I've been trying to solve the adventofcode day 5 question 2 (https://adventofcode.com/2017/day/5). It differs from the first question where if the item is bigger of equal to 3, it's decreased instead of increased by 1.
When running the implementation with the test data, it produces the correct outcome, so it seems that the implementation is perfect. Also, the recursive call looks to be in tail position, but it's still producing a stackoverflow exception.
The code looks like this
module AdventOfCode5 where
type Instruction = Int
type Position = Int
main :: IO ()
main = do
input <- readFile "day5input.txt"
let instructions = fmap (read :: String -> Instruction) $ lines input
_ <- putStrLn $ show $ computeResult (Prelude.length instructions) 0 (+1) $ instructions
return ()
main2 :: IO ()
main2 = do
input <- readFile "day5input.txt"
let instructions = fmap (read :: String -> Instruction) $ lines input
_ <- putStrLn $ show $ computeResult (Prelude.length instructions) 0 decAbove3AndIncBelow3 instructions
return ()
decAbove3AndIncBelow3 :: Int -> Int
decAbove3AndIncBelow3 x
| x >= 3 = x - 1
| otherwise = x + 1
computeResult :: Int -> Position -> (Int -> Int) -> [Instruction] -> Int
computeResult = takeStep' 0
where takeStep' :: Int -> Int -> Position -> (Int -> Int) -> [Instruction] -> Int
takeStep' count max pos changeInteger instructions
| pos >= max = count
| otherwise =
let
elementAtPos = instructions!!pos
newCount = count + 1
newPos = pos + elementAtPos
newInstructions = (take pos instructions) ++ ([(changeInteger elementAtPos)]) ++ (drop (pos + 1)) instructions
in
takeStep' newCount max newPos changeInteger newInstructions
The idea of the implementation is that you hold a counter and increment the counter for every iteration, in combination with altering the list of instructions with the updated version (where the Int -> Int is the function that knows how to update). You got a position where to look at and the recursion stops as soon as the position is larger than the list size (which i passed as input but could also be derived from the list of instructions).
Can anybody explain to me why this one is producing a stackoverflow?
There is a space leak in the first argument of takeStep', because it builds a thunk (... ((0 + 1) + 1) ...) + 1 instead of just evaluating the integer.
The stack may explode when that thunk gets evaluated.
Use seq to force count before continuing, e.g., count `seq` otherwise in the guard;
or compile with optimizations.
ghci is interpreting it, not compiling it. In particular, it doesn't perform the strictness analysis necessary to automatically fix that leak.
You can run this command to compile with optimizations (-O)
ghc -O -main-is AdventOfCode5.main2 AdventOfCode5.hs
(although even without optimizations compilation seems to reduce space usage enough to succeed.)
Related
I have written this function that computes Collatz sequences, and I see wildly varying times of execution depending on the spin I give it. Apparently it is related to something called "memoization", but I have a hard time understanding what it is and how it works, and, unfortunately, the relevant article on HaskellWiki, as well as the papers it links to, have all proven to not be easily surmountable. They discuss intricate details of the relative performance of highly layman-indifferentiable tree constructions, while what I miss must be some very basic, very trivial point that these sources neglect to mention.
This is the code. It is a complete program, ready to be built and executed.
module Main where
import Data.Function
import Data.List (maximumBy)
size :: (Integral a) => a
size = 10 ^ 6
-- Nail the basics.
collatz :: Integral a => a -> a
collatz n | even n = n `div` 2
| otherwise = n * 3 + 1
recollatz :: Integral a => a -> a
recollatz = fix $ \f x -> if (x /= 1)
then f (collatz x)
else x
-- Now, I want to do the counting with a tuple monad.
mocollatz :: Integral b => b -> ([b], b)
mocollatz n = ([n], collatz n)
remocollatz :: Integral a => a -> ([a], a)
remocollatz = fix $ \f x -> if x /= 1
then f =<< mocollatz x
else return x
-- Trivialities.
