Question

所以我在haskell有一个任务，我坚持我的逻辑，这就是我给出的问题：

graphColor查找图形的所有可能的颜色着色是节点到颜色的分配这样两个节点如果用一个连接就必须有不同的颜色边缘

graphColor采用以下参数

颜色列表
节点列表
边缘列表（即节点对）
部分着色（即，节点和颜色对的列表）

边缘示例

e1 :: [Edge]
e1 = [(1,3),(2,1),(2,5),(3,1),(3,2),(3,3),(3,4),(3,5),(5,6),(2,6)]

e2 :: [Edge]
e2 = [(1,2),(2,3),(3,4),(1,3),(2,4),(1,4)]

type Code = [(Char,Char)]

-- domain of our code

domain1 :: [Char]
domain1 = ['a'..'z']

-- associated range

range1 :: [Char]
range1 = ['q'..'z'] ++ ['a'..'p']

-- create a code out of our domain and range
code1 :: Code
code1 = zip domain1 range1

graphColor返回图表的所有完整色彩，如果有多个解决方案，则列表不需要包含所有这些，只要它包含至少一个。

练习1：

> take 1 (graphColor "abc" [1..6] e1 [])
[[(6,'c'),(5,'a'),(4,'a'),(3,'c'),(2,'b'),(1,'a')]]

练习2：

> take 1 (graphColor "abc" [1..4] e2 [])
[]

这是我到目前为止的代码，无论如何都会返回空列表。

graphColor :: [Color] -> [Node] -> [Edge] -> Coloring -> [Coloring]
-- Fill in your code here
graphColor colors nodes edges solution
  | length nodes == length solution = [solution]
  | otherwise = concat [graphColor colors nodes edges (newSol : solution) |
                             x <- nodes, y <- colors,
                             newSol <- [(x,y)],
                             notRepeat solution edges x y,
                             notElem newSol solution]
notRepeat :: Coloring -> [Edge] -> Node -> Color -> Bool
notRepeat solution edges x y 
  | x `elem` [z | (z,_) <- solution] = False
  |otherwise = helpHelper edges [z | (z,y) <- solution] x

helpHelper edges matchedNodes x
   = x `elem` [z | y <- matchedNodes, (z,y) <- edges] ++
     [z | y <- matchedNodes, (y,z) <- edges]

调试一个简单的haskell函数，将颜色放置到节点＆＃34; graphColor＆＃34;

练习1：

练习2：

0 个答案: