The partitions package provides efficient vectorized code to enumerate solutions to various integer equations. For example, we might note that
[ 5 = 4+1 = 3+2 = 3+1+1 = 2+2+1 = 2+1+1+1 = 1+1+1+1+1](https://latex.codecogs.com/png.latex?%0A5%20%3D%204%2B1%20%3D%203%2B2%20%3D%203%2B1%2B1%20%3D%202%2B2%2B1%20%3D%202%2B1%2B1%2B1%20%3D%201%2B1%2B1%2B1%2B1%0A " 5 = 4+1 = 3+2 = 3+1+1 = 2+2+1 = 2+1+1+1 = 1+1+1+1+1 ")
and we might want to list all seven in a consistent format (note here that each sum is written in nonincreasing order, so is considered to be the same as ).
You can install the released version of wedge from CRAN with:
partitions package in useTo enumerate the partitions of 5:
parts(5)
#>
#> [1,] 5 4 3 3 2 2 1
#> [2,] 0 1 2 1 2 1 1
#> [3,] 0 0 0 1 1 1 1
#> [4,] 0 0 0 0 0 1 1
#> [5,] 0 0 0 0 0 0 1(each column is padded with zeros). Of course, larger integers have many more partitions and in this case we can use summary():
summary(parts(16))
#>
#> [1,] 16 15 14 14 13 13 13 12 12 12 ... 3 2 2 2 2 2 2 2 2 1
#> [2,] 0 1 2 1 3 2 1 4 3 2 ... 1 2 2 2 2 2 2 2 1 1
#> [3,] 0 0 0 1 0 1 1 0 1 2 ... 1 2 2 2 2 2 2 1 1 1
#> [4,] 0 0 0 0 0 0 1 0 0 0 ... 1 2 2 2 2 2 1 1 1 1
#> [5,] 0 0 0 0 0 0 0 0 0 0 ... 1 2 2 2 2 1 1 1 1 1
#> [6,] 0 0 0 0 0 0 0 0 0 0 ... 1 2 2 2 1 1 1 1 1 1
#> [7,] 0 0 0 0 0 0 0 0 0 0 ... 1 2 2 1 1 1 1 1 1 1
#> [8,] 0 0 0 0 0 0 0 0 0 0 ... 1 2 1 1 1 1 1 1 1 1
#> [9,] 0 0 0 0 0 0 0 0 0 0 ... 1 0 1 1 1 1 1 1 1 1
#> [10,] 0 0 0 0 0 0 0 0 0 0 ... 1 0 0 1 1 1 1 1 1 1
#> [11,] 0 0 0 0 0 0 0 0 0 0 ... 1 0 0 0 1 1 1 1 1 1
#> [12,] 0 0 0 0 0 0 0 0 0 0 ... 1 0 0 0 0 1 1 1 1 1
#> [13,] 0 0 0 0 0 0 0 0 0 0 ... 1 0 0 0 0 0 1 1 1 1
#> [14,] 0 0 0 0 0 0 0 0 0 0 ... 1 0 0 0 0 0 0 1 1 1
#> [15,] 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 0 1 1
#> [16,] 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 0 0 1Sometimes we want to find the unequal partitions (that is, partitions without repeats):
summary(diffparts(16))
#>
#> [1,] 16 15 14 13 13 12 12 11 11 11 ... 8 8 7 7 7 7 7 6 6 6
#> [2,] 0 1 2 3 2 4 3 5 4 3 ... 5 4 6 6 5 5 4 5 5 4
#> [3,] 0 0 0 0 1 0 1 0 1 2 ... 2 3 3 2 4 3 3 4 3 3
#> [4,] 0 0 0 0 0 0 0 0 0 0 ... 1 1 0 1 0 1 2 1 2 2
#> [5,] 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 0 0 1Sometimes we have restrictions on the partition. For example, to enumerate the partitions of 9 into 5 parts we would use restrictedparts():
summary(restrictedparts(9,5))
#>
#> [1,] 9 8 7 6 5 7 6 5 4 5 ... 5 4 4 3 3 5 4 3 3 2
#> [2,] 0 1 2 3 4 1 2 3 4 2 ... 2 3 2 3 2 1 2 3 2 2
#> [3,] 0 0 0 0 0 1 1 1 1 2 ... 1 1 2 2 2 1 1 1 2 2
#> [4,] 0 0 0 0 0 0 0 0 0 0 ... 1 1 1 1 2 1 1 1 1 2
#> [5,] 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 1 1 1 1 1and if we want the partitions of 9 into parts not exceeding 5 we would use the conjugate of this:
summary(conjugate(restrictedparts(9,5)))
#>
#> [1,] 1 2 2 2 2 3 3 3 3 3 ... 4 4 4 4 4 5 5 5 5 5
