I'm using CUBLAS (Cuda Blas libraries) for matrix operations.
Is possible to use CUBLAS to achieve the exponentiation/root mean square of a matrix items?
I mean, having the 2x2 matrix
1 4
9 16
What I want is a function to elevate to a given value e.g. 2
1 16
81 256
and calculating the root mean square e.g.
1 2
3 4
Is this possible with CUBLAS? I can't find a function suitable to this goal, but I'll ask here first to begin coding my own kernel.
So this may well be something you do have to implement yourself, because the library won't do it for you. (There's probably some way to implement it some of it in terms of BLAS level 3 routines - certainly the squaring of the matrix elements - but it would involve expensive and otherwise unnecessary matrix-vector multiplications. And I still don't know how you'd do the squareroot operation). The reason is that these operations aren't really linear-algebra procedures; taking the square root of each matrix element doesn't really correspond to any fundamental linear algebra operation.
The good news is that these elementwise operations are very simple to implement in CUDA. Again, there are lots of tuning options one could play with for best performance, but one can get started fairly easily.
As with the matrix addition operations, you'll be treating the NxM matricies here as (N*M)-length vectors; the structure of the matrix doesn't matter for these elementwise operations. So you'll be passing in a pointer to the first element of the matrix and treating it as a single list of N*M numbers. (I'm going to assume you're using floats here, as you were talking about SGEMM and SAXPY earlier.)
The kernel, the actual bit of CUDA code which implements the operation, is quite simple. For now, each thread will compute the square (or squareroot) of one array element. (Whether this is optimal or not for performance is something you could test). So the kernels would look like the following. I'm assuming you're doing something like B_ij = (A_ij)^2; if you wanted to do the operation inplace, eg A_ij = (A_ij)^2, you could do that, too:
__global__ void squareElements(float *a, float *b, int N) {
/* which element does this compute? */
int tid = blockDim.x * blockIdx.x + threadIdx.x;
/* if valid, squre the array element */
if (tid < N)
b[tid] = (a[tid]*a[tid]);
}
__global__ void sqrtElements(float *a, float *b, int N) {
/* which element does this compute? */
int tid = blockDim.x * blockIdx.x + threadIdx.x;
/* if valid, sqrt the array element */
if (tid < N)
b[tid] = sqrt(a[tid]); /* or sqrtf() */
}
Note that if you're ok with very slightly increased error, the 'sqrtf()' function which has maximum error of 3 ulp (units in the last place) is significantly faster.
How you call these kernels will depend on the order in which you're doing things. If you've already made some CUBLAS calls on these matricies, you'll want to use them on the arrays which are already in GPU memory.
Related
I'm a total newbie to OpenCL.
I'm trying to code a reduction kernel that sums along one axis for a multi-dimensional array. I have stumbled upon that code which comes from here: https://tmramalho.github.io/blog/2014/06/16/parallel-programming-with-opencl-and-python-parallel-reduce/
__kernel void reduce(__global float *a, __global float *r, __local float *b) {
uint gid = get_global_id(0);
uint wid = get_group_id(0);
uint lid = get_local_id(0);
uint gs = get_local_size(0);
b[lid] = a[gid];
barrier(CLK_LOCAL_MEM_FENCE);
for(uint s = gs/2; s > 0; s >>= 1) {
if(lid < s) {
b[lid] += b[lid+s];
}
barrier(CLK_LOCAL_MEM_FENCE);
}
if(lid == 0) r[wid] = b[lid];
}
I don't understand the for loop part. I get that uint s = gs/2 means that we split the array in half, but then it is a complete mystery. Without understanding it, I can't really implement another version for taking the maximum of an array for instance, even less for multi-dimensional arrays.
Furthermore, as far as I understand, the reduce kernel needs to be rerun another time if "N is bigger than the number of cores in a single unit".
Could you give me further explanations on that whole piece of code? Or even guidance on how to implement it for taking the max of an array?
Complete code can be found here: https://github.com/tmramalho/easy-pyopencl/blob/master/008_localreduce.py
Your first question about the meaning of the for loop:
for(uint s = gs/2; s > 0; s >>= 1)
It means that you divide the local size gs by 2, and keep dividing by 2 (the shift part s >>= 1 is equivalent to s = s/2) while s > 0, in other words, until s = 1. This algorithm depends on your array's size being a power of 2, otherwise you'd have to deal with the excess of a power of 2 until you have reduced the whole array, or you'd have to fill your array with neutral values for the reduction until completing a power of 2 size.
Your second concern when N is bigger than the capacity of your GPU, you are right: you have to run your reduction in portions that fit and then merge the results.
Finally, when you ask for guidance on how to implement a reduction to get the max of an array, I would suggest the following:
For a simple reduction like max or sum, try using numpy, especially if you are dealing with programming the reduction by axis.
If you think that the GPU would give you an advantage, try first using pyopencl's Multidimensional Array functionality, e.g. max.
If the reduction is more math intensive, try using pyopencl's Parallel Algorithms, e.g. reduction
I think that the whole point of using pyopencl is to avoid dealing with the underlying GPU's architecture. Otherwise, it is easier to deal with CUDA or HIP directly instead of OpenCL.
I have a large genetic dataset (X, Y coordinates), of which I can easily know one dimension (X) during runtime.
I drafted the following code for a matrix class which allows to specify the size of one dimension, but leaves the other one dynamic by implementing std::vector. Each vector is new'd using unique_ptr, which is embedded in a C-style array, also with new and unique_ptr.
class Matrix
{
private:
typedef std::vector<Genotype> GenVec;
typedef std::unique_ptr<GenVec> upGenVec;
std::unique_ptr<upGenVec[]> m;
unsigned long size_;
public:
// ...
// construct
Matrix(unsigned long _size): m(new upGenVec[_size]), size_(_size)
{
for (unsigned long i = 0; i < this->size_; ++i)
this->m[i] = upGenVec(new GenVec);
}
};
My question:
Does it make sense to use this instead of std::vector< std::vector<Genotype> > ?
My reasoning behind this implementation is that I only require one dimension to be dynamic, while the other should be fixed. Using std::vectors could imply more memory allocation than needed. As I am working with data that would fill up estimated ~50GB of RAM, I would like to control memory allocation as much as I can.
Or, are there better solutions?
I won't cite any paragraphs from specification, but I'm pretty sure that std::vector memory overhead is fixed, i.e. it doesn't depend on number of elements it contains. So I'd say your solution with C-style array is actually worse memory-wise, because what you allocate, excluding actual data, is:
N * pointer_size (first dimension array)
N * vector_fixed_size (second dimension vectors)
In vector<vector<...>> solution what you allocate is:
1 * vector_fixed_size (first dimension vector)
N * vector_fixed_size (second dimension vectors)
I have a bunch of vectors (~500). I need to find triple products of all the combinations of the vectors in OpenCL. There are plenty of combination algorithms (r out of n things) in C++ but I am yet to find any implemented for GPU. I have seen quite a few parallel permutation algorithms in Cuda but I just want to know if there are any viable combination algorithms present?
I'll need to guess a bit here and there to answer your question.
I suppose you have an array V of n (~500) vectors. These vectors are all of same dimensionality m (probably m=3).
What you want is the component wise product of each 3 vectors vi, vj, vk where i,j,k in {0,..,n-1}.
Simple 3-dimensional example:
result[idx].x = V[i].x * V[j].x * V[k].x;
result[idx].y = V[i].y * V[j].y * V[k].y;
result[idx].z = V[i].z * V[j].z * V[k].z;
Now maybe your vectors are not 3-dimensional and maybe you don't want the component wise product but the sum of it (like in dot product), but I'm sure you're able to djust the code accordingly.
The real question here is how to compute all possible i,j,k and idx. Correct?
Now with CUDA you are in a very fortunate position. You can just launch n*n*n threads in a grid and therefore get i,j,k for free without having to think about ways to compute combinations or permutations at all. Just do the following:
dim3 grid, block;
block.x = n;
block.y = 1;
block z = 1;
grid.x = n;
grid.y = n;
grid.z = 1;
compute_product_kernel<<<grid, block>>>( V, result );
This way you'll launch n*n blocks of n threads. Computing i,j,k becomes trivial, computing idx is easy:
__device__ void compute_product_kernel( myVector* V, myVector* result)
{
int i = blockIdx.x;
int j = blockIdx.y;
int k = threadIdx.x;
int idx = i * gridDim.y * blockDim.x + j * blockDim.x + k;
...
}
Of course all of this only works because your n is within the limits of CUDA's block and grid range.
Two more things though:
Maybe you want permutations instead of combinations. You could do that by skipping every combination where any two of i,j,k are the same. But I'd recommend keeping them anyway because computing when to skip is probably more expensive that doing the actual work. Also I'd advise against using the permutation to save memory for result because it would save you less that 1% and make the calculation much more complex.
Are you sure you've got enough memory to actually do this? Storing the result requires n*n*n*m*sizeof(float) bytes. With n=500 and m=3 that would already be 1.5 GB. Is that really what you are looking for? Maybe the next step of your processing can be combined into the calculation so that storing the intermediate result is not neccessary.
How can I find the cube root of a number in an efficient way?
I think Newton-Raphson method can be used, but I don't know how to guess the initial solution programmatically to minimize the number of iterations.
This is a deceptively complex question. Here is a nice survey of some possible approaches.
In view of the "link rot" that overtook the Accepted Answer, I'll give a more self-contained answer focusing on the topic of quickly obtaining an initial guess suitable for superlinear iteration.
