This document covers the topic of finding combinations or permutations that meet a specific set of criteria. For example, retrieving all combinations of a vector that have a product between two bounds.

## Constraint Functions

There are 5 compiled constraint functions that can be utilized efficiently to test a given result.

1. sum
2. prod
3. mean
4. max
5. min

They are passed as strings to the constraintFun parameter. When these are employed without any other parameters being set, an additional column is added that represents the result of applying the given function to that combination/permutation. You can also set keepResults = TRUE (more on this later).

library(RcppAlgos)
options(width = 90)

packageVersion("RcppAlgos")
#> [1] '2.8.3'

cat(paste(capture.output(sessionInfo())[1:3], collapse = "\n"))
#> R version 4.3.1 (2023-06-16)
#> Platform: aarch64-apple-darwin20 (64-bit)
#> Running under: macOS Ventura 13.4.1

## base R using combn and FUN
combnSum = combn(20, 10, sum)
algosSum = comboGeneral(20, 10, constraintFun = "sum")

## Notice the additional column (i.e. the 11th column)
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
#> [1,]    1    2    3    4    5    6    7    8    9    10    55
#> [2,]    1    2    3    4    5    6    7    8    9    11    56
#> [3,]    1    2    3    4    5    6    7    8    9    12    57
#> [4,]    1    2    3    4    5    6    7    8    9    13    58
#> [5,]    1    2    3    4    5    6    7    8    9    14    59
#> [6,]    1    2    3    4    5    6    7    8    9    15    60

identical(as.integer(combnSum), algosSum[,11])
#> [1] TRUE

## Using parallel
paralSum = comboGeneral(20, 10, constraintFun = "sum", Parallel = TRUE)
identical(paralSum, algosSum)
#> [1] TRUE

library(microbenchmark)
microbenchmark(serial = comboGeneral(20, 10, constraintFun = "sum"),
parallel = comboGeneral(20, 10, constraintFun = "sum", Parallel = TRUE),
combnSum = combn(20, 10, sum), unit = "relative")
#> Warning in microbenchmark(serial = comboGeneral(20, 10, constraintFun = "sum"), : less
#> accurate nanosecond times to avoid potential integer overflows
#> Unit: relative
#>      expr        min         lq       mean     median         uq       max neval cld
#>    serial   3.448004   3.024937   2.742634   2.984405   2.930366  2.958093   100 a
#>  parallel   1.000000   1.000000   1.000000   1.000000   1.000000  1.000000   100  b
#>  combnSum 164.987777 143.760767 124.381205 139.771622 132.586909 65.974864   100   c

### Faster than rowSums and rowMeans

Finding row sums or row means is even faster than simply applying the highly efficient rowSums/rowMeans after the combinations have already been generated:

## Pre-generate combinations
combs = comboGeneral(25, 10)

## Testing rowSums alone against generating combinations as well as summing
microbenchmark(serial = comboGeneral(25, 10, constraintFun = "sum"),
parallel = comboGeneral(25, 10, constraintFun = "sum", Parallel = TRUE),
rowsums = rowSums(combs), unit = "relative")
#> Unit: relative
#>      expr      min       lq     mean   median       uq       max neval cld
#>    serial 3.549230 3.276643 2.958342 2.977668 2.869537 1.4453889   100 a
#>  parallel 1.000000 1.000000 1.000000 1.000000 1.000000 1.0000000   100  b
#>   rowsums 2.094864 1.955225 1.731866 1.712414 1.658254 0.4516231   100   c

all.equal(rowSums(combs),
comboGeneral(25, 10,
constraintFun = "sum",
Parallel = TRUE)[,11])
#> [1] TRUE

## Testing rowMeans alone against generating combinations as well as obtain row means
microbenchmark(serial = comboGeneral(25, 10, constraintFun = "mean"),
parallel = comboGeneral(25, 10, constraintFun = "mean", Parallel = TRUE),
rowmeans = rowMeans(combs), unit = "relative")
#> Unit: relative
#>      expr      min       lq     mean   median       uq       max neval cld
#>    serial 2.361449 2.413497 2.370115 2.386648 2.353974 1.4051019   100 a
#>  parallel 1.000000 1.000000 1.000000 1.000000 1.000000 1.0000000   100  b
#>  rowmeans 1.332882 1.343652 1.287887 1.325946 1.305233 0.3777631   100   c

all.equal(rowMeans(combs),
comboGeneral(25, 10,
constraintFun = "mean",
Parallel = TRUE)[,11])
#> [1] TRUE

In both cases above, RcppAlgos is doing double the work nearly twice as fast!!!

