# Reduction on CUDA

On this post I’m going to solve a common problem on CUDA, and that is a Reduction problem.

But using the algorithms posted by Mark Harris  “Optimizing Parallel Reduction in CUDA

We know on GPU is not very effective perform a reduction operation, and trying to make the most optimal code can be very difficult and not very intuitive.

The post by  Mark Harris  only gives the algorithm but is not quite clear on how to use it, I’m going to do that.

Let us solve the following integral this is not a calculus class, but in the old days, this was solved liked this the sum of all the areas of the rectangles gives the approximate solution to the integral and is not other than PI . And yes you are indeed right, this is a reduction problem.

So… how are we going to tackle this one?

first we need to calculate all the rectangle areas. Since one rectangle does not depend on any other rectangle, is perfect for our GPU to solve this one.

```__global__ void calcRectangles(float * rectangles,int n){
int i = blockIdx.x * blockDim.x + threadIdx.x;
float x;

if (i >= 0 && i < n){
x = (i + 0.5) / n;
rectangles[i] = 4.0 / (1.0 + x*x);
}
}
```

where in the input

float * rectangles is the array to store all the areas.

int n the precision, bigger the n the more accurate the solution will be.

this kernel will give us an array of float values, this result will be out input to the reduce kernel.

Now let us prepare for the real deal

for each reduction I can only use 32 threads, this is because of my graphics card limitation, son in order to make large reduction I needed to use recursive kernel calls.

For example, to reduce an array of 2^21 elements

in the image we can see how an array of 2^21 elements is reduced to an array of 2^16 elements.

Now the same process with this new array, resulting in a 2^11 array, until it has 31 elements or less.

```float calcReductionGeneral(int n, float * rectangles,int total , int type){

clock_t start; 	//vars to measure time
clock_t end;
float secondsTotal = 0;

if (n<32){ //If the lenght of the array is less than 32, we make the reduction in the host

start = clock();

float pi;
for (int i=0;i<n;i++)
pi = pi + rectangles[i];
pi = pi / total;
end = clock();
secondsTotal = ((float)(end - start) / CLOCKS_PER_SEC);
printf ("pi = %f " ,pi);

return secondsTotal;

}
else{ //reduction on GPU

float * resParcial = (float*)malloc(sizeof(float)*n/32); //partial solution, array of 32 floats

for (int i=0;i<n/32;i++){

/*i.e.   if we have 64 rectangles we are going to make 2 kernel calls
first call with 0..31 elemets
second call with 32..63 elements*/

float * inputParcial = (float*)malloc(sizeof(float)*32); //partial input to the kernel allocation
float * input_d;
gpuErrchk(cudaMalloc((void**)&input_d, 32*sizeof(float))); //GPU allocation
copia(inputParcial,rectangles,i);//copy the relative 32 elements of the rectangles to the partial input

/*  i.e  if we have 64 elements in the first loop iteration we copy the 0..31 elements and in the second iteration from 32..63 elements
*/
cudaMemcpy(input_d, inputParcial, sizeof(float)*32, cudaMemcpyHostToDevice);
//We move the parcial input to the GPU

float * output;
gpuErrchk(cudaMalloc((void**)&output, 32*sizeof(float)));	//allocating memory on GPU
float * resParcial132 = (float*)malloc(sizeof(float)*32);   //allocating memory for parcial result

start = clock();//we measure time
switch (type){  //calling the kernel, using the type parameter
case 0: reduce0 <<<1,32,32 >>>(input_d, output);	break;
case 1: reduce1 <<<1,32,32 >>>(input_d, output);	break;
case 2: reduce2 <<<1,32,32 >>>(input_d, output);	break;
default: break;
}
end = clock();
secondsTotal = secondsTotal + ((float)(end - start) / CLOCKS_PER_SEC);

gpuErrchk (cudaMemcpy(resParcial132,  output, sizeof(float)*32, cudaMemcpyDeviceToHost)); // we move the result to the host
//printf ("resParcial132 = %f" ,resParcial132);
resParcial[i]= resParcial132;

//clean up
free(inputParcial);
free(resParcial132);
cudaFree(input_d);
cudaFree(output);

}

return calcReductionGeneral((int)n/32,resParcial,total,type) + secondsTotal;  //recursive call.
}

}
```
```__global__ void reduce0(float *g_idata, float *g_odata) {
extern __shared__ float sdata[];
unsigned int i = blockIdx.x*blockDim.x + threadIdx.x;
sdata[tid] = g_idata[i];

// do reduction in shared mem

for
(unsigned int s = 1; s < blockDim.x; s *= 2) {
if (tid % (2 * s) == 0) {
sdata[tid] += sdata[tid + s];

}
}

// write result for this block to global mem
if (tid == 0) g_odata[blockIdx.x] = sdata;
}
```
```__global__ void reduce1(float *g_idata, float *g_odata) {
extern __shared__ float sdata[];
unsigned int i = blockIdx.x*blockDim.x + threadIdx.x;
sdata[tid] = g_idata[i];

for (unsigned int s=1; s < blockDim.x; s *= 2) {
int index = 2 * s * tid;
if (index < blockDim.x) { 			sdata[index] += sdata[index + s]; 		} 		__syncthreads(); 	} 	 	// write result for this block to global mem 	if (tid == 0) g_odata[blockIdx.x] = sdata; } ```
``` __global__ void reduce2(float *g_idata, float *g_odata) { 	extern __shared__ float sdata[]; 	// each thread loads one element from global to shared mem 	unsigned int tid = threadIdx.x; 	unsigned int i = blockIdx.x*blockDim.x + threadIdx.x; 	sdata[tid] = g_idata[i];       __syncthreads();         	for (unsigned int s=blockDim.x/2; s>0; s>>=1) {
if (tid < s) {
sdata[tid] += sdata[tid + s];
}
}

