How to optimise Haskell code to pass HackerRanks timed out test cases (Not for any ongoing competition, just me practicing) - algorithm

I been learning Haskell for around 4 months now and I have to say, the learning curve is definitely hard(scary also :p).
After solving about 15 easy questions, today I moved to my first medium difficulty problem on HackerRank https://www.hackerrank.com/challenges/climbing-the-leaderboard/problem.
It was 10 test cases and I am able to pass 6 of them, but the rest fail with timeout, now the interesting part is, I can already see a few parts that have potential for performance increase, for example, I am using nub to remove duplicated from a [Int], but still I am not able to build a mental model for algorithmic performance, the main cause of that being unsure about Haskell compiler will change my code and how laziness plays a role here.
import Data.List (nub)
getInputs :: [String] -> [String]
getInputs (_:r:_:p:[]) = [r, p]
findRating :: Int -> Int -> [Int] -> Int
findRating step _ [] = step
findRating step point (x:xs) = if point >= x then step else findRating (step + 1) point xs
solution :: [[Int]] -> [Int]
solution [rankings, points] = map (\p -> findRating 1 p duplicateRemovedRatings) points
where duplicateRemovedRatings = nub rankings
main :: IO ()
main = interact (unlines . map show . solution . map (map read . words) . getInputs . lines)
Test Case in GHCI
:l "solution"
let i = "7\n100 100 50 40 40 20 10\n4\n5 25 50 120"
solution i // "6\n4\n2\n1\n"
Specific questions I have:
Will the duplicateRemovedRankings variable be calculated once, or on each iteration of the map function call.
Like in imperative programming languages, I can verify the above question using some sort of printing mechanism, is there some equivalent way of doing the same with Haskell.
According to my current understanding, the complexity of this algorithm would be, I know nub is O(n^2)
findRating is O(n)
getInputs is O(1)
solution is O(n^2)
How can I reason about this and build a mental model for performance.
If this violates community guidelines, please comment and I will delete this. Thank you for the help :)

First, to answer your questions:
Yes, duplicateRemovedRankings is computed only once. No repeated computation.
To debug-trace, you can use trace and its friends (see the docs for examples and explanation). Yes, it can be used even in pure, non-IO code. But obviously, don't use it for "normal" output.
Yes, your understanding of complexity is correct.
Now, how to pass HackerRank's tricky tests.
First, yes, you're right that nub is O(N^2). However, in this particular case you don't have to settle for that. You can use the fact that the rankings come pre-sorted to get a linear version of nub. All you have to do is skip elements while they're equal to the next one:
betterNub (x:y:rest)
| x == y = betterNub (y:rest)
| otherwise = x : betterNub (y:rest)
betterNub xs = xs
This gives you O(N) for betterNub itself, but it's still not good enough for HackerRank, because the overall solution is still O(N*M) - for each game you are iterating over all rankings. No bueno.
But here you can get another improvement by observing that the rankings are sorted, and searching in a sorted list doesn't have to be linear. You can use a binary search instead!
To do this, you'll have to get yourself constant-time indexing, which can be achieved by using Array instead of list.
Here's my implementation (please don't judge harshly; I realize I probably got edge cases overengineered, but hey, it works!):
import Data.Array (listArray, bounds, (!))
findIndex arr p
| arr!end' > p = end' + 1
| otherwise = go start' end'
where
(start', end') = bounds arr
go start end =
let mid = (start + end) `div` 2
midValue = arr ! mid
in
if midValue == p then mid
else if mid == start then (if midValue < p then start else end)
else if midValue < p then go start mid
else go mid end
solution :: [[Int]] -> [Int]
solution [rankings, points] = map (\p -> findIndex duplicateRemovedRatings p + 1) points
where duplicateRemovedRatings = toArr $ betterNub rankings
toArr l = listArray (0, (length l - 1)) l
With this, you get O(log N) for the search itself, making the overall solution O(M * log N). And this seems to be good enough for HackerRank.
(note that I'm adding 1 to the result of findIndex - this is because the exercise requires 1-based index)