collatzLength :: Integral a => a -> Int
collatzLength x = (length . fst $ (remocollatz x)) + 1
collatzPairs :: Integral a => a -> [(a, Int)]
collatzPairs n = zip [1..n] (collatzLength <$> [1..n])
longestCollatz :: Integral a => a -> (a, Int)
longestCollatz n = maximumBy order $ collatzPairs n
where
order :: Ord b => (a, b) -> (a, b) -> Ordering
order x y = snd x `compare` snd y
main :: IO ()
main = print $ longestCollatz size
With ghc -O2 it takes about 17 seconds, without ghc -O2 -- about 22 seconds to deliver the length and the seed of the longest Collatz sequence starting at any point below size.
Now, if I make these changes:
diff --git a/Main.hs b/Main.hs
index c78ad95..9607fe0 100644
--- a/Main.hs
+++ b/Main.hs
## -1,6 +1,7 ##
module Main where
import Data.Function
+import qualified Data.Map.Lazy as M
import Data.List (maximumBy)
size :: (Integral a) => a
## -22,10 +23,15 ## recollatz = fix $ \f x -> if (x /= 1)
mocollatz :: Integral b => b -> ([b], b)
mocollatz n = ([n], collatz n)
-remocollatz :: Integral a => a -> ([a], a)
-remocollatz = fix $ \f x -> if x /= 1
- then f =<< mocollatz x
- else return x
+remocollatz :: (Num a, Integral b) => b -> ([b], a)
+remocollatz 1 = return 1
+remocollatz x = case M.lookup x (table mutate) of
+ Nothing -> mutate x
+ Just y -> y
+ where mutate x = remocollatz =<< mocollatz x
+
+table :: (Ord a, Integral a) => (a -> b) -> M.Map a b
+table f = M.fromList [ (x, f x) | x <- [1..size] ]
-- Trivialities.
-- Then it will take just about 4 seconds with ghc -O2, but I would not live long enough to see it complete without ghc -O2.
Looking at the details of cost centres with ghc -prof -fprof-auto -O2 reveals that the first version enters collatz about a hundred million times, while the patched one -- just about one and a half million times. This must be the reason of the speedup, but I have a hard time understanding the inner workings of this magic. My best idea is that we replace a portion of expensive recursive calls with O(log n) map lookups, but I don't know if it's true and why it depends so much on some godforsaken compiler flags, while, as I see it, such performance swings should all follow solely from the language.
Can I haz an explanation of what happens here, and why the performance differs so vastly between ghc -O2 and plain ghc builds?
P.S. There are two requirements to the achieving of automagical memoization highlighted elsewhere on Stack Overflow:
Make a function to be memoized a top-level name.
Make a function to be memoized a monomorphic one.
In line with these requirements, I rebuilt remocollatz as follows:
remocollatz :: Int -> ([Int], Int)
remocollatz 1 = return 1
remocollatz x = mutate x
mutate :: Int -> ([Int], Int)
mutate x = remocollatz =<< mocollatz x
Now it's as top level and as monomorphic as it gets. Running time is about 11 seconds, versus the similarly monomorphized table version:
remocollatz :: Int -> ([Int], Int)
remocollatz 1 = return 1
remocollatz x = case M.lookup x (table mutate) of
Nothing -> mutate x
Just y -> y
mutate :: Int -> ([Int], Int)
mutate = \x -> remocollatz =<< mocollatz x
table :: (Int -> ([Int], Int)) -> M.Map Int ([Int], Int)
table f = M.fromList [ (x, f x) | x <- [1..size] ]
-- Running in less than 4 seconds.
I wonder why the memoization ghc is supposedly performing in the first case here is almost 3 times slower than my dumb table.
Can I haz an explanation of what happens here, and why the performance differs so vastly between ghc -O2 and plain ghc builds?
Disclaimer: this is a guess, not verified by viewing GHC core output. A careful answer would do so to verify the conjectures outlined below. You can try peering through it yourself: add -ddump-simpl to your compilation line and you will get copious output detailing exactly what GHC has done to your code.