// write result for this block to global mem
if (tid == 0) g_odata[blockIdx.x] = sdata;
}

```

where reduce0, reduce1, and reduce2 are exactly the same algorithms used by Mark Harris.

Now, something to point out

pi = 3.141593 Time reduce0: 0.456775 seconds
pi = 3.141593 Time reduce1: 0.459540 seconds
pi = 3.141593 Time reduce2: 0.458745 seconds

the times of each algorithm for 4M elements is pretty much the same, where is this article “Optimizing Parallel Reduction in CUDA” the speedup from reduce0 to reduc1 is 2.33.

So… yeah.. I think they are selling a little bit of hot air in here, BUT, my implementation is very different from Nvidia’s, I try to use all the threads all the time and in their case they don’t, sticking to 128 threads because that is when the achieve peak performance and stuff like that.

The main structure of the program is the same, a tree based calling kernel function in a recursive manner and the results are very different still.

You can clone my entire code

# Multiplying two matrices in CUDA using BLAS…CUBLAS

On this post I’ll be making a simple program, multiplying two matrices using BLAS.

We already know the basics of CUDA, if not you can check me previous post here. To the matrices operation we are going to use the GEMM operation.

C = αop ( A ) op ( B ) + βC, Were α and β are scalar numbers α is going to be 1 and β is going to be 0.
In the CUDA toolkit documentation, the GEMM operation is declared as the following:

```cublasStatus_t cublasSgemm(cublasHandle_t handle,
cublasOperation_t transa,
cublasOperation_t transb,
int m,
int n,
int k,
const float *alpha,
const float *A,
int lda,
const float *B,
int ldb,
const float *beta,
float *C,
int ldc)
```

where

• M is number of rows of matrix op(A) and C.
• N is number of columns of matrix op(B) and C.
• K is number of columns of op(A) and rows of op(B).
• transa operation op(A) that is non- or (conj.) transpose.
• transb operation op(B) that is non- or (conj.) transpose.
• A array of dimensions lda x k with lda>=max(1,m) if transa == CUBLAS_OP_N and lda x m with lda>=max(1,k) otherwise.
• B array of dimension ldb x n with ldb>=max(1,k) if transa == CUBLAS_OP_N and ldb x k with ldb>=max(1,n) otherwise.
• Lda leading dimension of two-dimensional array used to store the matrix A.
• ldb leading dimension of two-dimensional array used to store matrix B.

Now let’s us go to the code:
Before starting let me declare some auxiliary functions just to make my code clearer.

