# How slow is Python really? (Or how fast is your language?)

I have this code which I have written in Python/NumPy

from __future__ import division
import numpy as np
import itertools

n = 6
iters = 1000
firstzero = 0
bothzero = 0
""" The next line iterates over arrays of length n+1 which contain only -1s and 1s """
for S in itertools.product([-1, 1], repeat=n+1):
"""For i from 0 to iters -1 """
for i in xrange(iters):
""" Choose a random array of length n.
Prob 1/4 of being -1, prob 1/4 of being 1 and prob 1/2 of being 0. """
F = np.random.choice(np.array([-1, 0, 0, 1], dtype=np.int8), size=n)
"""The next loop just makes sure that F is not all zeros."""
while np.all(F == 0):
F = np.random.choice(np.array([-1, 0, 0, 1], dtype=np.int8), size=n)
"""np.convolve(F, S, 'valid') computes two inner products between
F and the two successive windows of S of length n."""
FS = np.convolve(F, S, 'valid')
if FS[0] == 0:
firstzero += 1
if np.all(FS == 0):
bothzero += 1

print("firstzero: %i" % firstzero)
print("bothzero: %i" % bothzero)


It is counting the number of times the convolution of two random arrays, one which is one longer than the other, with a particular probability distribution, has a 0 at the first position or a 0 in both positions.

I had a bet with a friend who says Python is a terrible language to write code in that needs to be fast. It takes 9s on my computer. He says it could be made 100 times faster if written in a "proper language".

The challenge is to see if this code can indeed by made 100 times faster in any language of your choice. I will test your code and the fastest one week from now will win. If anyone gets below 0.09s then they automatically win and I lose.

Status

• Python. 30 times speed up by Alistair Buxon! Although not the fastest solution it is in fact my favourite.
• Octave. 100 times speed up by @Thethos.
• Rust. 500 times speed up by @dbaupp.
• C++. 570 times speed up by Guy Sirton.
• C. 727 times speed up by @ace.
• C++. Unbelievably fast by @Stefan.

The fastest solutions are now too fast to sensibly time. I have therefore increased n to 10 and set iters = 100000 to compare the best ones. Under this measure the fastest are.

• C. 7.5s by @ace.
• C++. 1s by @Stefan.

My Machine The timings will be run on my machine. This is a standard ubuntu install on an AMD FX-8350 Eight-Core Processor. This also means I need to be able to run your code.

Follow up posted As this competition was rather too easy to get a x100 speedup, I have posted a followup for those who want to exercise their speed guru expertise. See How Slow Is Python Really (Part II)?

# C++ bit magic

## 0.84ms with simple RNG, 1.67ms with c++11 std::knuth

0.16ms with slight algorithmic modification (see edit below)

The python implementation runs in 7.97 seconds on my rig. So this is 9488 to 4772 times faster depending on what RNG you choose.

#include <iostream>
#include <bitset>
#include <random>
#include <chrono>
#include <stdint.h>
#include <cassert>
#include <tuple>

#if 0
// C++11 random
std::random_device rd;
std::knuth_b gen(rd());

uint32_t genRandom()
{
return gen();
}
#else

uint32_t genRandom()
{
static uint32_t seed = std::random_device()();
auto oldSeed = seed;
seed = seed*1664525UL + 1013904223UL; // numerical recipes, 32 bit
return oldSeed;
}
#endif

#ifdef _MSC_VER
uint32_t popcnt( uint32_t x ){ return _mm_popcnt_u32(x); }
#else
uint32_t popcnt( uint32_t x ){ return __builtin_popcount(x); }
#endif

std::pair<unsigned, unsigned> convolve()
{
const uint32_t n = 6;
const uint32_t iters = 1000;
unsigned firstZero = 0;
unsigned bothZero = 0;

uint32_t S = (1 << (n+1));
// generate all possible N+1 bit strings
// 1 = +1
// 0 = -1
while ( S-- )
{
uint32_t s1 = S % ( 1 << n );
uint32_t s2 = (S >> 1) % ( 1 << n );
static_assert( n < 16, "packing of F fails when n > 16.");

for( unsigned i = 0; i < iters; i++ )
{
// generate random bit mess
uint32_t F;
do {
} while ( 0 == ((F % (1 << n)) ^ (F >> 16 )) );

// Assume F is an array with interleaved elements such that F[0] || F[16] is one element
// here MSB(F) & ~LSB(F) returns 1 for all elements that are positive
// and  ~MSB(F) & LSB(F) returns 1 for all elements that are negative
// this results in the distribution ( -1, 0, 0, 1 )
// to ease calculations we generate r = LSB(F) and l = MSB(F)

uint32_t r = F % ( 1 << n );
// modulo is required because the behaviour of the leftmost bit is implementation defined
uint32_t l = ( F >> 16 ) % ( 1 << n );

uint32_t posBits = l & ~r;
uint32_t negBits = ~l & r;
assert( (posBits & negBits) == 0 );

// calculate which bits in the expression S * F evaluate to +1
unsigned firstPosBits = ((s1 & posBits) | (~s1 & negBits));
// idem for -1
unsigned firstNegBits = ((~s1 & posBits) | (s1 & negBits));

if ( popcnt( firstPosBits ) == popcnt( firstNegBits ) )
{
firstZero++;

unsigned secondPosBits = ((s2 & posBits) | (~s2 & negBits));
unsigned secondNegBits = ((~s2 & posBits) | (s2 & negBits));

if ( popcnt( secondPosBits ) == popcnt( secondNegBits ) )
{
bothZero++;
}
}
}
}

return std::make_pair(firstZero, bothZero);
}

int main()
{
typedef std::chrono::high_resolution_clock clock;
int rounds = 1000;
std::vector< std::pair<unsigned, unsigned> > out(rounds);

// do 100 rounds to get the cpu up to speed..
for( int i = 0; i < 10000; i++ )
{
convolve();
}

auto start = clock::now();

for( int i = 0; i < rounds; i++ )
{
out[i] = convolve();
}

auto end = clock::now();
double seconds = std::chrono::duration_cast< std::chrono::microseconds >( end - start ).count() / 1000000.0;

#if 0
for( auto pair : out )
std::cout << pair.first << ", " << pair.second << std::endl;
#endif

std::cout << seconds/rounds*1000 << " msec/round" << std::endl;

return 0;
}


Compile in 64-bit for extra registers. When using the simple random generator the loops in convolve() run without any memory access, all variables are stored in the registers.

How it works: rather than storing S and F as in-memory arrays, it is stored as bits in an uint32_t.
For S, the n least significant bits are used where an set bit denotes an +1 and an unset bit denotes an -1.
F requires at least 2 bits to create an distribution of [-1, 0, 0, 1]. This is done by generating random bits and examining the 16 least significant (called r) and 16 most significant bits (called l). If l & ~r we assume that F is +1, if ~l & r we assume that F is -1. Otherwise F is 0. This generates the distribution we're looking for.

Now we have S, posBits with an set bit on every location where F == 1 and negBits with an set bit on every location where F == -1.

We can prove that F * S (where * denotes multiplication) evaluates to +1 under the condition (S & posBits) | (~S & negBits). We can also generate similar logic for all cases where F * S evaluates to -1. And finally, we know that sum(F * S) evaluates to 0 if and only if there is an equal amount of -1's and +1's in the result. This is very easy to calculate by simply comparing the number of +1 bits and -1 bits.

This implementation uses 32 bit ints, and the maximum n accepted is 16. It is possible to scale the implementation to 31 bits by modifying the random generate code, and to 63 bits by using uint64_t instead of uint32_t.

