Alternative exercise

Posted on 2017-06-16 by Oleg Grenrus

Haskell Programming from first principles - aka haskellbook "misses" a simple, yet hopefully insightful exercise about Alternative (introduced in Parser Combinators chapter). I have discovered this one, while working on a real world project:

Exercise Write a function with following signature:

count' :: Alternative f
       => Int  -- ^ minimum count
       -> Int  -- ^ maximum count
       -> f a
       -> f [a]

f could be a regular expression RE from regex-applicative package, in that case count' n m re should work as (re){n,m} in a traditional regexp syntax. In other words accept n to m occurrences of re.

lambda> "a" =~ count' 2 3 (sym 'a')

lambda> "aa" =~ count' 2 3 (sym 'a')
Just "aa"

lambda> "aaa" =~ count' 2 3 (sym 'a')
Just "aaa"

lambda> "aaaa" =~ count' 2 3 (sym 'a')

lambda> "ab" =~ count' 2 3 (sym 'a')

#Notes, hints and follow-ups

IF the exercise feels to hard to crack directly, there are few notes, hints and also few follow-ups.

#Negative numbers

If n < m, e.g. count' 2 1, you may return empty, or accept exactly 2 values. FWIW: JavaScript throws an exception: Invalid regular expression: /^a{3,2}$/: numbers out of order in {}, let's not do that.

Also in the following sub-exercises, treat negative numbers properly: write

foo n | n < 0 = ... -- not n == 0, or foo 0
      | otherwise = ... foo (n - 1)

This will save you from the infinite recursion when negative numbers are passed in.

#some and many

The definition of many and some in base library are slightly complicated:

-- | One or more.
some :: f a -> f [a]
some v = some_v
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v

-- | Zero or more.
many :: f a -> f [a]
many v = many_v
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v

Write many as a one-liner, using <$>, <*> and <|>.

Note: <$> and <*> bind tighter than <|>, so you can write expressions like fun <$> with <*> parsers <|> foo <$> bar <*> quux, which is grouped as (fun <$> with <*> parsers) <|> foo <$> bar <*> quux).

#Exact count

A small step towards the goal is to write simpler function first:

count :: Alternative f
      => Int  -- ^ exact count
      -> f a
      -> f [a]

That should accept the exact amount of parses. count n re = count' n n re(re){n} in traditional regex notation.

lambda> "aaa" =~ count 3 (sym 'a')
Just "aaa"


Another variant with the same type-signature (!):

upto :: Alternative f
     => Int  -- ^ up to this amount
     -> f a
     -> f [a]

This should accept upto some amount of parses. upto n re = count' 0 n re(re){n} in traditional regex notation.

#Divide and conquer

Now, when you have defined count and upto, you can define count' as simply as:

count' n m p = (++) <$> count n p <*> upto (m - n) p

GHC is smart, but not super-smart. Let's help it. Inline definition of count, removing the use of (++) from above definition.

#Order of alternatives

Virtually every parser combinator library is left-biased, in other word the left hand side of <|> is preferred. Explain what's the behavioral difference between

pure [] <|> some p
some p <|> pure []

Analyze your count', count and upto definitions, and try examples above with your backtracking parser-combinator library of choice (e.g. trifecta). Do they work as expected?

#Counting other Alternatives

There are also other Alternatives than parsers. For example [] and STM. Explain to your colleague what count' does for [] and STM. If there are other interesting Alternatives, where count' does something neat, please tell me! (e.g. via Twitter).


Fortunately, parser-combinators package by Mark Karpov provides count and count', so you could use them in your code, and check the source for one way of solving this exercise.

Site proudly generated by Hakyll