hmt-0.16: 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 :: 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 :: [[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