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I am comparing ways to perform equivalent matrix operations within Eigen, and am getting extraordinarily different runtimes, including some non-intuitive results.
I am comparing three mathematically equivalent forms of the matrix multiplication:
wx * transpose(data)
The three forms I'm comparing are:
result = wx * data.transpose() (straight multiply version)
result.noalias() = wx * data.transpose() (noalias version)
result = (data * wx.transpose()).transpose() (transposed version)
I am also testing using both Column Major and Row Major storage.
With column major storage, the transposed version is significantly faster (an order of magnitude) than both the straight multiply and the no alias version, which are both approximately equal in runtime.
With row major storage, the noalias and the transposed version are both significantly faster than the straight multiply in runtime.
I understand that Eigen uses lazy evaluation, and that the immediate results returned from an operation are often expression templates, and are not the intermediate values. I also understand that matrix * matrix operations will always produce a temporary when they are the last operation on the right hand side, to avoid aliasing issues, hence why I am attempting to speed things up through noalias().
My main questions:
Why is the transposed version always significantly faster, even (in the case of column major storage) when I explicitly state noalias so no temporaries are created?
Why does the (significant) difference in runtime only occur between the straight multiply and the noalias version when using column major storage?
The code I am using for this is below. It is being compiled using gcc 4.9.2, on a Centos 6 install, using the following command line.
g++ eigen_test.cpp -O3 -std=c++11 -o eigen_test -pthread -fopenmp -finline-functions
using Matrix = Eigen::Matrix<float, Eigen::Dynamic, Eigen::Dynamic, Eigen::ColMajor>;
// using Matrix = Eigen::Matrix<float, Eigen::Dynamic, Eigen::Dynamic, Eigen::RowMajor>;
int wx_rows = 8000;
int wx_cols = 1000;
int samples = 1;
// Eigen::MatrixXf matrix = Eigen::MatrixXf::Random(matrix_rows, matrix_cols);
Matrix wx = Eigen::MatrixXf::Random(wx_rows, wx_cols);
Matrix data = Eigen::MatrixXf::Random(samples, wx_cols);
Matrix result;
unsigned int iterations = 10000;
float sum = 0;
auto before = std::chrono::high_resolution_clock::now();
for (unsigned int ii = 0; ii < iterations; ++ii)
{
result = wx * data.transpose();
sum += result(result.rows() - 1, result.cols() - 1);
}
auto after = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(after - before).count();
std::cout << "original sum: " << sum << std::endl;
std::cout << "original time (ms): " << duration << std::endl;
std::cout << std::endl;
sum = 0;
before = std::chrono::high_resolution_clock::now();
for (unsigned int ii = 0; ii < iterations; ++ii)
{
result.noalias() = wx * data.transpose();
sum += result(wx_rows - 1, samples - 1);
}
after = std::chrono::high_resolution_clock::now();
duration = std::chrono::duration_cast<std::chrono::milliseconds>(after - before).count();
std::cout << "alias sum: " << sum << std::endl;
std::cout << "alias time (ms) : " << duration << std::endl;
std::cout << std::endl;
sum = 0;
before = std::chrono::high_resolution_clock::now();
for (unsigned int ii = 0; ii < iterations; ++ii)
{
result = (data * wx.transpose()).transpose();
sum += result(wx_rows - 1, samples - 1);
}
after = std::chrono::high_resolution_clock::now();
duration = std::chrono::duration_cast<std::chrono::milliseconds>(after - before).count();
std::cout << "new sum: " << sum << std::endl;
std::cout << "new time (ms) : " << duration << std::endl;
One half of the explanation is because, in the current version of Eigen, multi-threading is achieved by splitting the work over blocks of columns of the result (and the right-hand-side). With only 1 column, multi-threading does not take place. In the column-major case, this explain why cases 1 and 2 underperform. On the other hand, case 3 is evaluated as:
column_major_tmp.noalias() = data * wx.transpose();
result = column_major_tmp.transpose();
and since wx.transpose().cols() is huge, multi-threading is effective.
