Haskell中的嵌套列表推导

Question

I am following this Haskell tutorial and am on the higher order functions section. It defines a function called chain as:

chain :: (Integral a) => a -> [a]
chain 1 = [1]
chain n
    | even n = n:chain (n `div` 2)
    | odd n = n:chain (n * 3 + 1)

There is an exercise to find number of "chains" that have a length longer than 15. They do so like this:

numLongChains :: Int  
numLongChains = length (filter isLong (map chain [1..100]))  
    where isLong xs = length xs > 15

I am trying to come up with a list comprehension that instead of giving me the number of chains gives me a list of chains that are longer than 15 from [1..100]. My closest attempt so far looks like:

[ [ a | a <- chain b, length a > 15] | b <- [1..100]]

but I get:

<interactive>:9:14:
No instance for (Integral [a0]) arising from a use of `chain'
Possible fix: add an instance declaration for (Integral [a0])
In the expression: chain b
In a stmt of a list comprehension: a <- chain b
In the expression: [a | a <- chain b, length a > 15]

<interactive>:9:45:
    No instance for (Enum [a0])
      arising from the arithmetic sequence `1 .. 100'
    Possible fix: add an instance declaration for (Enum [a0])
    In the expression: [1 .. 100]
    In a stmt of a list comprehension: b <- [1 .. 100]
    In the expression:
      [[a | a <- chain b, length a > 15] | b <- [1 .. 100]]

<interactive>:9:46:
    No instance for (Num [a0]) arising from the literal `1'
    Possible fix: add an instance declaration for (Num [a0])
    In the expression: 1
    In the expression: [1 .. 100]
    In a stmt of a list comprehension: b <- [1 .. 100]

Am I even close? I do want to solve this problem using a nested comprehension for the sake of learning despite the possible better ways to approach this.

Answer 1

You're close. You're looking for:

[ a | b <- [1..10], let a = chain b, length a > 15 ]

The expression

[ [ a | a <- chain b, length a > 15] | b <- [1..100]]

has type:

Integral [a] => [[[a]]]

which is clearly not what you want, but because of Haskell's polymorphic numeric literals, it could possibly type check if the correct definition were in place.

In this case b is inferred to be a list of some type, and that is why you see the error:

No instance for (Integral [a0]) ...

Here is a complete run down on how the types are inferred.

1. from b <- [1..100] we can infer Enum b is a constraint
2. from the call chain b we can infer Integral b is a constraint
3. from a <- chain b we can infer a and b have the same type
4. from length a we can infer a is a list of something, eg a ~ [t]

So from (3) and (4) we can infer that b ~ [t] and so we need both Integral [t] and Enum [t] for some type t .

To further elaborate on (3), we know chain b is a list, eg [t] where t is the type of b . From a <- chain b we know a has the same type as the elements of chain b , ie the type of a is also t .