You write:
remocollatz x = {- ... -} table mutate {- ... -}
where mutate x = remocollatz =<< mocollatz x
The expression table mutate in fact does not depend on x; but it appears on the right-hand side of an equation that takes x as an argument. Consequently, without optimizations, this table is recomputed each time remocollatz is called (presumably even from inside the computation of table mutate).
With optimizations, GHC notices that table mutate does not depend on x, and floats it to its own definition, effectively producing:
fresh_variable_name = table mutate
where mutate x = remocollatz =<< mocollatz x
remocollatz x = case M.lookup x fresh_variable_name of
{- ... -}
The table is therefore computed just once for the entire program run.
don't know why it [the performance] depends so much on some godforsaken compiler flags, while, as I see it, such performance swings should all follow solely from the language.
Sorry, but Haskell doesn't work that way. The language definition tells clearly what the meaning of a given Haskell term is, but does not say anything about the runtime or memory performance needed to compute that meaning.
Another approach to memoization that works in some situations, like this one, is to use a boxed vector, whose elements are computed lazily. The function used to initialize each element can use other elements of the vector in its calculation. As long as the evaluation of an element of the vector doesn't loop and refer to itself, just the elements it recursively depends on will be evaluated. Once evaluated, an element is effectively memoized, and this has the further benefit that elements of the vector that are never referenced are never evaluated.
The Collatz sequence is a nearly ideal application for this technique, but there is one complication. The next Collatz value(s) in sequence from a value under the limit may be outside the limit, which would cause a range error when indexing the vector. I solved this by just iterating through the sequence until back under the limit and counting the steps to do so.
The following program takes 0.77 seconds to run unoptimized and 0.30 when optimized:
import qualified Data.Vector as V
limit = 10 ^ 6 :: Int
-- The Collatz function, which given a value returns the next in the sequence.
nextCollatz val
| odd val = 3 * val + 1
| otherwise = val `div` 2
-- Given a value, return the next Collatz value in the sequence that is less
-- than the limit and the number of steps to get there. For example, the
-- sequence starting at 13 is: [13, 40, 20, 10, 5, 16, 8, 4, 2, 1], so if
-- limit is 100, then (nextCollatzWithinLimit 13) is (40, 1), but if limit is
-- 15, then (nextCollatzWithinLimit 13) is (10, 3).
nextCollatzWithinLimit val = (firstInRange, stepsToFirstInRange)
where
firstInRange = head rest
stepsToFirstInRange = 1 + (length biggerThanLimit)
(biggerThanLimit, rest) = span (>= limit) (tail collatzSeqStartingWithVal)
collatzSeqStartingWithVal = iterate nextCollatz val
-- A boxed vector holding Collatz length for each index. The collatzFn used
-- to generate the value for each element refers back to other elements of
-- this vector, but since the vector elements are only evaluated as needed and
-- there aren't any loops in the Collatz sequences, the values are calculated
-- only as needed.
collatzVec :: V.Vector Int
collatzVec = V.generate limit collatzFn
where
collatzFn :: Int -> Int
collatzFn index
| index <= 1 = 1
| otherwise = (collatzVec V.! nextWithinLimit) + stepsToGetThere
where
(nextWithinLimit, stepsToGetThere) = nextCollatzWithinLimit index
main :: IO ()
main = do
-- Use a fold through the vector to find the longest Collatz sequence under
-- the limit, and keep track of both the maximum length and the initial
-- value of the sequence, which is the index.
let (maxLength, maxIndex) = V.ifoldl' accMaxLen (0, 0) collatzVec
accMaxLen acc#(accMaxLen, accMaxIndex) index currLen
| currLen <= accMaxLen = acc
| otherwise = (currLen, index)
putStrLn $ "Max Collatz length below " ++ show limit ++ " is "
++ show maxLength ++ " at index " ++ show maxIndex
all.
While trying to solve some programming quiz:
https://www.hackerrank.com/challenges/missing-numbers
, I came across with space leak.
Main function is difference, which implements multi-set difference.
I've found out that List ':' and Triples (,,) kept on heaps
with -hT option profiling. However, only big lists are difference's
two arguments, and it shrinks as difference keeps on tail recursion.