To understand the row-major case, you also need to know that internally matrix products is implemented for a column-major destination. If the destination is row-major, as in case 2, then the product is transposed, so what really happens is:
row_major_result.transpose().noalias() = data * wx.transpose();
and so we're back to case 3 but without temporary.
This is clearly a limitation of current Eigen's multi-threading implementation for highly unbalanced matrix sizes. Ideally threads should be spread on row-block and/or column-block depending on the size of the matrices at hand.
BTW, you should also compile with -march=native to let Eigen fully exploit your CPU (AVX, FMA, AVX512...).
Actually, I'm trying to subtract the background from this image. Apparently, I just want to subtract the green background and here is the code I'm using:
Mat img_object = imread(patternImageName);
Mat imageInHSV;
cvtColor(img_object, imageInHSV, CV_BGR2HSV);
Mat chan[3],imgThreshed, processed;
split( imageInHSV, chan );
Mat H = chan[0];
// compute statistics for Hue value
cv::Scalar mean, stddev;
cv::meanStdDev(H, mean, stddev);
// ensure we get 95% of all valid Hue samples (statistics 3*sigma rule)
float minHue = 80;
float maxHue = 95;
cout << "MinValue :" << mean[0] << " MaxHue:" << stddev[0] << endl;
cout << H << endl;
// STEP 2: detection phase
cv::inRange(H, cv::Scalar(minHue), cv::Scalar(maxHue), imgThreshed);
imshow("thresholded", imgThreshed);
I checked the values of the channel H to decide the minHue and maxHue so I choosed the interval of the most frequent values in the matrix which will definitely be the green one. But, I got this result which is obsiously not what I'm looking for because there is missing stuff in it. Any idea how to improve it? how to get better subtract the background from this kind of images?
I am not sure exactly what your goal is. However, I got much better result on your sample image from the two other channels (Saturation and lightness rather than hue) using the range of [mean-stddev,mean+stddev]. Averaging the results of all three channels shows some improvement:
using namespace std ;
using namespace cv ;
int main()
{
Mat img_object = imread("1.png");
Mat imageInHSV;
cvtColor(img_object, imageInHSV, CV_BGR2HSV);
Mat chan[3];
split( imageInHSV, chan );
Mat result ;
Mat threshImg[3] ;
for(int i=0 ; i<3 ; i++)
{
Mat H = chan[i];
// compute statistics for each channel
cv::Scalar mean, stddev;
cv::meanStdDev(H, mean, stddev);
// statistically 68% of data should be in this range
float minVal = mean[0]-stddev[0];
float maxVal = mean[0]+stddev[0];
cout << "MinValue :" << mean[0] << " MaxHue:" << stddev[0] << endl;
// Separating the dominant 68% which we guess should be the background.
cv::inRange(H, cv::Scalar(minVal), cv::Scalar(maxVal), threshImg[i]);
}
// averaging the results from three different channels (Hue, Saturation, lightness).
result = (threshImg[0]+threshImg[1]+threshImg[2])/3 ;
imwrite("thresholded_012.jpg", result) ;
}
Input image:
Output image:
I do not intend to use this for security purposes or statistical analysis. I need to create a simple random number generator for use in my computer graphics application. I don't want to use the term "random number generator", since people think in very strict terms about it, but I can't think of any other word to describe it.
it has to be fast.
it must be repeatable, given a particular seed.
Eg: If seed = x, then the series a,b,c,d,e,f..... should happen every time I use the seed x.
Most importantly, I need to be able to compute the nth term in the series in constant time.
It seems, that I cannot achieve this with rand_r or srand(), since these need are state dependent, and I may need to compute the nth in some unknown order.
I've looked at Linear Feedback Shift registers, but these are state dependent too.
So far I have this:
int rand = (n * prime1 + seed) % prime2
n = used to indicate the index of the term in the sequence. Eg: For
first term, n ==1
prime1 and prime2 are prime numbers where
prime1 > prime2
seed = some number which allows one to use the same function to
produce a different series depending on the seed, but the same series
for a given seed.
I can't tell how good or bad this is, since I haven't used it enough, but it would be great if people with more experience in this can point out the problems with this, or help me improve it..
EDIT - I don't care if it is predictable. I'm just trying to creating some randomness in my computer graphics.
