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Folding Operations Applied to Addition: Examples, Proofs, and Types

Explore foldr and foldl in infix notation with examples and proof of types for addition operations. Understand the recursion and base cases for folding.

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Folding Operations Applied to Addition: Examples, Proofs, and Types

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  1. Folding foldr and foldl applied to addition (using infix notation)

  2. First some examples • Then proofs of the types of foldr and foldl

  3. foldr (+) 0 [2,3,4] foldr (+) 0 [2,3,4] = (recursion)(+) 2 (foldr (+) 0 [3,4]) = (recursion)(+) 2 ( (+) 3 (+) (foldr (+) 0 [4]) ) = (recursion)(+) 2 ( (+) 3 ( (+) (4 (+) (foldr (+) 0 []) ))) = (base)(+) 2 ( (+) 3 ( (+) (4 (+) (0) ))) = (infix notation)This computes(2 + (3 + (4 + 0))) = 9

  4. foldl (+) 0 [2,3,4] foldl (+) 0 [2,3,4] = (recursion)foldl(+) ((+) 0 2) [3,4] =foldl(+) (2 [3,4]) = (recursion)foldl (+) ((+) 2 3) [4]) = foldl (+) (5 [4]) = (recursion) foldl (+) ((+) 5 4) [] = (base) foldl (+) 9 [] = 9This computes (((0(+)2)(+)3)(+)4) = 9

  5. type of foldl / foldr • As applied to (+), both functions have the type:

  6. type of foldl / foldr • As applied to (+), both functions have the type: (Int -> Int -> Int) -> (Int -> [Int] -> Int) • The lecture notes show that this is a special case. In general, foldr :: (a->b->b) - > b->[a]->b foldl :: (a->b->a) -> a->[b]->a

  7. First foldr, starting from its definition • The first question to ask is, what is the type of f • More specifically, what’s the type of the outcome of f? (Does this outcome have the same type as f’s 1st argument or the same type as f’s 2nd argument?)

  8. R S R,S,P,Q are types foldr f y [] = y P Q foldr f y (x:xs) = f x (foldr f y xs) ?? [a] a b f :: a  b  ??=b because Q=P=b y :: ?? ??=b because S=P Therefore, foldr :: (abb)b[a]b

  9. Now foldl, starting from its definition foldl f y [] = y P and Q are types foldl f y (x:xs) = foldl f (fyx) xs P Q=a f :: a -> b -> ? Within fyx, the first arg of f is y, so Type(y)=a. P = Qtherefore, Type(y)=P=a Therefore (comparing left and right), Type(fyx)=a. Therefore, ?=a Therefore, Type(foldl) = (a->b->a) -> a->[b]->a

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