## edit

The folowing convolve function:

std::pair<unsigned, unsigned> convolve()
{
const uint32_t n = 6;
const uint32_t iters = 1000;
unsigned firstZero = 0;
unsigned bothZero = 0;
static_assert( n < 16, "packing of F fails when n > 16.");

for( unsigned i = 0; i < iters; i++ )
{
// generate random bit mess
uint32_t F;
do {
} while ( 0 == ((F % (1 << n)) ^ (F >> 16 )) );

// Assume F is an array with interleaved elements such that F[0] || F[16] is one element
// here MSB(F) & ~LSB(F) returns 1 for all elements that are positive
// and  ~MSB(F) & LSB(F) returns 1 for all elements that are negative
// this results in the distribution ( -1, 0, 0, 1 )
// to ease calculations we generate r = LSB(F) and l = MSB(F)

uint32_t r = F % ( 1 << n );
// modulo is required because the behaviour of the leftmost bit is implementation defined
uint32_t l = ( F >> 16 ) % ( 1 << n );

uint32_t posBits = l & ~r;
uint32_t negBits = ~l & r;
assert( (posBits & negBits) == 0 );

uint32_t mask = posBits | negBits;
uint32_t totalBits = popcnt( mask );
// if the amount of -1 and +1's is uneven, sum(S*F) cannot possibly evaluate to 0
if ( totalBits & 1 )
continue;

uint32_t adjF = posBits & ~negBits;
uint32_t desiredBits = totalBits / 2;

uint32_t S = (1 << (n+1));
// generate all possible N+1 bit strings
// 1 = +1
// 0 = -1
while ( S-- )
{
// calculate which bits in the expression S * F evaluate to +1
auto secondBits = (S & ( mask << 1 ) ) ^ ( adjF << 1 );

bool a = desiredBits == popcnt( firstBits );
bool b = desiredBits == popcnt( secondBits );
firstZero += a;
bothZero += a & b;
}
}

return std::make_pair(firstZero, bothZero);
}


cuts the runtime to 0.160-0.161ms. Manual loop unroll (not pictured above) makes that 0.150. The less trivial n=10, iter=100000 case runs under 250ms. I'm sure i can get it under 50ms by leveraging additional cores but that's too easy.

This is done by making the inner loop branch free and swapping the F and S loop.
If bothZero is not required i can cut down the run time to 0.02ms by sparsely looping over all possible S arrays.

• Could you provide a gcc friendly version and also what your command line would be please? I am not sure I can test it currently. – user9206 May 1 '14 at 17:49
• I know nothing about this but google tells me that __builtin_popcount might be a replacement for _mm_popcnt_u32() . – user9206 May 1 '14 at 17:58
• Code updated, uses #ifdef switch to select the correct popcnt command. It compiles with -std=c++0x -mpopcnt -O2 and takes 1.01ms to run in 32 bit mode (i don't have a 64-bit GCC version at hand). – Stefan May 1 '14 at 18:06
• Could you make it print the output? I am not sure if it is actually doing anything currently :) – user9206 May 1 '14 at 18:26
• You are clearly a wizard. + 1 – BurntPizza May 2 '14 at 15:33

# Fortran 90+: 0.029 s0.003 s0.022 s 0.010 s

Damn straight you lost your bet! Not a drop of parallelization here too, just straight Fortran 90+.

EDIT I've taken Guy Sirton's algorithm for permuting the array S (good find :D). I apparently also had the -g -traceback compiler flags active which were slowing this code down to about 0.017s. Currently, I am compiling this as

ifort -fast -o convolve convolve_random_arrays.f90


For those who don't have ifort, you can use

gfortran -O3 -ffast-math -o convolve convolve_random_arrays.f90


EDIT 2: The decrease in run-time is because I was doing something wrong previously and got an incorrect answer. Doing it the right way is apparently slower. I still can't believe that C++ is faster than mine, so I'm probably going to spend some time this week trying to tweak the crap out of this to speed it up.

EDIT 3: By simply changing the RNG section using one based on BSD's RNG (as suggested by Sampo Smolander) and eliminating the constant divide by m1, I cut the run-time to the same as the C++ answer by Guy Sirton. Using static arrays (as suggested by Sharpie) drops the run-time to under the C++ run-time! Yay Fortran! :D

EDIT 4 Apparently this doesn't compile (with gfortran) and run correctly (incorrect values) because the integers are overstepping their limits. I've made corrections to ensure it works, but this requires one to have either ifort 11+ or gfortran 4.7+ (or another compiler that allows iso_fortran_env and the F2008 int64 kind).

Here's the code:

program convolve_random_arrays
use iso_fortran_env
implicit none
integer(int64), parameter :: a1 = 1103515245
integer(int64), parameter :: c1 = 12345
integer(int64), parameter :: m1 = 2147483648
real, parameter ::    mi = 4.656612873e-10 ! 1/m1
integer, parameter :: n = 6
integer :: p, pmax, iters, i, nil(0:1), seed
!integer, allocatable ::  F(:), S(:), FS(:)
integer :: F(n), S(n+1), FS(2)

!n = 6
!allocate(F(n), S(n+1), FS(2))
iters = 1000
nil = 0

!call init_random_seed()

S = -1
pmax = 2**(n+1)
do p=1,pmax
do i=1,iters
F = rand_int_array(n)
if(all(F==0)) then
do while(all(F==0))
F = rand_int_array(n)
enddo
endif

FS = convolve(F,S)

if(FS(1) == 0) then
nil(0) = nil(0) + 1
if(FS(2) == 0) nil(1) = nil(1) + 1
endif

enddo
call permute(S)
enddo

print *,"first zero:",nil(0)
print *," both zero:",nil(1)

contains
pure function convolve(x, h) result(y)
!x is the signal array
!h is the noise/impulse array
integer, dimension(:), intent(in) :: x, h
integer, dimension(abs(size(x)-size(h))+1) :: y
integer:: i, j, r
y(1) = dot_product(x,h(1:n-1))
y(2) = dot_product(x,h(2:n  ))
end function convolve

pure subroutine permute(x)
integer, intent(inout) :: x(:)
integer :: i

do i=1,size(x)
if(x(i)==-1) then
x(i) = 1
return
endif
x(i) = -1
enddo
end subroutine permute

function rand_int_array(i) result(x)
integer, intent(in) :: i
integer :: x(i), j
real :: y
do j=1,i
y = bsd_rng()
if(y <= 0.25) then
x(j) = -1
else if (y >= 0.75) then
x(j) = +1
else
x(j) = 0
endif
enddo
end function rand_int_array

function bsd_rng() result(x)
real :: x
integer(int64) :: b=3141592653
b = mod(a1*b + c1, m1)
x = real(b)*mi
end function bsd_rng
end program convolve_random_arrays


I suppose the question now is will you stop using slow-as-molasses Python and use fast-as-electrons-can-move Fortran ;).

• Wouldn't the case statement be faster than a generator function anyway? Unless you're expecting some kind of branch-prediction/cache-line/etc speedup? – Stop Harming Monica Apr 26 '14 at 20:48
• Speed should be compared on the same machine. What runtime did you get for the OP's code? – nbubis Apr 26 '14 at 21:59
• The C++ answer implements its own, very lightweight random number generator. Your answer used the default that comes with the compiler, which could be slower? – Sampo Smolander Apr 30 '14 at 14:26
• Also, the C++ example appears to be using statically allocated arrays. Try using fixed-length arrays that are set at compile time and see if it shaves any time off. – Sharpie Apr 30 '14 at 14:45
• @KyleKanos @Lembik the problem is that the integer assignment in fortran is not using implicitly the int64 specification, hence the numbers are int32 before any conversion is made. The code should be: integer(int64) :: b = 3141592653_int64 for all int64. This is part of the fortran standard and is expected by the programmer in a type-declared programming language. (notice that default settings of course can override this) – zeroth May 1 '14 at 10:35

Python 2.7 - 0.882s 0.283s

(OP's original: 6.404s)

Edit: Steven Rumbalski's optimization by precomputing F values. With this optimization cpython beats pypy's 0.365s.

import itertools
import operator
import random

n=6
iters = 1000
firstzero = 0
bothzero = 0

choicesF = filter(any, itertools.product([-1, 0, 0, 1], repeat=n))

for S in itertools.product([-1,1], repeat = n+1):
for i in xrange(iters):
F = random.choice(choicesF)
if not sum(map(operator.mul, F, S[:-1])):
firstzero += 1
if not sum(map(operator.mul, F, S[1:])):
bothzero += 1

print "firstzero", firstzero
print "bothzero", bothzero


OP's original code uses such tiny arrays there is no benefit to using Numpy, as this pure python implementation demonstrates. But see also this numpy implementation which is three times faster again than my code.