I believe Fyodor's answer is excellent for your first two and a half questions. For the final half, "How can I build a mental model for performance?", I can say that SPJ is an absolute master of writing highly technical papers in a way accessible to the smart but ignorant reader. The implementation book Implementing lazy functional languages on stock hardware is excellent and can serve as the basis of a mental execution model. There is also Okasaki's thesis, Purely functional data structures, which discusses a complementary and significantly higher-level approach to doing asymptotic complexity analyses. (Actually, I read his book, which apparently includes some extra content, so bear that in mind when deciding for yourself about this recommendation.)
Please don't be daunted by their length. I personally found they were actually actively fun to read; and the topic they cover is a big one, not compressible to a short/quick answer.

Related

Performance of Longest Substring Without Repeating Characters in Haskell

Upon reading this Python question and proposing a solution, I tried to solve the same challenge in Haskell.
I've come up with the code below, which seems to work. However, since I'm pretty new to this language, I'd like some help in understand whether the code is good performancewise.
lswrc :: String -> String
lswrc s = reverse $ fst $ foldl' step ("","") s
where
step ("","") c = ([c],[c])
step (maxSubstr,current) c
| c `elem` current = step (maxSubstr,init current) c
| otherwise = let candidate = (c:current)
longerThan = (>) `on` length
newMaxSubstr = if maxSubstr `longerThan` candidate
then maxSubstr
else candidate
in (newMaxSubstr, candidate)
Some points I think could be better than they are
I carry on a pair of strings (the longest tracked, and the current candidate) but I only need the former; thinking procedurally, there's no way to escape this, but maybe FP allows another approach?
I construct (c:current) but I use it only in the else; I could make a more complicated longerThan to add 1 to the lenght of its second argument, so that I can apply it to maxSubstr and current, and construct (c:current) in the else, without even giving it a name.
I drop the last element of current when c is in the current string, because I'm piling up the strings with :; I could instead pattern match when checking c against the string (as in c `elem` current#(a:as)), but then when adding the new character I should do current ++ [c], which I know is not as performant as c:current.
I use foldl' (as I know foldl doesn't really make sense); foldr could be an alternative, but since I don't see how laziness enters this problem, I can't tell which one would be better.
Running elem on every iteration makes your algorithm Ω(n^2) (for strings with no repeats). Running length on, in the worst case, every iteration makes your algorithm Ω(n^2) (for strings with no repeats). Running init a lot makes your algorithm Ω(n*sqrt(n)) (for strings that are sqrt(n) repetitions of a sqrt(n)-long string, with every other one reversed, and assuming an O(1) elem replacement).
A better way is to pay one O(n) cost up front to copy into a data structure with constant-time indexing, and to keep a set (or similar data structure) of seen elements rather than a flat list. Like this:
import Data.Set (Set)
import Data.Vector (Vector)
import qualified Data.Set as S
import qualified Data.Vector as V
lswrc2 :: String -> String
lswrc2 "" = ""
lswrc2 s_ = go S.empty 0 0 0 0 where
s = V.fromList s_
n = V.length s
at = V.unsafeIndex s
go seen lo hi bestLo bestHi
| hi == n = V.toList (V.slice bestLo (bestHi-bestLo+1) s)
-- it is probably faster (possibly asymptotically so?) to use findIndex
-- to immediately pick the correct next value of lo
| at hi `S.member` seen = go (S.delete (at lo) seen) (lo+1) hi bestLo bestHi
| otherwise = let rec = go (S.insert (at hi) seen) lo (hi+1) in
if hi-lo > bestHi-bestLo then rec lo hi else rec bestLo bestHi
This should have O(n*log(n)) worst-case performance (achieving that worst case on strings with no repeats). There may be ways that are better still; I haven't thought super hard about it.
On my machine, lswrc2 consistently outperforms lswrc on random strings. On the string ['\0' .. '\100000'], lswrc takes about 40s and lswrc2 takes 0.03s. lswrc2 can handle [minBound .. maxBound] in about 0.4s; I gave up after more than 20 minutes of letting lswrc chew on that list.