But the memory consumed by lists keeps increasing as program runs.
Triples is ephemeral array structure, used for bookkeeping the count of multiset's each element. But the memory consumed by triples also
keeps increasing, and I cannot find out why.
Though I've browsed similar 'space leak' questions in stackoverflow,
I couldn't grasp the idea. Surely I have much to study.
I appreciate any comments. Thank you.
p.s) executable is compiled with -O2 switch.
$ ./difference -hT < input04.txt
Stack space overflow: current size 8388608 bytes.
$ ghc --version
The Glorious Glasgow Haskell Compilation System, version 7.6.3
.
import Data.List
import Data.Array
-- array (non-zero-count, start-offset, array_data)
array_size=101
myindex :: Int -> Int -> Int
myindex key offset
| key >= offset = key - offset
| otherwise = key - offset + array_size
mylookup x (_,offset,arr) = arr ! idx
where idx = myindex x offset
addOrReplace :: Int -> Int -> (Int, Int, Array Int (Int,Int)) -> (Int, Int, Array Int (Int,Int))
addOrReplace key value (count,offset,arr) = (count', offset, arr // [(idx,(key,value))])
where idx = myindex key offset
(_,prev_value) = arr ! idx
count' = case (prev_value, value) of
(0,0) -> count
(0,_) -> count + 1
(_,0) -> count - 1
otherwise -> count
difference :: (Int,Int,Array Int (Int,Int)) -> [Int] -> [Int] -> [Int]
difference (count,offset,arr) [] []
| count == 0 = []
| otherwise = [ k | x <- [0..array_size-1], let (k,v) = (arr ! x), v /= 0]
difference m (x:xs) y = difference new_m xs y
where (_,v) = mylookup x m
new_m = addOrReplace x (v + 1) m
difference m [] (y:ys) = difference new_m [] ys
where (_,v) = mylookup y m
new_m = if v == 0
then m
else addOrReplace y (v - 1) m
main = do
n <- readLn :: IO Int
pp <- getLine
m <- readLn :: IO Int
qq <- getLine
let p = map (read :: String->Int) . words $ pp
q = map (read :: String->Int) . words $ qq
startArray = (0,head q, array (0,100) [(i,(0,0)) | i <- [0..100]] )
putStrLn . unwords . map show . sort $ difference startArray q p
[EDIT]
I seq'ed value and Array thanks to Carl's advice.
I attach heap diagram.
[original heap profiling]
[]1
[after seq'ing value v]
difference m (x:xs) y = difference new_m xs y
where (_,v) = mylookup x m
new_m = v `seq` addOrReplace x (v + 1) m
[after seq'ing value v and Array]
difference m (x:xs) y = new_m `seq` difference new_m xs y
where (_,v) = mylookup x m
new_m = v `seq` addOrReplace x (v + 1) m
I see three main problems with this code.
First (and not the cause of the memory use, but definitely the cause of generally poor performance) Array is horrible for this use case. O(1) lookups are useless when updates are O(n).
Speaking of, the values being stored in the Array aren't forced while difference is looping over its first input. They are thunks containing pointers to an unevaluated lookup in the previous version of the array. You can ensure that the value is evaluated at the same time the array is updated, in a variety of ways. When difference loops over its second input, it does this accidentally, in fact, by comparing the value against 0.
Third, difference doesn't even force the evaluation of the new arrays being created while traversing its first argument. Nothing requires the old array to be evaluated during that portion of the loop.
Both of those latter issues need to be resolved to fix the space leak. The first issue doesn't cause a space leak, just much higher overheads than needed.
TL:DR: Is there a way to do example 3 without passing an argument
I'm trying to understand the state monad in haskell (Control.Monad.State). I made an extremely simple function:
Example 1
example :: State Int Int
example = do
e <- get
put (e*5)
return e
This example works in ghci...
runState example 3
(3,15)
I modified it to be able to take arguments....