I also optimize by skipping the rest of the convolution if the first result isn't zero.

• With pypy this runs in about 0.5 seconds. – Alistair Buxton Apr 27 '14 at 12:41
• You get a much more convincing speedup if you set n = 10. I get 19s versus 4.6s for cpython versus pypy. – user9206 Apr 27 '14 at 16:34
• Another optimization would be to precompute the possiblities for F because there are only 4032 of them. Define choicesF = filter(any, itertools.product([-1, 0, 0, 1], repeat=n)) outside of the loops. Then in the innerloop define F = random.choice(choicesF). I get a 3x speedup with such an approach. – Steven Rumbalski Apr 28 '14 at 13:36
• How about compiling this in Cython? Then adding a few tactful static types? – Thane Brimhall Apr 28 '14 at 20:10
• Put everything in a function and call it at the end. That localizes the names, which also makes the optimzation proposed by @riffraff work. Also, move the creation of range(iters) out of the loop. Altogether, I get a speedup of about 7% over your very nice answer. – Reinstate Monica May 2 '14 at 22:27

# Rust: 0.011s

### Original Python: 8.3

A straight translation of the original Python.

extern crate rand;

use rand::Rng;

static N: uint = 6;
static ITERS: uint = 1000;

fn convolve<T: Num>(into: &mut [T], a: &[T], b: &[T]) {
// we want a to be the longest array
if a.len() < b.len() {
convolve(into, b, a);
return
}

assert_eq!(into.len(), a.len() - b.len() + 1);

for (n,place) in into.mut_iter().enumerate() {
for (x, y) in a.slice_from(n).iter().zip(b.iter()) {
*place = *place + *x * *y
}
}
}

fn main() {
let mut first_zero = 0;
let mut both_zero = 0;
let mut rng = rand::XorShiftRng::new().unwrap();

for s in PlusMinus::new() {
for _ in range(0, ITERS) {
let mut f = [0, .. N];
while f.iter().all(|x| *x == 0) {
for p in f.mut_iter() {
match rng.gen::<u32>() % 4 {
0 => *p = -1,
1 | 2 => *p = 0,
_ => *p = 1
}
}
}

let mut fs = [0, .. 2];
convolve(fs, s, f);

if fs[0] == 0 { first_zero += 1 }
if fs.iter().all(|&x| x == 0) { both_zero += 1 }
}
}

println!("{}\n{}", first_zero, both_zero);
}

/// An iterator over [+-]1 arrays of the appropriate length
struct PlusMinus {
done: bool,
current: [i32, .. N + 1]
}
impl PlusMinus {
fn new() -> PlusMinus {
PlusMinus { done: false, current: [-1, .. N + 1] }
}
}

impl Iterator<[i32, .. N + 1]> for PlusMinus {
fn next(&mut self) -> Option<[i32, .. N+1]> {
if self.done {
return None
}

let ret = self.current;

// a binary "adder", that just adds one to a bit vector (where
// -1 is the zero, and 1 is the one).
for (i, place) in self.current.mut_iter().enumerate() {
*place = -*place;
if *place == 1 {
break
} else if i == N {
// we've wrapped, so we want to stop after this one
self.done = true
}
}

Some(ret)
}
}

• Compiled with --opt-level=3
• My rust compiler is a recent nightly: (rustc 0.11-pre-nightly (eea4909 2014-04-24 23:41:15 -0700) to be precise)
• I got it to compile using the nightly version of rust. However I think the code is wrong. The output should be something close to firstzero 27215 bothzero 12086. Instead it gives 27367 6481 – user9206 Apr 27 '14 at 18:55
• @Lembik, whoops, got my as and bs mixed up in the convolution; fixed (doesn't change the runtime noticably). – huon Apr 28 '14 at 1:24
• It's a very nice demonstration of rust's speed. – user9206 May 1 '14 at 6:57

C++ (VS 2012) - 0.026s 0.015s

Python 2.7.6/Numpy 1.8.1 - 12s

Speedup ~x800.

The gap would be a lot smaller if the convolved arrays were very large...

#include <vector>
#include <iostream>
#include <ctime>

using namespace std;

static unsigned int seed = 35;

int my_random()
{
seed = seed*1664525UL + 1013904223UL; // numerical recipes, 32 bit

switch((seed>>30) & 3)
{
case 0: return 0;
case 1: return -1;
case 2: return 1;
case 3: return 0;
}
return 0;
}

bool allzero(const vector<int>& T)
{
for(auto x : T)
{
if(x!=0)
{
return false;
}
}
return true;
}

void convolve(vector<int>& out, const vector<int>& v1, const vector<int>& v2)
{
for(size_t i = 0; i<out.size(); ++i)
{
int result = 0;
for(size_t j = 0; j<v2.size(); ++j)
{
result += v1[i+j]*v2[j];
}
out[i] = result;
}
}

{
for(auto &x : v)
{
if(x==-1)
{
x = 1;
return;
}
x = -1;
}
}

void convolve_random_arrays(void)
{
const size_t n = 6;
const int two_to_n_plus_one = 128;
const int iters = 1000;
int bothzero = 0;
int firstzero = 0;

vector<int> S(n+1);
vector<int> F(n);
vector<int> FS(2);

time_t current_time;
time(&current_time);
seed = current_time;

for(auto &x : S)
{
x = -1;
}
for(int i=0; i<two_to_n_plus_one; ++i)
{
for(int j=0; j<iters; ++j)
{
do
{
for(auto &x : F)
{
x = my_random();
}
} while(allzero(F));
convolve(FS, S, F);
if(FS[0] == 0)
{
firstzero++;
if(FS[1] == 0)
{
bothzero++;
}
}
}
}
cout << firstzero << endl; // This output can slow things down
cout << bothzero << endl; // comment out for timing the algorithm
}


A few notes:

• The random function is being called in the loop so I went for a very light weight linear congruential generator (but generously looked at the MSBs).
• This is really just the starting point for an optimized solution.
• Didn't take that long to write...
• I iterate through all the values of S taking S[0] to be the "least significant" digit.

Add this main function for a self contained example:

int main(int argc, char** argv)
{
for(int i=0; i<1000; ++i) // run 1000 times for stop-watch
{
convolve_random_arrays();
}
}

• Indeed. The tiny size of the arrays in OP's code means using numpy is actually an order of magnitude slower than straight python. – Alistair Buxton Apr 27 '14 at 4:29
• Now x800 is what I am talking about! – user9206 Apr 27 '14 at 5:53
• Very nice! I've increased the speed-up on my code because of your advance function, so my code is now faster than yours :P (but very good competition!) – Kyle Kanos Apr 27 '14 at 15:18
• @lembik yes as Mat says. You need C++11 supprt and a main function. Let me know if you need more help to get this to run... – Guy Sirton Apr 27 '14 at 19:54
• I just tested this and could shave of another 20% by using plain arrays instead of std::vector.. – PlasmaHH Apr 28 '14 at 13:00

## C

Takes 0.015s on my machine, with OP's original code taking ~ 7.7s. Tried to optimize by generating the random array and convolving in the same loop, but it doesn't seem to make a lot of difference.

The first array is generated by taking an integer, write it out in binary, and change all 1 to -1 and all 0 to 1. The rest should be very straightforward.

Edit: instead of having n as an int, now we have n as a macro-defined constant, so we can use int arr[n]; instead of malloc.

