# Combinatorial Sampling

#### 10/12/2019

This document covers topics in generating random samples of combinations/permutations. It is encouraged to read General Combinatorics first.

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?

## Base R

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,]    4    5    7    8    9   11   13   14   17    18
[2,]    4    6    7   10   11   12   13   14   15    16
[3,]    1    3    4    7    9   11   14   15   17    20
[4,]    3    4    9   10   11   12   14   18   19    20
[5,]    2    4    5    6    8   11   12   16   17    20

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 Solutions

RcppAlgos provides three functions: comboSample, permuteSample, and comboGroupsSample for seamlessly attacking these types of problems. All functions provide the following:

• Easily generate random samples of combinations/permutations or partition of groups in parallel.
• You can pass a vector of specific indices 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 (i.e. combo/permuteGeneral and comboGroups)
• Useful when we need a reproducible set of random combinations/permutations or partitions of groups.
• 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,]    4    5    7    8    9   11   13   14   17    18
[2,]    4    6    7   10   11   12   13   14   15    16
[3,]    1    3    4    7    9   11   14   15   17    20
[4,]    3    4    9   10   11   12   14   18   19    20
[5,]    2    4    5    6    8   11   12   16   17    20

## 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,]    4    5    7    8    9   11   13   14   17    18
[2,]    4    6    7   10   11   12   13   14   15    16
[3,]    1    3    4    7    9   11   14   15   17    20
[4,]    3    4    9   10   11   12   14   18   19    20
[5,]    2    4    5    6    8   11   12   16   17    20

## fiveRndCombs produced above
identical(t(fiveRndCombs),
comboSample(20, 10, n = 5, seed = 42))
[1] TRUE

## Samples of Results with Repetition

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,]    3    3    3    6    6   10   10   10
[2,]    1    3    3    4    4    7    9   10
[3,]    3    7    7    7    9   10   10   10

permuteSample(10, 8, TRUE, n = 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]    8    7    8   10    1    2    7    6
[2,]    3    3    8   10    2    4    4    6
[3,]    3    7    8    4    2    9   10    4

comboSample(10, 12, freqs = 1:10, n = 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,]    1    2    3    4    5    5    6    8    8     9    10    10
[2,]    1    4    4    4    5    5    5    5    5     7     7     7
[3,]    2    3    4    5    5    6    7    7    7     7     7     7

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

## Specific Results with 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

## Using 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]
33554924331145   10    7   24    5   29    6   30   12   16    11
60218249947169   17   18   15   19   14    2    1    4    7    29
51084688265260   15    2   20   27    8   10   25   30    3    18

## Same output as above
permuteSample(30, 10, sampleVec = row.names(testInd))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]   10    7   24    5   29    6   30   12   16    11
[2,]   17   18   15   19   14    2    1    4    7    29
[3,]   15    2   20   27    8   10   25   30    3    18

## Parallel Computing and GMP Support

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

## Efficiency

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
3.526   0.065   3.604

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
1.482   0.006   1.491

## 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
2.409   0.002   0.310

## User Defined Functions

Again, just as with the general functions, you can pass a custom function to combo/permuteSample 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

## Sampling Partitions of Groups of Equal Size with 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,] A♠   2♥   6♠   10♦  10♥  J♣   8♠   10♣  3♠   Q♦   6♣   8♦
[2,] A♠   10♥  Q♦   10♦  2♥   8♦   8♠   J♣   6♣   10♣  3♠   6♠
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]
10939 A♠   2♥   6♠
3791  A♠   10♥  Q♦

, , Grp2

[,1] [,2] [,3]
10939 10♦  10♥  J♣
3791  10♦  2♥   8♦

, , Grp3

[,1] [,2] [,3]
10939 8♠   10♣  3♠
3791  8♠   J♣   6♣

, , Grp4

[,1] [,2] [,3]
10939 Q♦   6♣   8♦
3791  10♣  3♠   6♠

52 Levels: 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ ... A♣