Re: [Haskell-cafe] Haskell and memoization
Maybe not related, but does the following prove next is called once and only once. import qualified Data.ByteString as BS import qualified Data.ByteString.Char8 as BSC next = do nextcache - BS.readFile next.cache let nextint = readInt (BSC.unpack nextcache) BS.writeFile next.cache (BSC.pack (show (nextint+1))) return nextint readInt :: String - Int readInt = read I put a single character, 1 in the file next.cache when I run this through ghci, and call next several times, I always get a 1. Whereas in the file there is a 2. I see that next is a trial of creating a function which returns different things everytime its called, but it's in the IO monad, so that should be doable. When I re-run ghci, now it starts to give 2 everytime I call it. Does that mean, it doesn't bother to re-read the file while we are in the same process. Hope it relates to the OP's question in some way :) Best, 2009/12/16 Daniel Fischer daniel.is.fisc...@web.de Am Mittwoch 16 Dezember 2009 15:49:54 schrieb michael rice: Thanks all, OK, so this definition of fib fib 0 = 1 fib 1 = 1 fib n = fib (n-1) + fib (n-2) would involve a lot of recomputation for some large n, Where large can start as low as 20; 60 would be out of reach. which memoization would eliminate? Right. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Ozgur Akgun ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Sorry for the last mail, I now tried it and it returns the next value every time I call it. I was using an unsafeperformIO trick somewhere, and that fas the one resulting in the previously described behaviour. You can just ignore the previous mail. 2009/12/17 Ozgur Akgun ozgurak...@gmail.com Maybe not related, but does the following prove next is called once and only once. import qualified Data.ByteString as BS import qualified Data.ByteString.Char8 as BSC next = do nextcache - BS.readFile next.cache let nextint = readInt (BSC.unpack nextcache) BS.writeFile next.cache (BSC.pack (show (nextint+1))) return nextint readInt :: String - Int readInt = read I put a single character, 1 in the file next.cache when I run this through ghci, and call next several times, I always get a 1. Whereas in the file there is a 2. I see that next is a trial of creating a function which returns different things everytime its called, but it's in the IO monad, so that should be doable. When I re-run ghci, now it starts to give 2 everytime I call it. Does that mean, it doesn't bother to re-read the file while we are in the same process. Hope it relates to the OP's question in some way :) Best, 2009/12/16 Daniel Fischer daniel.is.fisc...@web.de Am Mittwoch 16 Dezember 2009 15:49:54 schrieb michael rice: Thanks all, OK, so this definition of fib fib 0 = 1 fib 1 = 1 fib n = fib (n-1) + fib (n-2) would involve a lot of recomputation for some large n, Where large can start as low as 20; 60 would be out of reach. which memoization would eliminate? Right. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Ozgur Akgun -- Ozgur Akgun ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Thanks all, OK, so this definition of fib fib 0 = 1 fib 1 = 1 fib n = fib (n-1) + fib (n-2) would involve a lot of recomputation for some large n, which memoization would eliminate? Michael --- On Wed, 12/16/09, michael rice nowg...@yahoo.com wrote: From: michael rice nowg...@yahoo.com Subject: Re: [Haskell-cafe] Haskell and memoization To: Daniel Fischer daniel.is.fisc...@web.de, Gregory Crosswhite gcr...@phys.washington.edu Cc: haskell-cafe@haskell.org Date: Wednesday, December 16, 2009, 12:58 AM Hi all, I think this (#3 below) is where I got the idea: http://en.wikipedia.org/wiki/Lazy_evaluation Excerpt: --- Lazy evaluation refers to how expressions are evaluated when they are passed as arguments to functions and entails the following three points:[1] 1. The expression is only evaluated if the result is required by the calling function, called delayed evaluation.[2] 2. The expression is only evaluated to the extent that is required by the calling function, called Short-circuit evaluation. 3. the expression is never evaluated more than once, called applicative-order evaluation.[3] --- So, I guess #3 doesn't apply to Haskell, or maybe I just misunderstood the meaning of the statement. I assumed that if f(p) = q (by some calculation) then that calculation would be replaced by q so the next time the function was called it could just return q, as occurs in memoization. Michael --- On Tue, 12/15/09, Gregory Crosswhite gcr...@phys.washington.edu wrote: From: Gregory Crosswhite gcr...@phys.washington.edu Subject: Re: [Haskell-cafe] Haskell and memoization To: Daniel Fischer daniel.is.fisc...@web.de Cc: haskell-cafe@haskell.org Date: Tuesday, December 15, 2009, 11:47 PM Hmm, you raise an On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote: Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite: Not even then, necessarily. And it's not always a good idea. f k = [1 .. 20^k] You raise a really good point here. One can force sharing, as I understand it, by using a let clause: n = let xs = f 20 in length (xs ++ xs) If I understand correctly, this should cause xs to be first evaluated, and then cached until the full length is computed, which in this case is obviously undesirable behavior. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -Inline Attachment Follows- ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Am Mittwoch 16 Dezember 2009 15:49:54 schrieb michael rice: Thanks all, OK, so this definition of fib fib 0 = 1 fib 1 = 1 fib n = fib (n-1) + fib (n-2) would involve a lot of recomputation for some large n, Where large can start as low as 20; 60 would be out of reach. which memoization would eliminate? Right. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