## Comparison Operators and limitConstraints

The standard 5 comparison operators (i.e. "<", ">", "<=", ">=", & "==") can be used in a variety of ways. In order for them to have any effect, they must be used in conjunction with constraintFun as well as limitConstraints. The latter is the value(s) that will be used for comparison. It can be passed as a single value or a vector of two numerical values. This is useful when one wants to find results that are between (or outside) of a given range.

### One Comparison Operator

First we will look at cases with only one comparison and one value for the limitConstraint.

## Generate some random data. N.B. Using R >= 4.0.0
set.seed(101)
myNums = sample(500, 20)

myNums
#>  [1] 329 313 430  95 209 442 351 317 444 315 246 355 128 131 288   9 352 489 354 244

## Find all 5-tuples combinations without repetition of myNums
## (defined above) such that the sum is equal to 1176.
p1 = comboGeneral(v = myNums, m = 5,
constraintFun = "sum",
comparisonFun = "==",
limitConstraints = 1176)

tail(p1)
#>       [,1] [,2] [,3] [,4] [,5]
#> [10,]   95  128  246  352  355
#> [11,]   95  128  288  313  352
#> [12,]   95  131  244  351  355
#> [13,]   95  131  244  352  354
#> [14,]   95  209  244  313  315
#> [15,]  128  131  246  317  354

## Authenticate with brute force
allCombs = comboGeneral(sort(myNums), 5)
identical(p1, allCombs[which(rowSums(allCombs) == 1176), ])
#> [1] TRUE

## How about finding combinations with repetition
## whose mean is less than or equal to 150.
p2 = comboGeneral(v = myNums, m = 5, TRUE,
constraintFun = "mean",
comparisonFun = "<=",
limitConstraints = 150)

## Again, we authenticate with brute force
allCombs = comboGeneral(sort(myNums), 5, TRUE)
identical(p2, allCombs[which(rowMeans(allCombs) <= 150), ])
#> [1] FALSE  ### <-- What? They should be the same

## N.B.
class(p2[1, ])
#> [1] "numeric"

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

## When mean is employed or it can be determined that integral
## values will not suffice for the comparison, we fall back to
## numeric types, thus all.equal should return TRUE
all.equal(p2, allCombs[which(rowMeans(allCombs) <= 150), ])
#> [1] TRUE

### Two Comparison Operators

Sometimes, we need to generate combinations/permutations such that when we apply a constraint function, the results are between (or outside) a given range. There is a natural two step process when finding results outside a range, however for finding results between a range, this two step approach could become computationally demanding. The underlying algorithms in RcppAlgos are optimized for both cases and avoids adding results that will eventually be removed.

Using two comparisons is easy. The first comparison operator is applied to the first limit and the second operator is applied to the second limit.

Note that in the examples below, we have keepResults = TRUE. This means an additional column will be added to the output that is the result of applying constraintFun to that particular combination.

## Get combinations such that the product is
## strictly between 3600 and 4000
comboGeneral(5, 7, TRUE, constraintFun = "prod",
comparisonFun = c(">","<"),          ## Find results > 3600 and < 4000
limitConstraints = c(3600, 4000),
keepResults = TRUE)
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,]    1    2    3    5    5    5    5 3750
#> [2,]    1    3    4    4    4    4    5 3840
#> [3,]    2    2    3    4    4    4    5 3840
#> [4,]    3    3    3    3    3    3    5 3645
#> [5,]    3    3    3    3    3    4    4 3888

# ## The above is the same as doing the following:
# comboGeneral(5, 7, TRUE, constraintFun = "prod",
#              comparisonFun = c("<",">"),          ## Note that the comparison vector
#              limitConstraints = c(4000, 3600),    ## and the limits have flipped
#              keepResults = TRUE)

