# Tag Info

0

Yes, this is entirely possible. As long as your test passwords are randomly distributed in the space, you will get the measure you are seeking. The hard part comes if the requirements on passwords allow most or very few of them to pass. You will detect that, but will not get a good measure. As an example, let us assume a character set of $26$ upper ...

0

The prime number is often stated as $\pi(x) \approx \frac{x}{\log x}$ meaning that the number of primes less than $x$ is that much. However we can have probabilistic interpretation as $\frac{\pi(x)}{x} = \frac{1}{\log x}$ so the odds of a randomly chosen number at random is $\frac{1}{\log x}$. But what exactly is $\log x$ it is more or less the number of ...

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This is the entropy of a mixture of (two) distributions, the probability of sampling from distribution number $i$ being proportional to the the sum $W_i$ of the weights of the elements of multiset $i$ (for $i=1,2$). The entropy is $\sum f(p)$ where $f(p)=-p\log p$ and $p$ runs through the probabilities of the different elements, so $\sum f(tp_1 + ... 0 On slide 12 it is also noted that the constructed set$\left\{ y_j \right\}$,$j=1, \dots, 2^{L/\varepsilon}$, of piecewise linear functions is also an$\varepsilon/2$-packing, meaning that $$\left\Vert y_i - y_j \right\Vert > \varepsilon/2 \quad \mbox{ for all }i\ne j.$$ Now, for any$\varepsilon/4$-cover of$\mathbf F$for each$y_j$we can find a ... 0 I think the issue is that while the Shannon entropy could be considered an "objective" property of the library, the "algorithmic complexity" of the program required to generate the content of the library is not an objective property, but depends on the specifics of the Turing machine available. Since the library has a finite number of unique texts within ... -1 Other answers have given a proof of the inequality. What i find problematic in these cases is not the actual proof of something already given (although it is important of course), but how did one actually reached this statement, which then it requires a proof? So how did one reached something like this in the first place: $$H_{\mathrm b}(p) \le 2 ... 2 Given I=(0,1), for any x\in I let g(x)=-x\log x and f(x)=g(x)+g(1-x). Obviously f\in C^{\infty}(I), \,f(x)=f(1-x), \,f>0 and$$ \lim_{x\to 0^+}f(x)=\lim_{x\to 1^-}f(x) = 0. $$We have f'(x)=\log(1-x)-\log(x), hence:$$\begin{eqnarray*} \frac{d}{dx}\frac{f(x)^2}{x(1-x)}&=&f(x)\cdot\frac{-x\log ... 0 This inequality is actually not trivial. Here is a proof sketch. We know that there are three equality cases, so to get rid of the ones at the ends we take$\frac{H(p)}{2\sqrt{p(1-p)}}$. By differentiating with respect to$p$and slowly analyzing the result (elementary algebra plus the inequality$x-\frac{1}{2}x^2 \le \ln(1+x) \le x$for$x \ge 0$) it is ... 1 Starting with the definition of differential entropy (assuming the random variables are continuous) we have: $$H(Y)=-\int_{\mathbb{Y}} p(y)\log p(y)\ dy$$ and the following expression for the conditional entropy$\$\begin{align} H(X|Y)&=-\int_{\mathbb{Y}}\int_{\mathbb{X}} p(x,y)\log p(x|y)\ dx\ dy\\&=\int_{\mathbb{Y}}\int_{\mathbb{X}} p(x,y)[\log ...

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