99 questions/Solutions/94
< 99 questions | Solutions
(***) Generate K-regular simple graphs with N nodes
In a K-regular graph all nodes have a degree of K; i.e. the number of edges incident in each node is K.
This solution generates all possible graphs with n nodes and n * k / 2 edges, filters the k regular graphs, then collects all non-isomorphic graphs using graph canonization. It is somewhat of a slow solution, taking >10 s to run regular 6 3
.
data Graph a = Graph [a] [(a, a)]
deriving (Show, Eq)
data Adjacency a = Adj [(a, [a])]
deriving (Show, Eq)
regular :: Int -> Int -> [Graph Int]
regular n k | r == 1 || n <= k || n < 0 || k < 0 = []
| otherwise =
map (adjToGraph . fst) $
foldr (\x xs -> if any ((==) (snd x) . snd) xs then xs else x : xs) [] $
zip a $ map canon a
where
a = filter (\(Adj a) -> all ((==) k . length . snd) a) $
map (graphToAdj . Graph [1..n]) $ perm e q
e = map (\xs -> (head xs, last xs)) $ perm [1..n] 2
(q, r) = (n * k) `quotRem` 2
perm n k = foldr (\x xs ->
[i : s | i <- n, s <- xs, i `notElem` s, asc (i : s)])
[[]] [1..k]
asc xs = all (uncurry (<)) $ zip xs $ tail xs
graphToAdj :: (Eq a) => Graph a -> Adjacency a
graphToAdj (Graph [] _) = Adj []
graphToAdj (Graph (x:xs) ys) = Adj ((x, ys >>= f) : zs)
where
f (a, b)
| a == x = [b]
| b == x = [a]
| otherwise = []
Adj zs = graphToAdj (Graph xs ys)
adjToGraph :: (Eq a) => Adjacency a -> Graph a
adjToGraph (Adj []) = Graph [] []
adjToGraph (Adj ((v, a):vs)) = Graph (v : xs) ((a >>= f) ++ ys)
where
f x = if (v, x) `elem` ys || (x, v) `elem` ys
then []
else [(v, x)]
Graph xs ys = adjToGraph (Adj vs)
canon :: (Eq a, Ord a) => Adjacency a -> String
canon (Adj a) = minimum $ map f $ perm n
where
n = length a
v = map fst a
perm n = foldr (\x xs -> [i : s | i <- [1..n], s <- xs, i `notElem` s]) [[]] [1..n]
f p = let n = zip v p
in show [(snd x,
sort id $ map (\x ->
snd $ head $ snd $ break ((==) x . fst) n) $ snd $ find a x)
| x <- sort snd n]
sort f n = foldr (\x xs -> let (lt, gt) = break ((<) (f x) . f) xs
in lt ++ [x] ++ gt) [] n
find a x = let (xs, ys) = break ((==) (fst x) . fst) a in head ys