HaskellForMaths-0.4.8: Combinatorics, group theory, commutative algebra, non-commutative algebra

Safe Haskell | None
Language | Haskell98

Math.Combinatorics.LatinSquares

Synopsis

- findLatinSqs :: Eq a => [a] -> [[[a]]]
- isLatinSq :: Ord a => [[a]] -> Bool
- isOneOfEach :: Ord a => [a] -> Bool
- incidenceGraphLS :: Ord t => [[t]] -> Graph (Int, Int, t)
- incidenceGraphLS' :: Eq a => [[a]] -> Graph (Int, Int)
- isOrthogonal :: (Ord a, Ord b) => [[a]] -> [[b]] -> Bool
- findMOLS :: (Eq a, Num a, Ord b) => a -> [[[b]]] -> [[[[b]]]]
- isMOLS :: Ord a => [[[a]]] -> Bool
- fromProjectivePlane :: (Ord k, Num k) => Design [k] -> [[[Int]]]
- isOA :: Ord b => (Int, Int) -> [[b]] -> Bool
- fromLS :: Foldable t => t [Int] -> [[Int]]
- fromMOLS :: Foldable t => [t [Int]] -> [[Int]]
- graphOA :: Ord a => [[a]] -> Graph [a]
- srgParamsOA :: Num t => (t, t) -> Maybe (t, t, t, t)

# Documentation

findLatinSqs :: Eq a => [a] -> [[[a]]] Source

isLatinSq :: Ord a => [[a]] -> Bool Source

isOneOfEach :: Ord a => [a] -> Bool Source

incidenceGraphLS :: Ord t => [[t]] -> Graph (Int, Int, t) Source

incidenceGraphLS' :: Eq a => [[a]] -> Graph (Int, Int) Source

isOrthogonal :: (Ord a, Ord b) => [[a]] -> [[b]] -> Bool Source

Are the two latin squares orthogonal?

findMOLS :: (Eq a, Num a, Ord b) => a -> [[[b]]] -> [[[[b]]]] Source

isMOLS :: Ord a => [[[a]]] -> Bool Source

Are the latin squares mutually orthogonal (ie each pair is orthogonal)?

fromProjectivePlane :: (Ord k, Num k) => Design [k] -> [[[Int]]] Source

MOLS from a projective plane

isOA :: Ord b => (Int, Int) -> [[b]] -> Bool Source

fromLS :: Foldable t => t [Int] -> [[Int]] Source

fromMOLS :: Foldable t => [t [Int]] -> [[Int]] Source

graphOA :: Ord a => [[a]] -> Graph [a] Source

srgParamsOA :: Num t => (t, t) -> Maybe (t, t, t, t) Source