How do I optimize these ocaml functions for dynamic interval scheduling? - performance

I have a program that solves the weighted interval scheduling problem using dynamic programming (and believe it or not, it isn't for homework). I've profiled it, and I seem to be spending most of my time filling M with p(...). Here are the functions:
let rec get_highest_nonconflicting prev count start =
match prev with
head :: tail ->
if head < start then
count
else
get_highest_nonconflicting tail (count - 1) start
| [] -> 0;;
let m_array = Array.make (num_genes + 1) 0;;
let rec fill_m_array ?(count=1) ?(prev=[]) gene_spans =
match gene_spans with
head :: tail -> m_array.(count) <-
get_highest_nonconflicting prev (count - 1) (get_start head);
fill_m_array tail ~prev:(get_stop (head) :: prev) ~count:(count+1);
| [] -> ();;
I can't really think of any ways to optimize this, and based on my knowledge of this algorithm, this seems to be the place that is likely to take up the most time. But this is also only my second OCaml program. So is there any way to optimize this?

There isn't anything obviously inefficient with your two functions. Did you expect your implementation to be faster, for instance with reference to an implementation in another language?
I was wondering about the lists passed to get_highest_nonconflicting. If you have reasons to expect that this function is often passed lists that are physically identical to previously passed lists (and this includes the sub-lists passed on recursive calls), a cache there could help.
If you expect lists that are equal but physically different, hash-consing (and then caching) could help.

Related

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

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.

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.

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.

F# performance bug?

let test1fun x = [for i in 1..x-> i]
let test2fun x y= [for i in 1..x
do for i in 1..y-> i]
let singlesearcher i =
let rec searcher j agg =
if j > i
then agg
else searcher (j+1) (i::agg)
searcher 1 []
let doublesearcher i j =
let rec searcher k l agg =
if k > i
then searcher 1 (l+1) agg
else if l > j
then agg
else searcher (k+1) l ((k,l)::agg)
searcher 1 1 []
executing the above with #time and 10000 for all inputs yields
list comprehension/singlesearcher-> negligable
cross product -> 320
list comprehension crossproduct -> 630
Why is the nested list comprehension more than twice the the functional version?
Yes. List comprehension is usually slower than directly using F# list or array. (On my machine, I also find similar timing with you.)
Let's look into how they are implemented. The list comprehension version is actually quite complicated:
a sequence/IEnumerable<int> is created using the comprehension syntax. This is just a lazy sequence, little time is spent here.
then this sequence is transformed into F# List by using something like Seq.toList. The actual time is spent here. There are a lot of HasNext MoveNext and switch (state) like code executed here. With so many function calls, you cannot expect it fast.
While the functional version doublesearcher is properly optimized into a tail recursion. This is a more direct version than list comprehension, and few function calls are used therein.
Usually we don't care these small performance difference for sequence, lists or arrays if the operation is not very critical. I think in your example, the generation is anyway one-time. The two time timing is not a big problem. For other cases, e.g. the dot product of two vectors, using arrays could save a lot of time because this operation is executed for a lot of times.

Recursion or Iteration?

Is there a performance hit if we use a loop instead of recursion or vice versa in algorithms where both can serve the same purpose? Eg: Check if the given string is a palindrome.
I have seen many programmers using recursion as a means to show off when a simple iteration algorithm can fit the bill.
Does the compiler play a vital role in deciding what to use?
Loops may achieve a performance gain for your program. Recursion may achieve a performance gain for your programmer. Choose which is more important in your situation!
It is possible that recursion will be more expensive, depending on if the recursive function is tail recursive (the last line is recursive call). Tail recursion should be recognized by the compiler and optimized to its iterative counterpart (while maintaining the concise, clear implementation you have in your code).
I would write the algorithm in the way that makes the most sense and is the clearest for the poor sucker (be it yourself or someone else) that has to maintain the code in a few months or years. If you run into performance issues, then profile your code, and then and only then look into optimizing by moving over to an iterative implementation. You may want to look into memoization and dynamic programming.
