PyCUDA May 2018

pycuda@tiker.net

3 participants
8 discussions

Re: [PyCUDA] Create gpuarrays on different GPUs

by Andreas Kloeckner

Zhangsheng Lai <dunno.noe(a)gmail.com> writes: > My 'can access' simply means that I'm able to access the values in the > variable in python by typing x1 or x2. My understanding is that if the > variables are stored on different GPUs, then I should be able to type x1 > and get its values when ctx1 is active and similarly, I can type x2 and get > the x2 values when ctx2 is active, not when ctx1 is active. You could measure bandwidths to between host/presumed gpu1/presumed gpu2 to ascertain where the data actually resides if you have doubts about that/don't trust the API. Andreas

2 years, 10 months

Re: [PyCUDA] Create gpuarrays on different GPUs

by Andreas Kloeckner

Zhangsheng Lai <dunno.noe(a)gmail.com> writes: > with the setup above, I tried to check by poping ctx2 and pushing ctx1, can > I access x1 and not x2 and vice versa, popping ctx1 and pushing ctx2, I can > access x2 and not x1. However, I realise that I can access x1 and x2 in > both contexts. Can you clarify what you mean by 'can access'? I'm guessing 'submit kernel launches with that pointer as an argument'? Andreas

2 years, 10 months

Create gpuarrays on different GPUs

by Zhangsheng Lai

Hi, I'm trying to create different GPU arrays on different GPUs. ``` import pycuda import pycuda.driver as cuda from pycuda.compiler import SourceModule import pycuda.curandom as curandom d = 2 ** 15 cuda.init() dev1 = cuda.Device(1) ctx1 = dev1.make_context() curng1 = curandom.XORWOWRandomNumberGenerator() x1 = curng1.gen_normal((d,d), dtype = np.float32) # so x1 is stored in GPU 1 memory ctx1.pop() # clearing ctx of GPU1 dev2 = cuda.Device(1) ctx2 = dev2.make_context() curng2 = curandom.XORWOWRandomNumberGenerator() x2 = curng2.gen_normal((d,d), dtype = np.float32) # so x2 is stored in GPU 2 ``` with the setup above, I tried to check by poping ctx2 and pushing ctx1, can I access x1 and not x2 and vice versa, popping ctx1 and pushing ctx2, I can access x2 and not x1. However, I realise that I can access x1 and x2 in both contexts. Thus I'm wondering if my assumptions of x1 stored in GPU1 and x2 stored in GPU2 are correct, or if it is actually the UVA and peer access that allows me to access both x1 and x2 even if only one of the two ctx is active. Thanks, Zhangsheng

2 years, 10 months

Re: [PyCUDA] I have been through the introduction, checked the faq, where to get help on pycuda -

by Andreas Kloeckner

Dear Anthony, "Anthony Pleticos" <anthonypleticos(a)unswalumni.com> writes: > I would like to know where people can go for 'assistance' in difficulties in > applying the pycuda. > > I could not find it in the > https://wiki.tiker.net/PyCuda/FrequentlyAskedQuestions and StackExchange > does not answer my specific issue, especially when running either the > tutorial and/or examples such as demo.py and hello_gpu in > C:\Python36\pycuda-2017.1.1\examples. > > > > I tried the follow step-by-step the tutorial at > https://documen.tician.de/pycuda/tutorial.html#where-to-go-from-here . > > The problem comes under the heading "Executing a Kernel" where I have a c++ > like module in the py file. > > mod = SourceModule(""" > > __global__ void method(args) > > { > > C++ like code > > } > > """) > > It happens under the sample code in your tutorial or the > pycuda-2017.1.1\examples I get the error message > > nvcc fatal : Value 'sm_21' is not defined for option 'gpu-architecture' Generally, the mailing list (cc'd, needs subscription to post) is a good place for requests like this. In your case, you seem to have fairly old GPU ("sm_21") that's no longer supported by your compiler (nvcc). Downgrading the CUDA toolkit may help. Andreas

2 years, 11 months

Example SparseSolve

by MarbHarmsen

I'm currently trying to build a simple FEA solver in python using an incomplete Cholesky decomposition preconditioned conjugate gradient method. I have exported an example stiffness matrix from my old (and slow) code into a symmetric .mtx file. This matrix is imported into the example code at https://wiki.tiker.net/PyCuda/Examples/SparseSolve. Running the code resulted in errors. The first part was solved by pymetis. However now a new problem appeared. I get a type error: "TypeError: No registered converter was able to produce a C++ rvalue of type int from this Python object of type numpy.int32". The full traceback: In [1]: runfile('/home/bram/Documenten/TUDelft/Thesis/Phyton/topopt/src/cgCUDATest.py', wdir='/home/bram/Documenten/TUDelft/Thesis/Phyton/topopt/src') starting... building... Traceback (most recent call last): File "<ipython-input-56-3395cc199316>", line 1, in <module> runfile('/home/bram/Documenten/TUDelft/Thesis/Phyton/topopt/src/cgCUDATest.py', wdir='/home/bram/Documenten/TUDelft/Thesis/Phyton/topopt/src') File "/home/bram/.anaconda3/lib/python3.6/site-packages/spyder/utils/site/sitecustomize.py", line 705, in runfile execfile(filename, namespace) File "/home/bram/.anaconda3/lib/python3.6/site-packages/spyder/utils/site/sitecustomize.py", line 102, in execfile exec(compile(f.read(), filename, 'exec'), namespace) File "/home/bram/Documenten/TUDelft/Thesis/Phyton/topopt/src/cgCUDATest.py", line 71, in <module> main_cg() File "/home/bram/Documenten/TUDelft/Thesis/Phyton/topopt/src/cgCUDATest.py", line 21, in main_cg spmv = PacketedSpMV(csr_mat, 'symmetric', csr_mat.dtype) File "/home/bram/.anaconda3/lib/python3.6/site-packages/pycuda-2017.1-py3.6-linux-x86_64.egg/pycuda/sparse/packeted.py", line 127, in __init__ xadj=adj_mat.indptr, adjncy=adj_mat.indices) File "/home/bram/.anaconda3/lib/python3.6/site-packages/pymetis/__init__.py", line 120, in part_graph return part_graph(nparts, xadj, adjncy, vweights, eweights, recursive) TypeError: No registered converter was able to produce a C++ rvalue of type int from this Python object of type numpy.int32 Do I need to change the python files in pymetis? -- Sent from: http://pycuda.2962900.n2.nabble.com/

