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.
There are 5 compiled constraint functions that can be utilized efficiently to test a given result.
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)
## 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")
#> Unit: relative
#> expr min lq mean median uq max neval cld
#> serial 3.177339 3.173798 3.054257 3.14192 2.948129 2.859115 100 a
#> parallel 1.000000 1.000000 1.000000 1.00000 1.000000 1.000000 100 a
#> combnSum 234.997953 208.891710 189.269384 201.36549 188.176221 176.208449 100 browSums and rowMeansFinding 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 3.18877 3.009145 2.592448 2.556819 2.446044 1.4423210 100 c
#> parallel 1.00000 1.000000 1.000000 1.000000 1.000000 1.0000000 100 a
#> rowsums 2.90689 2.636867 2.212371 2.069613 2.254097 0.6039177 100 b
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 3.063023 2.561861 2.411453 2.542078 2.529883 1.553592 100 c
#> parallel 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 100 a
#> rowmeans 1.759344 1.443430 1.348248 1.418801 1.404468 0.532154 100 b
all.equal(rowMeans(combs),
comboGeneral(25, 10,
constraintFun = "mean",
Parallel = TRUE)[,11])
#> [1] TRUEIn both cases above, RcppAlgos is doing double the work
nearly twice as fast!!!
limitConstraintsThe 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.
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] TRUESometimes, 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] 330toleranceWhen 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] 199If 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 6And 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 5.100000In 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 4.433333permuteGeneralTypically, 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 4.333333As 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.
In version 2.4.* it was possible to interrupt long
running jobs by taking advantage of
Rcpp::checkUserInterrupt. Since we no longer rely on
Rcpp in version 2.5.0, we are currently not
checking for user interruption. Our goal is to eventually include this
feature as it is very convenient, however it will take great care to
implement as it is non-trivial.
For now, 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.
set.seed(123)
s <- rnorm(1000)
## In previous versions, you could execute commands and interrupt
## them if it took a really long.
##
## Now, we use iterators.
iter <- comboIter(s, 20, TRUE,
constraintFun = "sum",
comparisonFun = "==",
limitConstraints = 0)
## Test one iteration to see if we need to relax the tolerance
system.time(iter@nextIter())
#> user system elapsed
#> 7.448 0.013 7.468
## 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 <- comboIter(s, 20, TRUE,
constraintFun = "sum",
comparisonFun = "==",
limitConstraints = 0,
tolerance = 1e-5)
system.time(relaxedIter@nextIter())
#> user system elapsed
#> 0.001 0.000 0.001Specialized 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 a wrapper of
comboGeneral with constraintFun = "sum" and
comparisonFun = "==". Instead of using the very general
limitConstraints parameter, we use target with
a default of max(abs(v)) as it seems more fitting for
partitions.
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 belowEverything 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 2Same 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 4freqs to Refine LengthWe 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 4freqsAs 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 4Note, 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
#> 0.092 0.051 0.143
## Using nThreads
system.time(partitionsGeneral(0:60, repetition = TRUE, nThreads=4))
#> user system elapsed
#> 0.116 0.103 0.056
## 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
#> 0.11 0.00 0.11
system.time(partitionsGeneral(0:120, freqs = c(120, rep(1, 120)), nThreads=4))
#> user system elapsed
#> 0.146 0.001 0.038