This document covers topics in generating random samples of combinations, permutations, partitions, compositions and partition of groups. It is encouraged to read General Combinatorics first.

At the heart of sampling is the ability to efficiently generate the *n ^{th}* lexicographical result. The algorithms in

`RcppAlgos`

are flexible and optimized, allowing for tackling this task with ease.To illustrate this in base R, let us consider getting 5 random combinations of the vector `1:20`

of length 10. How should we proceed?

A naive approach would be to generate all of the combinations using `combn`

and then call `sample`

:

naive <- function(v, m, n, s) { allCombs <- combn(v, m) set.seed(s) allCombs[, sample(ncol(allCombs), n)] } fiveRndCombs <- naive(20, 10, 5, 42) t(fiveRndCombs) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [1,] 1 3 5 10 11 14 15 16 18 20 #> [2,] 1 3 4 9 10 11 12 13 18 19 #> [3,] 2 3 4 6 9 10 12 13 15 19 #> [4,] 1 4 5 10 13 14 15 17 18 19 #> [5,] 1 3 4 5 7 8 13 15 18 19

This is okay for this small example (there are only `choose(20, 10) = 184756`

results), however what if we wanted to find one hundred thousand random combinations from the vector `1:100`

of length 20? Clearly, the approach above will not be feasible as there are far too many results to generate (`choose(100, 20) = 5.359834e+20`

). Furthermore, there are internal limitations on `sample`

. If we try to pass `choose(100, 20)`

, we will get an error:

sample(choose(100, 20), 5) #> Error in sample.int(x, size, replace, prob) : invalid first argument

We could also try calling `sample(100, 20)`

a bunch of times and hope we don't get duplicate combinations. This is neither promising nor elegant.

`RcppAlgos`

provides five functions: `comboSample`

, `permuteSample`

, `partitionsSample`

, `compositionsSample`

, and `comboGroupsSample`

for seamlessly attacking these types of problems. All functions provide the following:

- Easily generate random samples in parallel using the
`nThreads`

or the`Parallel`

parameters. - You can pass a vector of specific indices via
`sampleVec`

or rely on the internal sampling functions. We call`sample`

when the total number of results is small and for larger cases, the sampling is done in a very similar fashion to`urand.bigz`

from the`gmp`

package. - Consistent interface to their respective general functions (e.g.
`partitionsGeneral`

) - The
`seed`

parameter allows for generating reproducible samples. - If the gmp library is needed, the
`seed`

parameter must be set in order to have reproducible results (*E.g.*`set.seed()`

) has no effect in these cases).

`comboSample`

and `permuteSample`

Let's first look at the first problem above (i.e. getting 5 random combinations of the vector `1:20`

of length 10):

library(RcppAlgos) set.seed(42) comboSample(20, 10, n = 5) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [1,] 1 3 5 10 11 14 15 16 18 20 #> [2,] 1 3 4 9 10 11 12 13 18 19 #> [3,] 2 3 4 6 9 10 12 13 15 19 #> [4,] 1 4 5 10 13 14 15 17 18 19 #> [5,] 1 3 4 5 7 8 13 15 18 19 ## Use the seed argument directly to produce the same output comboSample(20, 10, n = 5, seed = 42) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [1,] 1 3 5 10 11 14 15 16 18 20 #> [2,] 1 3 4 9 10 11 12 13 18 19 #> [3,] 2 3 4 6 9 10 12 13 15 19 #> [4,] 1 4 5 10 13 14 15 17 18 19 #> [5,] 1 3 4 5 7 8 13 15 18 19 ## fiveRndCombs produced above identical(t(fiveRndCombs), comboSample(20, 10, n = 5, seed = 42)) #> [1] TRUE

Just like with `comboGeneral`

and `permuteGeneral`

, we can explore results with repetition.

comboSample(10, 8, TRUE, n = 3, seed = 84) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 2 5 5 7 9 9 9 9 #> [2,] 4 5 8 8 8 10 10 10 #> [3,] 2 6 6 6 6 6 9 9 permuteSample(10, 8, TRUE, n = 3) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 4 10 4 4 10 2 2 10 #> [2,] 1 4 5 10 5 5 2 2 #> [3,] 4 1 7 9 1 5 6 5 comboSample(10, 12, freqs = 1:10, n = 3) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] #> [1,] 2 2 3 5 5 6 6 6 7 8 9 10 #> [2,] 1 2 3 3 5 5 6 7 9 9 9 9 #> [3,] 1 2 5 5 5 6 6 9 10 10 10 10 permuteSample(10, 12, freqs = 1:10, n = 3, seed = 123) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] #> [1,] 2 8 7 4 8 9 10 10 7 1 8 2 #> [2,] 5 5 9 8 1 8 3 2 6 4 3 10 #> [3,] 10 3 8 8 4 8 8 6 10 6 3 8

`sampleVec`

We can also utilize `sampleVec`

to generate specific results.

