In my kernel, if a condition is met, I update an item of the output buffer
if (condition(input[i])) //?
output[i] = 1;
otherwise the output may stay the same, having value of 0.
The density of updates are quite unpredictable, depending on the input. Furthermore which output location will be updated is also not known. (i may force them though, in some cases)
My question is, is it better to write all items, to achieve coalescing, or do a selective write?
output[i] = condition(input[i]); //?
Would you mind discussing your statements?
Coalescing is achieved even if some threads in the warp do not participate in the load or store, as long as all participating threads satisfy the requirements of coalescing. So conditional writes should have no effect on memory throughput.
However, doing a conditional write may involve additional instructions due to involving a branch (this would probably explain, for example, the difference in performance measured by Eugene in his answer).
On my setup kernel that does conditional set (option 1) runs for 1.727 us and option 2 1.399 us. This is my code (setConditional is the faster one):
__global__ void conditionalSet(unsigned int* array) {
if ((threadIdx.x & 3) == 0) {
array[threadIdx.x] = 1;
}
}
__global__ void setConditional(unsigned int* array) {
array[threadIdx.x] = (threadIdx.x & 3) == 0 ? 1 : 0;
}
Related
Recently I have faced a problem when the program running the memory keep increasing, and when program is closed the memory would restore normal level. Obviously, it's a memory leak. After some work, I have located the code responsible, but I don't know why? The program's work flow is simple:
first use lidar api to get point cloud and image data;
then transport to next tbb flow graph to process these data;
finally use open3d api to visualzie them.
In the first step, the lidar itself's api use asio to asynchronously invoke some callback function to transport data, so I create some tbb concurrent_queue to store these data, and a align function to match cloud and image with timestamp. The problem is in the align function. In the function, I create a vector<shared_ptr<open3d::..::PointCloud>> and use iterator to store point cloud elements. However, I found when the function complete, the shared_ptr use count don't reduce . Similar but simpler example code like this:
std::pair<std::shared_ptr<int>, int> helper() {
auto a = std::make_shared<int>(90);
auto c = 100;
std::vector<std::pair<std::shared_ptr<int>, int>> container;
container.reserve(5);
auto iter = container.begin();
for (int i = 0; i < 3; i++) {
*iter = std::make_pair(a, c);
iter++;
}
return *(iter-1);
}
int main() {
auto b = helper();
std::cout << "shared_ptr use count: " << std::get<0>(b).use_count() << std::endl;
return 0;
}
Ubuntu 20.04 + gcc 9.4, the print result is shared_ptr use count: 4.
Why the vector can't be auto destroyed when function is completed? Hope someone kindly explain this problem.
Thanks #Retired Ninja! The root of the problem is vector.reserve just reserve capacity not physical space. So the vector space after reserve is 0. The following iterator operation is assumed point to some undifined memory. While the result can be transport to main function with no value error, the shared_ptr use count can't reduce to 1 after function call.
To solve the problem, One can just modify the reserve to resize, which can change physical space of the vector and iterator point to defined memory space. Or avoid use iterator, just use push_back and return back().
I am learning MQL4 language and am using this Code to plot a Simple moving Average, the Code works fine, but when I load it up on my MT4 it takes a lot of time, am I missing something ?
int start() // Special function start()
{
int i, // Bar index
n, // Formal parameter
Counted_bars; // Number of counted bars
// Sum of Low values for period
// --------------------------------------------------------------------
Counted_bars=IndicatorCounted(); // Number of counted bars
i=Bars-Counted_bars-1; // Index of the first uncounted
while(i>=0) // Loop for uncounted bars
{
Buf_0[i]=(iMA(Symbol(),PERIOD_M5,200,i,MODE_EMA,PRICE_HIGH,0);
i--; // Calculating index of the next bar
}
// --------------------------------------------------------------------
return; // Exit the special funct. start()
}
// --------------------------------------------------------------------
Q : am I missing something?
No, this is a standard feature to process all the Bars back, towards the earliest parts of the history.
If your intentions require a minimum setup-time, it is possible to "shorten" the re-painted part of the history to just, say, last week, not going all the way back all the Bars-number of Bars a few years back, as all that data have been stored in the OHLCV-history database.
