How well does it work? Profiling in Haskell

I’ve said it before, but I’ll say it again. As much as we’d like to think it’s the case, our Haskell code doesn’t work just because it compiles. This is why we have . But even if it passes our test suites this doesn’t mean it works as well as it could either. Sometimes we’ll realize that the code we wrote isn’t quite performant enough, so we’ll have to make improvements. test suites But improving our code can sometimes feel like taking shots in the dark. You’ll spend a great deal of time tweaking a certain piece. Then you’ll find you haven’t actually made much of a dent in the total run time of the application. Certain operations generally take longer, like database calls, network operations, and IO. So you can often have a decent idea of where to start. But it always helps to be sure. This is where benchmarking and profiling come in. We’re going to take a specific problem and learn how we can use some Haskell tools to zero in on the problem point. As a note, the tools we’ll use require you to be organizing your code using Stack or Cabal. If you’ve never used either of these before, you should check out our ! It’ll teach you the basics of creating a project with Stack. You’ll also learn the primary commands to use with Stack. It’s brand new and best of all FREE! Check it out! It’s our first course of any kind, so we’re looking for feedback! Stack Mini Course The Problem Our overarching problem for this article will be the “largest rectangle” problem. You can actually try to solve this problem yourself on under the name “John and Fences”. Imagine we have a series of vertical bars with varying heights placed next to each other. We want to find the area of the largest rectangle that we can draw over these bars that doesn’t include any empty space. Here’s a visualization of one such problem and solution: Hackerrank In this example, we have posts with heights . The largest rectangle we can form has an area of 12, as we see with the highlighted squares. [2,5,7,4,1,8] This problem is pretty neat and clean to solve with Haskell, as it lends itself to a recursive solution. First let’s define a couple newtypes to illustrate our concepts for this problem. We’ll use a compiler extension to derive the typeclass on our index type, as this will be useful later. Num {-# LANGUAGE GeneralizedNewtypeDeriving #-}...newtype FenceValues = FenceValues { unFenceValues :: [Int] }newtype FenceIndex = FenceIndex { unFenceIndex :: Int } deriving (Eq, Num, Ord)-- Left Index is inclusive, right index is non-inclusive newtype FenceInterval = FenceInterval { unFenceInterval :: (FenceIndex, FenceIndex) }newtype FenceSolution = FenceSolution { unFenceSolution :: Int } deriving (Eq, Show, Ord) Next, we’ll define our primary function. It will take our , a list of integers, and return our solution. FenceValues largestRectangle :: FenceValues -> FenceSolutionlargestRectangle values = ... It in turn will call our recursive helper function. This function will calculate the largest rectangle over a specific interval. We can solve it recursively by using smaller and smaller intervals. We’ll start by calling it on the interval of the whole list. largestRectangle :: FenceValues -> FenceSolutionlargestRectangle values = largestRectangleAtIndices values (FenceInterval (FenceIndex 0, FenceIndex (length (unFenceValues values)))) largestRectangleAtIndices :: FenceValues -> FenceInterval -> FenceSolutionlargestRectangleAtIndices = ... Now, to break this into recursive cases, we need some more information first. What we need is the index of the minimum height in this interval. One option is that we could make a rectangle spanning the whole interval with this height. i Any other “largest rectangle” won’t use this particular index. So we can then divide our problem into two more cases. In the first, we’ll find the largest rectangle on the interval to the left. In the second, we’ll look to the right. As your might realize, these two cases simply involve making recursive calls! Then we can easily compare their results. The only thing we need to add is a base case. Here are all these cases represented in code: largestRectangleAtIndices :: FenceValues -> FenceInterval -> FenceSolutionlargestRectangleAtIndices values interval@(FenceInterval (leftIndex, rightIndex)) = -- Base Case: Checks if left + 1 >= right if isBaseInterval interval then FenceSolution (valueAtIndex values leftIndex) -- Compare three cases else max (max middleCase leftCase) rightCase where -- Find the minimum height and its index (minIndex, minValue) = minimumHeightIndexValue