```
void init_matrix(float *M, int hM, int wM, float k)
{
int i,j;

for (i=0; i<hM; i++)
for (j=0; j<wM; j++)
if (i==j)
M[i*wM+j] = k*k*1.0f;
else
M[i*wM+j] = -k*1.0f;
}

```

This first function I just initialize a matrix with hM and wM dimensions the k value is just a random for the matrix to be populated with

```void print_matrix(float *M, int hM, int wM)
{
int i,j;

for (i=0; i<hM; i++){
for (j=0; j<wM; j++)
printf(&quot;%4.1f &quot;, M[i*wM+j]);
printf(&quot;\n&quot;);
}
}

```

this functions prints the matrix on screen

For the multiplication, let us declare a function, which accept 6 parameters, 2 inputs matrices, 1 output matrix, height and width of the first matrix and the width of the second matrix.

```void Mul(float* A, float* B, int hA, int wA, int wB, float* C){

//Now we have to allocate memory on GPU for the 3 matrices

size = hA * wA * sizeof(float);

float* Bd;
size = wA * wB * sizeof(float);
cudaMalloc((void**)&amp;Bd, size);
cudaMemcpy(Bd,B,size,cudaMemcpyHostToDevice);

// Allocate C on the device
float* Cd;
size = hA * wB * sizeof(float);
cudaMalloc((void**)&amp;Cd, size);

//declare the alfa and beta
const float alfa = 1;
const float beta = 0;
const float *alphac = &amp;alfa;
const float *betac = &amp;beta;

int lda=wA,ldb=hA,ldc=wA;

```

in order to use the CUBLAS methods first we need to create a handle and when we no loger need to use the CUBLAS method we should destroy it.

For the handle creation

```
cublasHandle_t handle;
cublasCreate(&amp;handle);

```

cublas gemm method

```
cublasSgemm(handle, CUBLAS_OP_N, CUBLAS_OP_N, wA, wB, hA, alphac, Ad,lda,Bd,ldb, betac, Cd, ldc);

```

Need to remember that CUBLAS method WORK ON GPU, that means, all input matrices must be on the device memory.

handle destruction

```cublasDestroy(handle);
```

bring output to host memory

```cudaMemcpy(C, Cd, size, cudaMemcpyDeviceToHost);
```

final touches

```cudaFree(Ad);
cudaFree(Bd);
cudaFree(Cd);
```

So the core of the program is finished, now we need a main method, initialise some matrices and test.

```
int main(int argc, char *argv[])
{
// Matrix variables
float *A, *B, *C;//Matrices declarations
int hA, wA, hB, wB;//declaration of dimensions

setbuf(stdout, NULL);

if (argc!=4){
printf(&quot;./exec hA hB/WA wB\n&quot;);
exit(-1);
}
//transform the input to integers
hA = atoi(argv);
hB = wA = atoi(argv);
wB = atoi(argv);

// Init A and B, malloc C
int size_A = wA * hA;
A = (float*)malloc(size_A*sizeof(float));
init_matrix(A, hA, wA,2.0);

int size_B = wB * hB;
B = (float*)malloc(size_B*sizeof(float));
init_matrix(B, hB, wB,1.0);

int size_C = wB * hA;
C = (float*)malloc(size_C*sizeof(float));

//matrix multiplication
Mul(A, B, hA, wA, wB, C);

//Print on console the 3 matrices
printf(&quot;\n\nMATRIX A\n&quot;);print_matrix(A, hA, wA);
printf(&quot;\n\nMATRIX B\n&quot;);print_matrix(B, hB, wB);
printf(&quot;\n\nMATRIX C\n&quot;);print_matrix(C, hA, wB);