Use a cryptographic block cipher in CTR mode. The Nth output is just encrypt(N). Not only does this give you the desired properties (O(1) computation of the Nth output); it also has strong non-predictability properties.
I stumbled on this a while back, looking for a solution for the same problem. Recently, I figured out how to do it in low-constant O(log(n)) time. While this doesn't quite match the O(1) requested by the author, It may be fast enough (a sample run, compiled with -O3, achieved performance of 1 billion arbitrary index random numbers, with n varying between 1 and 2^48, in 55.7s -- just shy of 18M numbers/s).
First, the theory behind the solution:
A common type of RNGs are Linear Congruential Generators, basically, they work as follows:
random(n) = (m*random(n-1) + b) mod p
Where m and b, and p are constants (see a reference on LCGs for how they are chosen). From this, we can devise the following using a bit of modular arithmetic:
random(0) = seed mod p
random(1) = m*seed + b mod p
random(2) = m^2*seed + m*b + b mod p
...
random(n) = m^n*seed + b*Sum_{i = 0 to n - 1} m^i mod p
= m^n*seed + b*(m^n - 1)/(m - 1) mod p
Computing the above can be a problem, since the numbers will quickly exceed numeric limits. The solution for the generic case is to compute m^n in modulo with p*(m - 1), however, if we take b = 0 (a sub-case of LCGs sometimes called Multiplicative congruential Generators), we have a much simpler solution, and can do our computations in modulo p only.
In the following, I use the constant parameters used by RANF (developed by CRAY), where p = 2^48 and g = 44485709377909. The fact that p is a power of 2 reduces the number of operations required (as expected):
#include <cassert>
#include <stdint.h>
#include <cstdlib>
class RANF{
// MCG constants and state data
static const uint64_t m = 44485709377909ULL;
static const uint64_t n = 0x0000010000000000ULL; // 2^48
static const uint64_t randMax = n - 1;
const uint64_t seed;
uint64_t state;
public:
// Constructors, which define the seed
RANF(uint64_t seed) : seed(seed), state(seed) {
assert(seed > 0 && "A seed of 0 breaks the LCG!");
}
// Gets the next random number in the sequence
inline uint64_t getNext(){
state *= m;
return state & randMax;
}
// Sets the MCG to a specific index
inline void setPosition(size_t index){
state = seed;
uint64_t mPower = m;
for (uint64_t b = 1; index; b <<= 1){
if (index & b){
state *= mPower;
index ^= b;
}
mPower *= mPower;
}
}
};
#include <cstdio>
void example(){
RANF R(1);
// Gets the number through random-access -- O(log(n))
R.setPosition(12345); // Goes to the nth random number
printf("fast nth number = %lu\n", R.getNext());
// Gets the number through standard, sequential access -- O(n)
R.setPosition(0);
for(size_t i = 0; i < 12345; i++) R.getNext();
printf("slow nth number = %lu\n", R.getNext());
}
While I presume the author has moved on by now, hopefully this will be of use to someone else.
If you're really concerned about runtime performance, the above can be made about 10x faster with lookup tables, at the cost of compilation time and binary size (it also is O(1) w.r.t the desired random index, as requested by OP)
In the version below, I used c++14 constexpr to generate the lookup tables at compile time, and got to 176M arbitrary index random numbers per second (doing this did however add about 12s of extra compilation time, and a 1.5MB increase in binary size -- the added time may be mitigated if partial recompilation is used).
class RANF{
// MCG constants and state data
static const uint64_t m = 44485709377909ULL;
static const uint64_t n = 0x0000010000000000ULL; // 2^48
static const uint64_t randMax = n - 1;
const uint64_t seed;
uint64_t state;
// Lookup table
struct lookup_t{
uint64_t v[3][65536];
constexpr lookup_t() : v() {
uint64_t mi = RANF::m;
for (size_t i = 0; i < 3; i++){
v[i][0] = 1;
uint64_t val = mi;
for (uint16_t j = 0x0001; j; j++){
v[i][j] = val;
val *= mi;
}
mi = val;
}
}
};
friend struct lookup_t;
public:
// Constructors, which define the seed
RANF(uint64_t seed) : seed(seed), state(seed) {
assert(seed > 0 && "A seed of 0 breaks the LCG!");
}
// Gets the next random number in the sequence
inline uint64_t getNext(){
state *= m;
return state & randMax;
}
// Sets the MCG to a specific index
// Note: idx.u16 indices need to be adapted for big-endian machines!