Lazily Tying the Knot for 1 Dimensional Dynamic Programming

Several years ago I took an algorithms course where we were giving the following problem (or one like it):
There is a building of n floors with an elevator that can only go up 2 floors at a time and down 3 floors at a time. Using dynamic programming write a function that will compute the number of steps it takes the elevator to get from floor i to floor j.
This is obviously easy using a stateful approach, you create an array n elements long and fill it up with the values. You could even use a technically non-stateful approach that involves accumulating a result as recursively passing it around. My question is how to do this in a non-stateful manner by using lazy evaluation and tying the knot.
I think I've devised the correct mathematical formula:
where i+2 and i-3 are within the allowed values.
Unfortunately I can't get it to terminate. If I put the i+2 case first and then choose an even floor I can get it to evaluate the even floors below the target level but that's it. I suspect that it shoots straight to the highest even floor for everything else, drops 3 levels, then repeats, forever oscillating between the top few floors.
So it's probably exploring the infinite space (or finite but with loops) in a depth first manner. I can't think of how to explore the space in a breadth first fashion without using a whole lot of data structures in between that effectively mimic a stateful approach.
Although this simple problem is disappointingly difficult I suspect that having seen a solution in 1 dimension I might be able to make it work for a 2 dimensional variation of the problem.
EDIT: A lot of the answers tried to solve the problem in a different way. The problem itself isn't interesting to me, the question is about the method used. Chaosmatter's approach of creating a minimal function which can compare potentially infinite numbers is possibly a step in the right direction. Unfortunately if I try to create a list representing a building with 100 floors the result takes too long to compute, since the solutions to sub problems are not reused.
I made an attempt to use a self-referencing data structure but it doesn't terminate, there is some kind of infinite loop going on. I'll post my code so you can understand what it is I'm going for. I'll change the accepted answer if someone can actually solve the problem using dynamic programming on a self-referential data structure using laziness to avoid computing things more than once.
levels = go [0..10]
where
go [] = []
go (x:xs) = minimum
[ if i == 7
then 0
else 1 + levels !! i
| i <- filter (\n -> n >= 0 && n <= 10) [x+2,x-3] ]
: go xs
You can see how 1 + levels !! i tries to reference the previously calculated result and how filter (\n -> n >= 0 && n <= 10) [x+2,x-3] tries to limit the values of i to valid ones. As I said, this doesn't actually work, it simply demonstrates the method by which I want to see this problem solved. Other ways of solving it are not interesting to me.
Since you're trying to solve this in two dimensions, and for other problems than the one described, let's explore some more general solutions. We are trying to solve the shortest path problem on directed graphs.
Our representation of a graph is currently something like a -> [a], where the function returns the vertices reachable from the input. Any implementation will additionally require that we can compare to see if two vertices are the same, so we'll need Eq a.
The following graph is problematic, and introduces almost all of the difficulty in solving the problem in general:
problematic 1 = [2]
problematic 2 = [3]
problematic 3 = [2]
problematic 4 = []
When trying to reach 4 from 1, there are is a cycle involving 2 and 3 that must be detected to determine that there is no path from 1 to 4.
Breadth-first search
The algorithm Will presented has, if applied to the general problem for finite graphs, worst case performance that is unbounded in both time and space. We can modify his solution to attack the general problem for graphs containing only finite paths and finite cycles by adding cycle detection. Both his original solution and this modification will find finite paths even in infinite graphs, but neither is able to reliably determine that there is no path between two vertices in an infinite graph.
acyclicPaths :: (Eq a) => (a->[a]) -> a -> a -> [[a]]
acyclicPaths steps i j = map (tail . reverse) . filter ((== j).head) $ queue
where
queue = [[i]] ++ gen 1 queue
gen d _ | d <= 0 = []
gen d (visited:t) = let r = filter ((flip notElem) visited) . steps . head $ visited
in map (:visited) r ++ gen (d+length r-1) t
shortestPath :: (Eq a) => (a->[a]) -> a -> a -> Maybe [a]
shortestPath succs i j = listToMaybe (acyclicPaths succs i j)
Reusing the step function from Will's answer as the definition of your example problem, we could get the length of the shortest path from floor 4 to 5 of an 11 story building by fmap length $ shortestPath (step 11) 4 5. This returns Just 3.
Let's consider a finite graph with v vertices and e edges. A graph with v vertices and e edges can be described by an input of size n ~ O(v+e). The worst case graph for this algorithm is to have one unreachable vertex, j, and the remaining vertexes and edges devoted to creating the largest number of acyclic paths starting at i. This is probably something like a clique containing all the vertices that aren't i or j, with edges from i to every other vertex that isn't j. The number of vertices in a clique with e edges is O(e^(1/2)), so this graph has e ~ O(n), v ~ O(n^(1/2)). This graph would have O((n^(1/2))!) paths to explore before determining that j is unreachable.