## What about finding combinations outside a range
outside = comboGeneral(5, 7, TRUE, constraintFun = "prod",
comparisonFun = c("<=",">="),
limitConstraints = c(3600, 4000),
keepResults = TRUE)

all(apply(outside[, -8], 1, prod) <= 3600
| apply(outside[, -8], 1, prod) >= 4000)
#> [1] TRUE

dim(outside)
#> [1] 325   8

## Note that we obtained 5 results when searching "between"
## 3600 and 4000. Thus we have: 325 + 5 = 330
comboCount(5, 7, T)
#> [1] 330

### Using tolerance

When the underlying type is numeric, round-off errors can occur. As stated in floating-point error mitigation:

“By definition, floating-point error cannot be eliminated, and, at best, can only be managed.”

Here is a great stackoverflow post that further illuminates this tricky topic:

For these reasons, the argument tolerance can be utilized to refine a given constraint. It is added to the upper limit and subtracted from the lower limit. The default value is sqrt(.Machine$double.eps) ~= 0.00000001490116. This default value is good and bad. For the good side: dim(comboGeneral(seq(0, 0.5, 0.05), 6, TRUE, constraintFun = "sum", comparisonFun = "==", limitConstraints = 1)) #> [1] 199 6 ## Confirm with integers and brute force allCbs = comboGeneral(seq(0L, 50L, 5L), 6, TRUE, constraintFun = "sum") sum(allCbs[, 7] == 100L) #> [1] 199 If we had a tolerance of zero, we would have obtained an incorrect result: ## We miss 31 combinations that add up to 1 dim(comboGeneral(seq(0, 0.5, 0.05), 6, TRUE, constraintFun = "sum", comparisonFun = "==", limitConstraints = 1, tolerance = 0)) #> [1] 168 6 And now for a less desirable result. The example below appears to give incorrect results. That is, we shouldn’t return any combination with a mean of 4.1 or 5.1. comboGeneral(c(2.1, 3.1, 5.1, 7.1), 3, T, constraintFun = "mean", comparisonFun = c("<", ">"), limitConstraints = c(5.1, 4.1), keepResults = TRUE) #> [,1] [,2] [,3] [,4] #> [1,] 2.1 3.1 7.1 4.100000 #> [2,] 2.1 5.1 5.1 4.100000 #> [3,] 2.1 5.1 7.1 4.766667 #> [4,] 3.1 3.1 7.1 4.433333 #> [5,] 3.1 5.1 5.1 4.433333 #> [6,] 3.1 5.1 7.1 5.100000 #> [7,] 5.1 5.1 5.1 5.100000 In the above example, the range that is actually tested against is c(4.0999999950329462, 5.1000000049670531). If you want to be absolutely sure you are getting the correct results, one must rely on integers as simple changes in arithmetic can throw off precision in floating point operations. comboGeneral(c(21, 31, 51, 71), 3, T, constraintFun = "mean", comparisonFun = c("<", ">"), limitConstraints = c(51, 41), keepResults = TRUE) / 10 #> [,1] [,2] [,3] [,4] #> [1,] 2.1 5.1 7.1 4.766667 #> [2,] 3.1 3.1 7.1 4.433333 #> [3,] 3.1 5.1 5.1 4.433333 ### Output Order with permuteGeneral Typically, when we call permuteGeneral, the output is in lexicographical order, however when we apply a constraint, the underlying algorithm checks against combinations only, as this is more efficient. If a particular combination meets a constraint, then all permutations of that vector also meet that constraint, so there is no need to check them. For this reason, the output isn’t in order. Observe: permuteGeneral(c(2, 3, 5, 7), 3, freqs = rep(2, 4), constraintFun = "mean", comparisonFun = c(">", "<"), limitConstraints = c(4, 5), keepResults = TRUE, tolerance = 0) #> [,1] [,2] [,3] [,4] #> [1,] 2 5 7 4.666667 ### <-- First combination that