hmt-0.15: Haskell Music Theory

Safe HaskellNone
LanguageHaskell98

Music.Theory.Permutations

Description

Permutation functions.

Synopsis

Documentation

factorial :: (Ord a, Num a) => a -> a Source

Factorial function.

(factorial 13,maxBound::Int)

nk_permutations :: Integral a => a -> a -> a Source

Number of k element permutations of a set of n elements.

(nk_permutations 4 3,nk_permutations 13 3) == (24,1716)

n_permutations :: Integral a => a -> a Source

Number of nk permutations where n == k.

map n_permutations [1..8] == [1,2,6,24,120,720,5040,40320]
n_permutations 16 `div` 1000000 == 20922789

permutation :: Eq a => [a] -> [a] -> Permute Source

Generate the permutation from p to q, ie. the permutation that, when applied to p, gives q.

apply_permutation (permutation [0,1,3] [1,0,3]) [0,1,3] == [1,0,3]

apply_permutation :: Eq a => Permute -> [a] -> [a] Source

Apply permutation f to p.

let p = permutation [1..4] [4,3,2,1]
in apply_permutation p [1..4] == [4,3,2,1]

apply_permutation_c :: Eq a => [[Int]] -> [a] -> [a] Source

Composition of apply_permutation and from_cycles.

apply_permutation_c [[0,3],[1,2]] [1..4] == [4,3,2,1]
apply_permutation_c [[0,2],[1],[3,4]] [1..5] == [3,2,1,5,4]
apply_permutation_c [[0,1,4],[2,3]] [1..5] == [2,5,4,3,1]
apply_permutation_c [[0,1,3],[2,4]] [1..5] == [2,4,5,1,3]

non_invertible :: Permute -> Bool Source

True if the inverse of p is p.

non_invertible (permutation [0,1,3] [1,0,3]) == True
let p = permutation [1..4] [4,3,2,1]
in non_invertible p == True && P.cycles p == [[0,3],[1,2]]

from_cycles :: [[Int]] -> Permute Source

Generate a permutation from the cycles c.

apply_permutation (from_cycles [[0,1,2,3]]) [1..4] == [2,3,4,1]

permutations_n :: Int -> [Permute] Source

Generate all permutations of size n.

map one_line (permutations_n 3) == [[1,2,3],[1,3,2]
                                   ,[2,1,3],[2,3,1]
                                   ,[3,1,2],[3,2,1]]

compose :: Permute -> Permute -> Permute Source

Composition of q then p.

let {p = from_cycles [[0,2],[1],[3,4]]
    ;q = from_cycles [[0,1,4],[2,3]]
    ;r = p `compose` q}
in apply_permutation r [1,2,3,4,5] == [2,4,5,1,3]

two_line :: Permute -> ([Int], [Int]) Source

Two line notation of p.

two_line (permutation [0,1,3] [1,0,3]) == ([1,2,3],[2,1,3])

one_line :: Permute -> [Int] Source

One line notation of p.

one_line (permutation [0,1,3] [1,0,3]) == [2,1,3]
map one_line (permutations_n 3) == [[1,2,3],[1,3,2]
                                   ,[2,1,3],[2,3,1]
                                   ,[3,1,2],[3,2,1]]

one_line_compact :: Permute -> String Source

Variant of one_line that produces a compact string.

one_line_compact (permutation [0,1,3] [1,0,3]) == "213"
let p = permutations_n 3
in unwords (map one_line_compact p) == "123 132 213 231 312 321"

multiplication_table :: Int -> [[Permute]] Source

Multiplication table of symmetric group n.

unlines (map (unwords . map one_line_compact) (multiplication_table 3))
==> 123 132 213 231 312 321
    132 123 312 321 213 231
    213 231 123 132 321 312
    231 213 321 312 123 132
    312 321 132 123 231 213
    321 312 231 213 132 123