I have taken Problem #12 from Project Euler as a programming exercise and to compare my (surely not optimal) implementations in C, Python, Erlang and Haskell. In order to get some higher execution times, I search for the first triangle number with more than 1000 divisors instead of 500 as stated in the original problem.
The result is the following:
C:
lorenzo@enzo:~/erlang$ gcc -lm -o euler12.bin euler12.c
lorenzo@enzo:~/erlang$ time ./euler12.bin
842161320
real 0m11.074s
user 0m11.070s
sys 0m0.000s
Python:
lorenzo@enzo:~/erlang$ time ./euler12.py
842161320
real 1m16.632s
user 1m16.370s
sys 0m0.250s
Python with PyPy:
lorenzo@enzo:~/Downloads/pypy-c-jit-43780-b590cf6de419-linux64/bin$ time ./pypy /home/lorenzo/erlang/euler12.py
842161320
real 0m13.082s
user 0m13.050s
sys 0m0.020s
Erlang:
lorenzo@enzo:~/erlang$ erlc euler12.erl
lorenzo@enzo:~/erlang$ time erl -s euler12 solve
Erlang R13B03 (erts-5.7.4) [source] [64-bit] [smp:4:4] [rq:4] [async-threads:0] [hipe] [kernel-poll:false]
Eshell V5.7.4 (abort with ^G)
1> 842161320
real 0m48.259s
user 0m48.070s
sys 0m0.020s
Haskell:
lorenzo@enzo:~/erlang$ ghc euler12.hs -o euler12.hsx
[1 of 1] Compiling Main ( euler12.hs, euler12.o )
Linking euler12.hsx ...
lorenzo@enzo:~/erlang$ time ./euler12.hsx
842161320
real 2m37.326s
user 2m37.240s
sys 0m0.080s
Summary:
- C: 100%
- Python: 692% (118% with PyPy)
- Erlang: 436% (135% thanks to RichardC)
- Haskell: 1421%
I suppose that C has a big advantage as it uses long for the calculations and not arbitrary length integers as the other three. Also it doesn’t need to load a runtime first (Do the others?).
Question 1:
Do Erlang, Python and Haskell lose speed due to using arbitrary length integers or don’t they as long as the values are less than MAXINT?
Question 2:
Why is Haskell so slow? Is there a compiler flag that turns off the brakes or is it my implementation? (The latter is quite probable as Haskell is a book with seven seals to me.)
Question 3:
Can you offer me some hints how to optimize these implementations without changing the way I determine the factors? Optimization in any way: nicer, faster, more "native" to the language.
EDIT:
Question 4:
Do my functional implementations permit LCO (last call optimization, a.k.a tail recursion elimination) and hence avoid adding unnecessary frames onto the call stack?
I really tried to implement the same algorithm as similar as possible in the four languages, although I have to admit that my Haskell and Erlang knowledge is very limited.
Source codes used:
#include <stdio.h>
#include <math.h>
int factorCount (long n)
{
double square = sqrt (n);
int isquare = (int) square;
int count = isquare == square ? -1 : 0;
long candidate;
for (candidate = 1; candidate <= isquare; candidate ++)
if (0 == n % candidate) count += 2;
return count;
}
int main ()
{
long triangle = 1;
int index = 1;
while (factorCount (triangle) < 1001)
{
index ++;
triangle += index;
}
printf ("%ld\n", triangle);
}
#! /usr/bin/env python3.2
import math
def factorCount (n):
square = math.sqrt (n)
isquare = int (square)
count = -1 if isquare == square else 0
for candidate in range (1, isquare + 1):
if not n % candidate: count += 2
return count
triangle = 1
index = 1
while factorCount (triangle) < 1001:
index += 1
triangle += index
print (triangle)
-module (euler12).
-compile (export_all).
factorCount (Number) -> factorCount (Number, math:sqrt (Number), 1, 0).
factorCount (_, Sqrt, Candidate, Count) when Candidate > Sqrt -> Count;
factorCount (_, Sqrt, Candidate, Count) when Candidate == Sqrt -> Count + 1;
factorCount (Number, Sqrt, Candidate, Count) ->
case Number rem Candidate of
0 -> factorCount (Number, Sqrt, Candidate + 1, Count + 2);
_ -> factorCount (Number, Sqrt, Candidate + 1, Count)
end.
nextTriangle (Index, Triangle) ->
Count = factorCount (Triangle),
if
Count > 1000 -> Triangle;
true -> nextTriangle (Index + 1, Triangle + Index + 1)
end.
solve () ->
io:format ("~p~n", [nextTriangle (1, 1) ] ),
halt (0).
factorCount number = factorCount' number isquare 1 0 - (fromEnum $ square == fromIntegral isquare)
where square = sqrt $ fromIntegral number
isquare = floor square
factorCount' number sqrt candidate count
| fromIntegral candidate > sqrt = count
| number `mod` candidate == 0 = factorCount' number sqrt (candidate + 1) (count + 2)
| otherwise = factorCount' number sqrt (candidate + 1) count
nextTriangle index triangle
| factorCount triangle > 1000 = triangle
| otherwise = nextTriangle (index + 1) (triangle + index + 1)
main = print $ nextTriangle 1 1
Using
GHC 7.0.3,gcc 4.4.6,Linux 2.6.29on an x86_64 Core2 Duo (2.5GHz) machine, compiling usingghc -O2 -fllvm -fforce-recompfor Haskell andgcc -O3 -lmfor C.-O3)-O2flag)factorCount'code isn’t explicitly typed and defaulting toInteger(thanks to Daniel for correcting my misdiagnosis here!). Giving an explicit type signature (which is standard practice anyway) usingIntand the time changes to 11.1 secondsfactorCount'you have needlessly calledfromIntegral. A fix results in no change though (the compiler is smart, lucky for you).modwhereremis faster and sufficient. This changes the time to 8.5 seconds.factorCount'is constantly applying two extra arguments that never change (number,sqrt). A worker/wrapper transformation gives us:That’s right, 7.95 seconds. Consistently half a second faster than the C solution. Without the
-fllvmflag I’m still getting8.182 seconds, so the NCG backend is doing well in this case too.Conclusion: Haskell is awesome.
Resulting Code
EDIT: So now that we’ve explored that, lets address the questions
In Haskell, using
Integeris slower thanIntbut how much slower depends on the computations performed. Luckily (for 64 bit machines)Intis sufficient. For portability sake you should probably rewrite my code to useInt64orWord64(C isn’t the only language with along).That was what I answered above. The answer was
-O2remnotmod(a frequently forgotten optimization) andYes, that wasn’t the issue. Good work and glad you considered this.