---
title: "Constraints, Partitions, and Compositions"
author: "Joseph Wood"
date: "01/31/2025"
output:
  rmarkdown::html_vignette:
    toc: true
    number_sections: false
vignette: >
  %\VignetteIndexEntry{Constraints, Partitions, and Compositions}
  %\VignetteEngine{knitr::rmarkdown}
  \usepackage[utf8]{inputenc}
---

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).

``` r
library(RcppAlgos)
options(width = 90)

packageVersion("RcppAlgos")
#> [1] '2.9.2'

cat(paste(capture.output(sessionInfo())[1:3], collapse = "\n"))
#> R version 4.4.2 (2024-10-31)
#> Platform: aarch64-apple-darwin20
#> Running under: macOS Sonoma 14.5

## 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)
head(algosSum)
#>      [,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
#>    serial   3.485355   3.291729   2.914214   3.197378   3.010128  1.716721   100
#>  parallel   1.000000   1.000000   1.000000   1.000000   1.000000  1.000000   100
#>  combnSum 171.180306 157.361966 133.895925 151.336700 139.596890 64.992480   100
```

### 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:

``` r
## 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
#>    serial 3.282834 3.118272 2.961110 2.873113 2.805536 3.622693   100
#>  parallel 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000   100
#>   rowsums 1.882470 1.857191 1.762444 1.684538 1.678611 1.722250   100

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
#>    serial 2.540918 2.273344 2.198220 2.233083 2.172085 1.593756   100
#>  parallel 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000   100
#>  rowmeans 1.520303 1.303887 1.271082 1.282142 1.248715 1.201293   100

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`.

``` r
## 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.

``` r
## 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](<https://en.wikipedia.org/wiki/Round-off_error>) can occur. As stated in [floating-point error mitigation](<https://en.wikipedia.org/wiki/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:

- [What is the correct/standard way to check if difference is smaller than machine precision?](<https://stackoverflow.com/q/59229545/4408538>)

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:

``` r
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:

``` r
## 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.

``` r
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.

``` r
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:

``` r
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 *2<sup>nd</sup>* through the *6<sup>th</sup>* entries are simply permutations of the *1<sup>st</sup>* entry. Similarly, entries *8* and *9* are permutations of the *7<sup>th</sup>* and entries *11* and *12* are permutations of the *10<sup>th</sup>*.

## Integer Partitions

Specialized algorithms are employed when it can be determined that we are looking for [integer partitions](<https://en.wikipedia.org/wiki/Partition_\(number_theory\)>).

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).

``` r
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.

``` r
## 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`.

``` r
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).

``` r
## 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*.

``` r
## 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

``` r
## 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.

``` r
## partitions of 60
partitionsCount(0:60, repetition = TRUE)
#> [1] 966467

## Single threaded
system.time(partitionsGeneral(0:60, repetition = TRUE))
#>    user  system elapsed 
#>   0.030   0.013   0.056

## Using nThreads
system.time(partitionsGeneral(0:60, repetition = TRUE, nThreads=4))
#>    user  system elapsed 
#>   0.032   0.021   0.014


## 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 
#>   0.022   0.005   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 
#>   0.246   0.082   0.329
```

## Integer Compositions

[Compositions](<https://en.wikipedia.org/wiki/Composition_\(combinatorics\)>) 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*

``` r
## 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*

``` r
## 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.

``` r
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

``` r
## 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.

``` r
## compositions of 25
system.time(compositionsGeneral(0:25, repetition = TRUE))
#>    user  system elapsed 
#>   1.649   0.085   1.738

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 
#>   1.819   0.099   0.514


## weak compositions of 12 usnig nThreads = 4
system.time(weakComp12 <- compositionsGeneral(0:12, repetition = TRUE,
                                              weak = TRUE, nThreads = 4))
#>    user  system elapsed 
#>   0.011   0.006   0.005

## And using permuteGeneral
system.time(weakPerm12 <- permuteGeneral(0:12, 12, repetition = TRUE,
                                         constraintFun = "sum",
                                         comparisonFun = "==",
                                         limitConstraints = 12))
#>    user  system elapsed 
#>   0.010   0.003   0.014

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 
#>   0.016   0.006   0.022

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.

``` r
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](<https://jwood000.github.io/RcppAlgos/articles/CombinatoricsIterators.html>)) 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.

``` 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 
#>   5.166   0.001   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       0       0
```