meets the criteria #> [2,] 2 7 5 4.666667 #> [3,] 5 2 7 4.666667 #> [4,] 5 7 2 4.666667 #> [5,] 7 2 5 4.666667 #> [6,] 7 5 2 4.666667 #> [7,] 3 3 7 4.333333 ### <-- Second combination that meets the criteria #> [8,] 3 7 3 4.333333 #> [9,] 7 3 3 4.333333 #> [10,] 3 5 5 4.333333 ### <-- Third combination that meets the criteria #> [11,] 5 3 5 4.333333 #> [12,] 5 5 3 4.333333 As you can see, the 2nd through the 6th entries are simply permutations of the 1st entry. Similarly, entries 8 and 9 are permutations of the 7th and entries 11 and 12 are permutations of the 10th. ## Integer Partitions Specialized algorithms are employed when it can be determined that we are looking for integer partitions. As of version 2.5.0, we now have added partitionsGeneral which is similar to comboGeneral with constraintFun = "sum" and comparisonFun = "==". Instead of using the very general limitConstraints parameter, we use target with a default of max(v) as it seems more fitting for partitions. ### Case 1: All Integer Partitions of N We need v = 0:N, repetition = TRUE. When we leave m = NULL, m is internally set to the length of the longest non-zero combination (this is true for all cases below). partitionsGeneral(0:5, repetition = TRUE) #> [,1] [,2] [,3] [,4] [,5] #> [1,] 0 0 0 0 5 #> [2,] 0 0 0 1 4 #> [3,] 0 0 0 2 3 #> [4,] 0 0 1 1 3 #> [5,] 0 0 1 2 2 #> [6,] 0 1 1 1 2 #> [7,] 1 1 1 1 1 ## Note that we could also use comboGeneral: ## comboGeneral(0:5, repetition = TRUE, ## constraintFun = "sum", ## comparisonFun = "==", limitConstraints = 5) ## ## The same goes for any of the examples below ### Case 2: Integer Partitions of N of Length m Everything is the same as above except for explicitly setting the desired length and deciding whether to include zero or not. ## Including zero partitionsGeneral(0:5, 3, repetition = TRUE) #> [,1] [,2] [,3] #> [1,] 0 0 5 #> [2,] 0 1 4 #> [3,] 0 2 3 #> [4,] 1 1 3 #> [5,] 1 2 2 ## Zero not included partitionsGeneral(5, 3, repetition = TRUE) #> [,1] [,2] [,3] #> [1,] 1 1 3 #> [2,] 1 2 2 ### Case 3: Integer Partitions of N into Distinct Parts Same as Case 1 & 2 except now we have repetition = FALSE. partitionsGeneral(0:10) #> [,1] [,2] [,3] [,4] #> [1,] 0 1 2 7 #> [2,] 0 1 3 6 #> [3,] 0 1 4 5 #> [4,] 0 2 3 5 #> [5,] 1 2 3 4 ## Zero not included and restrict the length partitionsGeneral(10, 3) #> [,1] [,2] [,3] #> [1,] 1 2 7 #> [2,] 1 3 6 #> [3,] 1 4 5 #> [4,] 2 3 5 ## Include zero and restrict the length partitionsGeneral(0:10, 3) #> [,1] [,2] [,3] #> [1,] 0 1 9 #> [2,] 0 2 8 #> [3,] 0 3 7 #> [4,] 0 4 6 #> [5,] 1 2 7 #> [6,] 1 3 6 #> [7,] 1 4 5 #> [8,] 2 3 5 ## partitions of 10 into distinct parts of every length lapply(1:4, function(x) { partitionsGeneral(10, x) }) #> [[1]] #> [,1] #> [1,] 10 #> #> [[2]] #> [,1] [,2] #> [1,] 1 9 #> [2,] 2 8 #> [3,] 3 7 #> [4,] 4 6 #> #> [[3]] #> [,1] [,2] [,3] #> [1,] 1 2 7 #> [2,] 1 3 6 #> [3,] 1 4 5 #> [4,] 2 3 5 #> #> [[4]] #> [,1] [,2] [,3] [,4] #> [1,] 1 2 3 4 #### Using freqs to Refine Length We can utilize the freqs argument to obtain more distinct partitions by allowing for repeated zeros. The super optimized algorithm will only be carried out if zero is included and the number of repetitions for every number except zero is one. For example, given v = 0:N and J >= 1, if freqs = c(J, rep(1, N)), then the super optimized algorithm will be