Comparing recursion to iteration is like comparing a phillips head screwdriver to a flat head screwdriver. For the most part you could remove any phillips head screw with a flat head, but it would just be easier if you used the screwdriver designed for that screw right?
Some algorithms just lend themselves to recursion because of the way they are designed (Fibonacci sequences, traversing a tree like structure, etc.). Recursion makes the algorithm more succinct and easier to understand (therefore shareable and reusable).
Also, some recursive algorithms use "Lazy Evaluation" which makes them more efficient than their iterative brothers. This means that they only do the expensive calculations at the time they are needed rather than each time the loop runs.
That should be enough to get you started. I'll dig up some articles and examples for you too.
Link 1: Haskel vs PHP (Recursion vs Iteration)
Here is an example where the programmer had to process a large data set using PHP. He shows how easy it would have been to deal with in Haskel using recursion, but since PHP had no easy way to accomplish the same method, he was forced to use iteration to get the result.
http://blog.webspecies.co.uk/2011-05-31/lazy-evaluation-with-php.html
Link 2: Mastering Recursion
Most of recursion's bad reputation comes from the high costs and inefficiency in imperative languages. The author of this article talks about how to optimize recursive algorithms to make them faster and more efficient. He also goes over how to convert a traditional loop into a recursive function and the benefits of using tail-end recursion. His closing words really summed up some of my key points I think:
"recursive programming gives the programmer a better way of organizing
code in a way that is both maintainable and logically consistent."
https://developer.ibm.com/articles/l-recurs/
Link 3: Is recursion ever faster than looping? (Answer)
Here is a link to an answer for a stackoverflow question that is similar to yours. The author points out that a lot of the benchmarks associated with either recursing or looping are very language specific. Imperative languages are typically faster using a loop and slower with recursion and vice-versa for functional languages. I guess the main point to take from this link is that it is very difficult to answer the question in a language agnostic / situation blind sense.
Is recursion ever faster than looping?
Recursion is more costly in memory, as each recursive call generally requires a memory address to be pushed to the stack - so that later the program could return to that point.
Still, there are many cases in which recursion is a lot more natural and readable than loops - like when working with trees. In these cases I would recommend sticking to recursion.
Typically, one would expect the performance penalty to lie in the other direction. Recursive calls can lead to the construction of extra stack frames; the penalty for this varies. Also, in some languages like Python (more correctly, in some implementations of some languages...), you can run into stack limits rather easily for tasks you might specify recursively, such as finding the maximum value in a tree data structure. In these cases, you really want to stick with loops.
Writing good recursive functions can reduce the performance penalty somewhat, assuming you have a compiler that optimizes tail recursions, etc. (Also double check to make sure that the function really is tail recursive---it's one of those things that many people make mistakes on.)
Apart from "edge" cases (high performance computing, very large recursion depth, etc.), it's preferable to adopt the approach that most clearly expresses your intent, is well-designed, and is maintainable. Optimize only after identifying a need.
Recursion is better than iteration for problems that can be broken down into multiple, smaller pieces.
For example, to make a recursive Fibonnaci algorithm, you break down fib(n) into fib(n-1) and fib(n-2) and compute both parts. Iteration only allows you to repeat a single function over and over again.
However, Fibonacci is actually a broken example and I think iteration is actually more efficient. Notice that fib(n) = fib(n-1) + fib(n-2) and fib(n-1) = fib(n-2) + fib(n-3). fib(n-1) gets calculated twice!
A better example is a recursive algorithm for a tree. The problem of analyzing the parent node can be broken down into multiple smaller problems of analyzing each child node. Unlike the Fibonacci example, the smaller problems are independent of each other.
So yeah - recursion is better than iteration for problems that can be broken down into multiple, smaller, independent, similar problems.
Your performance deteriorates when using recursion because calling a method, in any language, implies a lot of preparation: the calling code posts a return address, call parameters, some other context information such as processor registers might be saved somewhere, and at return time the called method posts a return value which is then retrieved by the caller, and any context information that was previously saved will be restored. the performance diff between an iterative and a recursive approach lies in the time these operations take.