## E.g. the below generates the 1st, 5th, 25th, 125th, and ## 625th lexicographical combinations comboSample(10, 8, TRUE, sampleVec = c(1, 5, 25, 125, 625)) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 1 1 1 1 1 1 1 1 #> [2,] 1 1 1 1 1 1 1 5 #> [3,] 1 1 1 1 1 1 3 8 #> [4,] 1 1 1 1 1 3 6 9 #> [5,] 1 1 1 1 5 6 10 10 ## Is the same as: comboGeneral(10, 8, TRUE)[5^(0:4), ] #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 1 1 1 1 1 1 1 1 #> [2,] 1 1 1 1 1 1 1 5 #> [3,] 1 1 1 1 1 1 3 8 #> [4,] 1 1 1 1 1 3 6 9 #> [5,] 1 1 1 1 5 6 10 10

`namedSample`

Have you ever wondered which lexicographical combinations/permutations are returned when sampling? No worries, simply set `namedSample = TRUE`

:

testInd <- permuteSample(30, 10, n = 3, seed = 100, namedSample = TRUE) testInd #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> 86626302070118 24 26 7 29 3 21 20 9 16 28 #> 15871916538841 5 12 21 9 6 3 14 23 4 20 #> 87932455980012 25 6 20 23 18 10 27 30 19 29 ## Same output as above permuteSample(30, 10, sampleVec = row.names(testInd)) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [1,] 24 26 7 29 3 21 20 9 16 28 #> [2,] 5 12 21 9 6 3 14 23 4 20 #> [3,] 25 6 20 23 18 10 27 30 19 29

Just like the `General`

counterparts, the sampling functions utilize GMP to allow for exploration of combinations/permutations of large vectors where the total number of results is enormous. They also offer parallel options using `Parallel`

or `nThreads`

.

## Uses min(stdThreadMax() - 1, 5) threads (in this case) permuteSample(500, 10, TRUE, n = 5, seed = 123, Parallel = TRUE) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [1,] 55 435 274 324 200 152 6 313 121 377 #> [2,] 196 166 331 154 443 329 155 233 354 442 #> [3,] 235 325 94 27 370 117 302 86 229 126 #> [4,] 284 104 464 104 207 127 117 9 390 414 #> [5,] 456 76 381 456 219 23 376 187 11 123 permuteSample(factor(state.abb), 15, n = 3, seed = 50, nThreads = 3) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] #> [1,] ME FL DE OK ND CA PA AL ID MO NM HI KY MT NJ #> [2,] AZ CA AL CT ME SD ID SC OK NH HI TN ND IA MT #> [3,] MD MO NC MT NH AL VA MA VT WV NJ NE MN MS MI #> 50 Levels: AK AL AR AZ CA CO CT DE FL GA HI IA ID IL IN KS KY LA MA MD ME MI MN ... WY permuteCount(factor(state.abb), 15) #> Big Integer ('bigz') : #> [1] 2943352142120754524160000

The algorithms are incredibly efficient and offer tremendous gains over the naive approach above:

## the function "naive" is defined above system.time(naive(25, 10, 5, 15)) #> user system elapsed #> 3.197 0.066 3.287 system.time(comboSample(25, 10, n = 5, seed = 15)) #> user system elapsed #> 0.002 0.000 0.001

Even when dealing with extremely large numbers, these algorithms are very fast. And using the parallel options have even greater effects than we saw with the general counterparts (typically around ~2-3 times faster with the general functions, whereas with the last example below with sampling we see a nearly 5 fold improvement).

## Lightning fast even with examples involving many results system.time(comboSample(2500, 100, n = 5, seed = 15)) #> user system elapsed #> 0.002 0.000 0.002 ## The total number of combinations has ~180 digits gmp::log10.bigz(comboCount(2500, 100)) #> [1] 180.9525 ## Still fast with larger samples system.time(comboSample(2500, 100, n = 1e4, seed = 157)) #> user system elapsed #> 1.124 0.008 1.142 ## Using Parallel/nThreads in these cases has an even greater effect system.time(comboSample(2500, 100, n = 1e4, seed = 157, nThreads = 8)) #> user system elapsed #> 2.032 0.005 0.268

Again, just as with the general functions, you can pass a custom function to `{combo|permute}Sample`

using the `FUN`

argument.

