb[1:5],
        some_data = data.frame(a = c('a', 'b'), b = c(10, 100))
    ),
    m = 2
)
#> [[1]]
#> [[1]]$numbers
#> [1] -1.9376332  0.2583997 -0.7198657 -0.8985872
#> 
#> [[1]]$states
#> [1] "AL" "AK" "AZ" "AR" "CA"
#> 
#> 
#> [[2]]
#> [[2]]$numbers
#> [1] -1.9376332  0.2583997 -0.7198657 -0.8985872
#> 
#> [[2]]$some_data
#>   a   b
#> 1 a  10
#> 2 b 100
#> 
#> 
#> [[3]]
#> [[3]]$states
#> [1] "AL" "AK" "AZ" "AR" "CA"
#> 
#> [[3]]$some_data
#>   a   b
#> 1 a  10
#> 2 b 100

This feature was inspired by ggrothendieck here: Issue 20.