From an implementation point of view, you really start noticing the difference when the time it takes to handle the calling context is comparable to the time it takes for your method to execute. If your recursive method takes longer to execute then the calling context management part, go the recursive way as the code is generally more readable and easy to understand and you won't notice the performance loss. Otherwise go iterative for efficiency reasons.
I believe tail recursion in java is not currently optimized. The details are sprinkled throughout this discussion on LtU and the associated links. It may be a feature in the upcoming version 7, but apparently it presents certain difficulties when combined with Stack Inspection since certain frames would be missing. Stack Inspection has been used to implement their fine-grained security model since Java 2.
http://lambda-the-ultimate.org/node/1333
There are many cases where it gives a much more elegant solution over the iterative method, the common example being traversal of a binary tree, so it isn't necessarily more difficult to maintain. In general, iterative versions are usually a bit faster (and during optimization may well replace a recursive version), but recursive versions are simpler to comprehend and implement correctly.
Recursion is very useful is some situations. For example consider the code for finding the factorial
int factorial ( int input )
{
int x, fact = 1;
for ( x = input; x > 1; x--)
fact *= x;
return fact;
}
Now consider it by using the recursive function
int factorial ( int input )
{
if (input == 0)
{
return 1;
}
return input * factorial(input - 1);
}
By observing these two, we can see that recursion is easy to understand.
But if it is not used with care it can be so much error prone too.
Suppose if we miss if (input == 0), then the code will be executed for some time and ends with usually a stack overflow.
In many cases recursion is faster because of caching, which improves performance. For example, here is an iterative version of merge sort using the traditional merge routine. It will run slower than the recursive implementation because of caching improved performances.
Iterative implementation
public static void sort(Comparable[] a)
{
int N = a.length;
aux = new Comparable[N];
for (int sz = 1; sz < N; sz = sz+sz)
for (int lo = 0; lo < N-sz; lo += sz+sz)
merge(a, lo, lo+sz-1, Math.min(lo+sz+sz-1, N-1));
}
Recursive implementation
private static void sort(Comparable[] a, Comparable[] aux, int lo, int hi)
{
if (hi <= lo) return;
int mid = lo + (hi - lo) / 2;
sort(a, aux, lo, mid);
sort(a, aux, mid+1, hi);
merge(a, aux, lo, mid, hi);
}
PS - this is what was told by Professor Kevin Wayne (Princeton University) on the course on algorithms presented on Coursera.
Using recursion, you're incurring the cost of a function call with each "iteration", whereas with a loop, the only thing you usually pay is an increment/decrement. So, if the code for the loop isn't much more complicated than the code for the recursive solution, loop will usually be superior to recursion.
Recursion and iteration depends on the business logic that you want to implement, though in most of the cases it can be used interchangeably. Most developers go for recursion because it is easier to understand.
It depends on the language. In Java you should use loops. Functional languages optimize recursion.
Recursion has a disadvantage that the algorithm that you write using recursion has O(n) space complexity.
While iterative aproach have a space complexity of O(1).This is the advantange of using iteration over recursion.
Then why do we use recursion?
See below.
Sometimes it is easier to write an algorithm using recursion while it's slightly tougher to write the same algorithm using iteration.In this case if you opt to follow the iteration approach you would have to handle stack yourself.
If you're just iterating over a list, then sure, iterate away.
A couple of other answers have mentioned (depth-first) tree traversal. It really is such a great example, because it's a very common thing to do to a very common data structure. Recursion is extremely intuitive for this problem.
Check out the "find" methods here:
http://penguin.ewu.edu/cscd300/Topic/BSTintro/index.html
Recursion is more simple (and thus - more fundamental) than any possible definition of an iteration. You can define a Turing-complete system with only a pair of combinators (yes, even a recursion itself is a derivative notion in such a system). Lambda calculus is an equally powerful fundamental system, featuring recursive functions. But if you want to define an iteration properly, you'd need much more primitives to start with.