The memory required by this function for this case is O((n^(1/2))!), since it only requires a constant increase in the queue for each path.
The time required by this function for this case is O((n^(1/2))! * n^(1/2)). Each time it expands a path, it must check that the new node isn't already in the path, which takes O(v) ~ O(n^(1/2)) time. This could be improved to O(log (n^(1/2))) if we had Ord a and used a Set a or similar structure to store the visited vertices.
For non-finite graphs, this function should only fail to terminate exactly when there doesn't exists a finite path from i to j but there does exist a non-finite path from i to j.
Dynamic Programming
A dynamic programming solution doesn't generalize in the same way; let's explore why.
To start with, we'll adapt chaosmasttter's solution to have the same interface as our breadth-first search solution:
instance Show Natural where
show = show . toNum
infinity = Next infinity
shortestPath' :: (Eq a) => (a->[a]) -> a -> a -> Natural
shortestPath' steps i j = go i
where
go i | i == j = Zero
| otherwise = Next . foldr minimal infinity . map go . steps $ i
This works nicely for the elevator problem, shortestPath' (step 11) 4 5 is 3. Unfortunately, for our problematic problem, shortestPath' problematic 1 4 overflows the stack. If we add a bit more code for Natural numbers:
fromInt :: Int -> Natural
fromInt x = (iterate Next Zero) !! x
instance Eq Natural where
Zero == Zero = True
(Next a) == (Next b) = a == b
_ == _ = False
instance Ord Natural where
compare Zero Zero = EQ
compare Zero _ = LT
compare _ Zero = GT
compare (Next a) (Next b) = compare a b
we can ask if the shortest path is shorter than some upper bound. In my opinion, this really shows off what's happening with lazy evaluation. problematic 1 4 < fromInt 100 is False and problematic 1 4 > fromInt 100 is True.
Next, to explore dynamic programming, we'll need to introduce some dynamic programming. Since we will build a table of the solutions to all of the sub-problems, we will need to know the possible values that the vertices can take. This gives us a slightly different interface:
shortestPath'' :: (Ix a) => (a->[a]) -> (a, a) -> a -> a -> Natural
shortestPath'' steps bounds i j = go i
where
go i = lookupTable ! i
lookupTable = buildTable bounds go2
go2 i | i == j = Zero
| otherwise = Next . foldr minimal infinity . map go . steps $ i
-- A utility function that makes memoizing things easier
buildTable :: (Ix i) => (i, i) -> (i -> e) -> Array i e
buildTable bounds f = array bounds . map (\x -> (x, f x)) $ range bounds
We can use this like shortestPath'' (step 11) (1,11) 4 5 or shortestPath'' problematic (1,4) 1 4 < fromInt 100. This still can't detect cycles...
Dynamic programming and cycle detection
The cycle detection is problematic for dynamic programming, because the sub-problems aren't the same when they are approached from different paths. Consider a variant of our problematic problem.
problematic' 1 = [2, 3]
problematic' 2 = [3]
problematic' 3 = [2]
problematic' 4 = []
If we are trying to get from 1 to 4, we have two options:
go to 2 and take the shortest path from 2 to 4
go to 3 and take the shortest path from 3 to 4
If we choose to explore 2, we will be faced with the following option:
go to 3 and take the shortest path from 3 to 4
We want to combine the two explorations of the shortest path from 3 to 4 into the same entry in the table. If we want to avoid cycles, this is really something slightly more subtle. The problems we faced were really:
go to 2 and take the shortest path from 2 to 4 that doesn't visit 1
go to 3 and take the shortest path from 3 to 4 that doesn't visit 1
After choosing 2
go to 3 and take the shortest path from 3 to 4 that doesn't visit 1 or 2
These two questions about how to get from 3 to 4 have two slightly different answers. They are two different sub-problems which can't fit in the same spot in a table. Answering the first question eventually requires determining that you can't get to 4 from 2. Answering the second question is straightforward.
We could make a bunch of tables for each possible set of previously visited vertices, but that doesn't sound very efficient. I've almost convinced myself that we can't do reach-ability as a dynamic programming problem using only laziness.
Breadth-first search redux
While working on a dynamic programming solution with reach-ability or cycle detection, I realized that once we have seen a node in the options, no later path visiting that node can ever be optimal, whether or not we follow that node. If we reconsider problematic':
If we are trying to get from 1 to 4, we have two options:
go to 2 and take the shortest path from 2 to 4 without visiting 1, 2, or 3
go to 3 and take the shortest path from 3 to 4 without visiting 1, 2, or 3
This gives us an algorithm to find the length of the shortest path quite easily:
-- Vertices first reachable in each generation
generations :: (Ord a) => (a->[a]) -> a -> [Set.Set a]
generations steps i = takeWhile (not . Set.null) $ Set.singleton i: go (Set.singleton i) (Set.singleton i)
where go seen previouslyNovel = let reachable = Set.fromList (Set.toList previouslyNovel >>= steps)
novel = reachable `Set.difference` seen
nowSeen = reachable `Set.union` seen
in novel:go nowSeen novel
lengthShortestPath :: (Ord a) => (a->[a]) -> a -> a -> Maybe Int
