CUDA+OpenCV 绘制朱利亚(Julia)集合图形

Julia集中的元素都是经过简单的迭代计算得到的，很适合用CUDA进行加速。对一个600*600的图像，需要进行360000次迭代计算，所以在CUDA中创建了600*600个线程块（block），每个线程块包含1个线程，并行执行360000次运行，图像的创建和显示通过OpenCV实现：

#include "cuda_runtime.h"
#include <highgui.hpp>

using namespace cv;

#define DIM 600   //图像长宽

struct cuComplex
{
	float   r;
	float   i;
	__device__ cuComplex(float a, float b) : r(a), i(b) {}
	__device__ float magnitude2(void)
	{
		return r * r + i * i;
	}
	__device__ cuComplex operator*(const cuComplex& a)
	{
		return cuComplex(r*a.r - i*a.i, i*a.r + r*a.i);
	}
	__device__ cuComplex operator+(const cuComplex& a)
	{
		return cuComplex(r + a.r, i + a.i);
	}
};

__device__ int julia(int x, int y)
{
	const float scale = 1.5;
	float jx = scale * (float)(DIM / 2 - x) / (DIM / 2);
	float jy = scale * (float)(DIM / 2 - y) / (DIM / 2);

	cuComplex c(0.25, 0.010);
	cuComplex a(jx, jy);

	int i = 0;
	for (i = 0; i < 200; i++)
	{
		a = a * a + c;
		if (a.magnitude2() > 1000)
			return 0;
	}
	return 1;
}

__global__ void kernel(unsigned char *ptr)
{
	// map from blockIdx to pixel position
	int x = blockIdx.x;
	int y = blockIdx.y;
	int offset = x + y * gridDim.x;

	// now calculate the value at that position
	int juliaValue = julia(x, y);
	ptr[offset * 3 + 0] = 0;
	ptr[offset * 3 + 1] = 0;
	ptr[offset * 3 + 2] = 255 * juliaValue;
}

// globals needed by the update routine
struct DataBlock
{
	unsigned char   *dev_bitmap;
};

int main(void)
{
	DataBlock   data;
	cudaError_t error;

	Mat image = Mat(DIM, DIM, CV_8UC3, Scalar::all(0));

	data.dev_bitmap = image.data;
	unsigned char    *dev_bitmap;

	error = cudaMalloc((void**)&dev_bitmap, 3 * image.cols*image.rows);
	data.dev_bitmap = dev_bitmap;

	dim3    grid(DIM, DIM);
	kernel << <grid, 1 >> > (dev_bitmap);

	error = cudaMemcpy(image.data, dev_bitmap,
		3 * image.cols*image.rows,
		cudaMemcpyDeviceToHost);

	error = cudaFree(dev_bitmap);

	imshow("CUDA For Julia | c(0.25, 0.010)", image);
	waitKey();
}

c(-0.8,0.156):

c(-0.85,0.06):

c(-0.305,0.60):

c(0.25,0.010):