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Can someone help me to clarify this equation: $$g^{*}=arg\left(\sum_{g=0}^{255}h(g) > \delta N\right)$$

  • where h(g) is the amount at gray level g
  • threshold $g^{*}$
  • $\delta=0.99$
  • N is total number of pixel

Everything is accorded to this doc (equation(14))

Btw, I'm only known of argmax, argmin

doudev
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  • The doc says $\delta = 0.1$. – Hoc Ngo Apr 17 '17 at 07:58
  • The value of $\delta$ has multipurpose. I misreferenced. But this is not what I concern about – doudev Apr 17 '17 at 08:01
  • I don't know what $\arg$ means in this context so I need to understand the physics of this formula. (Normally, $\arg z$ is the angle $\theta$ of the complex number $z$.) After staring at the formula for a while, I think I have a feeling that there is a typo. The correct formula may be: $$ g^* = \arg\left(\sum_{g=0}^{255} h(g)\right) > \delta N. \quad(1)$$ – Hoc Ngo Apr 17 '17 at 08:12
  • Here, think of $\arg$ as an operator which gives the argument of $f(x)$ when operating on $f$, i.e., $$ \arg f(x) = x. \quad(2)$$ In this context, $f$ is $h$. So the inverse of $\arg$ is $h$. Now apply $h$ to (1) gives $$ h(g*) = h(\arg(...)) > h(\delta N).\quad(3)$$ Note $h(\arg(u)) = u$. – Hoc Ngo Apr 17 '17 at 08:14
  • My wild guess, from a quick look at the application, is that they really mean the smallest $g^\in{0,1,\ldots,255}$ for which $\sum_{g=0}^{g^}h(g) > \delta N$. But if so, it is an amazingly clumsy way to express this. – Harald Hanche-Olsen Apr 17 '17 at 08:14
  • Continue from (3), consider only the left part of the inequality (3): $$h(g) = \sum_{g=0}^{255} h(g). \quad(4)$$ The summation on the RHS of (4) represents the total count of pixels of all gray levels. So can we find a gray level $g$ such that $h(g*)$ is the same as the total count as given on the RHS? And we want this total count to be greater than $\delta N$. That's all I can describe from the formula. – Hoc Ngo Apr 17 '17 at 08:30
  • If you think $\sum_{g=0}^{g*}$ is what the paper means, then it is not necessary to introduce $\arg$ into the formula. – Hoc Ngo Apr 17 '17 at 08:33
  • Actually, I also guess what the author meant for this formula, just like @Harald said. But the way he explained it makes me confused. – doudev Apr 17 '17 at 09:08

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