As for the code - no, recursive code is in fact much easier to understand and to maintain than a purely iterative one, since most data structures are recursive. Of course, in order to get it right one would need a language with a support for high order functions and closures, at least - to get all the standard combinators and iterators in a neat way. In C++, of course, complicated recursive solutions can look a bit ugly, unless you're a hardcore user of FC++ and alike.
I would think in (non tail) recursion there would be a performance hit for allocating a new stack etc every time the function is called (dependent on language of course).
it depends on "recursion depth".
it depends on how much the function call overhead will influence the total execution time.
For example, calculating the classical factorial in a recursive way is very inefficient due to:
- risk of data overflowing
- risk of stack overflowing
- function call overhead occupy 80% of execution time
while developing a min-max algorithm for position analysis in the game of chess that will analyze subsequent N moves can be implemented in recursion over the "analysis depth" (as I'm doing ^_^)
Recursion? Where do I start, wiki will tell you “it’s the process of repeating items in a self-similar way"
Back in day when I was doing C, C++ recursion was a god send, stuff like "Tail recursion". You'll also find many sorting algorithms use recursion. Quick sort example: http://alienryderflex.com/quicksort/
Recursion is like any other algorithm useful for a specific problem. Perhaps you mightn't find a use straight away or often but there will be problem you’ll be glad it’s available.
In C++ if the recursive function is a templated one, then the compiler has more chance to optimize it, as all the type deduction and function instantiations will occur in compile time. Modern compilers can also inline the function if possible. So if one uses optimization flags like -O3 or -O2 in g++, then recursions may have the chance to be faster than iterations. In iterative codes, the compiler gets less chance to optimize it, as it is already in the more or less optimal state (if written well enough).
In my case, I was trying to implement matrix exponentiation by squaring using Armadillo matrix objects, in both recursive and iterative way. The algorithm can be found here... https://en.wikipedia.org/wiki/Exponentiation_by_squaring.
My functions were templated and I have calculated 1,000,000 12x12 matrices raised to the power 10. I got the following result:
iterative + optimisation flag -O3 -> 2.79.. sec
recursive + optimisation flag -O3 -> 1.32.. sec
iterative + No-optimisation flag -> 2.83.. sec
recursive + No-optimisation flag -> 4.15.. sec
These results have been obtained using gcc-4.8 with c++11 flag (-std=c++11) and Armadillo 6.1 with Intel mkl. Intel compiler also shows similar results.
Mike is correct. Tail recursion is not optimized out by the Java compiler or the JVM. You will always get a stack overflow with something like this:
int count(int i) {
return i >= 100000000 ? i : count(i+1);
}
You have to keep in mind that utilizing too deep recursion you will run into Stack Overflow, depending on allowed stack size. To prevent this make sure to provide some base case which ends you recursion.
Using just Chrome 45.0.2454.85 m, recursion seems to be a nice amount faster.
Here is the code:
(function recursionVsForLoop(global) {
"use strict";
// Perf test
function perfTest() {}
perfTest.prototype.do = function(ns, fn) {
console.time(ns);
fn();
console.timeEnd(ns);
};
// Recursion method
(function recur() {
var count = 0;
global.recurFn = function recurFn(fn, cycles) {
fn();
count = count + 1;
if (count !== cycles) recurFn(fn, cycles);
};
})();
// Looped method
function loopFn(fn, cycles) {
for (var i = 0; i < cycles; i++) {
fn();
}
}
// Tests
var curTest = new perfTest(),
testsToRun = 100;
curTest.do('recursion', function() {
recurFn(function() {
console.log('a recur run.');
}, testsToRun);
});
curTest.do('loop', function() {
loopFn(function() {
console.log('a loop run.');
}, testsToRun);
});
})(window);
RESULTS
// 100 runs using standard for loop
100x for loop run.
Time to complete: 7.683ms
// 100 runs using functional recursive approach w/ tail recursion
100x recursion run.
Time to complete: 4.841ms
In the screenshot below, recursion wins again by a bigger margin when run at 300 cycles per test
If the iterations are atomic and orders of magnitude more expensive than pushing a new stack frame and creating a new thread and you have multiple cores and your runtime environment can use all of them, then a recursive approach could yield a huge performance boost when combined with multithreading. If the average number of iterations is not predictable then it might be a good idea to use a thread pool which will control thread allocation and prevent your process from creating too many threads and hogging the system.
