CODEWITHSUNDEEP
rcombinationspermutationcombinatoricspartition
dyluns
dyluns

Reputation: 155

Partition and permutation using R

I'm trying to find all possible partitions (including 0) of M into N parts using R. Then, I would like to permutate each outcome without replacement and without repeatition.

For example, for M=4 and N=2, I want to get:

[1,] 4 3 2 0 1
[2,] 0 1 2 4 3

Now, I can get:

[1,] 4 3 2
[2,] 0 1 2

using partitions::restrictedparts(4, 2, include.zero=TRUE). How should I continue?

To give some background of this question, I'm actually trying to find all possible outcomes of the number of occurence of each side after rolling a die for 60 times.

Upvotes: 2

Views: 154

Answers (1)

Joseph Wood
Joseph Wood

Reputation: 7597

There is a function available in the package that you are currently using (i.e. partitions) that does just what you are looking for. It is appropriately called compositions (See Composition (combinatorics) for more information).

partitions::compositions(4, 2)

[1,] 4 3 2 1 0
[2,] 0 1 2 3 4

Now, to solve your actual problem, we have:

myParts <- partitions::compositions(60, 6)  ## Note that include.zero is TRUE by default

dim(myParts)
[1]       6 8259888

And here is the output:

transMat <- t(as.matrix(myParts))
head(transMat)
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]   60    0    0    0    0    0
[2,]   59    1    0    0    0    0
[3,]   58    2    0    0    0    0
[4,]   57    3    0    0    0    0
[5,]   56    4    0    0    0    0
[6,]   55    5    0    0    0    0

tail(transMat)
           [,1] [,2] [,3] [,4] [,5] [,6]
[8259883,]    1    0    0    0    0   59
[8259884,]    0    1    0    0    0   59
[8259885,]    0    0    1    0    0   59
[8259886,]    0    0    0    1    0   59
[8259887,]    0    0    0    0    1   59
[8259888,]    0    0    0    0    0   60

Upvotes: 2

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