permuteSample(5000, 1000, n = 3, seed = 101, FUN = sd) #> [[1]] #> [1] 1431.949 #> #> [[2]] #> [1] 1446.859 #> #> [[3]] #> [1] 1449.272 ## Example using complex numbers myCplx <- as.complex(1:100 + rep(c(-1, 1), 50) * 1i) permuteSample(myCplx, 10, freqs = rep(1:5, 20), n = 3, seed = 101, FUN = function(x) { sqrt(sum(x)) }) #> [[1]] #> [1] 24.83948+0i #> #> [[2]] #> [1] 20.9285+0.04778i #> #> [[3]] #> [1] 22.20379+0.09007i

`partitionsSample`

The `partitionsSample`

function allows one to draw a random sample of partitions of a number. Many of the features present in `comboSample`

and `permuteSample`

are available in `partitionsSample`

.

## Use the seed parameter to obtain reproducible results partitionsSample(100, 8, TRUE, n = 3, seed = 42) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 1 1 3 3 4 20 23 45 #> [2,] 1 1 2 7 14 14 29 32 #> [3,] 2 10 11 11 16 16 16 18 ## Used namedSample to obtain the lexicographical indices partitionsSample(100, 8, TRUE, n = 3, seed = 42, namedSample = TRUE) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 61413 1 1 3 3 4 20 23 45 #> 54425 1 1 2 7 14 14 29 32 #> 623844 2 10 11 11 16 16 16 18 ## Use sampleVec to obtain specific results partitionsSample(100, 8, TRUE, sampleVec = c(61413, 54425, 623844)) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 1 1 3 3 4 20 23 45 #> [2,] 1 1 2 7 14 14 29 32 #> [3,] 2 10 11 11 16 16 16 18 partitionsCount(2500, 10) #> Big Integer ('bigz') : #> [1] 2621914835336941325 ## Algorithms are very efficient system.time(serial <- partitionsSample(2500, 10, n = 1e3, seed = 8128)) #> user system elapsed #> 5.220 0.010 5.235 ## Use nThreads for greater efficiency system.time(multi <- partitionsSample(2500, 10, n = 1e3, seed = 8128, nThreads = 8)) #> user system elapsed #> 10.723 0.022 1.358 identical(multi, serial) #> [1] TRUE ## Even works with non-standard setup partitionsSample(17 + (1:10) * 3, 10, TRUE, target = 320, n = 3, seed = 111) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [1,] 23 23 26 26 29 29 38 38 41 47 #> [2,] 26 26 26 29 29 29 32 41 41 41 #> [3,] 20 23 23 26 26 35 38 41 44 44

There are sampling algorithms available for most partition cases, but some cases are not covered. For example, with standard multisets, we are currently unable to *efficiently* generate the *n ^{th}* lexicographical result. Another example is when the source vector is not uniform (

Observe the following:

## No sampling algorithm available when the source vector is not uniform partitionsSample(c(1, 4, 6, 7, 10, seq(11, 100, 7)), 10, n = 1, target = 340) #> Error in partitionsSample(c(1, 4, 6, 7, 10, seq(11, 100, 7)), 10, n = 1, : #> Partition sampling not available for this case. ## As stated above, the standard multiset case doesn't work either partitionsSample(0:50, 6, freqs = rep(1:3, 17), n = 2) #> Error in partitionsSample(0:50, 6, freqs = rep(1:3, 17), n = 2) : #> Partition sampling not available for this case. ## If we use freqs to indicate that zeros can repeat, ## then we can obtain random samples partitionsSample(0:50, 6, freqs = c(50, rep(1, 50)), n = 3, seed = 222) #> [,1] [,2] [,3] [,4] [,5] [,6] #> [1,] 0 0 1 4 9 36 #> [2,] 0 0 0 0 17 33 #> [3,] 2 4 5 6 8 25 ## Even works when the vector is restricted in regards to the target partitionsSample(0:50, 6, freqs = c(50, rep(1, 50)), n = 3, seed = 222, target = 100) #> [,1] [,2] [,3] [,4] [,5] [,6] #> [1,] 0 1 6 15 29 49 #> [2,] 0 0 0 8 43 49 #> [3,] 4 7 17 19 22 31

There is ongoing research in this area and our goal is to eventually be able to cover the standard multiset case.

`compositionsSample`

The `compositionsSample`

function allows one to draw a random sample of compositions of a number. Many of the features present in `comboSample`

and `permuteSample`

are available in `compositionsSample`

.