lengthShortestPath steps i j = findIndex (Set.member j) $ generations steps i
As expected, lengthShortestPath (step 11) 4 5 is Just 3 and lengthShortestPath problematic 1 4 is Nothing.
In the worst case, generations requires space that is O(v*log v), and time that is O(v*e*log v).
The problem is that min needs to fully evaluate both calls to f,
so if one of them loops infinitly min will never return.
So you have to create a new type, encoding that the number returned by f is Zero or a Successor of Zero.
data Natural = Next Natural
| Zero
toNum :: Num n => Natural -> n
toNum Zero = 0
toNum (Next n) = 1 + (toNum n)
minimal :: Natural -> Natural -> Natural
minimal Zero _ = Zero
minimal _ Zero = Zero
minimal (Next a) (Next b) = Next $ minimal a b
f i j | i == j = Zero
| otherwise = Next $ minimal (f l j) (f r j)
where l = i + 2
r = i - 3
This code actually works.
standing on the floor i of n-story building, find minimal number of steps it takes to get to the floor j, where
step n i = [i-3 | i-3 > 0] ++ [i+2 | i+2 <= n]
thus we have a tree. we need to search it in breadth-first fashion until we get a node holding the value j. its depth is the number of steps. we build a queue, carrying the depth levels,
solution n i j = case dropWhile ((/= j).snd) queue
of [] -> Nothing
((k,_):_) -> Just k
where
queue = [(0,i)] ++ gen 1 queue
The function gen d p takes its input p from d notches back from its production point along the output queue:
gen d _ | d <= 0 = []
gen d ((k,i1):t) = let r = step n i1
in map (k+1 ,) r ++ gen (d+length r-1) t
Uses TupleSections. There's no knot tying here, just corecursion, i.e. (optimistic) forward production and frugal exploration. Works fine without knot tying because we only look for the first solution. If we were searching for several of them, then we'd need to eliminate the cycles somehow.
see also: https://en.wikipedia.org/wiki/Corecursion#Discussion
With the cycle detection:
solutionCD1 n i j = case dropWhile ((/= j).snd) queue
of [] -> Nothing
((k,_):_) -> Just k
where
step n i visited = [i2 | let i2=i-3, not $ elem i2 visited, i2 > 0]
++ [i2 | let i2=i+2, not $ elem i2 visited, i2 <=n]
queue = [(0,i)] ++ gen 1 queue [i]
gen d _ _ | d <= 0 = []
gen d ((k,i1):t) visited = let r = step n i1 visited
in map (k+1 ,) r ++
gen (d+length r-1) t (r++visited)
e.g. solution CD1 100 100 7 runs instantly, producing Just 31. The visited list is pretty much a copy of the instantiated prefix of the queue itself. It could be maintained as a Map, to improve time complexity (as it is, sol 10000 10000 7 => Just 3331 takes 1.27 secs on Ideone).
Some explanations seem to be in order.
First, there's nothing 2D about your problem, because the target floor j is fixed.
What you seem to want is memoization, as your latest edit indicates. Memoization is useful for recursive solutions; your function is indeed recursive - analyzing its argument into sub-cases, synthetizing its result from results of calling itself on sub-cases (here, i+2 and i-3) which are closer to the base case (here, i==j).
Because arithmetics is strict, your formula is divergent in the presence of any infinite path in the tree of steps (going from floor to floor). The answer by chaosmasttter, by using lazy arithmetics instead, turns it automagically into a breadth-first search algorithm which is divergent only if there's no finite paths in the tree, exactly like my first solution above (save for the fact that it's not checking for out-of-bounds indices). But it is still recursive, so indeed memoization is called for.
The usual way to approach it first, is to introduce sharing by "going through a list" (inefficient, because of sequential access; for efficient memoization solutions see hackage):
f n i j = g i
where
gs = map g [0..n] -- floors 1,...,n (0 is unused)
g i | i == j = Zero
| r > n = Next (gs !! l) -- assuming there's enough floors in the building
| l < 1 = Next (gs !! r)
| otherwise = Next $ minimal (gs !! l) (gs !! r)
where r = i + 2
l = i - 3
not tested.
My solution is corecursive. It needs no memoization (just needs to be careful with the duplicates), because it is generative, like the dynamic programming is too. It proceeds away from its starting case, i.e. the starting floor. An external accessor chooses the appropriate generated result.
It does tie a knot - it defines queue by using it - queue is on both sides of the equation. I consider it the simpler case of knot tying, because it is just about accessing the previously generated values, in disguise.
The knot tying of the 2nd kind, the more complicated one, is usually about putting some yet-undefined value in some data structure and returning it to be defined by some later portion of the code (like e.g. a back-link pointer in doubly-linked circular list); this is indeed not what my1 code is doing. What it does do is generating a queue, adding at its end and "removing" from its front; in the end it's just a difference list technique of Prolog, the open-ended list with its end pointer maintained and updated, the top-down list building of tail recursion modulo cons - all the same things conceptually. First described (though not named) in 1974, AFAIK.
1 based entirely on the code from Wikipedia.
Others have answered your direct question about dynamic programming. However, for this kind of problem I think the greedy approach works the best. It's implementation is very straightforward.
f i j :: Int -> Int -> Int
f i j = snd $ until (\(i,_) -> i == j)
(\(i,x) -> (i + if i < j then 2 else (-3),x+1))
(i,0)

How is this memoized DP table too slow for SPOJ?

SPOILERS: I'm working on http://www.spoj.pl/problems/KNAPSACK/ so don't peek if you don't want a possible solution spoiled for you.
The boilerplate:
import Data.Sequence (index, fromList)
import Data.MemoCombinators (memo2, integral)
main = interact knapsackStr
knapsackStr :: String -> String
knapsackStr str = show $ knapsack items capacity numItems
where [capacity, numItems] = map read . words $ head ls
ls = lines str
items = map (makeItem . words) $ take numItems $ tail ls
Some types and helpers to set the stage:
type Item = (Weight, Value)
type Weight = Int
type Value = Int
weight :: Item -> Weight
weight = fst
value :: Item -> Value
value = snd
makeItem :: [String] -> Item
makeItem [w, v] = (read w, read v)
And the primary function:
knapsack :: [Item] -> Weight -> Int -> Value
knapsack itemsList = go
where go = memo2 integral integral knapsack'
items = fromList $ (0,0):itemsList
knapsack' 0 _ = 0
knapsack' _ 0 = 0
knapsack' w i | wi > w = exclude
| otherwise = max exclude include
where wi = weight item
vi = value item
item = items `index` i
exclude = go w (i-1)
include = go (w-wi) (i-1) + vi
And this code works; I've tried plugging in the SPOJ sample test case and it produces the correct result. But when I submit this solution to SPOJ (instead of importing Luke Palmer's MemoCombinators, I simply copy and paste the necessary parts into the submitted source), it exceeds the time limit. =/
I don't understand why; I asked earlier about an efficient way to perform 0-1 knapsack, and I'm fairly convinced that this is about as fast as it gets: a memoized function that will only recursively calculate the sub-entries that it absolutely needs in order to produce the correct result. Did I mess up the memoization somehow? Is there a slow point in this code that I am missing? Is SPOJ just biased against Haskell?
I even put {-# OPTIONS_GHC -O2 #-} at the top of the submission, but alas, it didn't help. I have tried a similar solution that uses a 2D array of Sequences, but it was also rejected as too slow.
There's one major problem which really slows this down. It's too polymorphic. Type-specialized versions of functions can be much faster than polymorphic varieties, and for whatever reason GHC isn't inlining this code to the point where it can determine the exact types in use. When I change the definition of integral to:
integral :: Memo Int
integral = wrap id id bits
I get an approximately 5-fold speedup; I think it's fast enough to be accepted on SPOJ.
This is still significantly slower than gorlum0's solution however. I suspect the reason is because he's using arrays and you use a custom trie type. Using a trie will take much more memory and also make lookups slower due to extra indirections, cache misses, etc. You might be able to make up a lot of the difference if you strictify and unbox fields in IntMap, but I'm not sure that's possible. Trying to strictify fields in BitTrie creates runtime crashes for me.
Pure haskell memoizing code can be good, but I don't think it's as fast as doing unsafe things (at least under the hood). You might apply Lennart Augustsson's technique to see if it fares better at memoization.
The one thing that slows down Haskell is IO, The String type in Haskell gives UTF8 support which we don't need for SPOJ. ByteStrings are blazing fast so you might want to consider using them instead.

