Question

I am new to both Haskell and programming. My question about binding in pattern-matched, recursive functions. For instance, suppose I have a function which checks whether a given list (x:xs) is a sublist of another list, (y:ys). My initial thought, following the examples in my textbook, was:

sublist [] ys = True
sublist xs [] = False
sublist (x:xs) (y:ys)
   | x == y = sublist xs ys
   | x /= y = sublist (x:xs) ys

This works on test data, e.g.,

sublist [1, 2, 3] [1, 2, 4, 1, 2, 3]

where I expected it to fail. I expect it to fail, since

sublist [1, 2, 3] [1, 2, 4, 1, 2, 3]
   = sublist [2, 3] [2, 4, 1, 2, 3]
   = sublist [3] [4, 1, 2, 3]

at which point, I thought, [3] = 3:[] will be matched with (x:xs) in sublist, and [4, 1, 2, 3] will be matched with (y:ys) in sublist. How, then, is sublist working?

Edit: Thanks to everyone here, I think I’ve solved my problem. As noted, I was (“subconsciously”) wanting sublist to backtrack for me. Using the last answer (BMeph) posted as a guide, I decided to approach the problem differently, in order to solve the “binding problem,” i.e., the “backtracking” problem.

subseq :: (Eq a) => [a] -> [a] -> Bool
subseq [] _ = True
subseq _ [] = False
subseq (x:xs) (y:ys) =

-- subseq' decides whether the list bound to (x:xs) = M is a prefix of the list
-- bound to L = (y:ys); it recurses through L and returns a Bool value. subseq
-- recurses through M and L, returning a disjunction of Bool
-- values. Each recursive call to subseq passes M and ys to subseq', which
-- decides whether M is a prefix of the **current list bound to ys**.

   let subseq' :: (Eq a) => [a] -> [a] -> Bool
       subseq' [] _ = True
       subseq' _ [] = False
       subseq' (x:xs) (y:ys) = (x == y) && subseq' xs ys
          in subseq' (x:xs) (y:ys) || subseq (x:xs) ys

Answer 1

It is working because:

• [3] is matched as x:xs as 3:[],
• [4, 1, 2, 3] is matched as y:ys as 4:[1,2,3]
• 3/=4 so sublist (x:xs) ys is evaluated, which eventually is True

trace:

sublist [1, 2, 3] [1, 2, 4, 1, 2, 3]
   = sublist [2, 3] [2, 4, 1, 2, 3]
   = sublist [3] [4, 1, 2, 3]
   = sublist [3] [1, 2, 3]
   = sublist [3] [2, 3]
   = sublist [3] [3]
   = sublist [] [] = True