Haskell函数:: [Name]-> [[（（Name，Bool）]]

Question

Given the following:

type Name = String
envs :: [Name] -> [[(Name , Bool)]]

I have to implement 'envs' so that given a list of Names, it returns all possible combinations of Names and booleans

My attempt didn't return all possible combinations, this was my code:

envs xxs@(x:xs) = [[ (name, value) | name <- xxs  , value <- [False, True] ]]

the expected results for

envs ["P", "Q"]

are:

[ [ ("P",False)
  , ("Q",False)
  ]
, [ ("P",False)
  , ("Q",True)
  ]
, [ ("P",True)
  , ("Q",False)
  ]
, [ ("P",True)
  , ("Q",True)
  ]
]

But, mine were:

[[("P",True),("Q",True)],[("P",False),("Q",False)]]

So what is the proper implementation of the 'envs' function, the one that returns the expected result?

Answer 1

What you want to do is get all combinations of True and False for the number of names, which can be done easily with replicateM specialised to lists:

replicateM (length names) [False, True]

Eg if length names == 2 , then this produces:

[ [False, False]
, [False, True]
, [True, False]
, [True, True]
]

What remains is just to zip each of these with the names themselves:

envs names =
  [ zip names values
  | let n = length names
  , values <- replicateM n [False, True]
  ]

For envs ["P", "Q"] this produces:

[ [("P", False), ("Q", False)]
, [("P", False), ("Q", True)]
, [("P", True), ("Q", False)]
, [("P", True), ("Q", True)]
]

And it works for any number of inputs, even 0, for which your original implementation would fail since you don't match [] . It always returns 2 ⁿ assignments:

-- 2^0 == 1
envs [] == [[]]

-- 2^3 == 8
envs ["P", "Q", "R"] ==
  [ [("P", False), ("Q", False), ("R", False)]
  , [("P", False), ("Q", False), ("R", True)]
  , [("P", False), ("Q", True), ("R", False)]
  , [("P", False), ("Q", True), ("R", True)]
  , [("P", True), ("Q", False), ("R", False)]
  , [("P", True), ("Q", False), ("R", True)]
  , [("P", True), ("Q", True), ("R", False)]
  , [("P", True), ("Q", True), ("R", True)]
  ]