Tail-recursive pow() algorithm with memoization?

I'm looking for an algorithm to compute pow() that's tail-recursive and uses memoization to speed up repeated calculations.
Performance isn't an issue; this is mostly an intellectual exercise - I spent a train ride coming up with all the different pow() implementations I could, but was unable to come up with one that I was happy with that had these two properties.
My best shot was the following:
def calc_tailrec_mem(base, exp, cache_line={}, acc=1, ctr=0):
if exp == 0:
return 1
elif exp == 1:
return acc * base
elif exp in cache_line:
val = acc * cache_line[exp]
cache_line[exp + ctr] = val
return val
else:
cache_line[ctr] = acc
return calc_tailrec_mem(base, exp-1, cache_line, acc * base, ctr + 1)
It works, but it doesn't memoize the results of all calculations - only those with exponents 1..exp/2 and exp.
You'll get better performance if you use the successive squaring technique described in SICP section 1.2.4 Exponentiation. It doesn't use memoization, but the general approach is O(log n) instead of O(n), so you should still see an improvement.
I talk about the solution to the iterative process from exercise 1.16 here.
I don't think you're recording the correct thing in your cache, the mapping changed when you call it with different arguments.
I think you need to have a cache of (base,exp) -> pow(base,exp).
I understand what ctr is for, and why only half of what you expect is recorded.
Consider calc_tailrec_mem(2,4): First level, pow(2,1) is recorded as 2, the next level = calc_tailrec_mem(2,3,...), and pow(2,2) is recorded. The next level is calc_tailrec_mem(2,2,...), but that is already saved in the cache, so the recursion stops.
The function is very confusing because it's caching something completely different from what it's supposed to be calculating, due to the acculumator and ctr.
This is way too late, but anyone out there looking for the answer, here it is:
int powMem(int base,int exp){
//initializes once and for all
static map<int,int> memo;
//base case to stop the recursion
if(exp <= 1) return base;
//check if the value is already calculated before. If yes just return it.
if(memo.find(exp) != memo.end())
return memo[exp];
//else just find it and then store it in memo for further use.
int x = powMem(base,exp/2);
memo[exp] = x*x;
//return the answer
return memo[exp];
}
This uses the memo array - a map , to be exact - to store the already calculated values.

