# Combination and Permutation Basics

#### 08/14/2022

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("--------\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)

## 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).
#>      [,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

#>      [,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
#> 0.097   0.051   0.148

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

ht(comboGeneral(25, 12))
#>      [,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
#>   0.090   0.050   0.141

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

ht(permuteGeneral(13, 7))
#>      [,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))
#>      [,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))
#>      [,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)
#>      [,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))
#>      [,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 <- c(1,1,1,1,2,2,2,7,7,7,7,7)

system.time(test <- getPermsWithSpecificRepetition(a, 6))
#>   user  system elapsed
#>  2.477   0.069   2.553

### 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!

system.time(test2 <- permuteGeneral(unique(a), 6, freqs = rle(a)$lengths)) #> user system elapsed #> 0 0 0 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 ## Compared to combn using 4 threads microbenchmark(combn = combn(25, 10), serAlgos = comboGeneral(25, 10), parAlgos = comboGeneral(25, 10, nThreads = 4), times = 10, unit = "relative") #> Unit: relative #> expr min lq mean median uq max neval cld #> combn 117.041129 113.265819 41.953407 67.182638 22.416065 17.111740 10 b #> serAlgos 1.871675 1.802551 1.524188 2.206488 1.181195 1.097826 10 a #> parAlgos 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 10 a ## 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 3.027218 2.983932 2.830728 2.981056 2.885341 1.58061 100 b #> parallel 1.000000 1.000000 1.000000 1.000000 1.000000 1.00000 100 a ### 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
#>  67.482  55.057 123.033

## 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
#>  63.994  59.189  31.574

## 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
#>   0.009   0.000   0.010

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 5.884807 5.602342 4.886637 5.574073 5.327725 1.780681   100   b
#>    f2 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000   100  a

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) paste0(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 powerset
comboGeneral(c("", letters[1:3]), 3,
freqs = c(2, rep(1, 3)),
FUN = function(x) paste0(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