inline void setPosition(size_t index){
static constexpr auto lookup = lookup_t();
union { uint16_t u16[4]; uint64_t u64; } idx;
idx.u64 = index;
state = seed * lookup.v[0][idx.u16[0]] * lookup.v[1][idx.u16[1]] * lookup.v[2][idx.u16[2]];
}
};
Basically, what this does is splits the computation of, for example, m^0xAAAABBBBCCCC mod p, into (m^0xAAAA00000000 mod p)*(m^0xBBBB0000 mod p)*(m^CCCC mod p) mod p, and then precomputes tables for each of the values in the 0x0000 - 0xFFFF range that could fill AAAA, BBBB or CCCC.
RNG in a normal sense, have the sequence pattern like f(n) = S(f(n-1))
They also lost precision at some point (like % mod), due to computing convenience, therefore it is not possible to expand the sequence to a function like X(n) = f(n) = trivial function with n only.
This mean at best you have O(n) with that.
To target for O(1) you therefore need to abandon the idea of f(n) = S(f(n-1)), and designate a trivial formula directly so that the N'th number can be calculated directly without knowing (N-1)'th; this also render the seed meaningless.
So, you end up have a simple algebra function and not a sequence. For example:
int my_rand(int n) { return 42; } // Don't laugh!
int my_rand(int n) { 3*n*n + 2*n + 7; }
If you want to put more constraint to the generated pattern (like distribution), it become a complex maths problem.
However, for your original goal, if what you want is constant speed to get pseudo-random numbers, I suggest to pre-generate it with traditional RNG and access with lookup table.
EDIT: I noticed you have concern with a table size for a lot of numbers, however you may introduce some hybrid model, like a table of N entries, and do f(k) = g( tbl[k%n], k), which at least provide good distribution across N continue sequence.
This demonstrates an PRNG implemented as a hashed counter. This might appear to duplicate R.'s suggestion (using a block cipher in CTR mode as a stream cipher), but for this, I avoided using cryptographically secure primitives: for speed of execution and because security wasn't a desired feature.
If we were trying to create a secure stream cipher with your requirement that any emitted sequence be trivially repeatable, given knowledge of its index...
...then we could choose a secure hash algorithm (like SHA256) and a counter with a lot of bits (maybe 2048 -> sequence repeats every 2^2048 generated numbers without reseeding).
HOWEVER, the version I present here uses Bob Jenkins' famous hash function (simple and fast, but not secure) along with a 64-bit counter (which is as big as integers can get on my system, without needing custom incrementing code).
Code in main demonstrates that knowledge of the RNG's counter (seed) after initialization allows a PRNG sequence to be repeated, as long as we know how many values were generated leading up to the repetition point.
Actually, if you know the counter's value at any point in the output sequence, you will be able to retrieve all values generated previous to that point, AND all values which will be generated afterward. This only involves adding or subtracting ordinal differences to/from a reference counter value associated with a known point in the output sequence.
It should be pretty easy to adapt this class for use as a testing framework -- you could plug in other hash functions and change the counter's size to see what kind of impact there is on speed as well as the distribution of generated values (the only uniformity analysis I did was to look for patterns in the screenfuls of hexadecimal numbers printed by main()).