For example, in some languages, there are recursive multithreaded merge sort implementations.
But again, multithreading can be used with looping rather than recursion, so how well this combination will work depends on more factors including the OS and its thread allocation mechanism.
I found another differences between those approaches.
It looks simple and unimportant, but it has a very important role while you prepare for interviews and this subject arises, so look closely.
In short:
1) iterative post-order traversal is not easy - that makes DFT more complex
2) cycles check easier with recursion
Details:
In the recursive case, it is easy to create pre and post traversals:
Imagine a pretty standard question: "print all tasks that should be executed to execute the task 5, when tasks depend on other tasks"
Example:
//key-task, value-list of tasks the key task depends on
//"adjacency map":
Map<Integer, List<Integer>> tasksMap = new HashMap<>();
tasksMap.put(0, new ArrayList<>());
tasksMap.put(1, new ArrayList<>());
List<Integer> t2 = new ArrayList<>();
t2.add(0);
t2.add(1);
tasksMap.put(2, t2);
List<Integer> t3 = new ArrayList<>();
t3.add(2);
t3.add(10);
tasksMap.put(3, t3);
List<Integer> t4 = new ArrayList<>();
t4.add(3);
tasksMap.put(4, t4);
List<Integer> t5 = new ArrayList<>();
t5.add(3);
tasksMap.put(5, t5);
tasksMap.put(6, new ArrayList<>());
tasksMap.put(7, new ArrayList<>());
List<Integer> t8 = new ArrayList<>();
t8.add(5);
tasksMap.put(8, t8);
List<Integer> t9 = new ArrayList<>();
t9.add(4);
tasksMap.put(9, t9);
tasksMap.put(10, new ArrayList<>());
//task to analyze:
int task = 5;
List<Integer> res11 = getTasksInOrderDftReqPostOrder(tasksMap, task);
System.out.println(res11);**//note, no reverse required**
List<Integer> res12 = getTasksInOrderDftReqPreOrder(tasksMap, task);
Collections.reverse(res12);//note reverse!
System.out.println(res12);
private static List<Integer> getTasksInOrderDftReqPreOrder(Map<Integer, List<Integer>> tasksMap, int task) {
List<Integer> result = new ArrayList<>();
Set<Integer> visited = new HashSet<>();
reqPreOrder(tasksMap,task,result, visited);
return result;
}
private static void reqPreOrder(Map<Integer, List<Integer>> tasksMap, int task, List<Integer> result, Set<Integer> visited) {
if(!visited.contains(task)) {
visited.add(task);
result.add(task);//pre order!
List<Integer> children = tasksMap.get(task);
if (children != null && children.size() > 0) {
for (Integer child : children) {
reqPreOrder(tasksMap,child,result, visited);
}
}
}
}
private static List<Integer> getTasksInOrderDftReqPostOrder(Map<Integer, List<Integer>> tasksMap, int task) {
List<Integer> result = new ArrayList<>();
Set<Integer> visited = new HashSet<>();
reqPostOrder(tasksMap,task,result, visited);
return result;
}
private static void reqPostOrder(Map<Integer, List<Integer>> tasksMap, int task, List<Integer> result, Set<Integer> visited) {
if(!visited.contains(task)) {
visited.add(task);
List<Integer> children = tasksMap.get(task);
if (children != null && children.size() > 0) {
for (Integer child : children) {
reqPostOrder(tasksMap,child,result, visited);
}
}
result.add(task);//post order!
}
}
Note that the recursive post-order-traversal does not require a subsequent reversal of the result. Children printed first and your task in the question printed last. Everything is fine. You can do a recursive pre-order-traversal (also shown above) and that one will require a reversal of the result list.
Not that simple with iterative approach! In iterative (one stack) approach you can only do a pre-ordering-traversal, so you obliged to reverse the result array at the end:
List<Integer> res1 = getTasksInOrderDftStack(tasksMap, task);
Collections.reverse(res1);//note reverse!