Edit2: Instead of built-in rand() function, this now implements an xorshift PRNG. Also, a lot of conditional statements are removed when generating the random array.

Compile instructions:

gcc -O3 -march=native -fwhole-program -fstrict-aliasing -ftree-vectorize -Wall ./test.c -o ./test


Code:

#include <stdio.h>
#include <time.h>

#define n (6)
#define iters (1000)
unsigned int x,y=34353,z=57768,w=1564; //PRNG seeds

/* xorshift PRNG
* Taken from https://en.wikipedia.org/wiki/Xorshift#Example_implementation
* Used under CC-By-SA */
int myRand() {
unsigned int t;
t = x ^ (x << 11);
x = y; y = z; z = w;
return w = w ^ (w >> 19) ^ t ^ (t >> 8);
}

int main() {
int firstzero=0, bothzero=0;
int arr[n+1];
unsigned int i, j;
x=(int)time(NULL);

for(i=0; i< 1<<(n+1) ; i++) {
unsigned int tmp=i;
for(j=0; j<n+1; j++) {
arr[j]=(tmp&1)*(-2)+1;
tmp>>=1;
}
for(j=0; j<iters; j++) {
int randArr[n];
unsigned int k, flag=0;
int first=0, second=0;
do {
for(k=0; k<n; k++) {
randArr[k]=(1-(myRand()&3))%2;
flag+=(randArr[k]&1);
first+=arr[k]*randArr[k];
second+=arr[k+1]*randArr[k];
}
} while(!flag);
firstzero+=(!first);
bothzero+=(!first&&!second);
}
}
printf("firstzero %d\nbothzero %d\n", firstzero, bothzero);
return 0;
}

• I tested this. It is very fast (try n = 10) and gives correct looking output. Thank you. – user9206 Apr 27 '14 at 16:43
• This implementation doesn't follow the original because if the random vector is all zeros only the last element would be re-generated. In the original the entire vector would be. You need to enclose that loop in do{}while(!flag) or something to that effect. I don't expect it will change the run-time much (may make it faster). – Guy Sirton Apr 27 '14 at 22:28
• @Guy Sirton Notice that before the continue; statement I assigned -1 to k, so k will loop from 0 again. – ace_HongKongIndependence Apr 27 '14 at 22:55
• @ace ah! you're right. I was scanning too quickly and it looked like that was -= rather than =- :-) A while loop would be more readable. – Guy Sirton Apr 27 '14 at 23:01

# J

I don't expect to beat out any compiled languages, and something tells me it'd take a miraculous machine to get less than 0.09 s with this, but I'd like to submit this J anyway, because it's pretty slick.

NB. constants
num =: 6
iters =: 1000

NB. convolve
NB. take the multiplication table                */
NB. then sum along the NE-SW diagonals           +//.
NB. and keep the longest ones                    #~ [: (= >./) #/.
NB. operate on rows of higher dimensional lists  " 1
conv =: (+//. #~ [: (= >./) #/.) @: (*/) " 1