Example 2
example :: Int -> State Int Int
example n = do
e <- get
put (e*n)
return e
also works in ghci...
runState (example 5) 3
(3,15)
I made it recursive, counting the number of steps it takes for a computation to satisfy some condition
Example 3
example :: Int -> State Int Int
example n = do
e <- get
if (n /= 1)
then do
put (succ e)
example (next n)
else return (succ e)
next :: Int -> Int
next n
| even n = div n 2
| otherwise = 3*n+1
ghci
evalState (example 13) 0
10
My question is, is there a way to do the previous example without explicitly passing a value?
You can store n in the state along side of e, for example, something like:
example = do
(e,n) <- get
if n /= 1
then do put (succ e, next n); example
else return e
There is some overhead to using the State monad, so you should compare this with the alternatives.
For instance, a more Haskelly way of approaching this problem is compose list operations to compute the answer, e.g.:
collatz :: Int -> [Int]
collatz n = iterate next n
collatzLength n = length $ takeWhile (/= 1) $ collatz n
Task: "Sum the first 15,000,000 even numbers."
Haskell:
nats = [1..] :: [Int]
evens = filter even nats :: [Int]
MySum:: Int
MySum= sum $ take 15000000 evens
...but MySum takes ages. More precisely, about 10-20 times slower than C/C++.
Many times I've found, that a Haskell solution coded naturally is something like 10 times slower than C. I expected that GHC was a very neatly optimizing compiler and task such this don't seem that tough.
So, one would expect something like 1.5-2x slower than C. Where is the problem?
Can this be solved better?
This is the C code I'm comparing it with:
long long sum = 0;
int n = 0, i = 1;
for (;;) {
if (i % 2 == 0) {
sum += i;
n++;
}
if (n == 15000000)
break;
i++;
}
Edit 1: I really know, that it can be computed in O(1). Please, resist.
Edit 2: I really know, that evens are [2,4..] but the function even could be something else O(1) and need to be implemented as a function.
Lists are not loops
So don't be surprised if using lists as a loop replacement, you get slower code if the loop body is small.
nats = [1..] :: [Int]
evens = filter even nats :: [Int]
dumbSum :: Int
dumbSum = sum $ take 15000000 evens
sum is not a "good consumer", so GHC is not (yet) able to eliminate the intermediate lists completely.
If you compile with optimisations (and don't export nat), GHC is smart enough to fuse the filter with the enumeration,
Rec {
Main.main_go [Occ=LoopBreaker]
:: GHC.Prim.Int# -> GHC.Prim.Int# -> [GHC.Types.Int]
[GblId, Arity=1, Caf=NoCafRefs, Str=DmdType L]
Main.main_go =
\ (x_aV2 :: GHC.Prim.Int#) ->
let {
r_au7 :: GHC.Prim.Int# -> [GHC.Types.Int]
[LclId, Str=DmdType]
r_au7 =
case x_aV2 of wild_Xl {
__DEFAULT -> Main.main_go (GHC.Prim.+# wild_Xl 1);
9223372036854775807 -> n_r1RR
} } in
case GHC.Prim.remInt# x_aV2 2 of _ {
__DEFAULT -> r_au7;
0 ->
let {
wild_atm :: GHC.Types.Int
[LclId, Str=DmdType m]
wild_atm = GHC.Types.I# x_aV2 } in
let {
lvl_s1Rp :: [GHC.Types.Int]
[LclId]
lvl_s1Rp =
GHC.Types.:
# GHC.Types.Int wild_atm (GHC.Types.[] # GHC.Types.Int) } in
\ (m_aUL :: GHC.Prim.Int#) ->
case GHC.Prim.<=# m_aUL 1 of _ {
GHC.Types.False ->
GHC.Types.: # GHC.Types.Int wild_atm (r_au7 (GHC.Prim.-# m_aUL 1));
GHC.Types.True -> lvl_s1Rp
}
}
end Rec }
but that's as far as GHC's fusion takes it. You are left with boxing Ints and constructing list cells. If you give it a loop, like you give it to the C compiler,
module Main where
import Data.Bits
main :: IO ()
main = print dumbSum
dumbSum :: Int
dumbSum = go 0 0 1
where
go :: Int -> Int -> Int -> Int
go sm ct n
| ct >= 15000000 = sm
| n .&. 1 == 0 = go (sm + n) (ct+1) (n+1)
| otherwise = go sm ct (n+1)
you get the approximate relation of running times between the C and the Haskell version you expected.