It does compute the result of a function application more than once if you ask for it more than once, and that's why we need memoization. Dan On Tue, Dec 15, 2009 at 10:32 PM, michael rice nowg...@yahoo.com wrote: I'm looking here at the Fibonacci stuff: http://www.haskell.org/haskellwiki/Memoization Since (I've read) Haskell never computes the value of a function more than once, I don't understand the need for memoization. Enlighten me. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Haskell does not maintain a cache mapping function calls to their values, so if you have some function f and call it with, say, the argument 7 in two different places in your code, then it will re-evaluate the function at each point. The only time it will not do this is when it can see explicitly that the value will be shared, i.e. situations like g (f 7) (f 7) should only result in one evaluation of f 7 in simple cases, presuming the compiler is sufficiently smart. Cheers, Greg On Dec 15, 2009, at 7:32 PM, michael rice wrote: I'm looking here at the Fibonacci stuff: http://www.haskell.org/haskellwiki/Memoization Since (I've read) Haskell never computes the value of a function more than once, I don't understand the need for memoization. Enlighten me. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite: Haskell does not maintain a cache mapping function calls to their values, so if you have some function f and call it with, say, the argument 7 in two different places in your code, then it will re-evaluate the function at each point. The only time it will not do this is when it can see explicitly that the value will be shared, i.e. situations like g (f 7) (f 7) should only result in one evaluation of f 7 in simple cases, presuming the compiler is sufficiently smart. Not even then, necessarily. And it's not always a good idea. f k = [1 .. 20^k] g xs ys = genericLength (ys ++ xs) Finding out when to share is really hard. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Hmm, you raise an On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote: Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite: Not even then, necessarily. And it's not always a good idea. f k = [1 .. 20^k] You raise a really good point here. One can force sharing, as I understand it, by using a let clause: n = let xs = f 20 in length (xs ++ xs) If I understand correctly, this should cause xs to be first evaluated, and then cached until the full length is computed, which in this case is obviously undesirable behavior. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Hi all, I think this (#3 below) is where I got the idea: http://en.wikipedia.org/wiki/Lazy_evaluation Excerpt: --- Lazy evaluation refers to how expressions are evaluated when they are passed as arguments to functions and entails the following three points:[1] 1. The expression is only evaluated if the result is required by the calling function, called delayed evaluation.[2] 2. The expression is only evaluated to the extent that is required by the calling function, called Short-circuit evaluation. 3. the expression is never evaluated more than once, called applicative-order evaluation.[3] --- So, I guess #3 doesn't apply to Haskell, or maybe I just misunderstood the meaning of the statement. I assumed that if f(p) = q (by some calculation) then that calculation would be replaced by q so the next time the function was called it could just return q, as occurs in memoization. Michael --- On Tue, 12/15/09, Gregory Crosswhite gcr...@phys.washington.edu wrote: From: Gregory Crosswhite gcr...@phys.washington.edu Subject: Re: [Haskell-cafe] Haskell and memoization To: Daniel Fischer daniel.is.fisc...@web.de Cc: haskell-cafe@haskell.org Date: Tuesday, December 15, 2009, 11:47 PM Hmm, you raise an On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote: Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite: Not even then, necessarily. And it's not always a good idea. f k = [1 .. 20^k] You raise a really good point here. One can force sharing, as I understand it, by using a let clause: n = let xs = f 20 in length (xs ++ xs) If I understand correctly, this should cause xs to be first evaluated, and then cached until the full length is computed, which in this case is obviously undesirable behavior. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Am Mittwoch 16 Dezember 2009 05:47:20 schrieb Gregory Crosswhite: On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote: Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite: Not even then, necessarily. And it's not always a good idea. f k = [1 .. 