Floyd-Steinberg Algorithm

For (y=0; y<my; y++)

For (x=0; x<mx; x++){

iq[x][y]=quantize(I[x][y]);

e=i[x][y] - iq[x][y];

i[x+1][y] += 3/8*e;

i[x][y+1] += 3/8*e;

I[x+1][y+1] += 1/4*e;

Previous slide

Back to first slide

View graphic version