That "shortened"-part of the history will this way become as long as your needs can handle and not a single bar "longer".
Hooray, The Problem was solved.
BONUS PART :
Given your code instructs to work with EMA, not SMA, there is one more vector of attack onto the shortest possible time.
For EMA, any next Bar value will become a product of alfa * High[next] added to a previously known as EMA[next+1]
Where a constant alfa = 2. / ( N_period + 1 ) is known and constant across the whole run of the history processed.
This approach helped me gain about ~20-30 [us] FASTER processing for a 20-cell Price-vector, when using this algorithmic shortcut on an array of float32-values compared to cell-by-cell processing. Be sure to benchmark the code for your use-case and may polish further tricks with using different call-signatures of iHigh() instead of accessing an array of High[]-s for any potential speedups, if in utmost need to shave-off any further [us] possible.
I am trying to learn how to use the Sparse Coding algorithm with the mlpack library. When I call Encode() on my instance of mlpack::sparse_coding:SparseCoding, I get the error
[WARN] There are 63 inactive atoms. They will be reinitialized randomly.
error: solve(): solution not found
Is it simply that the algorithm cannot learn a latent representation of the data. Or perhaps it is my usage? The relevant section follows
EDIT: One line was modified to fix an unrelated error, but the original error remains.
double* Application::GetSparseCodes(arma::mat* trainingExample, int atomCount)
{
double* latentRep = new double[atomCount];
mlpack::sparse_coding::SparseCoding<mlpack::sparse_coding::DataDependentRandomInitializer> sc(*trainingExample, Utils::ATOM_COUNT, 1.0);
sc.Encode(Utils::MAX_ITERATIONS);
arma::mat& latentRepMat = sc.Codes();
for (int i = 0; i < atomCount; i++)
latentRep[i] = latentRepMat.at(i, 0);
return latentRep;
}
Some relevant parameters
const static int IMAGE_WIDTH = 20;
const static int IMAGE_HEIGHT = 20;
const static int PIXEL_COUNT = IMAGE_WIDTH * IMAGE_HEIGHT;
const static int ATOM_COUNT = 64;
const static int MAX_ITERATIONS = 100000;
This could be one of a handful of issues but given the description it's a little difficult to tell which of these it is (or if it is something else entirely). However, these three ideas should provide a good place to start:
Matrices in mlpack are column-major. That means each observation should represent a column. If you use mlpack::data::Load() to load, e.g., a CSV file (which are generally one row per observation), it will automatically transpose the dataset. SparseCoding will act oddly if you pass it transposed data. See also http://www.mlpack.org/doxygen.php?doc=matrices.html.
If there are 63 inactive atoms, then only one atom is actually active (given that ATOM_COUNT is 64). This means that the algorithm has found that the best way to represent the dictionary (at a given step) uses only one atom. This could happen if the matrix you are passing consists of all zeros.
mlpack will provide verbose output, which may also be helpful for debugging. Usually this is used by using mlpack's CLI class to parse command-line input, but you can enable verbose output with mlpack::Log::Info.ignoreInput = false. You may obtain a lot of output that way, but it will give a better look at what is going on...
The mlpack project has its own mailing list where you may be likely to get a quicker or more comprehensive response, by the way.
I'm having trouble here. I launch two kernels , check if some value is the one expected (memcpy to the host), if it is I stop, if it isn't I launch the two kernels again.
the first kernel:
__global__ void aco_step(const KPDeviceData* data)
{
int obj = threadIdx.x;
int ant = blockIdx.x;
int id = threadIdx.x + blockIdx.x * blockDim.x;
*(data->added) = 1;
while(*(data->added) == 1)
{
*(data->added) = 0;
//check if obj fits
int fits = (data->obj_weights[obj] + data->weight[ant] <= data->max_weight);
fits = fits * !(getElement(data->selections, data->selections_pitch, ant, obj));
if(obj == 0)
printf("ant %d going..\n", ant);
__syncthreads();
...
The code goes on after this. But that printf never gets printed, that syncthreads is there just for debugging purposes.