#include <iostream>
#include <iomanip>
#include <ctime>
using namespace std;
class CHashedCounterRng {
static unsigned JenkinsHash(const void *input, unsigned len) {
unsigned hash = 0;
for(unsigned i=0; i<len; ++i) {
hash += static_cast<const unsigned char*>(input)[i];
hash += hash << 10;
hash ^= hash >> 6;
}
hash += hash << 3;
hash ^= hash >> 11;
hash += hash << 15;
return hash;
}
unsigned long long m_counter;
void IncrementCounter() { ++m_counter; }
public:
unsigned long long GetSeed() const {
return m_counter;
}
void SetSeed(unsigned long long new_seed) {
m_counter = new_seed;
}
unsigned int operator ()() {
// the next random number is generated here
const auto r = JenkinsHash(&m_counter, sizeof(m_counter));
IncrementCounter();
return r;
}
// the default coontructor uses time()
// to seed the counter
CHashedCounterRng() : m_counter(time(0)) {}
// you can supply a predetermined seed here,
// or after construction with SetSeed(seed)
CHashedCounterRng(unsigned long long seed) : m_counter(seed) {}
};
int main() {
CHashedCounterRng rng;
// time()'s high bits change very slowly, so look at low digits
// if you want to verify that the seed is different between runs
const auto stored_counter = rng.GetSeed();
cout << "initial seed: " << stored_counter << endl;
for(int i=0; i<20; ++i) {
for(int j=0; j<8; ++j) {
const unsigned x = rng();
cout << setfill('0') << setw(8) << hex << x << ' ';
}
cout << endl;
}
cout << endl;
cout << "The last line again:" << endl;
rng.SetSeed(stored_counter + 19 * 8);
for(int j=0; j<8; ++j) {
const unsigned x = rng();
cout << setfill('0') << setw(8) << hex << x << ' ';
}
cout << endl << endl;
return 0;
}
In the last week i have been programming some 2-dimensional convolutions with FFTW, by passing to the frequency domain both signals, multiplying, and then coming back.
Surprisingly, I am getting the correct result only when input size is less than a fixed number!
I am posting some working code, in which i take simple initial constant matrixes of value 2 for the input, and 1 for the filter on the spatial domain. This way, the result of convolving them should be a matrix of the average of the first matrix values, i.e., 2, since it is constant. This is the output when I vary the sizes of width and height from 0 to h=215, w=215 respectively; If I set h=216, w=216, or greater, then the output gets corrupted!! I would really appreciate some clues about where could I be making some mistake. Thank you very much!
#include <fftw3.h>
int main(int argc, char* argv[]) {
int h=215, w=215;
//Input and 1 filter are declared and initialized here
float *in = (float*) fftwf_malloc(sizeof(float)*w*h);
float *identity = (float*) fftwf_malloc(sizeof(float)*w*h);
for(int i=0;i<w*h;i++){
in[i]=5;
identity[i]=1;
}
//Declare two forward plans and one backward
fftwf_plan plan1, plan2, plan3;
//Allocate for complex output of both transforms
fftwf_complex *inTrans = (fftwf_complex*) fftw_malloc(sizeof(fftwf_complex)*h*(w/2+1));
fftwf_complex *identityTrans = (fftwf_complex*) fftw_malloc(sizeof(fftwf_complex)*h*(w/2+1));
//Initialize forward plans
plan1 = fftwf_plan_dft_r2c_2d(h, w, in, inTrans, FFTW_ESTIMATE);
plan2 = fftwf_plan_dft_r2c_2d(h, w, identity, identityTrans, FFTW_ESTIMATE);
//Execute them
fftwf_execute(plan1);
fftwf_execute(plan2);
//Multiply in frequency domain. Theoretically, no need to multiply imaginary parts; since signals are real and symmetric
//their transform are also real, identityTrans[i][i] = 0, but i leave here this for more generic implementation.
for(int i=0; i<(w/2+1)*h; i++){
inTrans[i][0] = inTrans[i][0]*identityTrans[i][0] - inTrans[i][1]*identityTrans[i][1];
inTrans[i][1] = inTrans[i][0]*identityTrans[i][1] + inTrans[i][1]*identityTrans[i][0];
}
//Execute inverse transform, store result in identity, where identity filter lied.
plan3 = fftwf_plan_dft_c2r_2d(h, w, inTrans, identity, FFTW_ESTIMATE);
fftwf_execute(plan3);
//Output first results of convolution(in, identity) to see if they are the average of in.