used, however if freqs = c(J, 2, rep(1, N - 1)), the general algorithm will be used. It should be noted that the general algorithms are still highly optimized so one should not fear using it. A pattern that is guaranteed to retrieve all distinct partitions of N is to set v = 0:N and freqs = c(N, rep(1, N)) (the extra zeros will be left off). ## Obtain all distinct partitions of 10 partitionsGeneral(0:10, freqs = c(10, rep(1, 10))) ## Same as c(3, rep(1, 10)) #> [,1] [,2] [,3] [,4] #> [1,] 0 0 0 10 #> [2,] 0 0 1 9 #> [3,] 0 0 2 8 #> [4,] 0 0 3 7 #> [5,] 0 0 4 6 #> [6,] 0 1 2 7 #> [7,] 0 1 3 6 #> [8,] 0 1 4 5 #> [9,] 0 2 3 5 #> [10,] 1 2 3 4 #### Caveats Using freqs As noted in Case 1, if m = NULL, the length of the output will be determined by the longest non-zero combination that sums to N. ## m is NOT NULL and output has at most 2 zeros partitionsGeneral(0:10, 3, freqs = c(2, rep(1, 10))) #> [,1] [,2] [,3] #> [1,] 0 0 10 #> [2,] 0 1 9 #> [3,] 0 2 8 #> [4,] 0 3 7 #> [5,] 0 4 6 #> [6,] 1 2 7 #> [7,] 1 3 6 #> [8,] 1 4 5 #> [9,] 2 3 5 ## m is NULL and output has at most 2 zeros partitionsGeneral(0:10, freqs = c(2, rep(1, 10))) #> [,1] [,2] [,3] [,4] #> [1,] 0 0 1 9 #> [2,] 0 0 2 8 #> [3,] 0 0 3 7 #> [4,] 0 0 4 6 #> [5,] 0 1 2 7 #> [6,] 0 1 3 6 #> [7,] 0 1 4 5 #> [8,] 0 2 3 5 #> [9,] 1 2 3 4 ### Case 4: Integer Partitions of N into Parts of Varying Multiplicity ## partitions of 12 into 4 parts where each part can ## be used a specific number of times (e.g. 2 or 3) partitionsGeneral(12, 4, freqs = rep(2:3, 6)) #> [,1] [,2] [,3] [,4] #> [1,] 1 1 2 8 #> [2,] 1 1 3 7 #> [3,] 1 1 4 6 #> [4,] 1 1 5 5 #> [5,] 1 2 2 7 #> [6,] 1 2 3 6 #> [7,] 1 2 4 5 #> [8,] 1 3 3 5 #> [9,] 1 3 4 4 #> [10,] 2 2 2 6 #> [11,] 2 2 3 5 #> [12,] 2 2 4 4 #> [13,] 2 3 3 4 ## Efficiency Generating Partitions Note, as of version 2.5.0, one can generate partitions in parallel using the nThreads argument. ## partitions of 60 partitionsCount(0:60, repetition = TRUE) #> [1] 966467 ## Single threaded system.time(partitionsGeneral(0:60, repetition = TRUE)) #> user system elapsed #> 0.025 0.012 0.038 ## Using nThreads system.time(partitionsGeneral(0:60, repetition = TRUE, nThreads=4)) #> user system elapsed #> 0.032 0.022 0.015 ## partitions of 120 into distinct parts partitionsCount(0:120, freqs = c(120, rep(1, 120))) #> [1] 2194432 system.time(partitionsGeneral(0:120, freqs = c(120, rep(1, 120)))) #> user system elapsed #> 0.020 0.006 0.026 system.time(partitionsGeneral(0:120, freqs = c(120, rep(1, 120)), nThreads=4)) #> user system elapsed #> 0.023 0.006 0.008 ## partitions of 100 into parts of 15 with specific multiplicity partitionsCount(100, 15, freqs = rep(4:8, 20)) #> [1] 6704215 ## Over 6 million in just over a second! system.time(partitionsGeneral(100, 15, freqs = rep(4:8, 20))) #> user system elapsed #> 0.341 0.092 0.433 ## Integer Compositions Compositions are related to integer partitions, however order matters. With RcppAlgos, we generate standard compositions with compositionsGeneral. Currently, the composition algorithms are limited to a subset of cases of compositions with repetiion. The output with compositionGeneral will be in lexicographical order. When we set weak = TRUE, we will obtain weak compositions, which allow for zeros to be a part of the sequence (E.g. c(0, 0, 5), c(0, 5, 0), c(5, 0, 0) are weak