Will multithreading provide improved performance?

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I am new to programming in general so please keep that in mind when you answer my question.

I have a program that takes a large 3D array (1 billion elements) and sums up elements along the various axis to produce a 2D array of a projection of each side of the data. The problem here is that it is very ram intensive as the program is constantly fetching information from the ram, both reading and writing.

The question is, will i gain any performance increases if i multithread the program or will I end up running into a RAM access bottleneck? When i say multithreading, i only mean multithreading for 2 or 4 cores, no more.

If it helps, my current computer configuration is 2.4ghz core2 quad, 1033 fsb, 4gb ram at 667mhz.

Thanks in advance,

-Faken

Edit:

It seems to me that people here are much more interested in this question that I had first expected. I'll expand the question and post some code for those who are interested.

First of all, a little background on me so you understand where I'm coming from. I am a mechanical engineering graduate student who some how managed to pick a topic that pretty much had nothing to do with mechanical engineering. I have taken 1 course in introductory java (forced) approximately 5 years ago and have never touched programming until about a month ago when i started my thesis in earnest. I have also taken (again forced, still don't know why) a course in electronics and computer engineering, we dealt with micro-controllers (8-bit), their inner workings, and some ASM coding for them. Other than that, I know next to nothing about programming.

Here is the code:

int dim = 1000;
int steps = 7 //ranges from 1 to  255

for (int stage = 1; stage < steps; stage++)
for (int j = 0; j < dim; j++)
    for (int i = 0; i < dim; i++)
    {
        sum = 0;
        for (int k = 0; k < dim; k++)
            if (partMap[(((i * dim) + k) * dim) + j] >= stage)
                sum++;

        projection[(j*dim) + i] = sum;
    }

This section of code operates on the z-axis only. The main data, due to the way it was constructed, has a strange addressing system but you don't need to worry about that. There is also other code for doing the projections of other sides of the cube but they do very different things.


There is only one way to optimize code: figure out what you're doing that's slow, and do less of it. A special case of "doing less of it" is to do something else instead that's faster.

So first of all, here's what I'm doing based on your posted code:

#include <fstream>
#include <sstream>
using std::ios_base;

template<typename Iterator, typename Value>
void iota(Iterator start, Iterator end, Value val) {
    while (start != end) {
        *(start++) = val++;
    }
}

int main() {

    const int dim = 1000;
    const int cubesize = dim*dim*dim;
    const int squaresize = dim*dim;
    const int steps = 7; //ranges from 1 to  255
    typedef unsigned char uchar;

    uchar *partMap = new uchar[cubesize];
    // dummy data. I timed this separately and it takes about
    // a second, so I won't worry about its effect on overall timings.
    iota(partMap, partMap + cubesize, uchar(7));
    uchar *projection = new uchar[squaresize];

    for (int stage = 1; stage < steps; stage++) {
        for (int j = 0; j < dim; j++) {
                for (int i = 0; i < dim; i++)
                {
                        int sum = 0;
                        for (int k = 0; k < dim; k++)
                            if (partMap[(((i * dim) + k) * dim) + j] >= stage)
                                sum++;

                        projection[(j*dim) + i] = sum;
                }
        }

        std::stringstream filename;
        filename << "results" << stage << ".bin";
        std::ofstream file(filename.str().c_str(),
            ios_base::out | ios_base::binary | ios_base::trunc);
        file.write((char *)projection, squaresize);
    }

    delete[] projection;
    delete[] partMap;
}

(Edit: just noticed that "projection" should be an array of int, not uchar. My bad. This will make a difference to some of the timings, but hopefully not too big of one.)

Then I copied result*.bin to gold*.bin, so I can check my future changes as follows:

$ make big -B CPPFLAGS="-O3 -pedantic -Wall" && time ./big; for n in 1 2 3 4 5
6; do diff -q results$n.bin gold$n.bin; done
g++  -O3 -pedantic -Wall   big.cpp   -o big

real    1m41.978s
user    1m39.450s
sys     0m0.451s

OK, so 100 seconds at the moment.

So, speculating that it's striding through the billion-item data array that's slow, let's try only going through once, instead of once per stage:

    uchar *projections[steps];
    for (int stage = 1; stage < steps; stage++) {
         projections[stage] = new uchar[squaresize];
    }

    for (int j = 0; j < dim; j++) {
            for (int i = 0; i < dim; i++)
            {
                    int counts[256] = {0};
                    for (int k = 0; k < dim; k++)
                            counts[partMap[(((i * dim) + k) * dim) + j]]++;

                    int sum = 0;
                    for (int idx = 255; idx >= steps; --idx) {
                        sum += counts[idx];
                    }
                    for (int stage = steps-1; stage > 0; --stage) {
                        sum += counts[stage];
                        projections[stage][(j*dim) + i] = sum;
                    }
            }
    }

    for (int stage = 1; stage < steps; stage++) {
        std::stringstream filename;
        filename << "results" << stage << ".bin";
        std::ofstream file(filename.str().c_str(),
            ios_base::out | ios_base::binary | ios_base::trunc);
        file.write((char *)projections[stage], squaresize);
    }

    for (int stage = 1; stage < steps; stage++) delete[] projections[stage];
    delete[] partMap;