## Use the seed parameter to obtain reproducible results compositionsSample(100, 8, TRUE, n = 3, seed = 42) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 5 4 34 31 7 3 1 15 #> [2,] 3 21 8 6 12 43 6 1 #> [3,] 6 6 1 36 17 18 10 6 ## Used namedSample to obtain the lexicographical indices compositionsSample(100, 8, TRUE, n = 3, seed = 42, namedSample = TRUE) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 4024715585 5 4 34 31 7 3 1 15 #> 2756281572 3 21 8 6 12 43 6 1 #> 4873365553 6 6 1 36 17 18 10 6 ## Use sampleVec to obtain specific results compositionsSample(100, 8, TRUE, sampleVec = c(4024715585, 2756281572, 4873365553)) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> [1,] 5 4 34 31 7 3 1 15 #> [2,] 3 21 8 6 12 43 6 1 #> [3,] 6 6 1 36 17 18 10 6 compositionsCount(2500, 10, TRUE) #> Big Integer ('bigz') : #> [1] 10324518903611974891453624 ## Algorithms are very efficient... ## The below retrieves 10,000 compositions in under a second system.time(serial <- compositionsSample(2500, 10, TRUE, n = 1e4, seed = 8128)) #> user system elapsed #> 0.506 0.002 0.507 ## Use nThreads for greater efficiency system.time(multi <- compositionsSample(2500, 10, TRUE, n = 1e4, seed = 8128, nThreads = 8)) #> user system elapsed #> 1.154 0.004 0.152 identical(multi, serial) #> [1] TRUE ## Sample weak compositions compositionsSample(0:100, 8, repetition = TRUE, weak = TRUE, seed = 245659, n = 3, namedSample = TRUE) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 21547195040 23 0 0 8 5 22 25 17 #> 7649268748 5 0 13 36 42 4 0 0 #> 7598223523 4 54 21 4 6 0 8 3

Currently, there are only sampling algorithms for most cases of compositions with repetition. There is ongoing work to expand these algorithms in the future.

`comboGroupsSample`

Just as we can generate random samples of combinations and permutations, we are also able to generate random samples of partitions of groups of equal size. There are many problems that present in this manner. Below, we examine one involving playing cards.

Let's say we have 4 players and each player is to have 3 cards a piece. Given that the deck is shuffled, the dealer then distrubutes 12 cards.

What possible hands can each player have?

See Creating A Deck Of Cards In R Without Using While And Double For Loop (Credit to @MichaelChirico)

cards <- c(2:10, "J", "Q", "K", "A") suits <- c("♠", "♥", "♦", "♣") deck <- paste0(rep(cards, length(suits)), #card values rep(suits, each = length(cards))) #suits set.seed(1738) shuffled <- factor(deck[sample(52)], levels = deck) ## Here are 3 possibilities comboGroupsSample(shuffled[1:12], numGroups = 4, n = 2, seed = 13) #> Grp1 Grp1 Grp1 Grp2 Grp2 Grp2 Grp3 Grp3 Grp3 Grp4 Grp4 Grp4 #> [1,] 8♦ 3♥ 5♦ 9♦ J♠ 7♥ 8♠ K♦ 10♦ A♦ J♥ 3♦ #> [2,] 8♦ K♦ 10♦ 9♦ J♥ 3♥ J♠ 8♠ 3♦ A♦ 5♦ 7♥ #> 52 Levels: 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ ... A comboGroupsSample(shuffled[1:12], numGroups = 4, retType = "3Darray", n = 2, seed = 13, namedSample = TRUE) #> , , Grp1 #> #> [,1] [,2] [,3] #> 13784 8♦ 3♥ 5♦ #> 9152 8♦ K♦ 10♦ #> #> , , Grp2 #> #> [,1] [,2] [,3] #> 13784 9♦ J♠ 7♥ #> 9152 9♦ J♥ 3♥ #> #> , , Grp3 #> #> [,1] [,2] [,3] #> 13784 8♠ K♦ 10♦ #> 9152 J♠ 8♠ 3♦ #> #> , , Grp4 #> #> [,1] [,2] [,3] #> 13784 A♦ J♥ 3♦ #> 9152 A♦ 5♦ 7♥ #> #> 52 Levels: 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ ... A♣♣

Ranking is the complement of sampling. That is, given a combination (or permutation/partition), determine which lexicographical result it is. As an example, consider all of the combinations of 5 choose 3:

comboGeneral(5, 3) #> [,1] [,2] [,3] #> [1,] 1 2 3 #> [2,] 1 2 4 #> [3,] 1 2 5 #> [4,] 1 3 4 #> [5,] 1 3 5 #> [6,] 1 4 5 #> [7,] 2 3 4 #> [8,] 2 3 5 #> [9,] 2 4 5 #> [10,] 3 4 5

We can see that the rank of the combination: `c(2, 3, 4)`

is 7. That is, `c(2, 3, 4)`

is the 7^{th} combination of 5 choose 3.

