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Date: Mon, 11 Nov 1996 17:30:57 GMTServer: NCSA/1.5Content-type: text/htmlLast-modified: Mon, 23 Oct 1995 20:29:37 GMTContent-length: 4603<html><head><title>CS 766 HW #2 Hints and FAQs</title></head><body><H1>CS 766 HW #2 Hints and FAQs</H1><p><hr><UL><LI><B>How are the weights in the mask w defined?</B><BR>The weights used should be based on the values a=.4, b=.25, and c=0.05.These should be scaled so that integer arithmetic, not floating point,is used in the convolution operation. A good 1D convolution mask is:1 4 6 4 1. The result must be rescaled by multiplying by 1/16 to obtain theright answer. More correctly, if the result of convolving the above maskat a pixel is x, then the rescaled and rounded result should becomputed as (x + 8)/16 since division truncates.<P><LI><B>What does it mean to subtract two images?</B><BR>This is a pixel-wise difference operation resulting in another image.<P><LI><B>How can I create two input images that approximately matchin scale so that the splining creates a cool image?</B><BR><B>vscale</B> can be useful for initially scaling yourtwo input images so that the splined result is more interesting.<P><LI><B>Since the Laplacian images have both positive and negative values,with a range of -255 to 255, what is the best way to store these images?</B><BR>In practice most pixels will have values which are close to 0, so itwill not usually introduce much error if you truncate the Laplacian valuesthat are less than -128 and greater than 127 and store the image as a byteper pixel (Vista type SBtye). <P><LI><B>How should integer arithmetic operations be performed so as toavoid losing precision?</B><BR>Whenever you do an integer division operation, be sure to computea rounded result by doing (a + (b/2))/b instead of a/b.<P><LI><B>How are the windows from the two input images determined when theimages are arbitrary sizes?</B><BR>After aligning the center pixels of the two input images, there is some n x marea where the two overlap. Make the n x m left image by using the n x mwindow of the first image that is centered on its center point. Similarly,construct an n x m right image from the second input image. You can eitheruse these n x m images as input to your programs (stopping at a level ofthe pyramid where n or m = 1), or else by first padding the n x m imagesto be of size (2**k + 1) x (2**k + 1), although padding will likely introducesome artifacts. <P><LI><B>How is the Expand operation defined, and why is there a 4 in theformula?</B><BR>Maybe I can answer your question about EXPAND by showing two implementations.An "incorrect" implementation is:<BR><pre> tmp = 0 for m = -2 to 2 for n = -2 to 2 tmp = tmp + w(m,n) * image((row+m)/2, (col+n)/2) store tmp</pre><p>instead of a "correct" version like:<BR><pre> tmp = 0 for m = -2 to 2 for n = -2 to 2 if (row+m mod 2 = 0 and col+n mod 2 = 0) then tmp = tmp + w(m,n) * image((row+m)/2, (col+n)/2) store 4*tmp</pre><p>In the correct version, suppose row and column are both even, say 10.Then you actually use a 3x3 window on the input image [row values: (10-2)/2,10/2, (10+2)/2 and column values (10-2)/2, 10/2 and (10+2)/2]. Thus only nineof the 25 weights in the kernel are used corresponding to the x's below:<br><pre> x o x o x o o o o o x o x o x o o o o o x o x o x</pre>and if you consider the weights of those nine locations, they turn out to be:<br><pre> .0025 .0200 .0025 .0200 .1600 .0200 .0025 .0200 .0025 </pre>summing up to 0.25 exactly. And thus you should multiply your result by 4 sothat the effective sum of weights is 1.<p>In the incorrect version (still assuming row and column to be even), you areeffectively truncating (row+m)/2, (col+n)/2 for m,n = -2..2. For instance thecomputed row index is same for m = 0 and m = 1 [row/2 = (row+1)/2]. Thus you are still using a 3x3 window on your input image, but some values in this window are repeated. The effective weights for the window are:<br><pre> .0900 .1950 .0150 .1950 .4225 .0325 .0150 .0325 .0025 </pre>summing up to 1.0. This weight window is obviously very different fromBurt-Adelson's weights.For row odd, column even etc, you can similarly compute the effective weightmatrices and see that the results are not similar for the two algorithms.<p>The "correct" version tries to interpolate pixel values in the larger (output)image by looking at a neighborhood in the smaller (input) image. It makessense to use a symmetric Gaussian-like weight matrix to do that.</UL></body></html>⌨️ 快捷键说明
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