NB. main program
S  =: > , { (num+1) # < _1 1                NB. all {-1,1}^(num+1)
F  =: (3&= - 0&=) (iters , num) ?@$4 NB. iters random arrays of length num FS =: ,/ S conv/ F NB. make a convolution table FB =: +/ ({. , *./)"1 ] 0 = FS NB. first and both zero ('first zero ',:'both zero ') ,. ":"0 FB NB. output results  This takes about 0.5 s on a laptop from the previous decade, only about 20x as fast as the Python in the answer. Most of the time is spent in conv because we write it lazily (we compute the entire convolution) and in full generality. Since we know things about S and F, we can speed things up by making specific optimizations for this program. The best I've been able to come up with is conv =: ((num, num+1) { +//.)@:(*/)"1—select specifically the two numbers that correspond from the diagonal sums to the longest elements of the convolution—which approximately halves the time. • J is always worth submitting, man :) – Vitaly Dyatlov Apr 30 '14 at 17:50 Perl - 9.3X faster...830% improvement On my ancient netbook, the OP's code takes 53 seconds to run; Alistair Buxton's version takes about 6.5 seconds, and the following Perl version takes about 5.7 seconds. use v5.10; use strict; use warnings; use Algorithm::Combinatorics qw( variations_with_repetition ); use List::Util qw( any sum ); use List::MoreUtils qw( pairwise ); my$n         = 6;
my $iters = 1000; my$firstzero = 0;
my $bothzero = 0; my$variations = variations_with_repetition([-1, 1], $n+1); while (my$S = $variations->next) { for my$i (1 .. $iters) { my @F; until (@F and any {$_ } @F)
{
@F = map +((-1,0,0,1)[rand 4]), 1..$n; } # The pairwise function doesn't accept array slices, # so need to copy into a temp array @S0 my @S0 = @$S[0..$n-1]; unless (sum pairwise {$a * $b } @F, @S0) {$firstzero++;
my @S1 = @$S[1..$n];  # copy again :-(
$bothzero++ unless sum pairwise {$a * $b } @F, @S1; } } } say "firstzero ",$firstzero;
say "bothzero ", $bothzero;  Python 2.7 - numpy 1.8.1 with mkl bindings - 0.086s (OP's original: 6.404s) (Buxton's pure python: 0.270s) import numpy as np import itertools n=6 iters = 1000 #Pack all of the Ses into a single array S = np.array( list(itertools.product([-1,1], repeat=n+1)) ) # Create a whole array of test arrays, oversample a bit to ensure we # have at least (iters) of them F = np.random.rand(int(iters*1.1),n) F = ( F < 0.25 )*-1 + ( F > 0.75 )*1 goodrows = (np.abs(F).sum(1)!=0) assert goodrows.sum() > iters, "Got very unlucky" # get 1000 cases that aren't all zero F = F[goodrows][:iters] # Do the convolution explicitly for the two # slots, but on all of the Ses and Fes at the # same time firstzeros = (F[:,None,:]*S[None,:,:-1]).sum(-1)==0 secondzeros = (F[:,None,:]*S[None,:,1:]).sum(-1)==0 firstzero_count = firstzeros.sum() bothzero_count = (firstzeros * secondzeros).sum() print "firstzero", firstzero_count print "bothzero", bothzero_count  As Buxton points out, OP's original code uses such tiny arrays there is no benefit to using Numpy. This implementation leverages numpy by doing all of the F and S cases at once in an array oriented way. This combined with mkl bindings for python leads to a very fast implementation. Note also that just loading the libraries and starting the interpreter takes 0.076s so the actual computation is taking ~ 0.01 seconds, similar to the C++ solution. • What are mkl bindings and how do I get them on ubuntu? – user9206 Apr 30 '14 at 17:53 • Running python -c "import numpy; numpy.show_config()" will show you if your version of numpy is compiled against blas/atlas/mkl, etc. ATLAS is a free accelerated math package that numpy can be linked against, Intel MKL you usually have to pay for (unless you're an academic) and can be linked to numpy/scipy. – alemi Apr 30 '14 at 18:57 • For an easy way, use the anaconda python distribution and use the accelerate package. Or use the enthought distribution. – alemi Apr 30 '14 at 19:00 • If you're on windows, just download numpy from here. Pre-compiled numpy installers linked against MKL. – Fake Name May 2 '14 at 10:20 # MATLAB 0.024s Computer 1 • Original Code: ~ 3.3 s • Alistar Buxton's Code: ~ 0.51 s • Alistar Buxton's new Code: ~0.25 s • Matlab Code: ~ 0.024 s (Matlab already running) Computer 2 • Original Code: ~ 6.66 s • Alistar Buxton's Code: ~ 0.64 s • Alistar Buxton's new Code: ? • Matlab: ~ 0.07 s (Matlab already running) • Octave: ~ 0.07 s I decided to give the oh so slow Matlab a try. If you know how, you can get rid of most of the loops (in Matlab), which makes it quite fast. However, the memory requirements are higher than for looped solutions but this will not be an issue if you don't have very large arrays... function call_convolve_random_arrays tic convolve_random_arrays toc end function convolve_random_arrays n = 6; iters = 1000; firstzero = 0; bothzero = 0; rnd = [-1, 0, 0, 1]; S = -1 *ones(1, n + 1); IDX1 = 1:n; IDX2 = IDX1 + 1; for i = 1:2^(n + 1) F = rnd(randi(4, [iters, n])); sel = ~any(F,2); while any(sel) F(sel, :) = rnd(randi(4, [sum(sel), n])); sel = ~any(F,2); end sum1 = F * S(IDX1)'; sel = sum1 == 0; firstzero = firstzero + sum(sel); sum2 = F(sel, :) * S(IDX2)'; sel = sum2 == 0; bothzero = bothzero + sum(sel); S = permute(S); end fprintf('firstzero %i \nbothzero %i \n', firstzero, bothzero); end function x = permute(x) for i=1:length(x) if(x(i)==-1) x(i) = 1; return end x(i) = -1; end end  Here is what I do: • use Kyle Kanos function to permute through S • calculate all n * iters random numbers at once • map 1 to 4 to [-1 0 0 1] • use Matrix multiplication (elementwise sum(F * S(1:5)) is equal to matrix multiplication of F * S(1:5)' • for bothzero: only calculate members that fullfill the first condition I assume you don't have matlab, which is too bad as I really would have liked to see how it compares... (The function can be slower the first time you run it.) • Well I have octave if you can make it work for that...? – user9206 Apr 28 '14 at 12:12 • I can give it a try - I never worked with octave, though. – mathause Apr 28 '14 at 12:43 • Ok, I can run it as is in octave if i put the code in a file named call_convolve_random_arrays.m and then call it from octave. – mathause Apr 28 '14 at 18:29 • Does it need some more code to actually get it to do anything? When I do "octave call_convolve_random_arrays.m" it doesn't output anything. See bpaste.net/show/JPtLOCeI3aP3wc3F3aGf – user9206 Apr 28 '14 at 18:33 • sorry, try to open octave and run it then. It should display firstzero, bothzero and execution time. – mathause Apr 28 '14 at 18:34 # Julia: 0.30 s # Op's Python: 21.36 s (Core2 duo) 71x speedup function countconv() n = 6 iters = 1000 firstzero = 0 bothzero = 0 cprod= Iterators.product(fill([-1,1], n+1)...) F=Array(Float64,n); P=[-1. 0. 0. 1.] for S in cprod Sm=[S...] for i = 1:iters F=P[rand(1:4,n)] while all(F==0) F=P[rand(1:4,n)] end if dot(reverse!(F),Sm[1:end-1]) == 0 firstzero += 1 if dot(F,Sm[2:end]) == 0 bothzero += 1 end end end end return firstzero,bothzero end  I did some modifications of Arman's Julia answer: First of all, I wrapped it in a function, as global variables make it hard for Julia's type inference and JIT: A global variable can change its type at any time, and must be checked every operation. Then, I got rid of the anonymous functions and array comprehensions. They aren't really necessary, and are still pretty slow. Julia is faster with lower-level abstractions right now. There's lots more ways to make it faster, but this does a decent job. • Are you measuring the time in the REPL or running the whole file from command line? – Aditya Apr 29 '14 at 5:30 • both from the REPL. – user20768 Apr 29 '14 at 17:35 Ok I am posting this just because I feel Java needs to be represented here. I am terrible with other languages and I confess to not understand the problem exactly, so I will need some help to fix this code. I stole most of the code ace's C example, and then borrowed some snippets from others. I hope that isn't a faux pas... One thing I would like to point out is that languages that optimize at run time need to be run several/many times to get up to full speed. I think it is justified to take the fully optimized speed (or at least the average speed) because most things you are concerned with running fast will be run a bunch of times. The code still needs to be fixed, but I ran it anyways to see what times I would get. Here are the results on an Intel(R) Xeon(R) CPU E3-1270 V2 @ 3.50GHz on Ubuntu running it 1000 times: server:/tmp# time java8 -cp . Tester firstzero 40000 bothzero 20000 first run time: 41 ms last run time: 4 ms real 0m5.014s user 0m4.664s sys 0m0.268s Here is my crappy code: public class Tester { public static void main( String[] args ) { long firstRunTime = 0; long lastRunTime = 0; String testResults = null; for( int i=0 ; i<1000 ; i++ ) { long timer = System.currentTimeMillis(); testResults = new Tester().runtest(); lastRunTime = System.currentTimeMillis() - timer; if( i ==0 ) { firstRunTime = lastRunTime; } } System.err.println( testResults ); System.err.println( "first run time: " + firstRunTime + " ms" ); System.err.println( "last run time: " + lastRunTime + " ms" ); } private int x,y=34353,z=57768,w=1564; public String runtest() { int n = 6; int iters = 1000; //#define iters (1000) //PRNG seeds /* xorshift PRNG * Taken from https://en.wikipedia.org/wiki/Xorshift#Example_implementation * Used under CC-By-SA */ int firstzero=0, bothzero=0; int[] arr = new int[n+1]; int i=0, j=0; x=(int)(System.currentTimeMillis()/1000l); for(i=0; i< 1<<(n+1) ; i++) { int tmp=i; for(j=0; j<n+1; j++) { arr[j]=(tmp&1)*(-2)+1; tmp>>=1; } for(j=0; j<iters; j++) { int[] randArr = new int[n]; int k=0; long flag = 0; int first=0, second=0; do { for(k=0; k<n; k++) { randArr[k]=(1-(myRand()&3))%2; flag+=(randArr[k]&1); first+=arr[k]*randArr[k]; second+=arr[k+1]*randArr[k]; } } while(allzero(randArr)); if( first == 0 ) { firstzero+=1; if( second == 0 ) { bothzero++; } } } } return ( "firstzero " + firstzero + "\nbothzero " + bothzero + "\n" ); } private boolean allzero(int[] arr) { for(int x : arr) { if(x!=0) { return false; } } return true; } public int myRand() { long t; t = x ^ (x << 11); x = y; y = z; z = w; return (int)( w ^ (w >> 19) ^ t ^ (t >> 8)); } }  And I tried running the python code after upgrading python and installing python-numpy but I get this: server:/tmp# python tester.py Traceback (most recent call last): File "peepee.py", line 15, in <module> F = np.random.choice(np.array([-1,0,0,1], dtype=np.int8), size = n) AttributeError: 'module' object has no attribute 'choice'  • Comments: Never use currentTimeMillis for benchmarking (use the nano version in System) and 1k runs may not be enough to get the JIT involved (1.5k for client and 10k for server would be the defaults, although you call myRand often enough that that will be JITed which should cause some functions up the callstack to be compiled which may work here).Last but not least the weak PNRG is cheating, but so does the C++ solution and others, so I guess that's not too unfair. – Voo May 4 '14 at 20:25 • On windows you need to avoid currentTimeMillis, but for linux for all but very fine granularity measurements you don't need nano time, and the call to get nano time is much more expensive than millis. So I very much disagree that you should NEVER use it. – Chris Seline May 5 '14 at 20:58 • So you're writing Java code for one particular OS and JVM implementation? Actually I'm not sure which OS you're using, because I just checked in my HotSpot dev tree and Linux uses gettimeofday(&time, NULL) for milliSeconds which isn't monotonical and doesn't give any accuracy guarantees (so on some platforms/kernels exactly the same problems as the currentTimeMillis Windows implementation - so either that one's fine too or neither is). nanoTime on the other hand uses clock_gettime(CLOCK_MONOTONIC, &tp) which clearly is also the right thing to use when benchmarking on Linux. – Voo May 5 '14 at 23:38 • It has never caused an issue for me since I have been coding java on any Linux distro or kernel. – Chris Seline May 7 '14 at 0:35 Golang version 45X of python on my machine on below Golang codes: package main import ( "fmt" "time" ) const ( n = 6 iters = 1000 ) var ( x, y, z, w = 34353, 34353, 57768, 1564 //PRNG seeds ) /* xorshift PRNG * Taken from https://en.wikipedia.org/wiki/Xorshift#Example_implementation * Used under CC-By-SA */ func myRand() int { var t uint t = uint(x ^ (x << 11)) x, y, z = y, z, w w = int(uint(w^w>>19) ^ t ^ (t >> 8)) return w } func main() { var firstzero, bothzero int var arr [n + 1]int var i, j int x = int(time.Now().Unix()) for i = 0; i < 1<<(n+1); i = i + 1 { tmp := i for j = 0; j < n+1; j = j + 1 { arr[j] = (tmp&1)*(-2) + 1 tmp >>= 1 } for j = 0; j < iters; j = j + 1 { var randArr [n]int var flag uint var k, first, second int for { for k = 0; k < n; k = k + 1 { randArr[k] = (1 - (myRand() & 3)) % 2 flag += uint(randArr[k] & 1) first += arr[k] * randArr[k] second += arr[k+1] * randArr[k] } if flag != 0 { break } } if first == 0 { firstzero += 1 if second == 0 { bothzero += 1 } } } } println("firstzero", firstzero, "bothzero", bothzero) }  and the below python codes copyed from above: import itertools import operator import random n=6 iters = 1000 firstzero = 0 bothzero = 0 choicesF = filter(any, itertools.product([-1, 0, 0, 1], repeat=n)) for S in itertools.product([-1,1], repeat = n+1): for i in xrange(iters): F = random.choice(choicesF) if not sum(map(operator.mul, F, S[:-1])): firstzero += 1 if not sum(map(operator.mul, F, S[1:])): bothzero += 1 print "firstzero", firstzero print "bothzero", bothzero  and the time below: $time python test.py
firstzero 27349
bothzero 12125