It's a bit faster:

$ make big -B CPPFLAGS="-O3 -pedantic -Wall" && time ./big; for n in 1 2 3 4 5
6; do diff -q results$n.bin gold$n.bin; done
g++  -O3 -pedantic -Wall   big.cpp   -o big

real    1m15.176s
user    1m13.772s
sys     0m0.841s

Now, steps is quite small in this example, so we're doing a lot of unnecessary work with the "counts" array. Without even profiling, I'm guessing that counting to 256 twice (once to clear the array and once to sum it) is quite significant compared with counting to 1000 (to run along our column). So let's change that:

    for (int j = 0; j < dim; j++) {
            for (int i = 0; i < dim; i++)
            {
                    // steps+1, not steps. I got this wrong the first time,
                    // which at least proved that my diffs work as a check
                    // of the answer...
                    int counts[steps+1] = {0};
                    for (int k = 0; k < dim; k++) {
                        uchar val = partMap[(((i * dim) + k) * dim) + j];
                        if (val >= steps)
                            counts[steps]++;
                        else counts[val]++;
                    }

                    int sum = counts[steps];
                    for (int stage = steps-1; stage > 0; --stage) {
                        sum += counts[stage];
                        projections[stage][(j*dim) + i] = sum;
                    }
            }
    }

Now we're only using as many buckets as we actually need.

$ make big -B CPPFLAGS="-O3 -pedantic -Wall" && time ./big; for n in 1 2 3 4 5
6; do diff -q results$n.bin gold$n.bin; done
g++  -O3 -pedantic -Wall   big.cpp   -o big

real    0m27.643s
user    0m26.551s
sys     0m0.483s

Hurrah. The code is nearly 4 times as fast as the first version, and produces the same results. All I've done is change what order the maths is done: we haven't even looked at multi-threading or prefetching yet. And I haven't attempted any highly technical loop optimisation, just left it to the compiler. So this can be considered a decent start.

However it's still taking an order of magnitude longer than the 1s which iota runs in. So there are probably big gains still to find. One main difference is that iota runs over the 1d array in sequential order, instead of leaping about all over the place. As I said in my first answer, you should aim to always use sequential order on the cube.

So, let's make a one-line change, switching the i and j loops:

            for (int i = 0; i < dim; i++)
    for (int j = 0; j < dim; j++) {

This still isn't sequential order, but it does mean we're focussing on one million-byte slice of our cube at a time. A modern CPU has at least 4MB cache, so with a bit of luck we'll only hit main memory for any given part of the cube once in the entire program. With even better locality we could reduce the traffic in and out of L1 cache, too, but main memory is the slowest.

How much difference does it make?

$ make big -B CPPFLAGS="-O3 -pedantic -Wall" && time ./big; for n in 1 2 3 4 5
6; do diff -q results$n.bin gold$n.bin; done
g++  -O3 -pedantic -Wall   big.cpp   -o big

real    0m8.221s
user    0m4.507s
sys     0m0.514s

Not bad. In fact, this change alone brings the original code from 100s to 20s. So this is responsible for a factor of 5, and everything else I did is responsible for another factor of 5 (I think the difference between 'user' and 'real' time in the above is mostly accounted for by the fact my virus scanner is running, which it wasn't earlier. 'user' is how much time the program occupied a CPU, 'real' includes time spent suspended, either waiting on I/O or giving another process time to run).

Of course, my bucket sort relies on the fact that whatever we're doing with the values in each column is commutative and associative. Reducing the number of buckets only worked because large values are all treated the same. This might not be true for all your operations, so you'll have to look at the inner loop of each one in turn to figure out what to do with it.

And the code is a bit more complicated. Instead of running over the data doing "blah" for each stage, we're computing all the stages at the same time in a single run over the data. If you start doing row and column computations in a single pass, as I recommended in my first answer, this will get worse. You may have to start breaking your code into functions to keep it readable.

Finally, a lot of my performance gain came from an optimisation for the fact that "steps" is small. With steps=100, I get:

$ make big -B CPPFLAGS="-O3 -pedantic -Wall" && time ./big; for n in 1 2 3 4 5
6; do diff -q results$n.bin gold$n.bin; done
g++  -O3 -pedantic -Wall   big.cpp   -o big

real    0m22.262s
user    0m10.108s
sys     0m1.029s

This isn't so bad. With steps=100 the original code probably takes about 1400 seconds, although I'm not going to run it to prove that. But it's worth remembering that I haven't completely taken away the time dependency on "steps", just made it sub-linear.