The figure shown above is a binary images I in which white piİxels have values of one and black pixels have values of zero. First compute derivatives of the image intensities in the x and y directions...


The figure shown above is a binary images I in which white piİxels have values of one and black pixels have values of zero. First compute derivatives of the image intensities in the x and y directions<br>following the numerical way to obtain Ix and Iy : Ix(i,j)= (I(i+1,j)-I(i-1,j))/2 and Iy(i,j) = (I(i,j+1)-I(i,j-1))/2. For the computation of Ix(i,j) on pixels in the first column<br>Ix(i, j) = I(i+1,j) - I(i,j) or of Iy(i,j) on pixels in the first row Iy(i,j) = I(i,j+1) - I(i,j). For the computation of Ix(i,j) on pixels in the last column Ix(i,j) = I(i,j) - I(i-1,j) or<br>of Iy(i,j) on pixels in the last row Iy(i,j) = I(i,j) - I(i,j-1).<br>Calculate the response R of the Harris corner detector at each pixel based on Ix and ly, assuming (i) a value of k=0.05 in the calculation, (ii) that products of derivatives are summed over an equally<br>weighted 3 by 3 pixel window around each pixel, (iii) that the image is padded with zeros when computing at pixels on the edge.<br>ANSWER:<br>R=<br>(to 2 decimal places)<br>

Extracted text: The figure shown above is a binary images I in which white piİxels have values of one and black pixels have values of zero. First compute derivatives of the image intensities in the x and y directions following the numerical way to obtain Ix and Iy : Ix(i,j)= (I(i+1,j)-I(i-1,j))/2 and Iy(i,j) = (I(i,j+1)-I(i,j-1))/2. For the computation of Ix(i,j) on pixels in the first column Ix(i, j) = I(i+1,j) - I(i,j) or of Iy(i,j) on pixels in the first row Iy(i,j) = I(i,j+1) - I(i,j). For the computation of Ix(i,j) on pixels in the last column Ix(i,j) = I(i,j) - I(i-1,j) or of Iy(i,j) on pixels in the last row Iy(i,j) = I(i,j) - I(i,j-1). Calculate the response R of the Harris corner detector at each pixel based on Ix and ly, assuming (i) a value of k=0.05 in the calculation, (ii) that products of derivatives are summed over an equally weighted 3 by 3 pixel window around each pixel, (iii) that the image is padded with zeros when computing at pixels on the edge. ANSWER: R= (to 2 decimal places)

Jun 09, 2022
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