The "survey" by metamerist (Wayback link) provided some timing comparisons for various starting value/iteration combinations (both Newton and Halley methods are included). Its references are to works by W. Kahan, "Computing a Real Cube Root", and by K. Turkowski, "Computing the Cube Root".
metamarist updates the DEC-VAX era bit-fiddling technique of W. Kahan with this snippet, which "assumes 32-bit integers" and relies on IEEE 754 format for doubles "to generate initial estimates with 5 bits of precision":
inline double cbrt_5d(double d)
{
const unsigned int B1 = 715094163;
double t = 0.0;
unsigned int* pt = (unsigned int*) &t;
unsigned int* px = (unsigned int*) &d;
pt[1]=px[1]/3+B1;
return t;
}
The code by K. Turkowski provides slightly more precision ("approximately 6 bits") by a conventional powers-of-two scaling on float fr, followed by a quadratic approximation to its cube root over interval [0.125,1.0):
/* Compute seed with a quadratic qpproximation */
fr = (-0.46946116F * fr + 1.072302F) * fr + 0.3812513F;/* 0.5<=fr<1 */
and a subsequent restoration of the exponent of two (adjusted to one-third). The exponent/mantissa extraction and restoration make use of math library calls to frexp and ldexp.
Comparison with other cube root "seed" approximations
To appreciate those cube root approximations we need to compare them with other possible forms. First the criteria for judging: we consider the approximation on the interval [1/8,1], and we use best (minimizing the maximum) relative error.
That is, if f(x) is a proposed approximation to x^{1/3}, we find its relative error:
error_rel = max | f(x)/x^(1/3) - 1 | on [1/8,1]
The simplest approximation would of course be to use a single constant on the interval, and the best relative error in that case is achieved by picking f_0(x) = sqrt(2)/2, the geometric mean of the values at the endpoints. This gives 1.27 bits of relative accuracy, a quick but dirty starting point for a Newton iteration.
A better approximation would be the best first-degree polynomial:
f_1(x) = 0.6042181313*x + 0.4531635984
This gives 4.12 bits of relative accuracy, a big improvement but short of the 5-6 bits of relative accuracy promised by the respective methods of Kahan and Turkowski. But it's in the ballpark and uses only one multiplication (and one addition).
Finally, what if we allow ourselves a division instead of a multiplication? It turns out that with one division and two "additions" we can have the best linear-fractional function:
f_M(x) = 1.4774329094 - 0.8414323527/(x+0.7387320679)
which gives 7.265 bits of relative accuracy.
At a glance this seems like an attractive approach, but an old rule of thumb was to treat the cost of a FP division like three FP multiplications (and to mostly ignore the additions and subtractions). However with current FPU designs this is not realistic. While the relative cost of multiplications to adds/subtracts has come down, in most cases to a factor of two or even equality, the cost of division has not fallen but often gone up to 7-10 times the cost of multiplication. Therefore we must be miserly with our division operations.
static double cubeRoot(double num) {
double x = num;
if(num >= 0) {
for(int i = 0; i < 10 ; i++) {
x = ((2 * x * x * x) + num ) / (3 * x * x);
}
}
return x;
}
It seems like the optimization question has already been addressed, but I'd like to add an improvement to the cubeRoot() function posted here, for other people stumbling on this page looking for a quick cube root algorithm.
The existing algorithm works well, but outside the range of 0-100 it gives incorrect results.
Here's a revised version that works with numbers between -/+1 quadrillion (1E15). If you need to work with larger numbers, just use more iterations.
static double cubeRoot( double num ){
boolean neg = ( num < 0 );
double x = Math.abs( num );
for( int i = 0, iterations = 60; i < iterations; i++ ){
x = ( ( 2 * x * x * x ) + num ) / ( 3 * x * x );
}
if( neg ){ return 0 - x; }
return x;
}
Regarding optimization, I'm guessing the original poster was asking how to predict the minimum number of iterations for an accurate result, given an arbitrary input size. But it seems like for most general cases the gain from optimization isn't worth the added complexity. Even with the function above, 100 iterations takes less than 0.2 ms on average consumer hardware. If speed was of utmost importance, I'd consider using pre-computed lookup tables. But this is coming from a desktop developer, not an embedded systems engineer.
I have two large vectors, I am trying to do some sort of element multiplication, where an even-numbered element in the first vector is multiplied by the next odd-numbered element in the second vector... and where the odd-numbered element in the first vector is multiplied by the preceding even-numbered element in the second vector.
For example:
vector 1 is V1(1) V1(2) V1(3) V1(4)
vector 2 is V2(1) V2(2) V2(3) V2(4)
V1(1) * V2(2)
V1(3) * V2(4)
V1(2) * V2(1)
V1(4) * V2(3)
I have written Cuda code to do this (Pds has the elements of the first vector in shared memory, Nds the second Vector):
// instead of % 2, checking the first bit to decide if a number
// is odd/even is faster
if ((tx & 0x0001) == 0x0000)
Nds[tx+1] = Pds[tx] * Nds[tx+1];
else
Nds[tx-1] = Pds[tx] * Nds[tx-1];
__syncthreads();
Is there anyway to further accelerate this code or avoid divergence?
You should be able to eliminate the branch like this:
int tx_index = tx ^ 1; // equivalent to: tx_index = (tx & 1) ? tx - 1 : tx + 1
Nds[tx_index] = Pds[tx] * Nds[tx_index];
This is an old post, may be someone finds my answer useful. If in your code tx is threadIdx, then you have branching or warp divergence. You must avoid divergence in blocks, because it serializes the process. It means that the threads with even indices will run first, and then threads with odd indices will run. If tx is threadIdx, try to change your algorithm such that branching depends on blockIdx.