Just as we saw before, we could easily produce a brute force approach that would work well with small cases, but would become unmanagemable very quickly. For example:

naive_rank <- function(v, m, comb) { comb <- as.integer(comb) which(apply(combn(v, m), 2, function(x) identical(x, comb))) } naive_rank(5, 3, 2:4) #> [1] 7 ## Larger example comb = comboSample(25, 12, sampleVec = 2e6)[1, ] system.time(print(naive_rank(25, 12, comb))) #> [1] 2000000 #> user system elapsed #> 13.703 0.172 13.901

Similar to the sampling problem, `RcppAlgos`

provides four functions: `comboRank`

, `permuteRank`

, `partitionsRank`

, and `compositionsRank`

(currently there is not a ranking function for `comboGroups`

). These functions are very similar to their sampling counterparts.

For both problems presented above, here is how you would attack them with `comboRank`

:

comboRank(2:4, v = 5) #> [1] 7 ## Since order doesn't matter with combinations, 4:2 should return 7 as well comboRank(4:2, v = 5) #> [1] 7 ## comb was provided above system.time(print(comboRank(comb, v = 25))) #> [1] 2000000 #> user system elapsed #> 0 0 0

All that is needed is the original vector that was used to produce the results and whether or not repetition is used via the `repetition`

or `freqs`

arguments. The width is determined automatically by the input.

A neat feature of the ranking functions is the ability to rank multiple inputs at once. We can either pass a single vector, multiple vectors, and we can even pass matrices.

`comboRank`

combs = comboSample(50, 8, n = 10, seed = 123, namedSample = TRUE) combs #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 161401295 3 4 6 13 40 44 48 50 #> 120710262 2 6 21 29 32 36 42 45 #> 220886245 3 22 23 25 35 36 48 50 #> 442099291 10 11 13 20 21 24 36 45 #> 334689371 6 9 13 20 28 41 43 48 #> 189241370 3 7 18 26 29 31 43 50 #> 168175018 3 4 20 21 27 42 44 49 #> 105740881 2 4 24 27 40 46 48 49 #> 36321127 1 5 10 15 18 32 40 46 #> 292830028 5 8 16 19 34 35 46 47 comboRank(combs, v = 50) #> [1] 161401295 120710262 220886245 442099291 334689371 #> [6] 189241370 168175018 105740881 36321127 292830028

`permuteRank`

perms_len_5 = permuteSample(100, 5, freqs = rep(1:5, 20), n = 3, seed = 987, namedSample = TRUE) perms_len_5 #> [,1] [,2] [,3] [,4] [,5] #> 3474930553 36 47 93 7 32 #> 5793832271 60 12 27 39 99 #> 797663634 9 16 23 3 35 perms_len_8 = permuteSample(100, 8, freqs = rep(1:5, 20), n = 3, seed = 123, namedSample = TRUE) perms_len_8 #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 6897937625624040 73 96 48 62 19 39 60 31 #> 6937176899580966 74 37 42 44 34 80 18 94 #> 5771073522599470 62 7 60 69 14 60 7 30 ## Note you can name the inputs permuteRank(p5 = perms_len_5, p8 = perms_len_8, v = 100, freqs = rep(1:5, 20)) #> $p5 #> [1] 3474930553 5793832271 797663634 #> #> $p8 #> Big Integer ('bigz') object of length 3: #> [1] 6897937625624040 6937176899580966 5771073522599470

`partitionsRank`

parts = partitionsSample(50, 8, target = 100, repetition = TRUE, n = 3, seed = 42, namedSample = TRUE) parts #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 61413 1 1 3 11 12 21 23 28 #> 54425 1 1 3 6 7 13 29 40 #> 623844 3 4 4 7 7 15 28 32 partitionsRank(parts, v = 50, target = 100, repetition = TRUE) #> [1] 61413 54425 623844

`compositionsRank`

comps = compositionsSample(50, 8, repetition = TRUE, n = 3, seed = 42, namedSample = TRUE) comps #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #> 36761573 4 6 2 1 13 3 15 6 #> 21025945 2 11 13 8 2 3 7 4 #> 71927012 11 4 17 1 2 5 5 5 compositionsRank(comps, v = 50, repetition = TRUE) #> [1] 36761573 21025945 71927012

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