20^k] You raise a really good point here. One can force sharing, as I understand it, by using a let clause: n = let xs = f 20 in length (xs ++ xs) If I understand correctly, this should cause xs to be first evaluated, It is evaluated during the calculation of the length. But whereas in n = let xs = f 20 ys = f 10 in length (xs ++ ys) every cell of xs can be immediately garbage collected - so it will only take insanely long to get a result -, in the above example the scond argument of (++) holds on to xs, so they can't be garbage collected (I wouldn't bet any body parts on that, an implementation is allowed to recalculate a named entity on every occurrence, but it's the expected behaviour) and you will run into memory problems. The good part of it is that it finishes much faster :) and then cached until the full length is computed, which in this case is obviously undesirable behavior. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
#3 is true for Haskell, it's just that when your function call appears in two different places, it is counted as two different expressions. Each separate expression will only be evaluated once, though. This is what is really meant, since the alternative --- i.e., no function ever being called more than once for a given set of arguments --- is way too cumbersome to be worth doing in practice for any language. Laziness really means that if you have, say, f x = (g 7) + x then g 7 need only be evaluated at the first call to f, and then after that it can be cached. In some circumstances, if we had f x = (g 7) + x h x = (g 7) * x Then maybe the compiler would decide not only to evaluate each (g 7) expression once, but also that the two expression should be merged into references to a single shared expression. However, this is not required for laziness; the only requirement there is that each expression separately only be evaluated once. Cheers, Greg On Dec 15, 2009, at 9:58 PM, michael rice wrote: Hi all, I think this (#3 below) is where I got the idea: http://en.wikipedia.org/wiki/Lazy_evaluation Excerpt: --- Lazy evaluation refers to how expressions are evaluated when they are passed as arguments to functions and entails the following three points:[1] 1. The expression is only evaluated if the result is required by the calling function, called delayed evaluation.[2] 2. The expression is only evaluated to the extent that is required by the calling function, called Short-circuit evaluation. 3. the expression is never evaluated more than once, called applicative-order evaluation.[3] --- So, I guess #3 doesn't apply to Haskell, or maybe I just misunderstood the meaning of the statement. I assumed that if f(p) = q (by some calculation) then that calculation would be replaced by q so the next time the function was called it could just return q, as occurs in memoization. Michael --- On Tue, 12/15/09, Gregory Crosswhite gcr...@phys.washington.edu wrote: From: Gregory Crosswhite gcr...@phys.washington.edu Subject: Re: [Haskell-cafe] Haskell and memoization To: Daniel Fischer daniel.is.fisc...@web.de Cc: haskell-cafe@haskell.org Date: Tuesday, December 15, 2009, 11:47 PM Hmm, you raise an On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote: Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite: Not even then, necessarily. And it's not always a good idea. f k = [1 .. 20^k] You raise a really good point here. One can force sharing, as I understand it, by using a let clause: n = let xs = f 20 in length (xs ++ xs) If I understand correctly, this should cause xs to be first evaluated, and then cached until the full length is computed, which in this case is obviously undesirable behavior. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Haskell and memoization
Am Mittwoch 16 Dezember 2009 07:22:42 schrieb Gregory Crosswhite: #3 is true for Haskell, it's just that when your function call appears in two different places, it is counted as two different expressions. Each separate expression will only be evaluated once, though. This is what is really meant, since the alternative --- i.e., no function ever being called more than once for a given set of arguments --- is way too cumbersome to be worth doing in practice for any language. Laziness really means that if you have, say, f x = (g 7) + x then g 7 need only be evaluated at the first call to f, and then after that it can be cached. In some circumstances, if we had f x = (g 7) + x h x = (g 7) * x Then maybe the compiler would decide not only to evaluate each (g 7) expression once, but also that the two expression should be merged into references to a single shared expression. However, this is not required for laziness; the only requirement there is that each expression separately only be evaluated once. And, strictly speaking, Haskell is non-strict, not lazy. Thus, if an implementation decides to evaluate g 7 thrice in f x = (x,x,x) r = f (g 7) it doesn't violate the specs. Cheers, Greg ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