Algorithm for multidimensional optimization / root-finding / something

I have five values, A, B, C, D and E.
Given the constraint A + B + C + D + E = 1, and five functions F(A), F(B), F(C), F(D), F(E), I need to solve for A through E such that F(A) = F(B) = F(C) = F(D) = F(E).
What's the best algorithm/approach to use for this? I don't care if I have to write it myself, I would just like to know where to look.
EDIT: These are nonlinear functions. Beyond that, they can't be characterized. Some of them may eventually be interpolated from a table of data.
There is no general answer to this question. A solver finding the solution to any equation does not exist. As Lance Roberts already says, you have to know more about the functions. Just a few examples
If the functions are twice differentiable, and you can compute the first derivative, you might try a variant of Newton-Raphson
Have a look at the Lagrange Multiplier Method for implementing the constraint.
If the function F is continuous (which it probably is, if it is an interpolant), you could also try the Bisection Method, which is a lot like binary search.
Before you can solve the problem, you really need to know more about the function you're studying.
As others have already posted, we do need some more information on the functions. However, given that, we can still try to solve the following relaxation with a standard non-linear programming toolbox.
min k
st.
A + B + C + D + E = 1
F1(A) - k = 0
F2(B) - k = 0
F3(C) -k = 0
F4(D) - k = 0
F5(E) -k = 0
Now we can solve this in any manner we wish, such as penalty method
min k + mu*sum(Fi(x_i) - k)^2
st
A+B+C+D+E = 1
or a straightforward SQP or interior-point method.
More details and I can help advise as to a good method.
m
The functions are all monotonically increasing with their argument. Beyond that, they can't be characterized. The approach that worked turned out to be:
1) Start with A = B = C = D = E = 1/5
2) Compute F1(A) through F5(E), and recalculate A through E such that each function equals that sum divided by 5 (the average).
3) Rescale the new A through E so that they all sum to 1, and recompute F1 through F5.
4) Repeat until satisfied.
It converges surprisingly fast - just a few iterations. Of course, each iteration requires 5 root finds for step 2.
One solution of the equations
A + B + C + D + E = 1
F(A) = F(B) = F(C) = F(D) = F(E)
is to take A, B, C, D and E all equal to 1/5. Not sure though whether that is what you want ...
Added after John's comment (thanks!)
Assuming the second equation should read F1(A) = F2(B) = F3(C) = F4(D) = F5(E), I'd use the Newton-Raphson method (see Martijn's answer). You can eliminate one variable by setting E = 1 - A - B - C - D. At every step of the iteration you need to solve a 4x4 system. The biggest problem is probably where to start the iteration. One possibility is to start at a random point, do some iterations, and if you're not getting anywhere, pick another random point and start again.
Keep in mind that if you really don't know anything about the function then there need not be a solution.
ALGENCAN (part of TANGO) is really nice. There are Python bindings, too.
http://www.ime.usp.br/~egbirgin/tango/codes.php - " general nonlinear programming that does not use matrix manipulations at all and, so, is able to solve extremely large problems with moderate computer time. The general algorithm is of Augmented Lagrangian type ... "
http://pypi.python.org/pypi/TANGO%20Project%20-%20ALGENCAN/1.0
Google OPTIF9 or ALLUNC. We use these for general optimization.
You could use standard search technic as the others mentioned. There are a few optimization you could make use of it while doing the search.
First of all, you only need to solve A,B,C,D because 1-E = A+B+C+D.
Second, you have F(A) = F(B) = F(C) = F(D), then you can search for A. Once you get F(A), you could solve B, C, D if that is possible. If it is not possible to solve the functions, you need to continue search each variable, but now you have a limited range to search for because A+B+C+D <= 1.
If your search is discrete and finite, the above optimizations should work reasonable well.
I would try Particle Swarm Optimization first. It is very easy to implement and tweak. See the Wiki page for it.

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