2 years, 11 months

Re: [PyCUDA] Invalid resource handle error

by Andreas Kloeckner

Zhangsheng Lai <dunno.noe(a)gmail.com> writes: > Hi, > > I'm trying to do some updates to a state which is a binary array. gputid is > a GPU thread class (https://wiki.tiker.net/PyCuda/Examples/MultipleThreads) > and it stores the state and the index of the array to be updated in another > class which can be accessed with gputid.mp.x_gpu and gputid.mp.neuron_gpu > respectively. Below is my kernel that takes in the gputid and performs the > update of the state. However, it the output of the code is not consistent > as it runs into errors and executes perfectly when i run it multiple times. > The error msg makes no sense to me: > > File "/root/anaconda3/lib/python3.6/site-packages/pycuda/driver.py", line > 447, in function_prepared_call > func._set_block_shape(*block) > pycuda._driver.LogicError: cuFuncSetBlockShape failed: invalid resource > handle I think the right way to interpret this is that if you cause an on-device segfault, the GPU context dies, and all the handles of objects contained in it (including the function) become invalid. HTH, Andreas

2 years, 11 months

Invalid resource handle error

by Zhangsheng Lai

Hi, I'm trying to do some updates to a state which is a binary array. gputid is a GPU thread class (https://wiki.tiker.net/PyCuda/Examples/MultipleThreads) and it stores the state and the index of the array to be updated in another class which can be accessed with gputid.mp.x_gpu and gputid.mp.neuron_gpu respectively. Below is my kernel that takes in the gputid and performs the update of the state. However, it the output of the code is not consistent as it runs into errors and executes perfectly when i run it multiple times. The error msg makes no sense to me: File "/root/anaconda3/lib/python3.6/site-packages/pycuda/driver.py", line 447, in function_prepared_call func._set_block_shape(*block) pycuda._driver.LogicError: cuFuncSetBlockShape failed: invalid resource handle My code: def local_update(gputid): mod = SourceModule(""" __global__ void local_update(int *x_gpu, float *n_gpu) { int tid = threadIdx.x + blockDim.x * blockIdx.x; if (tid == (int)(n_gpu[0])) { x_gpu[tid] = 1 - x_gpu[tid]; } } """) gputid.ctx.push() x_gpu = gputid.mp.x_gpu n_gpu = gputid.mp.neuron_gpu func = mod.get_function("local_update") func.prepare("PP") grid = (1,1) block = (gputid.mp.d,1,1) func.prepared_call(grid, block, x_gpu, n_gpu) gputid.ctx.pop() print ('1Pain')

2 years, 11 months

PyCUDA ImportError

by MarbHarmsen

My goal is to speed up my python FEA (finite elements analysis) with my quadro GPU. I however have issues when I import pycuda.autoinit or pycuda.driver into my python code. See the example from my Console: **code In [6] import pycuda.autoinit Traceback (most recent call last): File "<ipython-input-7-78816ba4a0fc>", line 1, in <module> import pycuda.autoinit File "/home/bram/.anaconda3/lib/python3.6/site-packages/pycuda-2017.1-py3.6-linux-x86_64.egg/pycuda/autoinit.py", line 2, in <module> import pycuda.driver as cuda File "/home/bram/.anaconda3/lib/python3.6/site-packages/pycuda-2017.1-py3.6-linux-x86_64.egg/pycuda/driver.py", line 5, in <module> from pycuda._driver import * # noqa ImportError: libcurand.so.8.0: cannot open shared object file: No such file or directory /***code Some details of my setup: - HP Zbook Studio G3 (Quadro M1000M) Ubuntu 18.04 - Cuda 9.1 (.run installer)(I added the path variables to ~/.bashrc - nvidia-driver-390 as driver pycuda 2017.1 (from anaconda) I've tried the solutions proposed by people encountering similar issues when using tensorflow-gpu: It was proposed to make a softlink from libcurand.se.8.0 to the libcurand.se.9.1 using the terminal: "user@device:~$ sudo ln -s libcublas.so.9.1 libcublas.so.8.0" This did not help however. I've checked the installation of CUDA by running a simple vectorAdd example in Exlips. That worked without any issues and when profiling it showed that the gpu was working as expected. I probably made a mistake somewhere and tell me if you need more information -- Sent from: http://pycuda.2962900.n2.nabble.com/

2 years, 11 months

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PyCUDA May 2018