The "added" variable was shared, but since shared memory is a PITA and usually throws bugs in the code, i just removed it for now. This "added" variable isn't the smartest thing to do but it's faster than the alternative, which is checking if any variable within an array is some value on the host and deciding to keep iterating or not.
The getElement, simply does the matrix memory calculation with the pitch to access the right position and returns the element there:
int* el = (int*) ((char*)mat + row * pitch) + col;
return *el;
The obj_weights array has the right size, n*sizeof(int). So does the weight array, ants*sizeof(float). So they aren't out of bounds.
The kernel after this one has a printf right on the beginning, and it doesn't get printed either and after the printf it sets a variable on the device memory, and this memory is copied to the CPU after the kernel finished, and it isn't the right value when I print it in the CPU code. So I think this kernel is doing something illegal and the second one doesn't even get launched.
I'm testing some instances, when I launch 8 blocks and 512 threads, it runs OK. 32 blocks, 512 threads, OK. But 8 blocks and 1024 threads, and this happens, the kernel doesn't work, neither 32 blocks and 1024 threads.
Am I doing something wrong? Memory access? Am I launching too many threads?
edit: tried removing the "added" variable and the while loop, so it should execute just once. Still doesnt work, nothing gets printed, even if the printf is right after the three initial lines and the next kernel also doesn't print anything.
edit: another thing, I'm using a GTX 570, so the "Maximum number of threads per block" is 1024 according to http://en.wikipedia.org/wiki/CUDA. Maybe I'll just stick with 512 maximum or check on how higher I can put this value.
__syncthreads() inside conditional code is only allowed if the condition evaluates identically on all threads of a block.
In your case the condition suffers a race condition and is nondeterministic, so it most probably evaluates to different results for different threads.
printf() output is only displayed after the kernel finishes successfully. In this case it doesn't due to the problem mentioned above, so the output never shows up. You could have figured out this by testing the return codes all CUDA function calls for errors.
in a simple MPI program I have used a column wise division of a large matrix.
How can I order the output so that each matrix appears next to the other ordered ?
I have tried this simple code the effect is quite different from the wanted:
for(int i=0;i<10;i++)
{
for(int k=0;k<numprocs;k++)
{
if (my_id==k){
for(int j=1;j<10;j++)
printf("%d",data[i][j]);
}
MPI_Barrier(com);
}
if(my_id==0)
printf("\n");
}
Seems that each process has his own stdout and so is impossible to have ordered lines output without sending all the data to one master which will print out. Is my guess true ? Or what I'm doing wrong ?
You guessed right. The MPI standard does not specify how stdout from different nodes should be collected for printing at the originating process. It is often the case that when multiple processes are doing prints the output will get merged in an unspecified way. fflush doesn't help.
If you want the output ordered in a certain way, the most portable method would be to send the data to the master process for printing.
For example, in pseudocode:
if (rank == 0) {
print_col(0);
for (i = 1; i < comm_size; i++) {
MPI_Recv(buffer, .... i, ...);
print_col(i);
}
} else {
MPI_Send(data, ..., 0, ...);
}
Another method which can sometimes work would be to use barries to lock step processes so that each process prints in turn. This of course depends on the MPI Implementation and how it handles stdout.
for(i = 0; i < comm_size; i++) {
MPI_Barrier(MPI_COMM_WORLD);
if (i == rank) {
printf(...);
}
}
Of course, in production code where the data is too large to print sensibly anyway, data is eventually combine by having each process writing to a separate file and merged separately, or using MPI I/O (defined in the MPI2 standards) to coordinate parallel writes.
I produced ordered output to a file before using the exact same method. You could try printing to a temporary file, printing the contents of said file and then deleting it.
Have the root processor do all of the printing. Use MPI_Send/MPI_Recv or MPI_Gather (or whatever) to send the data in turn from each processor to the root.
To solve this problem you can use short sleep. I use and then it works in 99%
printf("text nr 1\n");
MPI_Barrier(MPI_COMM_WORLD);
usleep(100);
printf("text nr 2\n");
It's not very elegant but works.