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Bilateral filter [closed]

How do I implement a bilateral filter, given a gaussian filter?

like image 468
Jack Avatar asked Apr 17 '11 19:04

Jack


1 Answers

A simple bilateral filter can be defined as

Inew (x,y) = Summation(j=y-n/2; j<=y+n/2)Summation(i=x-m/2; j<=x+m/2)w(i,j,x,y)I(i,j)

where common low-pass filter, such as a Gaussian filter, has a weight w(i,j,x,y) based on the distance from the center of the kernel (x,y) to each pixel (i,j). For the bilateral filter, the weight is determined based on two distances: an image space distance and a colorht space distance.

A simple C implementation is below

void convolution(uchar4 *_in, uchar4 *_out, int width, int height, int ~ halfkernelsize, float id, float cd)
{ 
  int kernelDim = 2*halfkernelsize+1;
for(int y=0; y float sumWeight = 0; unsigned int ctrIdx = y*width + x; float ctrPix[3]; ctrPix[0] = _in[ctrIdx].x; ctrPix[1] = _in[ctrIdx].y; ctrPix[2] = _in[ctrIdx].z; // neighborhood of current pixel int kernelStartX, kernelEndX, kernelStartY, kernelEndY; kernelStartX = x-halfkernelsize; kernelEndX = x+halfkernelsize; kernelStartY = y-halfkernelsize; kernelEndY = y+halfkernelsize; for(int j= kernelStartY; j<= kernelEndY; j++) { for(int i= kernelStartX; i<= kernelEndX; i++) { unsigned int idx = max(0, min(j, height-1))*width + max(0, min(i,width-1)); float curPix[3]; curPix[0] = _in[idx].x; curPix[1] = _in[idx].y; curPix[2] = _in[idx].z; float currWeight; // define bilateral filter kernel weights float imageDist = sqrt( (float)((i-x)*(i-x) + (j-y)*(j-y)) ); float colorDist = sqrt( (float)( (curPix[0] - ctrPix[0])*(curPix[0] - ctrPix[0]) + (curPix[1] - ctrPix[1])*(curPix[1] - ctrPix[1]) + (curPix[2] - ctrPix[2])*(curPix[2] - ctrPix[2]) ) ); currWeight = 1.0f/(exp((imageDist/id)*(imageDist/id)*0.5)*exp((colorDist/cd)*(colorDist/cd)*0.5)); sumWeight += currWeight; _sum[0] += currWeight*curPix[0]; _sum[1] += currWeight*curPix[1]; _sum[2] += currWeight*curPix[2]; } } _sum[0] /= sumWeight; _sum[1] /= sumWeight; _sum[2] /= sumWeight; _out[ctrIdx].x = (int)(floor(_sum[0])); _out[ctrIdx].y = (int)(floor(_sum[1])); _out[ctrIdx].z = (int)(floor(_sum[2])); _out[ctrIdx].w = _in[ctrIdx].w; } }

}

like image 89
rOrlig Avatar answered Oct 21 '22 12:10

rOrlig