real    0m0.477s
user    0m0.461s
sys 0m0.014s

$time ./hf firstzero 27253 bothzero 12142 real 0m0.011s user 0m0.008s sys 0m0.002s  • have you thought about using "github.com/yanatan16/itertools" ? also would you say this would work nice in multiple goroutines ? – ymg Apr 30 '14 at 17:02 # C# 0.135s C# based on Alistair Buxton's plain python: 0.278s Parallelised C#: 0.135s Python from the question: 5.907s Alistair's plain python: 0.853s I'm not actually certain this implementation is correct - its output is different, if you look at the results down at the bottom. There's certainly more optimal algorithms. I just decided to use a very similar algorithm to the Python one. ### Single-threaded C using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConvolvingArrays { static class Program { static void Main(string[] args) { int n=6; int iters = 1000; int firstzero = 0; int bothzero = 0; int[] arraySeed = new int[] {-1, 1}; int[] randomSource = new int[] {-1, 0, 0, 1}; Random rand = new Random(); foreach (var S in Enumerable.Repeat(arraySeed, n+1).CartesianProduct()) { for (int i = 0; i < iters; i++) { var F = Enumerable.Range(0, n).Select(_ => randomSource[rand.Next(randomSource.Length)]); while (!F.Any(f => f != 0)) { F = Enumerable.Range(0, n).Select(_ => randomSource[rand.Next(randomSource.Length)]); } if (Enumerable.Zip(F, S.Take(n), (f, s) => f * s).Sum() == 0) { firstzero++; if (Enumerable.Zip(F, S.Skip(1), (f, s) => f * s).Sum() == 0) { bothzero++; } } } } Console.WriteLine("firstzero {0}", firstzero); Console.WriteLine("bothzero {0}", bothzero); } // itertools.product? // http://ericlippert.com/2010/06/28/computing-a-cartesian-product-with-linq/ static IEnumerable<IEnumerable<T>> CartesianProduct<T> (this IEnumerable<IEnumerable<T>> sequences) { IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() }; return sequences.Aggregate( emptyProduct, (accumulator, sequence) => from accseq in accumulator from item in sequence select accseq.Concat(new[] { item })); } } }  ### Parallel C#: using System; using System.Collections.Concurrent; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading; using System.Threading.Tasks; namespace ConvolvingArrays { static class Program { static void Main(string[] args) { int n=6; int iters = 1000; int firstzero = 0; int bothzero = 0; int[] arraySeed = new int[] {-1, 1}; int[] randomSource = new int[] {-1, 0, 0, 1}; ConcurrentBag<int[]> results = new ConcurrentBag<int[]>(); // The next line iterates over arrays of length n+1 which contain only -1s and 1s Parallel.ForEach(Enumerable.Repeat(arraySeed, n + 1).CartesianProduct(), (S) => { int fz = 0; int bz = 0; ThreadSafeRandom rand = new ThreadSafeRandom(); for (int i = 0; i < iters; i++) { var F = Enumerable.Range(0, n).Select(_ => randomSource[rand.Next(randomSource.Length)]); while (!F.Any(f => f != 0)) { F = Enumerable.Range(0, n).Select(_ => randomSource[rand.Next(randomSource.Length)]); } if (Enumerable.Zip(F, S.Take(n), (f, s) => f * s).Sum() == 0) { fz++; if (Enumerable.Zip(F, S.Skip(1), (f, s) => f * s).Sum() == 0) { bz++; } } } results.Add(new int[] { fz, bz }); }); foreach (int[] res in results) { firstzero += res[0]; bothzero += res[1]; } Console.WriteLine("firstzero {0}", firstzero); Console.WriteLine("bothzero {0}", bothzero); } // itertools.product? // http://ericlippert.com/2010/06/28/computing-a-cartesian-product-with-linq/ static IEnumerable<IEnumerable<T>> CartesianProduct<T> (this IEnumerable<IEnumerable<T>> sequences) { IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() }; return sequences.Aggregate( emptyProduct, (accumulator, sequence) => from accseq in accumulator from item in sequence select accseq.Concat(new[] { item })); } } // http://stackoverflow.com/a/11109361/1030702 public class ThreadSafeRandom { private static readonly Random _global = new Random(); [ThreadStatic] private static Random _local; public ThreadSafeRandom() { if (_local == null) { int seed; lock (_global) { seed = _global.Next(); } _local = new Random(seed); } } public int Next() { return _local.Next(); } public int Next(int maxValue) { return _local.Next(maxValue); } } }  ## Test output: ### Windows (.NET) The C# is much faster on Windows. Probably because .NET is faster than mono. User and sys timing doesn't seem to work (used git bash for timing). $ time /c/Python27/python.exe numpypython.py
firstzero 27413
bothzero 12073

real    0m5.907s
user    0m0.000s
sys     0m0.000s
$time /c/Python27/python.exe plainpython.py firstzero 26983 bothzero 12033 real 0m0.853s user 0m0.000s sys 0m0.000s$ time ConvolvingArrays.exe
firstzero 28526
bothzero 6453

real    0m0.278s
user    0m0.000s
sys     0m0.000s
$time ConvolvingArraysParallel.exe firstzero 28857 bothzero 6485 real 0m0.135s user 0m0.000s sys 0m0.000s  ### Linux (mono) bob@phoebe:~/convolvingarrays$ time python program.py
firstzero 27059
bothzero 12131

real    0m11.932s
user    0m11.912s
sys     0m0.012s
bob@phoebe:~/convolvingarrays$mcs -optimize+ -debug- program.cs bob@phoebe:~/convolvingarrays$ time mono program.exe
firstzero 28982
bothzero 6512

real    0m1.360s
user    0m1.532s
sys     0m0.872s
bob@phoebe:~/convolvingarrays$mcs -optimize+ -debug- parallelprogram.cs bob@phoebe:~/convolvingarrays$ time mono parallelprogram.exe
firstzero 28857
bothzero 6496

real    0m0.851s
user    0m2.708s
sys     0m3.028s

• I don't think the code is correct as you say. The outputs are not right. – user9206 Apr 27 '14 at 16:43
• @Lembik Yea. I'd appreciate it if someone could tell me where it's wrong, though - I can't figure it out (only having a minimal understanding of what it's supposed to do doesn't help). – Bob Apr 28 '14 at 1:40
• Would be interesting to see how this does with .NET Native blogs.msdn.com/b/dotnet/archive/2014/04/02/… – Rick Minerich Apr 30 '14 at 16:33
• @Lembik I've just gone over all of it, as far as I can tell it should be identical to the other Python solution... now I'm really confused. – Bob Apr 30 '14 at 23:44