for(int i=0;i<h/h+4;i++){
for(int j=0;j<w/w+4;j++){
std::cout<<"After convolution, component (" << i <<","<< j << ") is " << identity[j+i*w]/(w*h*w*h) << endl;
}
}std::cout<<endl;
//Compute average of data
float sum=0.0;
for(int i=0; i<w*h;i++)
sum+=in[i];
std::cout<<"Mean of input was " << (float)sum/(w*h) << endl;
std::cout<< endl;
fftwf_destroy_plan(plan1);
fftwf_destroy_plan(plan2);
fftwf_destroy_plan(plan3);
return 0;
}
Your problem has nothing to do with fftw ! It comes from this line :
std::cout<<"After convolution, component (" << i <<","<< j << ") is " << identity[j+i*w]/(w*h*w*h) << endl;
if w=216 and h=216 then `w*h*w*h=2 176 782 336. The higher limit for signed 32bit integer is 2 147 483 647. You are facing an overflow...
Solution is to cast the denominator to float.
std::cout<<"After convolution, component (" << i <<","<< j << ") is " << identity[j+i*w]/(((float)w)*h*w*h) << endl;
The next trouble that you are going to face is this one :
float sum=0.0;
for(int i=0; i<w*h;i++)
sum+=in[i];
Remember that a float has 7 useful decimal digits. If w=h=4000, the computed average will be lower than the real one. Use a double or write two loops and sum on the inner loop (localsum) before summing the outer loop (sum+=localsum) !
Bye,
Francis
For one of the projects I'm doing right now, I need to look at the performance (amongst other things) of different concurrent enabled programming languages.
At the moment I'm looking into comparing stackless python and C++ PThreads, so the focus is on these two languages, but other languages will probably be tested later. Ofcourse the comparison must be as representative and accurate as possible, so my first thought was to start looking for some standard concurrent/multi-threaded benchmark problems, alas I couldn't find any decent or standard, tests/problems/benchmarks.
So my question is as follows: Do you have a suggestion for a good, easy or quick problem to test the performance of the programming language (and to expose it's strong and weak points in the process)?
Surely you should be testing hardware and compilers rather than a language for concurrency performance?
I would be looking at a language from the point of view of how easy and productive it is in terms of concurrency and how much it 'insulates' the programmer from making locking mistakes.
EDIT: from past experience as a researcher designing parallel algorithms, I think you will find in most cases the concurrent performance will depend largely on how an algorithm is parallelised, and how it targets the underlying hardware.
Also, benchmarks are notoriously unequal; this is even more so in a parallel environment. For instance, a benchmark that 'crunches' very large matrices would be suited to a vector pipeline processor, whereas a parallel sort might be better suited to more general purpose multi core CPUs.
These might be useful:
Parallel Benchmarks
NAS Parallel Benchmarks
Well, there are a few classics, but different tests emphasize different features. Some distributed systems may be more robust, have more efficient message-passing, etc. Higher message overhead can hurt scalability, since it the normal way to scale up to more machines is to send a larger number of small messages. Some classic problems you can try are a distributed Sieve of Eratosthenes or a poorly implemented fibonacci sequence calculator (i.e. to calculate the 8th number in the series, spin of a machine for the 7th, and another for the 6th). Pretty much any divide-and-conquer algorithm can be done concurrently. You could also do a concurrent implementation of Conway's game of life or heat transfer. Note that all of these algorithms have different focuses and thus you probably will not get one distributed system doing the best in all of them.
I'd say the easiest one to implement quickly is the poorly implemented fibonacci calculator, though it places too much emphasis on creating threads and too little on communication between those threads.
Surely you should be testing hardware
and compilers rather than a language
for concurrency performance?
No, hardware and compilers are irrelevant for my testing purposes. I'm just looking for some good problems that can test how well code, written in one language, can compete against code from another language. I'm really testing the constructs available in the specific languages to do concurrent programming. And one of the criteria is performance (measured in time).
Some of the other test criteria I'm looking for are:
how easy is it to write correct code; because as we all know concurrent programming is harder then writing single threaded programs
what is the technique used to to concurrent programming: event-driven, actor based, message parsing, ...
how much code must be written by the programmer himself and how much is done automatically for him: this can also be tested with the given benchmark problems
what's the level of abstraction and how much overhead is involved when translated back to machine code
So actually, I'm not looking for performance as the only and best parameter (which would indeed send me to the hardware and the compilers instead of the language itself), I'm actually looking from a programmers point of view to check what language is best suited for what kind of problems, what it's weaknesses and strengths are and so on...
