Recent content by daviddoria

Yes, that is exactly what I am observing. You got it :)

I am confused how to interpret the result of preforming a normalized correlation with a constant vector. Since you have to divide by the standard devation of both vectors (reference: http://en.wikipedia.org/wiki/Cross-correlation#Normalized_cross-correlation ) , if one of them is constant (say a...

@chiro - what kind of statistical techniques? Are you suggesting just computing a Euclidean style distance between local FFT components or something like that? Doesn't this ONLY contain frequency information, and no information in the spatial domain (i.e. the colors don't have to match at all)...

Yep, that's what I want :) Of course there isn't a perfect one, but I thought someone here might have an idea of a better one than a simple Euclidean distance. Assume everything is quite low resolution. For instance, this picture of the entire house/grass/road is ~500x500. Yes, only...

Hi Stephen, The values are just the RGB pixel values. It does very well in most cases, but when it fails (like the case I described), it fails miserably. Yes, simple and fast is definitely a requirement. I would like to avoid this at all costs :) I hope this clarifies some...

I am trying to match little square patches in an image. You can imagine that these patches have been "vectorized" in that the values are reordered consistently into a 1D array. At first glance, it seems reasonable to simply do a Euclidean distance style comparison of two of these arrays to get a...

I think it works if you just use the law of cosines. See attached.

Gah, you are right. However, this requires I have p. What if I don't have p?

I guess another way to say it is: "I need a difference function which will produce 'x' for f(p,d1) and also produce 'x' for f(d2,d1)"

I guess even better would be some transform of each distance, so that: f(d1)-f(p)=f(d2)-f(d1) of course f() may not be exactly the same function, it may depending on the position (i.e. it could be f(d1)-f(p)=g(d2)-g(d1) or something like that).

See the image in the attached document. I am looking for a function which will make f(d1-p)=f(d2-d1)=f(d3-d2) (see the very last part of the document) I thought it would be as simple as dividing by the angle between the lines, but that doesn't seem to work. Is it reasonable to do this...

This is for mathematics students. Certainly if you study mathematics as your "field" then you should know these things. But engineers need not know them, and certainly people studying/practicing non-technical disciplines need not study them in high school.

Haha Manheis, as I was reading this thread I was thinking "how has no one posted the Worlfram TED talk??". I don't work for Wolfram either, but he is definitely right. Frankly, I am quite shocked that people here do not support this concept. The people who will build the next-gen Maple style...

We all know \int \frac{1}{x} dx = ln(x) + c but if you try to apply the power rule for integration: \int x^n dx = \frac{x^{n+1}}{n+1} + c you get \int x^{-1} dx = \frac{x^0}{0} What can you learn from this/what does this mean? David

So they are just giving the first solution? I'm confused why it doesn't say, -1, e^{i\pi / 3}, e^{-i\pi / 3} ?