I was trying to solve ITA Software's "Word Nubmers" puzzle using a brute force approach. It looks like my Haskell version is more than 10 times slower than a C#/C++ version.
The answer
Thanks to Bryan O'Sullivan's answer, I was able to "correct" my program to acceptable performance. You can read his code which is much cleaner than mine. I am going to outline the key points here.
Int
isInt64
on Linux GHC x64. Unless youunsafeCoerce
, you should just useInt
. This saves you from having tofromIntegral
. DoingInt64
on Windows 32-bit GHC is just darn slow, avoid it. (This is in fact not GHC's fault. As mentioned in my blog post below, 64 bit integers in 32-bit programs is slow in general (at least in Windows))-fllvm
or-fvia-C
for performance.- Prefer
quotRem
todivMod
,quotRem
already suffices. That gave me 20% speed up. - In general, prefer
Data.Vector
toData.Array
as an "array" - Use the wrapper-worker pattern liberally.
The above points were enough to give me about 100% boost over my original version.
In my blog post, I have detailed a step-by-step illustrated example of how I turned the original program to match Bryan's program. There are other points mentioned there as well.
The original question
(This may sound like a "could you do the work for me" post, but I argue that such a concrete example would be very instructive since profiling Haskell performance is often seen as a myth)
(As noted in the comments, I think I have misinterpreted the problem. But who cares, we can focus on performance in a different problem)
Here's a my version of a quick recap of the problem:
A wordNumber is defined as
wordNumber 1 = "one"
wordNumber 2 = "onetwo"
wordNumber 3 = "onethree"
wordNumber 15 = "onetwothreefourfivesixseveneightnineteneleventwelvethirteenfourteenfifteen"
...
Problem: Find the 51-billion-th letter of (wordNumber Infinity); assume that letter is found at 'wordNumber x', also find 'sum [1..x]'
From an imperative perspective, a naive algorithm would be to have 2 counters, one for sum of numbers and one for sum of lengths. Keep counting the length of each wordNumber and "break" to return the result.
The imperative brute-force approach is implemented in C# here: http://ideone.com/JjCb3. It takes about 1.5 minutes to find the answer on my computer. There is also an C++ implementation that runs in 45 seconds on my computer.
Then I implemented a brute-force Haskell version: http://ideone.com/ngfFq. It cannot finish the calculation in 5 minutes on my machine. (Irony: it's has more lines than the C# version)
Here is the -p
profile of the Haskell program: http://hpaste.org/49934
Question: How to make it perform comparatively to the C# version? Are there obvious mistakes I am making?
(Note: I am fully aware that brute-forcing it is not the correct solution to this problem. I am mainly interested in making the Haskell version perform comparatively to the C# version. Right now it is at least 5x slower so obviously I am missing something obvious)
(Note 2: It does not seem to be space leaking. The program runs with constant memory (about 2MB) on my computer)
(Note 3: I am compiling with `ghc -O2 WordNumber.hs)
To make the question more reader friendly, I include the "gist" of the two versions.
// C#
long sumNum = 0;
long sumLen = 0;
long target = 51000000000;
long i = 1;
for (; i < 999999999; i++)
{
// WordiLength(1) = 3 "one"
// WordiLength(101) = 13 "onehundredone"
long newLength = sumLen + WordiLength(i);
if (newLength >= target)
break;
sumNum += i;
sumLen = newLength;
}
Console.WriteLine(Wordify(i)[Convert.ToInt32(target - sumLen - 1)]);
-
-- Haskell
-- This has become totally ugly during my squeeze for
-- performance
-- Tail recursive
-- n-th number (51000000000 in our problem) -> accumulated result -> list of 'zipped' left to try
-- accumulated has the format (sum of numbers, current lengths of the whole chain, the current number)
solve :: Int64 -> (Int64, Int64, Int64) -> [(Int64, Int64)] -> (Int64, Int64, Int64)
solve !n !acc@(!sumNum, !sumLen, !curr) ((!num, !len):xs)
| sumLen' >= n = (sumNum', sumLen, num)
| otherwise = solve n (sumNum', sumLen', num) xs
where
sumNum' = sumNum + num
sumLen' = sumLen + len
-- wordLength 1 = 3 "one"
-- wordLength 101 = 13 "onehundredone"
wordLength :: Int64 -> Int64
-- wordLength = ...
solution :: Int64 -> (Int64, Char)
solution !x =
let (sumNum, sumLen, n) = solve x (0,0,1) (map (\n -> (n, wordLength n)) [1..])
in (sumNum, (wordify n) !! (fromIntegral $ x - sumLen - 1))
wordLength'
. It allocates memory for no apparent reason. I have no idea why, or how to rewrite it so it doesn't. I have tried (not too hard) but with no success so far. – n. 1.8e9-where's-my-share m.