Bare in mind that this is just a small project and the tests are therefore to be kept small as well. (rigorous testing of everything is therefore not feasible)
I have decided to use the Mandelbrot set (the escape time algorithm to be more precise) to benchmark the different languages.
It fits me quite well as the original algorithm can easily be implemented and creating the multi threaded variant from it is not that much work.
below is the code I currently have. It is still a single threaded variant, but I'll update it as soon as I'm satisfied with the result.
#include <cstdlib> //for atoi
#include <iostream>
#include <iomanip> //for setw and setfill
#include <vector>
int DoThread(const double x, const double y, int maxiter) {
double curX,curY,xSquare,ySquare;
int i;
curX = x + x*x - y*y;
curY = y + x*y + x*y;
ySquare = curY*curY;
xSquare = curX*curX;
for (i=0; i<maxiter && ySquare + xSquare < 4;i++) {
ySquare = curY*curY;
xSquare = curX*curX;
curY = y + curX*curY + curX*curY;
curX = x - ySquare + xSquare;
}
return i;
}
void SingleThreaded(int horizPixels, int vertPixels, int maxiter, std::vector<std::vector<int> >& result) {
for(int x = horizPixels; x > 0; x--) {
for(int y = vertPixels; y > 0; y--) {
//3.0 -> so we always have -1.5 -> 1.5 as the window; (x - (horizPixels / 2) will go from -horizPixels/2 to +horizPixels/2
result[x-1][y-1] = DoThread((3.0 / horizPixels) * (x - (horizPixels / 2)),(3.0 / vertPixels) * (y - (vertPixels / 2)),maxiter);
}
}
}
int main(int argc, char* argv[]) {
//first arg = length along horizontal axis
int horizPixels = atoi(argv[1]);
//second arg = length along vertical axis
int vertPixels = atoi(argv[2]);
//third arg = iterations
int maxiter = atoi(argv[3]);
//fourth arg = threads
int threadCount = atoi(argv[4]);
std::vector<std::vector<int> > result(horizPixels, std::vector<int>(vertPixels,0)); //create and init 2-dimensional vector
SingleThreaded(horizPixels, vertPixels, maxiter, result);
//TODO: remove these lines
for(int y = 0; y < vertPixels; y++) {
for(int x = 0; x < horizPixels; x++) {
std::cout << std::setw(2) << std::setfill('0') << std::hex << result[x][y] << " ";
}
std::cout << std::endl;
}
}
I've tested it with gcc under Linux, but I'm sure it works under other compilers/Operating Systems as well. To get it to work you have to enter some command line arguments like so:
mandelbrot 106 500 255 1
the first argument is the width (x-axis)
the second argument is the height (y-axis)
the third argument is the number of maximum iterations (the number of colors)
the last ons is the number of threads (but that one is currently not used)
on my resolution, the above example gives me a nice ASCII-art representation of a Mandelbrot set. But try it for yourself with different arguments (the first one will be the most important one, as that will be the width)
Below you can find the code I hacked together to test the multi threaded performance of pthreads. I haven't cleaned it up and no optimizations have been made; so the code is a bit raw.
