question from Yet Another Haskell Tutorial

I have a question about a problem in Yet Another Haskell Tutorial (problem 7.1). My answers seems to disagree with Hal's, and it fact something looks wrong in Hal's answer (maybe it's an error in his paper). The problem is to write the following function in "point-free" style: func5 f list = foldr (\x y -> f(y,x)) 0 list Here's how I approached it. It's easy to drop 'list' by the eta-reduction (using partial application): func5 f = foldr (\x y -> f(y,x)) 0 But how to get rid of f? First I reasoned that f is a function of a tuple. That is, it is not curried. So to curry it: func5 f = foldr (\x y -> (curry f) y x) 0 The second argument to foldr is obviously flipping the arugments, so func5 f = foldr (flip $ curry f) 0 Now I want to use the eta-reduction again, but I have to transform this expression into something that takes f as its last argument instead of the 0. I can use flip again, this time on foldr: func5 f = flip foldr 0 (flip $ curry f) Now f can be dropped: THE FINAL ANSWER: func5 = flip foldr 0 . flip. curry This works, in a couple of my test cases. Now Hal gives this as the answer: func5 = foldr (uncurry $ flip f) 0 The first problem is that there's no argument f, though he refers to it. So maybe he meant func5 f = foldr (uncurry $ flip f) 0 But more problems. He's applying flip to f, but f takes only one argument (a 2-tuple). Then he's uncurry-ing it, but I thought it needed currying, not uncurry-ing. Can anyone figure out what Hal is up to, or does it look like a simple mistake? Thanks, Mike

I have a question about a problem in Yet Another Haskell Tutorial (problem 7.1). My answers seems to disagree with Hal's, and it fact something looks wrong in Hal's answer (maybe it's an error in his paper). The problem is to write the following function in "point-free" style:
func5 f list = foldr (\x y -> f(y,x)) 0 list
Here's how I approached it. It's easy to drop 'list' by the eta-reduction (using partial application):
func5 f = foldr (\x y -> f(y,x)) 0
But how to get rid of f? First I reasoned that f is a function of a tuple. That is, it is not curried. So to curry it:
func5 f = foldr (\x y -> (curry f) y x) 0
The second argument to foldr is obviously flipping the arugments, so
func5 f = foldr (flip $ curry f) 0
Now I want to use the eta-reduction again, but I have to transform
Am Sonntag 29 März 2009 08:33:13 schrieb Michael Mossey: this
expression into something that takes f as its last argument instead of the 0. I can use flip again, this time on foldr:
func5 f = flip foldr 0 (flip $ curry f)
Now f can be dropped:
THE FINAL ANSWER: func5 = flip foldr 0 . flip. curry
This works, in a couple of my test cases.
As it should, since it is correct. A simple way to check for sanity of the results is: Prelude> :t flip foldr 0 . flip. curry flip foldr 0 . flip. curry :: (Num c) => ((c, b) -> c) -> [b] -> c Prelude> :t \f list -> foldr (\x y -> f (y,x)) 0 list \f list -> foldr (\x y -> f (y,x)) 0 list :: (Num b) => ((b, a) -> b) -> [a] -> b Prelude> :t \f -> foldr (uncurry $ flip f) 0 \f -> foldr (uncurry $ flip f) 0 :: (Num b1) => (b -> a -> b1 -> b1) -> [(a, b)] -> b1 So you see that your result has the correct type, while Hal's hasn't.
Now Hal gives this as the answer:
func5 = foldr (uncurry $ flip f) 0
The first problem is that there's no argument f, though he refers to it. So maybe he meant
func5 f = foldr (uncurry $ flip f) 0
But more problems. He's applying flip to f, but f takes only one argument (a 2-tuple). Then he's uncurry-ing it, but I thought it needed currying, not uncurry-ing.
Can anyone figure out what Hal is up to, or does it look like a simple mistake?
Simple mistake. YAHT was never systematically proofread to catch all such mistakes, so there are several still in. Nevertheless, it is an excellent tutorial.
Thanks, Mike

Daniel Fischer wrote:
A simple way to check for sanity of the results is:
Prelude> :t flip foldr 0 . flip. curry flip foldr 0 . flip. curry :: (Num c) => ((c, b) -> c) -> [b] -> c
Prelude> :t \f list -> foldr (\x y -> f (y,x)) 0 list \f list -> foldr (\x y -> f (y,x)) 0 list :: (Num b) => ((b, a) -> b) -> [a] -> b
Prelude> :t \f -> foldr (uncurry $ flip f) 0 \f -> foldr (uncurry $ flip f) 0 :: (Num b1) => (b -> a -> b1 -> b1) -> [(a, b)] -> b1
So you see that your result has the correct type, while Hal's hasn't.
Can anyone figure out what Hal is up to, or does it look like a simple mistake?
Simple mistake. YAHT was never systematically proofread to catch all such mistakes, so there are several still in. Nevertheless, it is an excellent tutorial.
Thanks for the information, glad to know I was correct. I agree that YAHT is excellent... I am working from several books and tutorials simultaneously to get various perspectives, and YAHT covers much of the same material as the books, but in a more compact form, with good exercises. -Mike

On Sun, 29 Mar 2009 14:45:22 +0200, Daniel Fischer
Simple mistake. YAHT was never systematically proofread to catch all such mistakes, so there are several still in. Nevertheless, it is an excellent tutorial.
Thank you both, Michael and Daniel. I have compiled the information in this thread into a new page, "Errata," in the YAHT Wikibook, covering discrepancies between the original PDF and corrected HTML versions of YAHT: Haskell/YAHT/Errata - Wikibooks, collection of open-content textbooks http://en.wikibooks.org/wiki/Haskell/YAHT/Errata Please feel free to add to it or to correct it as would benefit students of Haskell. -- Benjamin L. Russell -- Benjamin L. Russell / DekuDekuplex at Yahoo dot com http://dekudekuplex.wordpress.com/ Translator/Interpreter / Mobile: +011 81 80-3603-6725 "Furuike ya, kawazu tobikomu mizu no oto." -- Matsuo Basho^
participants (3)
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Benjamin L.Russell
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Daniel Fischer
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Michael Mossey