System.out.println(res1);
private static List<Integer> getTasksInOrderDftStack(Map<Integer, List<Integer>> tasksMap, int task) {
List<Integer> result = new ArrayList<>();
Set<Integer> visited = new HashSet<>();
Stack<Integer> st = new Stack<>();
st.add(task);
visited.add(task);
while(!st.isEmpty()){
Integer node = st.pop();
List<Integer> children = tasksMap.get(node);
result.add(node);
if(children!=null && children.size() > 0){
for(Integer child:children){
if(!visited.contains(child)){
st.add(child);
visited.add(child);
}
}
}
//If you put it here - it does not matter - it is anyway a pre-order
//result.add(node);
}
return result;
}
Looks simple, no?
But it is a trap in some interviews.
It means the following: with the recursive approach, you can implement Depth First Traversal and then select what order you need pre or post(simply by changing the location of the "print", in our case of the "adding to the result list"). With the iterative (one stack) approach you can easily do only pre-order traversal and so in the situation when children need be printed first(pretty much all situations when you need start print from the bottom nodes, going upwards) - you are in the trouble. If you have that trouble you can reverse later, but it will be an addition to your algorithm. And if an interviewer is looking at his watch it may be a problem for you. There are complex ways to do an iterative post-order traversal, they exist, but they are not simple. Example:https://www.geeksforgeeks.org/iterative-postorder-traversal-using-stack/
Thus, the bottom line: I would use recursion during interviews, it is simpler to manage and to explain. You have an easy way to go from pre to post-order traversal in any urgent case. With iterative you are not that flexible.
I would use recursion and then tell: "Ok, but iterative can provide me more direct control on used memory, I can easily measure the stack size and disallow some dangerous overflow.."
Another plus of recursion - it is simpler to avoid / notice cycles in a graph.
Example (preudocode):
dft(n){
mark(n)
for(child: n.children){
if(marked(child))
explode - cycle found!!!
dft(child)
}
unmark(n)
}
It may be fun to write it as recursion, or as a practice.
However, if the code is to be used in production, you need to consider the possibility of stack overflow.
Tail recursion optimization can eliminate stack overflow, but do you want to go through the trouble of making it so, and you need to know you can count on it having the optimization in your environment.
Every time the algorithm recurses, how much is the data size or n reduced by?
If you are reducing the size of data or n by half every time you recurse, then in general you don't need to worry about stack overflow. Say, if it needs to be 4,000 level deep or 10,000 level deep for the program to stack overflow, then your data size need to be roughly 24000 for your program to stack overflow. To put that into perspective, a biggest storage device recently can hold 261 bytes, and if you have 261 of such devices, you are only dealing with 2122 data size. If you are looking at all the atoms in the universe, it is estimated that it may be less than 284. If you need to deal with all the data in the universe and their states for every millisecond since the birth of the universe estimated to be 14 billion years ago, it may only be 2153. So if your program can handle 24000 units of data or n, you can handle all data in the universe and the program will not stack overflow. If you don't need to deal with numbers that are as big as 24000 (a 4000-bit integer), then in general you don't need to worry about stack overflow.
However, if you reduce the size of data or n by a constant amount every time you recurse, then you can run into stack overflow when n becomes merely 20000. That is, the program runs well when n is 1000, and you think the program is good, and then the program stack overflows when some time in the future, when n is 5000 or 20000.
So if you have a possibility of stack overflow, try to make it an iterative solution.
As far as I know, Perl does not optimize tail-recursive calls, but you can fake it.
sub f{
my($l,$r) = #_;
if( $l >= $r ){
return $l;
} else {
# return f( $l+1, $r );
#_ = ( $l+1, $r );
goto &f;
}
}
When first called it will allocate space on the stack. Then it will change its arguments, and restart the subroutine, without adding anything more to the stack. It will therefore pretend that it never called its self, changing it into an iterative process.
Note that there is no "my #_;" or "local #_;", if you did it would no longer work.
