Haskell递归

Question

I've been working on adding a counter to variables in a given expression and have the following test case:

prop_add1 = add (And (Var "a") (And (Var "b") (Var "a"))) ==
And (Var "a1") (And (Var "b2") (Var "a1"))

I have been using Pattern Matching and Recursion to try and find a solution. Although I have come a little stuck, I have tried adding the original variables to a list and then using the list to determine the variable name to be output but I couldn't figure out how to correctly implement it and my output just adds a 1 to the end of all variables.

I'm wondering is there a better/easier solution to this?

My Attempt so far:

add :: Expr -> Expr
add T = T
add (Var x) = Var (x ++ show (check2(check x)))
add (And e1 e2) = And (add e1) (add e2)
add (Not e1) = Not (add e1)

check :: Variable -> [Variable]
check p = [p]

check2 :: [Variable] -> Int
check2 p = length (union p p)

Answer 1

You need to keep an environment mapping variable names to numbers. Something like

add :: Expr -> Expr
add expr = fst $ addWithEnv emptyEnv expr
  where
    emptyEnv = []
    addWithEnv env (And e1 e2)
        = case addWithEnv env e1 of
            (e1', env') ->
              case addWithEnv env' e2 of
                (e2', env'') -> (And e1' e2', env'')
    addWithEnv env (Var name)
        = case lookup name env of
            Just k -> -- stopping here,it's homework

I hope I've left enough for you to fill in.

Update:

In your attempt, you don't keep track of which variables you have already seen, so every variable seems to be the first, and every time a '1' is appended. Numbering items is a stateful computation, you must have a record of which variables have been seen so far to assign previously seen variables the old number and know which number to assign the next not-yet-seen variable. So you must carry that record around in the worker. If you already know about the Monad class and how to use that, you can implicitly carry it around using the State monad, otherwise you have to carry it around explicitly. Then add becomes a wrapper that calls the worker with an initially empty state (before the numbering/renaming starts, no variable has yet been seen). The worker then looks at the subexpressions of the given expression (if any) and renames variables and updates the state when a new variable is encountered.

So in the sketch above, we have

addWithEnv :: [(String,Int)] -> Expr -> (Expr, [(String,Int)])

since we cannot mutate the state, we have to return the new state along with the renamed expression. Now you have to define what the result shall be for each type of expression,

addWithEnv env T = ??
addWithEnv env (Var name) = ??
addWithEnv env (Add e1 e2) = ??
addWithEnv env (Not e) = ??

The T case of course does no renaming and doesn't update the environment. A Var has either been seen before, in which case the environment remains unchanged, or not, in which case it is added to the environment. A Not e has the same influence on the environment as e , and an And e1 e2 has the combined effects of first e1 then e2 on the environment.