#> [2,] 1 1 2 2 2 1 2 2 2 3 ... 2 2 3 3 4 1 2 2 3 4
#> [3,] 1 1 1 2 2 1 1 2 2 1 ... 1 2 1 2 1 1 1 2 1 0
#> [4,] 1 1 1 1 2 1 1 1 2 1 ... 1 1 1 0 0 1 1 0 0 0
#> [5,] 1 1 1 1 1 1 1 1 0 1 ... 1 0 0 0 0 1 0 0 0 0
#> [6,] 1 1 1 1 0 1 1 0 0 0 ... 0 0 0 0 0 0 0 0 0 0
#> [7,] 1 1 1 0 0 1 0 0 0 0 ... 0 0 0 0 0 0 0 0 0 0
#> [8,] 1 1 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 0 0 0
#> [9,] 1 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 0 0 0Sometimes we have restrictions on each element of a partition and in this case we would use blockparts():
summary(blockparts(1:6,10))
#>
#> [1,] 1 1 1 1 0 1 1 1 0 1 ... 0 1 0 0 0 1 0 0 0 0
#> [2,] 2 2 2 1 2 2 2 1 2 2 ... 0 0 1 0 0 0 1 0 0 0
#> [3,] 3 3 2 3 3 3 2 3 3 1 ... 2 0 0 1 0 0 0 1 0 0
#> [4,] 4 3 4 4 4 2 3 3 3 4 ... 0 1 1 1 2 0 0 0 1 0
#> [5,] 0 1 1 1 1 2 2 2 2 2 ... 2 2 2 2 2 3 3 3 3 4
#> [6,] 0 0 0 0 0 0 0 0 0 0 ... 6 6 6 6 6 6 6 6 6 6which would show all solutions to , .
Above we considered and to be the same partition, but if these are considered to be distinct, we need the compositions, not partitions:
compositions(4)
#>
#> [1,] 4 1 2 1 3 1 2 1
#> [2,] 0 3 2 1 1 2 1 1
#> [3,] 0 0 0 2 0 1 1 1
#> [4,] 0 0 0 0 0 0 0 1A set of 4 elements, WLOG , may be partitioned into subsets in a number of ways and these are enumerated with the setparts() function:
setparts(4)
#>
#> [1,] 1 1 1 1 2 1 1 1 1 1 1 2 2 2 1
#> [2,] 1 1 1 2 1 2 1 2 2 1 2 1 1 3 2
#> [3,] 1 2 1 1 1 2 2 1 3 2 1 3 1 1 3
#> [4,] 1 1 2 1 1 1 2 2 1 3 3 1 3 1 4In the above, column 2 3 1 1 would correspond to the set partition .
Knuth deals with multisets (that is, a generalization of the concept of set, in which elements may appear more than once) and gives an algorithm for enumerating a multiset. His simplest example is the permutations of :
multiset(c(1,2,2,3))
#>
#> [1,] 1 1 1 2 2 2 2 2 2 3 3 3
#> [2,] 2 2 3 1 1 2 2 3 3 1 2 2
#> [3,] 2 3 2 2 3 1 3 1 2 2 1 2
#> [4,] 3 2 2 3 2 3 1 2 1 2 2 1It is possible to answer questions such as the permutations of the word “pepper”:
library("magrittr")
"pepper" %>%
strsplit("") %>%
unlist %>%
match(letters) %>%
multiset %>%
apply(2,function(x){x %>% `[`(letters,.) %>% paste(collapse="")})
#> [1] "eepppr" "eepprp" "eeprpp" "eerppp" "epeppr" "epeprp" "eperpp" "eppepr"
#> [9] "epperp" "eppper" "epppre" "epprep" "epprpe" "eprepp" "eprpep" "eprppe"
#> [17] "ereppp" "erpepp" "erppep" "erpppe" "peeppr" "peeprp" "peerpp" "pepepr"
#> [25] "peperp" "pepper" "peppre" "peprep" "peprpe" "perepp" "perpep" "perppe"
#> [33] "ppeepr" "ppeerp" "ppeper" "ppepre" "pperep" "pperpe" "pppeer" "pppere"
#> [41] "pppree" "ppreep" "pprepe" "pprpee" "preepp" "prepep" "preppe" "prpeep"
#> [49] "prpepe" "prppee" "reeppp" "repepp" "reppep" "repppe" "rpeepp" "rpepep"
#> [57] "rpeppe" "rppeep" "rppepe" "rpppee"For more detail, see the package vignettes
vignette("partitionspaper")
vignette("setpartitions")
vignette("scrabble")