我正在尝试生成所有符合规范的独特有向图：
- 每个节点必须恰好有 2 个输入
- 并且允许任意多个输出到图中的其他节点
我目前的解决方案很慢。例如，对于 6 个节点，算法需要 1.5 天才能到达我认为完成的位置，但它可能还要再检查几天。
n我的节点图算法：
生成
n的全长字符串0，其中一个符号是 a1，例如，对于 n=3,[[0,0,1], [0,1,0], [1,0,0]]。这些可以被认为是来自单位矩阵的行。生成所有可能的
n * n矩阵，其中每一行是所有可能的组合step 1. + step 1.
这是连接矩阵，其中每个单元代表从column-index到的连接row-index
因此，对于 n=3，这些是可能的：
[0,1,0] + [1,0,0] = [1,1,0][1,0,0] + [1,0,0] = [2,0,0]
这些代表节点的输入，通过将步骤 1 添加到自身，结果将始终代表 2 个输入。
例如：
A B C
A' [[0,1,1],
B' [0,2,0],
C' [1,1,0]]
因此B，每个C连接A一次：B -> A', C -> A'，
并B连接到自身两次：B => B'
- 我只想要唯一的，所以对于生成的每个连接矩阵，我只能保留它，如果它不与已经看到的图同构。
这一步很昂贵。我需要通过遍历同构图的每个排列，对它们进行排序，并将第一个视为“规范形式”，将图转换为“规范形式”。
如果有人潜入测试其中任何一个，这里是n节点的唯一图的计数：
2 - 6
3 - 44
4 - 475
5 - 6874
6 - 109,934 (I think, it's not done running yet but I haven't found a new graph in >24 hrs.)
7 - I really wanna know!
可能的优化：
既然我要生成要测试的图表，有没有办法在不测试的情况下将它们排除在外，因为它们与已经看到的图形同构？
有更快的图同构算法吗？
我认为这与“Nauty”有关，我在论文中读过其他一些，但我还没有实现它们的专业知识（或带宽）。
这是一个可演示的连接矩阵，可以在 graphonline.ru 上绘制，以显示自连接和到同一节点的 2 个连接：
1, 0, 0, 0, 0, 1,
1, 0, 0, 0, 1, 0,
0, 1, 0, 1, 0, 0,
0, 1, 2, 0, 0, 0,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0,
这是haskell中的代码，如果你想玩它，但我更关心的是让算法正确（例如修剪搜索空间），而不是实现：
-- | generate all permutations of length n given symbols from xs
npermutations :: [a] -> Int -> [[a]]
npermutations xs size = mapM (const xs) [1..size]
identity :: Int -> [[Int]]
identity size = scanl
(\xs _ -> take size $ 0 : xs) -- keep shifting right
(1 : (take (size - 1) (repeat 0))) -- initial, [1,0,0,...]
[1 .. size-1] -- correct size
-- | return all possible pairings of [Column]
columnPairs :: [[a]] -> [([a], [a])]
columnPairs xs = (map (\x y -> (x,y)) xs)
<*> xs
-- | remove duplicates
rmdups :: Ord a => [a] -> [a]
rmdups = rmdups' Set.empty where
rmdups' _ [] = []
rmdups' a (b : c) = if Set.member b a
then rmdups' a c
else b : rmdups' (Set.insert b a) c
-- | all possible patterns for inputting 2 things into one node.
-- eg [0,1,1] means cells B, and C project into some node
-- [0,2,0] means cell B projects twice into one node
binaryInputs :: Int -> [[Int]]
binaryInputs size = rmdups $ map -- rmdups because [1,0]+[0,1] is same as flipped
(\(x,y) -> zipWith (+) x y)
(columnPairs $ identity size)
transposeAdjMat :: [[Int]] -> [[Int]]
transposeAdjMat ([]:_) = []
transposeAdjMat m = (map head m) : transposeAdjMat (map tail m)
-- | AdjMap [(name, inbounds)]
data AdjMap a = AdjMap [(a, [a])] deriving (Show, Eq)
addAdjColToMap :: Int -- index
-> [Int] -- inbound
-> AdjMap Int
-> AdjMap Int
addAdjColToMap ix col (AdjMap xs) =
let conns = foldl (\c (cnt, i) -> case cnt of
1 -> i:c
2 -> i:i:c
_ -> c
)
[]
(zip col [0..]) in
AdjMap ((ix, conns) : xs)
adjMatToMap :: [[Int]] -> AdjMap Int
adjMatToMap cols = foldl
(\adjMap@(AdjMap nodes) col -> addAdjColToMap (length nodes) col adjMap)
(AdjMap [])
cols
-- | a graph's canonical form : http://mfukar.github.io/2015/09/30/haskellxiii.html
-- very expensive algo, of course
canon :: (Ord a, Enum a, Show a) => AdjMap a -> String
canon (AdjMap g) = minimum $ map f $ Data.List.permutations [1..(length g)]
where
-- Graph vertices:
vs = map fst g
-- Find, via brute force on all possible orderings (permutations) of vs,
-- a mapping of vs to [1..(length g)] which is minimal.
-- For example, map [1, 5, 6, 7] to [1, 2, 3, 4].
-- Minimal is defined lexicographically, since `f` returns strings:
f p = let n = zip vs p
in (show [(snd x, sort id $ map (\x -> snd $ head $ snd $ break ((==) x . fst) n)
$ snd $ take_edge g x)
| x <- sort snd n])
-- Sort elements of N in ascending order of (map f N):
sort f n = foldr (\x xs -> let (lt, gt) = break ((<) (f x) . f) xs
in lt ++ [x] ++ gt) [] n
-- Get the first entry from the adjacency list G that starts from the given node X
-- (actually, the vertex is the first entry of the pair, hence `(fst x)`):
take_edge g x = head $ dropWhile ((/=) (fst x) . fst) g
-- | all possible matrixes where each node has 2 inputs and arbitrary outs
binaryMatrixes :: Int -> [[[Int]]]
binaryMatrixes size = let columns = binaryInputs size
unfiltered = mapM (const columns) [1..size] in
fst $ foldl'
(\(keep, seen) x -> let can = canon . adjMatToMap $ x in
(if Set.member can seen
then keep
else id $! x : keep
, Set.insert can seen))
([], Set.fromList [])
unfiltered