This sort of algorithm is not what GHC has been taught to optimise well, there are bigger fish to fry elsewhere before the limited manpower is put into these optimisations.
The problem why list fusion can't work here is actually rather subtle. Say we define the right RULE to fuse the list away:
import GHC.Base
sum2 :: Num a => [a] -> a
sum2 = sum
{-# NOINLINE [1] sum2 #-}
{-# RULES "sum" forall (f :: forall b. (a->b->b)->b->b).
sum2 (build f) = f (+) 0 #-}
(The short explanation is that we define sum2 as an alias of sum, which we forbid GHC to inline early, so the RULE has a chance to fire before sum2 gets eliminated. Then we look for sum2 directly next to the list-builder build (see definition) and replace it by direct arithmetic.)
This has mixed success, as it yields the following Core:
Main.$wgo =
\ (w_s1T4 :: GHC.Prim.Int#) ->
case GHC.Prim.remInt# w_s1T4 2 of _ {
__DEFAULT ->
case w_s1T4 of wild_Xg {
__DEFAULT -> Main.$wgo (GHC.Prim.+# wild_Xg 1);
15000000 -> 0
};
0 ->
case w_s1T4 of wild_Xg {
__DEFAULT ->
case Main.$wgo (GHC.Prim.+# wild_Xg 1) of ww_s1T7 { __DEFAULT ->
GHC.Prim.+# wild_Xg ww_s1T7
};
15000000 -> 15000000
}
}
Which is nice, completely fused code - with the sole problem that we have a call to $wgo in a non-tail-call position. This means that we aren't looking at a loop, but actually at a deeply recursive function, with predictable program results:
Stack space overflow: current size 8388608 bytes.
The root problem here is that the Prelude's list fusion can only fuse right folds, and computing the sum as a right fold directly causes the excessive stack consumption.
The obvious fix would be to use a fusion framework that can actually deal with left folds, such as Duncan's stream-fusion package, which actually implements sum fusion.
Another solution would be to hack around it - and implement the left fold using a right fold:
main = print $ foldr (\x c -> c . (+x)) id [2,4..15000000] 0
This actually produces close-to-perfect code with current versions of GHC. On the other hand, this is generally not a good idea as it relies on GHC being smart enough to eliminate the partially applied functions. Already adding a filter into the chain will break that particular optimization.
Sum first 15,000,000 even numbers:
{-# LANGUAGE BangPatterns #-}
g :: Integer -- 15000000*15000001 = 225000015000000
g = go 1 0 0
where
go i !a c | c == 15000000 = a
go i !a c | even i = go (i+1) (a+i) (c+1)
go i !a c = go (i+1) a c
ought to be the fastest.
If you want to be sure to traverse the list only once, you can write the traversal explicitly:
nats = [1..] :: [Int]
requiredOfX :: Int -> Bool -- this way you can write a different requirement
requiredOfX x = even x
dumbSum :: Int
dumbSum = dumbSum' 0 0 nats
where dumbSum' acc 15000000 _ = acc
dumbSum' acc count (x:xs)
| requiredOfX x = dumbSum' (acc + x) (count + 1) xs
| otherwise = dumbSum' acc (count + 1) xs
First, you can be clever as young Gauss was and compute the sum in O(1).
Fun stuff aside, your Haskell solution uses lists. I'm quite sure your C/C++ solution doesn't. (Haskell lists are very easy to use so one is tempted to use them even in cases where it might not be appropriate.) Try benchmarking this:
sumBy2 :: Integer -> Integer
sumBy2 = f 0
where
f result n | n <= 1 = result
| otherwise = f (n + result) (n - 2)
Compile it using GHC with -O2 argument. This function is tail-recursive so compiler can implement it very efficiently.
