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i want to implement this transformation formula on any image, give me code, links , tutorials related to it ?
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how i implement this formula on such kind of images friends plz help
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Image Analyst
2013 年 11 月 29 日
I don't know what x and y are. If they are coordinates then this t value does not depend on the image pixel intensities at all, which strikes me as strange.
10 件のコメント
Image Analyst
2013 年 11 月 30 日
I hope my code helped you. It does what you want, even though I don't understand what you want represents, or why you want it. Does this solve everything for you now?
Roger Stafford
2013 年 11 月 29 日
編集済み: Roger Stafford
2013 年 11 月 29 日
Chitresh, I can tell you what kind of mathematical problem your expression is a solution to. Let us have n x and y pairs
x = [x1,x2,x3,...,xn]
y = [y1,y2,y3,...,yn]
and suppose we want to find the line of regression through them - that is, to find constants a and b such that
sum((a*x+b-y).^2)
is a minimum. This is a least squares problem and the solution for the constant b is
b = (sum(x.^2)*sum(y)-sum(x)*sum(x.*y))/(n*sum(x.^2)-sum(x)^2)
which is precisely your expression.
I can only guess how this might apply to the fingerprint image you display. If the xi and yi pairs refer to pixel coordinates of the black areas, and if the coordinate origin is at the lower left corner of the image, then b would be the vertical distance above the bottom at the left side where the line of regression crosses. On the other hand if the origin is located in the center of the image, b would be the vertical distance above or below this origin where the regression line crosses a vertical line through that origin. I would have thought the expression for a would have at least as much significance in the image as with b, since it would give the slope of that line of regression.
There are probably several other possible meanings that the x and y pairs might possess with regard to the image. I think that is a puzzle you will have to crack. Note that when you do solve your puzzle, the above expression shows you how to implement it in matlab in one line.
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