## Haskell: ~2000x speedup per core

Compile with 'ghc -O3 -funbox-strict-fields -threaded -fllvm', and run with '+RTS -Nk' where k is the number of cores on your machine.

import Control.Parallel.Strategies
import Data.Bits
import Data.List
import Data.Word
import System.Random

n = 6 :: Int
iters = 1000 :: Int

data G = G !Word !Word !Word !Word deriving (Eq, Show)

gen :: G -> (Word, G)
gen (G x y z w) = let t  = x xor (x shiftL 11)
w' = w xor (w shiftR 19) xor t xor (t shiftR 8)
in (w', G y z w w')

mask = (.&.) $(2 ^ n) - 1 gen_nonzero :: G -> (Word, G) gen_nonzero g = let (x, g') = gen g a = mask x in if a == 0 then gen_nonzero g' else (a, g') data F = F {zeros :: !Word, posneg :: !Word} deriving (Eq, Show) gen_f :: G -> (F, G) gen_f g = let (a, g') = gen_nonzero g (b, g'') = gen g' in (F a$ mask b, g'')

inner :: Word -> F -> Int
inner s (F zs pn) = let s' = complement $s xor pn ones = s' .&. zs negs = (complement s') .&. zs in popCount ones - popCount negs specialised_convolve :: Word -> F -> (Int, Int) specialised_convolve s f@(F zs pn) = (inner s f', inner s f) where f' = F (zs shiftL 1) (pn shiftL 1) ss :: [Word] ss = [0..2 ^ (n + 1) - 1] main_loop :: [G] -> (Int, Int) main_loop gs = foldl1' (\(fz, bz) (fz', bz') -> (fz + fz', bz + bz')) . parMap rdeepseq helper$ zip ss gs
where helper (s, g) = go 0 (0, 0) g
where go k u@(fz, bz) g = if k == iters
then u
else let (f, g') = gen_f g
v = case specialised_convolve s f
of (0, 0) -> (fz + 1, bz + 1)
(0, _) -> (fz + 1, bz)
_      -> (fz, bz)
in go (k + 1) v g'

seed :: IO G
seed = do std_g <- newStdGen
let [x, y, z, w] = map fromIntegral $take 4 (randoms std_g :: [Int]) return$ G x y z w

main :: IO ()
main = (sequence $map (const seed) ss) >>= print . main_loop  • So with 4 cores it's over 9000?! There's no way that could be right. – Cees Timmerman May 6 '14 at 15:03 • Amdahl's law states the parallelization speedup is not linear to the number of parallel processing units. instead they only provide dimishing returns – xaedes Jul 11 at 7:39 • @xaedes The speedup seems essentially linear for low numbers of cores – user1502040 Jul 12 at 9:18 # F# solution Runtime is 0.030s when compiled to x86 on the CLR Core i7 4 (8) @ 3.4 Ghz I have no idea if the code is correct. • Functional optimization (inline fold) -> 0.026s • Building via Console Project -> 0.022s • Added a better algorithm for generation of the permutation arrays -> 0.018s • Mono for Windows -> 0.089s • Running Alistair's Python script -> 0.259s let inline ffoldi n f state = let mutable state = state for i = 0 to n - 1 do state <- f state i state let product values n = let p = Array.length values Array.init (pown p n) (fun i -> (Array.zeroCreate n, i) |> ffoldi n (fun (result, i') j -> result.[j] <- values.[i' % p] result, i' / p ) |> fst ) let convolute signals filter = let m = Array.length signals let n = Array.length filter let len = max m n - min m n + 1 Array.init len (fun offset -> ffoldi n (fun acc i -> acc + filter.[i] * signals.[m - 1 - offset - i] ) 0 ) let n = 6 let iters = 1000 let next = let arrays = product [|-1; 0; 0; 1|] n |> Array.filter (Array.forall ((=) 0) >> not) let rnd = System.Random() fun () -> arrays.[rnd.Next arrays.Length] let signals = product [|-1; 1|] (n + 1) let firstzero, bothzero = ffoldi signals.Length (fun (firstzero, bothzero) i -> let s = signals.[i] ffoldi iters (fun (first, both) _ -> let f = next() match convolute s f with | [|0; 0|] -> first + 1, both + 1 | [|0; _|] -> first + 1, both | _ -> first, both ) (firstzero, bothzero) ) (0, 0) printfn "firstzero %i" firstzero printfn "bothzero %i" bothzero  ## Ruby Ruby (2.1.0) 0.277s Ruby (2.1.1) 0.281s Python (Alistair Buxton) 0.330s Python (alemi) 0.097s n = 6 iters = 1000 first_zero = 0 both_zero = 0 choices = [-1, 0, 0, 1].repeated_permutation(n).select{|v| [0] != v.uniq} def convolve(v1, v2) [0, 1].map do |i| r = 0 6.times do |j| r += v1[i+j] * v2[j] end r end end [-1, 1].repeated_permutation(n+1) do |s| iters.times do f = choices.sample fs = convolve s, f if 0 == fs[0] first_zero += 1 if 0 == fs[1] both_zero += 1 end end end end puts 'firstzero %i' % first_zero puts 'bothzero %i' % both_zero  # thread wouldnt be complete without PHP ## 6.6x faster ### PHP v5.5.9 - 1.223 0.646 sec; ### vs ### Python v2.7.6 - 8.072 sec <?php$n = 6;
$iters = 1000;$firstzero = 0;
$bothzero = 0;$x=time();
$y=34353;$z=57768;
$w=1564; //PRNG seeds function myRand() { global$x;
global $y; global$z;
global $w;$t = $x ^ ($x << 11);
$x =$y; $y =$z; $z =$w;
return $w =$w ^ ($w >> 19) ^$t ^ ($t >> 8); } function array_cartesian() {$_ = func_get_args();
if (count($_) == 0) return array();$a = array_shift($_); if (count($_) == 0)
$c = array(array()); else$c = call_user_func_array(__FUNCTION__, $_);$r = array();
foreach($a as$v)
foreach($c as$p)
$r[] = array_merge(array($v), $p); return$r;
}

function rand_array($a,$n)
{
$r = array(); for($i = 0; $i <$n; $i++)$r[] = $a[myRand()%count($a)];
return $r; } function convolve($a, $b) { // slows down /*if(count($a) < count($b)) return convolve($b,$a);*/$result = array();
$w = count($a) - count($b) + 1; for($i = 0; $i <$w; $i++){$r = 0;
for($k = 0;$k < count($b);$k++)
$r +=$b[$k] *$a[$i +$k];
$result[] =$r;
}
return $result; }$cross = call_user_func_array('array_cartesian',array_fill(0,$n+1,array(-1,1))); foreach($cross as $S) for($i = 0; $i <$iters; $i++){ while(true) {$F = rand_array(array(-1,0,0,1), $n); if(in_array(-1,$F) || in_array(1, $F)) break; }$FS = convolve($S,$F);
if(0==$FS[0])$firstzero += 1;
if(0==$FS[0] && 0==$FS[1]) $bothzero += 1; } echo "firstzero$firstzero\n";
echo "bothzero $bothzero\n";  • Used a custom random generator (stolen from C answer), PHP one sucks and numbers dont match • convolve function simplified a bit to be more fast • Checking for array-with-zeros-only is very optimized too (see $F and $FS checkings). Outputs: $ time python num.py
firstzero 27050
bothzero 11990

