On Fri, 23 Apr 2021, Galaxy Being wrote:

I'm in Bird's Thinking Functionally with Haskell and he has this code to transpose a matrix based on a list of row lists transpose :: [[a]] -> [[a]] transpose [xs] = [[x] | x <- xs] transpose (xs:xss) = zipWith (:) xs (transpose xss)

then he says transpose can be rewritten with this pattern

transpose [] = ...

what could be the rest of it? The answer he gives is

transpose2 :: [[a]] -> [[a]] transpose2 [] = repeat []   transpose2 (xs:xss) = zipWith (:) xs (transpose2 xss)

where repeat [] gives an infinite list of repetitions. And, he says, note that

transpose [xs] =  zipWith (;) xs (repeat []) = [[x] | x <- xs]

I suppose I get this last equation, but I don't understand repeat in transpose2. Can someone explain this to me?

With, say transpose [] = [], the (zipWith (:)) would be shortened to the empty list. In contrast to that, (repeat []) provides as many list ends as needed.

Think about it like this: [image: image.png] The current elements are zipped with the tails (which are created with transpose). This has to bottom out with empty tails (repeat []). On Sat, Apr 24, 2021 at 2:32 AM Galaxy Being wrote:

I'm in Bird's *Thinking Functionally with Haskell *and he has this code to transpose a matrix based on a list of row lists

transpose :: [[a]] -> [[a]] transpose [xs] = [[x] | x <- xs] transpose (xs:xss) = zipWith (:) xs (transpose xss)

then he says transpose can be rewritten with this pattern

transpose [] = ...

what could be the rest of it? The answer he gives is

transpose2 :: [[a]] -> [[a]] transpose2 [] = repeat [] transpose2 (xs:xss) = zipWith (:) xs (transpose2 xss)

where repeat [] gives an infinite list of repetitions. And, he says, note that

transpose [xs] = zipWith (;) xs (repeat []) = [[x] | x <- xs]

I suppose I get this last equation, but I don't understand repeat in transpose2. Can someone explain this to me?

LB

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