"Is there a performance hit if we use a loop instead of
recursion or vice versa in algorithms where both can serve the same purpose?"
Usually yes if you are writing in a imperative language iteration will run faster than recursion, the performance hit is minimized in problems where the iterative solution requires manipulating Stacks and popping items off of a stack due to the recursive nature of the problem. There are a lot of times where the recursive implementation is much easier to read because the code is much shorter,
so you do want to consider maintainability. Especailly in cases where the problem has a recursive nature. So take for example:
The recursive implementation of Tower of Hanoi:
def TowerOfHanoi(n , source, destination, auxiliary):
if n==1:
print ("Move disk 1 from source",source,"to destination",destination)
return
TowerOfHanoi(n-1, source, auxiliary, destination)
print ("Move disk",n,"from source",source,"to destination",destination)
TowerOfHanoi(n-1, auxiliary, destination, source)
Fairly short and pretty easy to read. Compare this with its Counterpart iterative TowerOfHanoi:
# Python3 program for iterative Tower of Hanoi
import sys
# A structure to represent a stack
class Stack:
# Constructor to set the data of
# the newly created tree node
def __init__(self, capacity):
self.capacity = capacity
self.top = -1
self.array = [0]*capacity
# function to create a stack of given capacity.
def createStack(capacity):
stack = Stack(capacity)
return stack
# Stack is full when top is equal to the last index
def isFull(stack):
return (stack.top == (stack.capacity - 1))
# Stack is empty when top is equal to -1
def isEmpty(stack):
return (stack.top == -1)
# Function to add an item to stack.
# It increases top by 1
def push(stack, item):
if(isFull(stack)):
return
stack.top+=1
stack.array[stack.top] = item
# Function to remove an item from stack.
# It decreases top by 1
def Pop(stack):
if(isEmpty(stack)):
return -sys.maxsize
Top = stack.top
stack.top-=1
return stack.array[Top]
# Function to implement legal
# movement between two poles
def moveDisksBetweenTwoPoles(src, dest, s, d):
pole1TopDisk = Pop(src)
pole2TopDisk = Pop(dest)
# When pole 1 is empty
if (pole1TopDisk == -sys.maxsize):
push(src, pole2TopDisk)
moveDisk(d, s, pole2TopDisk)
# When pole2 pole is empty
else if (pole2TopDisk == -sys.maxsize):
push(dest, pole1TopDisk)
moveDisk(s, d, pole1TopDisk)
# When top disk of pole1 > top disk of pole2
else if (pole1TopDisk > pole2TopDisk):
push(src, pole1TopDisk)
push(src, pole2TopDisk)
moveDisk(d, s, pole2TopDisk)
# When top disk of pole1 < top disk of pole2
else:
push(dest, pole2TopDisk)
push(dest, pole1TopDisk)
moveDisk(s, d, pole1TopDisk)
# Function to show the movement of disks
def moveDisk(fromPeg, toPeg, disk):
print("Move the disk", disk, "from '", fromPeg, "' to '", toPeg, "'")
# Function to implement TOH puzzle
def tohIterative(num_of_disks, src, aux, dest):
s, d, a = 'S', 'D', 'A'
# If number of disks is even, then interchange
# destination pole and auxiliary pole
if (num_of_disks % 2 == 0):
temp = d
d = a
a = temp
total_num_of_moves = int(pow(2, num_of_disks) - 1)
# Larger disks will be pushed first
for i in range(num_of_disks, 0, -1):
push(src, i)
for i in range(1, total_num_of_moves + 1):
if (i % 3 == 1):
moveDisksBetweenTwoPoles(src, dest, s, d)
else if (i % 3 == 2):
moveDisksBetweenTwoPoles(src, aux, s, a)
else if (i % 3 == 0):
moveDisksBetweenTwoPoles(aux, dest, a, d)
# Input: number of disks
num_of_disks = 3
# Create three stacks of size 'num_of_disks'
# to hold the disks
src = createStack(num_of_disks)
dest = createStack(num_of_disks)
aux = createStack(num_of_disks)
tohIterative(num_of_disks, src, aux, dest)