real    0m8.072s
user    0m8.037s
sys 0m0.024s
$time php num.php firstzero 27407 bothzero 12216 real 0m1.223s user 0m1.210s sys 0m0.012s  Edit. Second version of script works for for just 0.646 sec: <?php$n = 6;
$iters = 1000;$firstzero = 0;
$bothzero = 0;$x=time();
$y=34353;$z=57768;
$w=1564; //PRNG seeds function myRand() { global$x;
global $y; global$z;
global $w;$t = $x ^ ($x << 11);
$x =$y; $y =$z; $z =$w;
return $w =$w ^ ($w >> 19) ^$t ^ ($t >> 8); } function array_cartesian() {$_ = func_get_args();
if (count($_) == 0) return array();$a = array_shift($_); if (count($_) == 0)
$c = array(array()); else$c = call_user_func_array(__FUNCTION__, $_);$r = array();
foreach($a as$v)
foreach($c as$p)
$r[] = array_merge(array($v), $p); return$r;
}

function convolve($a,$b)
{
// slows down
/*if(count($a) < count($b))
return convolve($b,$a);*/
$result = array();$w = count($a) - count($b) + 1;
for($i = 0;$i < $w;$i++){
$r = 0; for($k = 0; $k < count($b); $k++)$r += $b[$k] * $a[$i + $k];$result[] = $r; } return$result;
}

$cross = call_user_func_array('array_cartesian',array_fill(0,$n+1,array(-1,1)));

$choices = call_user_func_array('array_cartesian',array_fill(0,$n,array(-1,0,0,1)));

foreach($cross as$S)
for($i = 0;$i < $iters;$i++){
while(true)
{
$F =$choices[myRand()%count($choices)]; if(in_array(-1,$F) || in_array(1, $F)) break; }$FS = convolve($S,$F);
if(0==$FS[0]){$firstzero += 1;
if(0==$FS[1])$bothzero += 1;
}
}

echo "firstzero $firstzero\n"; echo "bothzero$bothzero\n";


# Q, 0.296 seg

n:6; iter:1000  /parametrization (constants)
c:n#0           /auxiliar constant (sequence 0 0.. 0 (n))
A:B:();         /A and B accumulates results of inner product (firstresult, secondresult)

/S=sequence with all arrays of length n+1 with values -1 and 1
S:+(2**m)#/:{,/x#/:-1 1}'m:|n(2*)\1

f:{do[iter; F:c; while[F~c; F:n?-1 0 0 1]; A,:+/F*-1_x; B,:+/F*1_x];} /hard work
f'S               /map(S,f)
N:~A; +/'(N;N&~B) / ~A is not A (or A=0) ->bitmap.  +/ is sum (population over a bitmap)
/ +/'(N;N&~B) = count firstResult=0, count firstResult=0 and secondResult=0


Q is a collection oriented language (kx.com)

Code rewrited to explote idiomatic Q, but no other clever optimizations

Scripting languages optimize programmer time, not execution time

• Q is not the best tool for this problem

First coding attempt = not a winner, but reasonable time (approx. 30x speedup)

• quite competitive among interpreters
• stop and choose another problem

NOTES.-

• program uses default seed (repeteable execs) To choose another seed for random generator use \S seed
• Result is given as a squence of two ints, so there is a final i-suffix at second value 27421 12133i -> read as (27241, 12133)
• Time not counting interpreter startup. \t sentence mesures time consumed by that sentence
• Very interesting thank you. – user9206 May 22 '16 at 8:42

# Julia: 12.149 6.929 s

Despite their claims to speed, the initial JIT compilation time holds us back!

Note that the following Julia code is effectively a direct translation of the original Python code (no optimisations made) as a demonstration that you can easily transfer your programming experience to a faster language ;)

require("Iterators")

n = 6
iters = 1000
firstzero = 0
bothzero = 0

for S in Iterators.product(fill([-1,1], n+1)...)
for i = 1:iters
F = [[-1 0 0 1][rand(1:4)] for _ = 1:n]
while all((x) -> round(x,8) == 0, F)
F = [[-1 0 0 1][rand(1:4)] for _ = 1:n]
end
FS = conv(F, [S...])
if round(FS[1],8) == 0
firstzero += 1
end
if all((x) -> round(x,8) == 0, FS)
bothzero += 1
end
end
end

println("firstzero ", firstzero)
println("bothzero ", bothzero)


# Edit

Running with n = 8 takes 32.935 s. Considering that the complexity of this algorithm is O(2^n), then 4 * (12.149 - C) = (32.935 - C), where C is a constant representing the JIT compilation time. Solving for C we find that C = 5.2203, suggesting that actual execution time for n = 6 is 6.929 s.

• How about increasing n to 8 to see if Julia comes into its own then? – user9206 Apr 27 '14 at 6:11
• This ignores many of the performance tips here: julia.readthedocs.org/en/latest/manual/performance-tips. See also the other Julia entry which does significantly better. The submission is appreciated though :-) – StefanKarpinski Apr 30 '14 at 15:08

# Rust, 6.6 ms, 1950x speedup

Pretty much a direct translation of Alistair Buxton's code to Rust. I considered making use of multiple cores with rayon (fearless concurrency!), but this didn't improve the performance, probably because it's very fast already.

extern crate itertools;
extern crate rand;
extern crate time;

use itertools::Itertools;
use rand::{prelude::*, prng::XorShiftRng};
use std::iter;
use time::precise_time_ns;

fn main() {
let start = precise_time_ns();

let n = 6;
let iters = 1000;
let mut first_zero = 0;
let mut both_zero = 0;
let choices_f: Vec<Vec<i8>> = iter::repeat([-1, 0, 0, 1].iter().cloned())
.take(n)
.multi_cartesian_product()
.filter(|i| i.iter().any(|&x| x != 0))
.collect();
// xorshift RNG is faster than default algorithm designed for security
// rather than performance.
let mut rng = XorShiftRng::from_entropy();
for s in iter::repeat(&[-1, 1]).take(n + 1).multi_cartesian_product() {
for _ in 0..iters {
let f = rng.choose(&choices_f).unwrap();
if f.iter()
.zip(&s[..s.len() - 1])
.map(|(a, &b)| a * b)
.sum::<i8>() == 0
{
first_zero += 1;
if f.iter().zip(&s[1..]).map(|(a, &b)| a * b).sum::<i8>() == 0 {
both_zero += 1;
}
}
}
}
println!("first_zero = {}\nboth_zero = {}", first_zero, both_zero);

println!("runtime {} ns", precise_time_ns() - start);
}


And Cargo.toml, as I use external dependencies:

[package]
name = "how_slow_is_python"
version = "0.1.0"

[dependencies]
itertools = "0.7.8"
rand = "0.5.3"
time = "0.1.40"


Speed comparison:

$time python2 py.py firstzero: 27478 bothzero: 12246 12.80user 0.02system 0:12.90elapsed 99%CPU (0avgtext+0avgdata 23328maxresident)k 0inputs+0outputs (0major+3544minor)pagefaults 0swaps$ time target/release/how_slow_is_python
first_zero = 27359
both_zero = 12162
runtime 6625608 ns
0.00user 0.00system 0:00.00elapsed 100%CPU (0avgtext+0avgdata 2784maxresident)k
0inputs+0outputs (0major+189minor)pagefaults 0swaps


6625608 ns is about 6.6 ms. This means 1950 times speedup. There are many optimizations possible here, but I was going for readability rather than performance. One possible optimization would be use arrays instead of vectors for storing choices, as they will always have n elements. It's also possible to use RNG other than XorShift, as while Xorshift is faster than the default HC-128 CSPRNG, it's slowest than naivest of PRNG algorithms.