values interval -- Case 1: Use the minimum index middleCase = FenceSolution $ (intervalSize interval) * minValue -- Recursive call #1 leftCase = largestRectangleAtIndices values (FenceInterval (leftIndex, minIndex)) -- Guard against case where there is no "right" interval rightCase = if minIndex + 1 == rightIndex then FenceSolution (maxBound :: Int) -- Supply a "fake" solution that we'll ignore -- Recursive call #2 else largestRectangleAtIndices values (FenceInterval (minIndex + 1, rightIndex)) And just like that, we’re actually almost finished. The only sticking point here is a few helper functions. Three of these are simple: valueAtIndex :: FenceValues -> FenceIndex -> IntvalueAtIndex values index = (unFenceValues values) !! (unFenceIndex index) isBaseInterval :: FenceInterval -> BoolisBaseInterval (FenceInterval (FenceIndex left, FenceIndex right)) = left + 1 >= right intervalSize :: FenceInterval -> IntintervalSize (FenceInterval (FenceIndex left, FenceIndex right)) = right - left Now we have to determine the minimum on this interval. Let’s do this in the most naive way, by scanning the whole interval with a fold. minimumHeightIndexValue :: FenceValues -> FenceInterval -> (FenceIndex, Int)minimumHeightIndexValue values (FenceInterval (FenceIndex left, FenceIndex right)) = foldl minTuple (FenceIndex (-1), maxBound :: Int) valsInInterval where valsInInterval :: [(FenceIndex, Int)] valsInInterval = drop left (take right (zip (FenceIndex [0..]) (unFenceValues values))) minTuple :: (FenceIndex, Int) -> (FenceIndex, Int) -> (FenceIndex, Int) minTuple old@(_, heightOld) new@(_, heightNew) = if heightNew < heightOld then new else old And now we’re done! Benchmarking our Code Now, this is a neat little algorithmic solution, but we want to know if our code is efficient. We need to know if it will scale to larger input values. We can find the answer to this question by writing benchmarks. Benchmarks are a feature we can use in conjunction with Cabal and Stack. They work a lot like test suites. But instead of proving the correctness of our code, they’ll show us how fast our code runs under various circumstances. We’ll use the library to do this. We’ll start by adding a section in our .cabal file for this benchmark: Criterion benchmark fences-benchmarks type: exitcode-stdio-1.0 hs-source-dirs: benchmark main-is: fences-benchmark.hs build-depends: base , FencesExample , criterion , random default-language: Haskell2010 Now we’ll make our file , make it a module and add a function. We’ll generate 6 lists, increasing in size by a factor of 10 each time. Then we’ll create a benchmark group and call the function on each situation. fences-benchmark.hs Main main bench module Main where import Criterionimport Criterion.Main (defaultMain)import System.Random import Lib main :: IO ()main = do [l1, l2, l3, l4, l5, l6] <- mapM randomList [1, 10, 100, 1000, 10000, 100000] defaultMain [ bgroup "fences tests" [ bench "Size 1 Test" $ whnf largestRectangle l1 , bench "Size 10 Test" $ whnf largestRectangle l2 , bench "Size 100 Test" $ whnf largestRectangle l3 , bench "Size 1000 Test" $ whnf largestRectangle l4 , bench "Size 10000 Test" $ whnf largestRectangle l5 , bench "Size 100000 Test" $ whnf largestRectangle l6 ] ] -- Generate a list of a particular sizerandomList :: Int -> IO FenceValuesrandomList n = FenceValues (sequence $ replicate n (randomRIO (1, 10000 :: Int))) We’d normally run these benchmarks with (or if you’re not using Stack). But we can also compile our code with the flag. This will automatically create a profiling report with more information about our code. Note using profiling requires re-compiling ALL the dependencies to use profiling as well. So you don't want to switch back and forth a lot. stack bench cabal bench --profile >> stack bench --profileBenchmark fences-benchmarks: RUNNING...benchmarking fences tests/Size 1 Testtime 47.79 ns (47.48 ns .. 48.10 ns) 1.000 R² (0.999 R² .. 1.000 R²)mean 47.78 ns (47.48 ns .. 48.24 ns)std dev 1.163 ns (817.2 ps .. 1.841 ns)variance introduced by outliers: 37% (moderately inflated) benchmarking fences tests/Size 10 Testtime 3.324 μs (3.297 μs .. 3.356 μs) 0.999 R² (0.999 R² .. 1.000 R²)mean 3.340 μs (3.312 μs .. 3.368 μs)std dev 98.52 ns (79.65 ns .. 127.2 ns)variance introduced by outliers: 38% (moderately inflated) benchmarking fences tests/Size 100 Testtime 107.3 μs (106.3 μs .. 108.2 μs) 0.999 R² (0.999 R² .. 0.999 R²)mean 107.2 μs (106.3 μs .. 108.4 μs)std dev 3.379 μs (2.692 μs .. 4.667 μs)variance introduced by outliers: 30% (moderately inflated) benchmarking fences tests/Size 1000 Testtime 8.724 ms (8.596 ms .. 