How to find connected object in a binary image in VIVADO HLS? - opencv

Question

I have a thresholded binary image as shown below:
I want to find all connected object in the image.
The code will take an input image stream and gives no. of connected components as output.
I have already implemented it in C where matrices are stored and can be accessed directly by A[][] format. But in HLS images come in as a stream, which I have converted in hls::Mat. I am not sure whether I can perform an element-wise operation on Mat and whether mat is available for offline operations just like a matrix.
I am not able to figure out how to implement the C code in Vivado HLS.
unsigned char paddedImg[FULL_HD_HEIGHT+2][FULL_HD_WIDTH+2] = {0};
void connectedComponents(unsigned char* Image,int width, int height, int
connectType, int* noOfComponents, char* list)
{
unsigned char west, north,northWest, northEast,south,east,southEast,southWest = 0;
int label = 0;
int min = 0;
for(int i=0; i<height; i++)
{
for(int j=0; j<width; j++)
{
paddedImg[i+1][j+1] = Image[(i*width) + j];
}
}
for(int i=1; i<(height+1); i++)
{
for(int j=1; j<(width+1); j++)
{
west = paddedImg[i][j-1];
northWest = paddedImg[i-1][j-1];
north = paddedImg[i-1][j];
northEast = paddedImg[i-1][j+1];
if(paddedImg[i][j] != 0)
{
min = 25500;
if(connectType == 8)
{
if( west != 0 || north != 0 || northWest != 0 || northEast != 0)
{
if(west < min && west != 0) min = west;
if(northWest < min && northWest != 0) min = northWest;
if(north < min && north != 0) min = north;
if(northEast < min && northEast != 0) min = northEast;
paddedImg[i][j] = min;
}
else
{
paddedImg[i][j] = ++label;
}
}
else if(connectType == 4)
{
if( west != 0 || north != 0)
{
if(west < min && west != 0) min = west;
if(north < min && north != 0) min = north;
paddedImg[i][j] = min;
}
else
{
paddedImg[i][j] = ++label;
}
}
else
{
printf("invalid connect type\n");
}
}
}
}
for(int i=1; i<height+1; i++)
{
for(int j=1; j<width+1; j++)
{
if(paddedImg[i][j] != 0)
{
if(connectType == 8)
{
west = paddedImg[i][j-1];
northWest = paddedImg[i-1][j-1];
north = paddedImg[i-1][j];
northEast = paddedImg[i-1][j+1];
east = paddedImg[i][j+1];
southEast = paddedImg[i+1][j+1];
south = paddedImg[i+1][j];
southWest = paddedImg[i+1][j-1];
min = paddedImg[i][j];
if(west < min && west != 0) min = west;
if(northWest < min && northWest != 0) min = northWest;
if(north < min && north != 0) min = north;
if(northEast < min && northEast != 0) min = northEast;
if(east < min && east != 0) min = east;
if(southEast < min && southEast != 0) min = southEast;
if(south < min && south != 0) min = south;
if(southWest < min && southWest != 0) min = southWest;
if(west != 0) paddedImg[i][j-1] = min;
if(northWest != 0) paddedImg[i-1][j-1] = min;
if(north != 0) paddedImg[i-1][j] = min;
if(northEast != 0) paddedImg[i-1][j+1] = min;
if(east != 0) paddedImg[i][j+1] = min;
if(southEast != 0) paddedImg[i+1][j+1] = min;
if(south != 0) paddedImg[i+1][j] = min;
if(southWest != 0) paddedImg[i+1][j-1] = min;
}
else if(connectType == 4)
{
west = paddedImg[i][j-1];
north = paddedImg[i-1][j];
east = paddedImg[i][j+1];
south = paddedImg[i+1][j];
min = paddedImg[i][j];
if(west < min && west != 0) min = west;
if(north < min && north != 0) min = north;
if(east < min && east != 0) min = east;
if(south < min && south != 0) min = south;
if(west != 0) paddedImg[i][j-1] = min;
if(north != 0) paddedImg[i-1][j] = min;
if(east != 0) paddedImg[i][j+1] = min;
if(south != 0) paddedImg[i+1][j] = min;
}
else
{
printf("invalid connect type\n");
}
}
}
}
for(int i=0; i<height; i++)
{
for(int j=0; j<width; j++)
{
Image[i*width + j] = paddedImg[i+1][j+1];
}
}
*noOfComponents = min;
}

Answer 1

The algorithm you describe seem to require that the entire image be loaded into FPGA memory to allow random access to elements. If the image is small enough, you can do that by reading it from the stream into memory beforehand.
If the image is large or you don't want to buffer the entire image, the paper FPGA implementation of a Single Pass Connected Components Algorithm proposes an algorithm for finding connected component in a single pass.