the code to save the calculated mandelbrot set as a bitmap is not mine, you can find it here
#include <cstdlib> //for atoi
#include <iostream>
#include <iomanip> //for setw and setfill
#include <vector>
#include "bitmap_Image.h" //for saving the mandelbrot as a bmp
#include <pthread.h>
pthread_mutex_t mutexCounter;
int sharedCounter(0);
int percent(0);
int horizPixels(0);
int vertPixels(0);
int maxiter(0);
//doesn't need to be locked
std::vector<std::vector<int> > result; //create 2 dimensional vector
void *DoThread(void *null) {
double curX,curY,xSquare,ySquare,x,y;
int i, intx, inty, counter;
counter = 0;
do {
counter++;
pthread_mutex_lock (&mutexCounter); //lock
intx = int((sharedCounter / vertPixels) + 0.5);
inty = sharedCounter % vertPixels;
sharedCounter++;
pthread_mutex_unlock (&mutexCounter); //unlock
//exit thread when finished
if (intx >= horizPixels) {
std::cout << "exited thread - I did " << counter << " calculations" << std::endl;
pthread_exit((void*) 0);
}
//set x and y to the correct value now -> in the range like singlethread
x = (3.0 / horizPixels) * (intx - (horizPixels / 1.5));
y = (3.0 / vertPixels) * (inty - (vertPixels / 2));
curX = x + x*x - y*y;
curY = y + x*y + x*y;
ySquare = curY*curY;
xSquare = curX*curX;
for (i=0; i<maxiter && ySquare + xSquare < 4;i++){
ySquare = curY*curY;
xSquare = curX*curX;
curY = y + curX*curY + curX*curY;
curX = x - ySquare + xSquare;
}
result[intx][inty] = i;
} while (true);
}
int DoSingleThread(const double x, const double y) {
double curX,curY,xSquare,ySquare;
int i;
curX = x + x*x - y*y;
curY = y + x*y + x*y;
ySquare = curY*curY;
xSquare = curX*curX;
for (i=0; i<maxiter && ySquare + xSquare < 4;i++){
ySquare = curY*curY;
xSquare = curX*curX;
curY = y + curX*curY + curX*curY;
curX = x - ySquare + xSquare;
}
return i;
}
void SingleThreaded(std::vector<std::vector<int> >& result) {
for(int x = horizPixels - 1; x != -1; x--) {
for(int y = vertPixels - 1; y != -1; y--) {
//3.0 -> so we always have -1.5 -> 1.5 as the window; (x - (horizPixels / 2) will go from -horizPixels/2 to +horizPixels/2
result[x][y] = DoSingleThread((3.0 / horizPixels) * (x - (horizPixels / 1.5)),(3.0 / vertPixels) * (y - (vertPixels / 2)));
}
}
}
void MultiThreaded(int threadCount, std::vector<std::vector<int> >& result) {
/* Initialize and set thread detached attribute */
pthread_t thread[threadCount];
pthread_attr_t attr;
pthread_attr_init(&attr);
pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
for (int i = 0; i < threadCount - 1; i++) {
pthread_create(&thread[i], &attr, DoThread, NULL);
}
std::cout << "all threads created" << std::endl;
for(int i = 0; i < threadCount - 1; i++) {
pthread_join(thread[i], NULL);
}
std::cout << "all threads joined" << std::endl;
}
int main(int argc, char* argv[]) {
//first arg = length along horizontal axis
horizPixels = atoi(argv[1]);
//second arg = length along vertical axis
vertPixels = atoi(argv[2]);
//third arg = iterations
maxiter = atoi(argv[3]);
//fourth arg = threads
int threadCount = atoi(argv[4]);
result = std::vector<std::vector<int> >(horizPixels, std::vector<int>(vertPixels,21)); // init 2-dimensional vector
if (threadCount <= 1) {
SingleThreaded(result);
} else {
MultiThreaded(threadCount, result);
}
//TODO: remove these lines
bitmapImage image(horizPixels, vertPixels);
for(int y = 0; y < vertPixels; y++) {
for(int x = 0; x < horizPixels; x++) {
image.setPixelRGB(x,y,16777216*result[x][y]/maxiter % 256, 65536*result[x][y]/maxiter % 256, 256*result[x][y]/maxiter % 256);
//std::cout << std::setw(2) << std::setfill('0') << std::hex << result[x][y] << " ";
}
std::cout << std::endl;
}
image.saveToBitmapFile("~/Desktop/test.bmp",32);
}
good results can be obtained using the program with the following arguments:
mandelbrot 5120 3840 256 3
that way you will get an image that is 5 * 1024 wide; 5 * 768 high with 256 colors (alas you will only get 1 or 2) and 3 threads (1 main thread that doesn't do any work except creating the worker threads, and 2 worker threads)
Since the benchmarks game moved to a quad-core machine September 2008, many programs in different programming languages have been re-written to exploit quad-core - for example, the first 10 mandelbrot programs.