Now the first one is way easier to read because suprise suprise shorter code is usually easier to understand than code that is 10 times longer. Sometimes you want to ask yourself is the extra performance gain really worth it? The amount of hours wasted debugging the code. Is the iterative TowerOfHanoi faster than the Recursive TowerOfHanoi? Probably, but not by a big margin. Would I like to program Recursive problems like TowerOfHanoi using iteration? Hell no. Next we have another recursive function the Ackermann function:
Using recursion:
if m == 0:
# BASE CASE
return n + 1
elif m > 0 and n == 0:
# RECURSIVE CASE
return ackermann(m - 1, 1)
elif m > 0 and n > 0:
# RECURSIVE CASE
return ackermann(m - 1, ackermann(m, n - 1))
Using Iteration:
callStack = [{'m': 2, 'n': 3, 'indentation': 0, 'instrPtr': 'start'}]
returnValue = None
while len(callStack) != 0:
m = callStack[-1]['m']
n = callStack[-1]['n']
indentation = callStack[-1]['indentation']
instrPtr = callStack[-1]['instrPtr']
if instrPtr == 'start':
print('%sackermann(%s, %s)' % (' ' * indentation, m, n))
if m == 0:
# BASE CASE
returnValue = n + 1
callStack.pop()
continue
elif m > 0 and n == 0:
# RECURSIVE CASE
callStack[-1]['instrPtr'] = 'after first recursive case'
callStack.append({'m': m - 1, 'n': 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
continue
elif m > 0 and n > 0:
# RECURSIVE CASE
callStack[-1]['instrPtr'] = 'after second recursive case, inner call'
callStack.append({'m': m, 'n': n - 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
continue
elif instrPtr == 'after first recursive case':
returnValue = returnValue
callStack.pop()
continue
elif instrPtr == 'after second recursive case, inner call':
callStack[-1]['innerCallResult'] = returnValue
callStack[-1]['instrPtr'] = 'after second recursive case, outer call'
callStack.append({'m': m - 1, 'n': returnValue, 'indentation': indentation + 1, 'instrPtr': 'start'})
continue
elif instrPtr == 'after second recursive case, outer call':
returnValue = returnValue
callStack.pop()
continue
print(returnValue)
And once again I will argue that the recursive implementation is much easier to understand. So my conclusion is use recursion if the problem by nature is recursive and requires manipulating items in a stack.
I'm going to answer your question by designing a Haskell data structure by "induction", which is a sort of "dual" to recursion. And then I will show how this duality leads to nice things.
We introduce a type for a simple tree:
data Tree a = Branch (Tree a) (Tree a)
| Leaf a
deriving (Eq)
We can read this definition as saying "A tree is a Branch (which contains two trees) or is a leaf (which contains a data value)". So the leaf is a sort of minimal case. If a tree isn't a leaf, then it must be a compound tree containing two trees. These are the only cases.
Let's make a tree:
example :: Tree Int
example = Branch (Leaf 1)
(Branch (Leaf 2)
(Leaf 3))
Now, let's suppose we want to add 1 to each value in the tree. We can do this by calling:
addOne :: Tree Int -> Tree Int
addOne (Branch a b) = Branch (addOne a) (addOne b)
addOne (Leaf a) = Leaf (a + 1)
First, notice that this is in fact a recursive definition. It takes the data constructors Branch and Leaf as cases (and since Leaf is minimal and these are the only possible cases), we are sure that the function will terminate.
What would it take to write addOne in an iterative style? What will looping into an arbitrary number of branches look like?
Also, this kind of recursion can often be factored out, in terms of a "functor". We can make Trees into Functors by defining:
instance Functor Tree where fmap f (Leaf a) = Leaf (f a)
fmap f (Branch a b) = Branch (fmap f a) (fmap f b)
and defining:
addOne' = fmap (+1)
We can factor out other recursion schemes, such as the catamorphism (or fold) for an algebraic data type. Using a catamorphism, we can write:
addOne'' = cata go where
go (Leaf a) = Leaf (a + 1)
go (Branch a b) = Branch a b

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