Update: If you want it using even function, it's possible:
sumBy2 :: Integer -> Integer
sumBy2 = f 0
where
f result n | n <= 0 = result
| even n = f (n + result) (n - 1)
| otherwise = f result (n - 1)
You can also easily make the filtering function a parameter:
sumFilter :: (Integral a) => (a -> Bool) -> a -> a
sumFilter filtfn = f 0
where
f result n | n <= 0 = result
| filtfn n = f (n + result) (n - 1)
| otherwise = f result (n - 1)
Strict version works much faster:
foldl' (+) 0 $ take 15000000 [2, 4..]
Another thing to note is that nats and evens are so-called Constant Applicative Forms, or CAFs for short. Basically, those correspond to top-level definitions without any arguments. CAFs are a bit of an odd duck, for instance being the reason for the Dreaded Monomorphism Restriction; I'm not sure the language definition even allows CAFs to be inlined.
In my mental model of how Haskell executes, by the time dumbSum returns a value, evens will be evaluated to look something like 2:4: ... : 30000000 : <thunk> and nats to 1:2: ... : 30000000 : <thunk>, where the <thunk>s represent something that's not been looked at yet. If my understanding is correct, these allocations of : do have to happen and can't be optimized away.
So one way of speeding things up without altering your code too much would be to simply write:
dumbSum :: Int
dumbSum = sum . take 15000000 . filter even $ [1..]
or
dumbSum = sum $ take 15000000 evens where
nats = [1..]
evens = filter even nats
On my machine, compiled with -O2, that alone seems to result in a roughly 30% speedup.
I'm no GHC connaisseur (I've never even profiled a Haskell program!), so I could be wildly off the mark, though.
--Standered Diviation
module SD where
diviation:: IO()
diviation =
do
putStrLn ("Enter Students Marks")
marks <- getLine
let m = ( read marks)::[Float]
let x = sum' m
let mean = (fromIntegral x)/(fromIntegral $ length )
let len = (read (length(m)))::Float
let divia = divi mean l
let std = map (^2) divia
let stdd = xx length(m-1) m
let final = map sqrt stdd
let tot = sum final
if(m==[])
then
putStrLn("empty List" ++ show(tot))
else do
putStrLn("The Standered Divation is" ++ (show(tot)))
sum' :: (Num a) => [a] -> a
sum' = foldl (+) 0
avg::Float->Float->Float
avg a b = (fromIntegral a)/(fromIntegral b)
divi::Float->[Float]->[Float]
divi a xs = [x-a | x <- xs]
xx::Float->[Float]->[Float]
xx a xs = [x/a|x<-xs]
i can not figure out what's wrong with this program. is shows an error like this
ERROR file:.\SD.hs:11 - Type error in application
*** Expression : read (length m)
*** Term : length m
*** Type : Int
can you guys please point me out problem in this program, thank you
* Does not match : [Char]
There are a few things going on in your code that you might want take a look at. You have:
let stdd = xx length(m-1) m
which will not typecheck. I think you meant:
let stdd = xx (length m-1) m
(Edit: actually, that won't check either due to the type signature for xx. (Why not?)) This line:
let mean = (fromIntegral x)/(fromIntegral $ length )
is a Num divided by a function.
In this line:`
let std = map (^2) divia
what is the type of divia, and what is the type of (^2)?
Finally, what actually happens in the case of the empty list, or even a singleton list?
As an aside, you might want to consider which parts of your program really need to live inside main. As it stands, you're
Printing a line
Reading input
Computing the standard deviation
Printing the result of that computation
Why not factor out a standardDev function? That might make your main function shorter and clearer. Writing pure functions also allows you to test your functions in the REPL more conveniently. When I code, I find it very helpful to build short, obviously correct functions, compose them in order to obtain tge desired behavior, and dump the result into the IO monad only at the very last moment.
The function read has type String -> 'a ; your error tells you that length m has type Int, while read is waiting for a String. You might want to use genericLength from Data.List:
let len = Data.List.genericLength m :: Float