//free the allocated memory
free(A)
free(B)
free(C)
return (1);
}

```

to compile remember to add the flags -lcudart -lcublas.

An output example of 4×4 matrices multiplication is the full code can be found on my git repository, feel free to clone

# Grids, threads… THE ULTIMATE GUIDE

So most users struggle a lot with this topic, on CUDA is important to make a good use of how many grids and threads we need to use so let’s start:

## What are grids blocks and thread blocks?

In CUDA we have to define how our grid is going to be and how many threads are going to be inside a grid.

The kernels are executed in grids just like the following image:

## Identify the dimension of the problem

There are multiple ways to do this, but let’s try to make it simple.

On CUDA or GPU programming in general we have from 1 to 3 dimensions, values different than those will give runtime errors.

So for example, if we have an array, our problem is a one-dimensional problem, a matrix is a two-dimensional problem and a cube/parallelepiped is a three-dimensional problem.

### One dimensional problems

These are the easiest ones, we only need to work on the X axis, and basically our problem is reduced to the length of the array.

Let’s say we have an Array with length 1Million:

we have to decide how many grids blocks we want and how many threads per block we want.

First we have to remember, on most devices the maximum threads per block available is 512 or 1024, something bigger than this will give an error.

After deciding the threads per blocks (I’m going to use 1024) the grid is ceil(1M/1024).

Our declaration and call to the kernel will look something like this

```
dim3 gridBlocks(ceil(1000000/1024),1,1);

```

this seems pretty straight forward, the Y and Z axis is 1 since we are only working on the X axis.

## Two dimensional problems

These might be the most common, we need to work on the X and Y axis.

Let us divide this into two examples,

A square and a rectangle, the first being an exclusive case of the second one, when both sides of the rectangle are the same length, is a square (Duh!?).

### Square:

Let us assume we have a square image with A x A pixels, se we need to create a grid that minimizes the number of threads in order to use them the most efficient way

First we need to set the size of our grid, this is a problem specific matter but we can set some various examples.

For very small values of A (A >8 and A <64):

We can divide the square into 4 sub-squares, for example if A = 16 we have the following layout

So our grid will be 2 x 2 and each grid block will have 8 x 8 threads

We can code that as follow:

```dim3 gridBlocks(2,2,1);
```

For medium sizes (A>64 and A < 512):

We can divide the square into 16-sub squares

This is 32 x 32 I’m using a small grid just for illustration purposes

We can code this as follow

```dim3 gridBlocks(4,4,1); // 4 x 4 grid
```

This examples are just illustrations, is not by any means a rule, remember the number of grids and threads is a specific matter of the problem itself.

### Rectangle

So let us assume we have an image that is shaped like a rectangle with A x B pixels and A != B

Assuming A and B big enough for illustration purposes, I will set two cases, A > B and B > A

#### A < B:

We get the following layout

This can be coded as follow

```dim3 gridBlocks(4,2,1); // 4 x 2 grid
```

#### A > B:

We get the following layout

This can be coded as follow