compositions of 5). As the Wikipedia article points out, we can increase the number of zeros indefinitely when weak = TRUE. For more general cases, we can make use of permuteGeneral, keeping in mind that the output will not be in lexicographical order. Another consideration with permuteGeneral is that when we include zero, we will always obtain weak compositions. With that in mind, generating compositions with RcppAlgos is easy, flexible, and quite efficient. ### Case 5: All Compositions of N ## See Case 1 compositionsGeneral(0:3, repetition = TRUE) #> [,1] [,2] [,3] #> [1,] 0 0 3 #> [2,] 0 1 2 #> [3,] 0 2 1 #> [4,] 1 1 1 ## Get weak compositions compositionsGeneral(0:3, repetition = TRUE, weak = TRUE) #> [,1] [,2] [,3] #> [1,] 0 0 3 #> [2,] 0 1 2 #> [3,] 0 2 1 #> [4,] 0 3 0 #> [5,] 1 0 2 #> [6,] 1 1 1 #> [7,] 1 2 0 #> [8,] 2 0 1 #> [9,] 2 1 0 #> [10,] 3 0 0 ## Get weak compositions with width > than target tail(compositionsGeneral(0:3, 10, repetition = TRUE, weak = TRUE)) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [215,] 2 0 0 0 0 1 0 0 0 0 #> [216,] 2 0 0 0 1 0 0 0 0 0 #> [217,] 2 0 0 1 0 0 0 0 0 0 #> [218,] 2 0 1 0 0 0 0 0 0 0 #> [219,] 2 1 0 0 0 0 0 0 0 0 #> [220,] 3 0 0 0 0 0 0 0 0 0 ## With permuteGeneral, we always get weak compositions, just ## not in lexicographical order permuteGeneral(0:3, repetition = TRUE, constraintFun = "sum", comparisonFun = "==", limitConstraints = 3) #> [,1] [,2] [,3] #> [1,] 0 0 3 #> [2,] 0 3 0 #> [3,] 3 0 0 #> [4,] 0 1 2 #> [5,] 0 2 1 #> [6,] 1 0 2 #> [7,] 1 2 0 #> [8,] 2 0 1 #> [9,] 2 1 0 #> [10,] 1 1 1 tail(permuteGeneral(0:3, 10, repetition = TRUE, constraintFun = "sum", comparisonFun = "==", limitConstraints = 3)) #> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] #> [215,] 1 1 0 0 0 0 0 1 0 0 #> [216,] 1 1 0 0 0 0 1 0 0 0 #> [217,] 1 1 0 0 0 1 0 0 0 0 #> [218,] 1 1 0 0 1 0 0 0 0 0 #> [219,] 1 1 0 1 0 0 0 0 0 0 #> [220,] 1 1 1 0 0 0 0 0 0 0 ### Case 6: Compositions of N of Length m ## See Case 2. N.B. weak = TRUE has no effect compositionsGeneral(6, 3, repetition = TRUE) #> [,1] [,2] [,3] #> [1,] 1 1 4 #> [2,] 1 2 3 #> [3,] 1 3 2 #> [4,] 1 4 1 #> [5,] 2 1 3 #> [6,] 2 2 2 #> [7,] 2 3 1 #> [8,] 3 1 2 #> [9,] 3 2 1 #> [10,] 4 1 1 ### Case 7: Compositions of N into Distinct Parts We must use permuteGeneral here. compositionsGeneral(6, 3) #> Error: Currently, there is no composition algorithm for this case. #> Use permuteCount, permuteIter, permuteGeneral, permuteSample, or #> permuteRank instead. ## See Case 3 permuteGeneral(6, 3, constraintFun = "sum", comparisonFun = "==", limitConstraints = 6) #> [,1] [,2] [,3] #> [1,] 1 2 3 #> [2,] 1 3 2 #> [3,] 2 1 3 #> [4,] 2 3 1 #> [5,] 3 1 2 #> [6,] 3 2 1 ### Case 8: Integer Compositions of N into Parts of Varying Multiplicity ## compositions of 5 into 3 parts where each part can ## be used a maximum of 2 times. permuteGeneral(5, 3, freqs = rep(2, 5), constraintFun = "sum", comparisonFun = "==", limitConstraints = 5) #> [,1] [,2] [,3] #> [1,] 1 1 3 #> [2,] 1 3 1 #> [3,] 3 1 1 #> [4,] 1 2 2 #> [5,] 2 1 2 #> [6,] 2 2 1 ## Efficiency Generating Partitions and Compositions With compositionGeneral we are able to take advantage of parallel computation. With permuteGeneral, the parallel options have no effect when generating compositions. ## compositions of 25 system.time(compositionsGeneral(0:25, repetition = TRUE)) #> user system elapsed #> 1.711 0.098 1.809 compositionsCount(0:25, repetition=TRUE) #> [1] 16777216 ## Use multiple threads for greater efficiency. Generate ## over 16 million compositions in under a second! system.time(compositionsGeneral(0:25, repetition = TRUE, nThreads = 4)) #> user system elapsed #> 1.871 0.103 0.547 ## weak compositions of 12 usnig nThreads = 4 system.time(weakComp12 <- compositionsGeneral(0:12, repetition = TRUE, weak = TRUE, nThreads = 4)) #> user system elapsed #> 0.012 0.007 0.005 ## And using permuteGeneral system.time(weakPerm12 <- permuteGeneral(0:12, 12, repetition = TRUE, constraintFun = "sum", comparisonFun = "==", limitConstraints = 12)) #> user system elapsed #> 0.011 0.003 0.015 dim(weakPerm12) #> [1] 1352078 12 identical(weakPerm12[do.call(order, as.data.frame(weakPerm12)), ], weakComp12) #> [1] TRUE ## General compositions with varying multiplicities system.time(comp25_gen <- permuteGeneral(25, 10, freqs = rep(4:8, 5), constraintFun = "sum", comparisonFun = "==", limitConstraints = 25)) #> user system elapsed #> 0.025 0.006 0.031 dim(comp25_gen) #> [1] 946092 10 ## Safely Interrupt Execution with cpp11::check_user_interrupt Some of these operations can take some time, especially when you are in the exploratory phase and you don’t have that much information about what type of solution you will obtain. For this reason, we have added the ability to interrupt execution. Under the hood, we call cpp11::check_user_interrupt() once every second to check if the user has requested for the process to be interrupted. Note that we only check for user interruptions when we cannot determine the number of results up front. This means that if we initiate a process that will take a long time or exhaust all of the available memory (e.g. we forget to put an upper limit on the number of results, relax the tolerance, etc.), we can simply hit Ctrl + c, or esc if using RStudio, to stop execution. set.seed(123) s = rnorm(1000) ## Oops!! We forgot to limit the output/put a loose tolerance ## There are as.numeric(comboCount(s, 20, T)) ~= 4.964324e+41 ## This will either take a long long time, or all of your ## memory will be consumed!!! ## ## No problem... simply hit Ctrl + c or if in RStudio, hit esc ## or hit the "Stop" button ## ## system.time(testInterrupt <- partitionsGeneral(s, 20, TRUE, target = 0)) ## Timing stopped at: 1.029 0.011 1.04 ## ### Note about Interrupting Execution Generally, we encourage user to use iterators (See Combinatorial Iterators in RcppAlgos) as they offer greater flexibility. For example, with iterators it is easy to avoid resource consuming calls by only fetching a few results at a time. Here is an example of how to investigate difficult problems due to combinatorial explosion without fear of having to restart R. ## We use "s" defined above iter = partitionsIter(s, 20, TRUE, target = 0) ## Test one iteration to see if we need to relax the tolerance system.time(iter@nextIter()) #> user system elapsed #> 5.163 0.018 5.182 ## 8 seconds per iteration is a bit much... Let's loosen things ## a little by increasing the tolerance from sqrt(.Machine$double.eps)
## ~= 1.49e-8 to 1e-5.
relaxedIter = partitionsIter(s, 20, TRUE, target = 0, tolerance = 1e-5)

system.time(relaxedIter@nextIter())
#>    user  system elapsed
#>   0.001   0.000   0.001