8.865 ms) 0.998 R² (0.997 R² .. 0.999 R²)mean 8.638 ms (8.560 ms .. 8.723 ms)std dev 228.8 μs (193.6 μs .. 272.8 μs) benchmarking fences tests/Size 10000 Testtime 909.2 ms (899.3 ms .. 914.1 ms) 1.000 R² (1.000 R² .. 1.000 R²)mean 915.1 ms (914.6 ms .. 915.8 ms)std dev 620.1 μs (136.0 as .. 664.8 μs)variance introduced by outliers: 19% (moderately inflated) benchmarking fences tests/Size 100000 Testtime 103.9 s (91.11 s .. 117.3 s) 0.997 R² (0.997 R² .. 1.000 R²)mean 107.3 s (103.7 s .. 109.4 s)std dev 3.258 s (0.0 s .. 3.702 s)variance introduced by outliers: 19% (moderately inflated) Benchmark fences-benchmarks: FINISH So when we run this, we’ll find something…troubling. It takes a looong time to run the final benchmark on size 100000. On average, this case takes over 100 seconds…more than a minute and a half! We can further take note of how the average run time increases based on the size of the case. Let’s pare down the data a little bit: Size 1: 47.78 nsSize 10: 3.340 μs (increased ~70x)Size 100: 107.2 μs (increased ~32x)Size 1000: 8.638 ms (increased ~81x)Size 10000: 915.1 ms (increased ~106x)Size 100000: 107.3 s (increased ~117x) Each time we increase the size of the problem by a factor of 10, the time spent increased by a factor closer to 100! This suggests our run time is (check out if you are unfamiliar with Big-O notation). We’d like to do better. O(n^2) this guide Determining the Problem So we want to figure out why our code isn’t performing very well. Luckily, we already profiled our benchmark!. This outputs a specific file that we can look at, called . It has some very interesting results: fences-benchmark.prof COST CENTRE %time %allocminimumHeightIndexValue.valsInInterva 69.8 99.7valueAtIndex 29.3 0.0 We see that we have two big culprits taking a lot of time. First, there is our function that determines the minimum between a specific interval. The report is even more specific, calling out the specific offending part of the function. We spend a lot of time getting the different values for a specific interval. In second place, we have . This means we also spend a lot of time getting values out of our list. valueAtIndex First let’s be glad we’ve factored our code well. If we had written our entire solution in one big function, we wouldn’t have any leads here. This makes it much easier for us to analyze the problem. When examining the code, we see why both of these functions could produce behavior. O(n^2) Due to the number of recursive calls we make, we’ll call each of these functions times. Then when we call , we use the operator on our linked list. This takes time. Scanning the whole interval for the minimum height has the same effect. In the worst case, we have to look at every element in the list! I’m hand waving a bit here, but that is the basic result. When we call these pieces times, we get time total. O(n) valueAtIndex (!!) O(n) O(n) O(n) O(n^2) Cliff Hanger Ending We can actually solve this problem in time, a dramatic improvement over the current . But we’ll have to improve our data structures to accomplish this. First, we’ll store our values so that we can go from the index to the element in sub-linear time. This is easy. Second, we have to determine the index containing the minimum element within an arbitrary interval. This is a bit trickier to do in sub-linear time. We'll need a more advanced data structure. O(n log n) O(n^2) To find out how we solve these problems, you’ll have to wait for part 2 of this series! Come back next week to the ! Monday Morning Haskell blog As a reminder, we’ve just published a FREE . It’ll teach you the basics of laying out a project and running commands on it using the Stack tool. You should enroll in the to sign up! This is our first course of any kind, so we would love to get some feedback! Once you know about Stack, it’ll be a lot easier to try this problem out for yourself! mini-course on Stack Monday Morning Haskell Academy In addition to Stack, recursion also featured pretty heavily in our solution here. If you’ve done any amount of functional programming you’ve seen recursion in action. But if you want to solidify your knowledge, you should download our FREE ! It has two chapters worth of content on recursion and it has 10 practice problems you can work through! Recursion Workbook Never programmed in Haskell? No problem! You can download our and it’ll help you get Haskell installed. It’ll also point you in the direction of some great learning resources as well. Take a look! 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