```dim3 gridBlocks(2,4,1); // 4 x 2 grid
```

# First steps on CUDA

On this post I’m going to use the programming language “C” for our first steps in CUDA. Remember that “C++” is also compatible.

To start we have to know the basic structure of any CUDA program: is like a regular program but in this one we have to communicate with our Nvidia GPU, keeping in mind memory assignment, memory transactions and calling the kernel itself.

For example, we want to create an array of 10 float elements in “C”. It would be something like this:

This memory is allocated in our HOST memory, is not visible to our GPU. Now lets  do the same procedure in a GPU.

## cudaMalloc

This directive will let us allocate memory in our GPU global memory. On the contrary as before, it is only visible from our GPU, therefore we cannot use this in our HOST program.

On the Nvidia Documentation we can check how a cudaMalloc works.

cudaError_t cudaMalloc(void** devPtr, size_t size): applying this procedure we can allocate an array of 10 float elements in our GPU memory.

Now, lets say we populate our array in our HOST and we want to move that array to our GPU for massive parallel computations. How can we do that?

## cudaMemcpy

This directive will let us transfer memory from our host to our GPU and vice-versa.

On the Nvidia Documentation we can check how a cudaMemcpy works.

cudaError_t cudaMemcpy (void* dst, const void* src, size_t count, cudaMemcpyKind kind), where:

void* dst is our destination.

const void* src is our source.

size_t count is the total amount of memory to transfer.

cudaMemcpyKind kind is the kind of copy we want to make, it can be from our host to our device or from our device to our host. For a better explanation click here.

For our example it will go like this: Allocate host memory and populate array h_array Allocate device memory d_array and then use cudaMemcpy to copy the memory from h_array (HOST) to d_array (DEVICE) using cudaMemcpyHostToDevice

Now we have our memory in our device ready.

So let’s recap a little. We know how to allocate and transfer memory, so now it’s time for the fun part, the kernel.

Basically the kernel is the procedure that is going to be executed in our GPU.

## Kernel, grids, blocks and threads

So lets say we want our device to add a number ‘p’ to each element in the array.

This seems like a problem that our GPU can easily take care of. For example, “N” threads and each thread can make 1 addition, so instead of “N” iterations in a loop, it would only be 1 instruction.

### How to create a kernel procedure

The kernel procedure must be declared with the directive __global__. This indicates that is a kernel and it’s going to be executed in the GPU.

Our kernel will look something like this:

__global__ void addPtoArray(float* d_array,int lenght,int p)

{

Implementation of the kernel here:

}

And to invoke our kernel we have to use the triple chevron, define grids and threads and finally the name of the kernel.

It’s better explained with an image:

In order to call a kernel, we have to specify how many threads are going to be executed. We can manage the threads in a grid like the previous image.

The grid and blocks are specified using an structure called dim3. Each dim3 structure has 3 values (X,Y and Z), so we can create a 1d,2d or 3d grid.

Inside the kernel we have information about the grid and thread that is being executed.

blockIdx is the index of the block. In the image example, the block highlighted in green is blockIdx.x=2 and blockIdx.y=1

blockDim is the dimension of the grid, in the image example is blockDim.x=4 and blockDim.y=4 Going back to our example the kernel will look something like this:

Now we are set to work that GPU!

Our code will look something like this:

```#include<stdio.h>;
#include <stdlib.h>;
#include <sys/time.h>;
//CUDA
#include <cuda.h>;

//kernel implementation
__global__ void addPtoArray(float *d_array, int lenght,int p){
int i;
i = blockIdx.x * blockDim.x + threadIdx.x; // We calculate the current thread of our context
if (i<lenght)//if the thread is inside the lenght of our array
d_array[i]+=p;
}

int main(int argc, char *argv[])
{
float *h_array, *d_array,*h_hostResult;
int p = 20;
int lenght=5;

// Mallocs CPU
h_array = (float *)malloc(sizeof(float)*lenght);
for (int i=0; i<lenght; i++){ h_array[i] = i;}    /* Mallocs GPU */    cudaMalloc((void **) &amp;amp;amp;amp;amp;amp;amp;amp;amp;d_array,sizeof(float)*lenght);   /* CPU-&amp;amp;amp;amp;amp;amp;amp;amp;gt;GPU */
cudaMemcpy(d_array, h_array, sizeof(float)*lenght, cudaMemcpyHostToDevice);

/*****************/
/*****************/
dim3 GridBlocks(256);

/* GPU->CPU */
h_hostResult = (float *)malloc(sizeof(float)*lenght);
cudaMemcpy(h_hostResult,d_array,sizeof(float)*lenght,cudaMemcpyDeviceToHost);

/************/
/* Results */
/************/
for (int i=0; i<lenght; i++)
{
printf("h_array[%i] = %f \n",i, h_array[i]);
printf("h_hostResult[%i] = %f \n",i, h_hostResult[i]);

}

/* Free CPU */
free(h_hostResult);
free(h_array);
/* Free GPU */
cudaFree(d_array);
return(CUDA_SUCCESS);
}
```

Save this code on a file, let’s say addPtoArray.cu