Kernal Blas

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Kernal Blas

Team Members

1. Shan Ariel Sioson
2. Joseph Pham

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Progress

Assignment 1

Calculation of Pi

For this assessment, we used code found at helloacm.com

In this version, the value of PI is calculated using the Monte Carlo method. This method states:

1. A circle with radius r in a squre with side length 2r
2. The area of the circle is Πr² and the area of the square is 4r²
3. The ratio of the area of the circle to the area of the square is: Πr² / 4r² = Π / 4
4. If you randomly generate N points inside the square, approximately N * Π / 4 of those points (M) should fall inside the circle.
5. Π is then approximated as:

• N * Π / 4 = M
• Π / 4 = M / N
• Π = 4 * M / N

For simplicity the radius of the circle is 1. With this, we randomly generate points within the area and count the number of times each point falls within the circle, between 0 and 1. We then calculate the ratio and multiply by 4 which will give us the approximation of Π.

When we run the program we see:

1st Run

1000            3.152 - took - 0 millisecs
10000           3.1328 - took - 0 millisecs
100000          3.14744 - took - 9 millisecs
1000000         3.141028 - took - 96 millisecs
10000000        3.1417368 - took - 998 millisecs
100000000       3.1419176 - took - 10035 millisecs

The first column represents the "stride" or the number of digits of pi we are calculating to.

With this method, we can see that the accuracy of the calculation is slightly off. This is due to the randomization of points within the circle. Running the program again will give us slightly different results.

2nd Run

1000            3.12 - took - 0 millisecs
10000           3.1428 - took - 0 millisecs
100000          3.13528 - took - 9 millisecs
1000000         3.143348 - took - 106 millisecs
10000000        3.1414228 - took - 1061 millisecs
100000000       3.14140056 - took - 8281 millisecs

The hotspot within the code lies in the loops. There iteration for the loops is O(n²)

Potential Speedup:

Using Gustafson's Law:

P = 50% of the code
P = 0.50
n = ?

S(n)= n - ( 1 - P ) ∙ ( n - 1 )
Sn = n - ( 1 - .5 ) ∙ ( n - 1 )
Sn =

Data Compression

For this suggested project we used the LZW algorithm from one of the suggested projects.

The compress dictionary copies the file into a new file with a .LZW filetype which uses ASCII

The function that compresses the file initializing with ASCII

void compress(string input, int size, string filename) {
	unordered_map<string, int> compress_dictionary(MAX_DEF);
	//Dictionary initializing with ASCII
	for (int unsigned i = 0; i < 256; i++) {
		compress_dictionary[string(1, i)] = i;
	}
	string current_string;
	unsigned int code;
	unsigned int next_code = 256;
	//Output file for compressed data
	ofstream outputFile;
	outputFile.open(filename + ".lzw");

	for (char& c : input) {
		current_string = current_string + c;
		if (compress_dictionary.find(current_string) == compress_dictionary.end()) {
			if (next_code <= MAX_DEF)
				compress_dictionary.insert(make_pair(current_string, next_code++));
			current_string.erase(current_string.size() - 1);
			outputFile << convert_int_to_bin(compress_dictionary[current_string]);
			current_string = c;
		}
	}
	if (current_string.size())
		outputFile << convert_int_to_bin(compress_dictionary[current_string]);
	outputFile.close();
}

The results of the compression from:

Parallelizing

From one of the suggested improvements in the algorithm post link. A potential improvement is changing from char& c to a const char in the for loop

	for (char& c : input) {

since char& c is not being modified.

Assignment 2

In order to parallelize the code from above, we decided to use a kernel to handle the calculations. The logic remains the same but the results are much faster.
Offloading to the GPU results in a pi calculation time to be reduced

Kernel code used

__global__ void calculate(float *sum, int nbin, float step, int nthreads, int nblocks) {
	int i;
	float x;
	int idx = blockIdx.x * blockDim.x + threadIdx.x;  // Sequential thread index across the blocks
	for (i = idx; i< nbin; i += nthreads*nblocks) {
		x = (i + 0.5)*step;
		sum[idx] += 4.0 / (1.0 + x*x);
	}
}

Main function

// Main routine that executes on the host
int main(int argc, char** argv) {
	// interpret command-line argument
	if (argc != 2) {
		std::cerr << argv[0] << ": invalid number of arguments\n";
		return 1;
	}
	float n = std::atoi(argv[1]);
	int nblocks = 30;

	steady_clock::time_point ts, te;
	dim3 dimGrid(nblocks, 1, 1);  // Grid dimensions
	dim3 dimBlock(ntpb, 1, 1);  // Block dimensions
	float *sumHost, *sumDev;  // Pointer to host & device arrays

	float step = 1.0 / n;  // Step size
	size_t size = nblocks*ntpb * sizeof(float);  //Array memory size
	sumHost = (float *)malloc(size);  //  Allocate array on host
	cudaMalloc((void **)&sumDev, size);  // Allocate array on device
										 // Initialize array in device to 0
	cudaMemset(sumDev, 0, size);
	// initialization
	std::srand(std::time(nullptr));

	ts = steady_clock::now();

	// Do calculation on device
	calculate << <dimGrid, dimBlock >> > (sumDev, n, step, ntpb, nblocks); // call CUDA kernel
	te = steady_clock::now();

	cudaMemcpy(sumHost, sumDev, size, cudaMemcpyDeviceToHost);

	for (tid = 0; tid<ntpb*nblocks; tid++)
		pi += sumHost[tid];
	pi *= step;

	// Print results
	printf("Number of  iterations= %f\nPI = %f\n", n,pi);
	reportTime("Pi calculation took ", te - ts);



	// Cleanup
	free(sumHost);
	cudaFree(sumDev);

	return 0;
}

Results CPU vs GPU



As we can see above, the more iterations, the more accurate the calculation of PI.
The CPU's results drastically change as we increase the iteration 10x.
However, the parallelized results seem to stay accurate throughout the iterations.
As the iteration passes 100 million, we have a bit of memory leaks which results in inaccurate results.