I'm trying to debug NTLM authentication issue. One of my ideas was to capture the network traffic and look thougth it. In my case NTLM authentication is going over non-stardart port (6901). Of course, Wireshark can't detect it. But there is no NTLM (NTLMSSP) protocol in the list in Decode as menu. I can't do like . Is there a way to ask Wireshark to decode traffic as NTLM? Or I need to modify captured traffic, e.g. change TCP port or somehow another?
I am running Wireshark on my linux machine and i have simple python web server running on localhost:8080  I am capturing only local port 8080. When i open browser on localhost:8080 wireshark captures huge amount of tcp communication (as page loads with all resources) but there`s no HTTP only TCP...   Why i cannot capture HTTP on custom port ?
Apologies in advance for the long question.  My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$   t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.    Why is this so?     Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$.  Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$.  If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$.  If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.  I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$   T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$.  Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$   I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align}   I_{ba}  &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &amp;= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.  How do others feel about this argument?  I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.  I imagine that in GR a similar argument could be made by invoking local flatness of the metric. Which queue for an unmarried EU/non-EU couple travelling through UK Immigration Control together? How to show that Klein four-group is a normal subgroup of the alternating group $A_4$ I want to show that the Klein four-group is a normal subgroup of the alternating group $A_4$.   I am using the information in , that shows explicitly $A_4$, and Klein four-group as a subgroup.  I know that there is the direct way, by definition, but is there a way that does not require actually multiplying so many permutations ? What are these auxiliary file descriptors? While poking around /proc/self folder for my mksh shell,  I found a peculiar thing: in /proc/self/fd/* there are all the standard file descriptors (0 for stdin, 1 for stdout, and 2 stderr) some file descriptors, but also some extra ones - 24, 25, 3. And I technically can list them with a glob in the shell:  $ for fd in /proc/self/fd/* ; do echo $fd ; done                                                       /proc/self/fd/0 /proc/self/fd/1 /proc/self/fd/2 /proc/self/fd/24 /proc/self/fd/25 /proc/self/fd/3   But when I try to stat them or use find on them, they are reported as non-existent.  $ find  /proc/self/fd/*                                                                                /proc/self/fd/0 /proc/self/fd/1 /proc/self/fd/2 find: ‘/proc/self/fd/24’: No such file or directory find: ‘/proc/self/fd/25’: No such file or directory /proc/self/fd/3   Same occurs in bash , but just with just one auxiliary file descriptor.  $ for fd in /proc/self/fd/* ; do echo $fd; done /proc/self/fd/0 /proc/self/fd/1 /proc/self/fd/2 /proc/self/fd/255 /proc/self/fd/3  $ find /proc/self/fd/* /proc/self/fd/0 /proc/self/fd/1 /proc/self/fd/2 find: ‘/proc/self/fd/255’: No such file or directory /proc/self/fd/3   Question is: what are those additional file descriptors? What is their purpose? How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? Is there a known mathematical equation to find the nth prime? I've solved for it making a computer program, but was wondering there was a mathematical equation that you could use to solve for the nth prime? I am familiar with the set-theoretical definition of functions.   A function $f : A \rightarrow B$ from a set $A$ to a set $B$ is a subset $f \subseteq A \times B$ of the Cartesian product $A \times B$ such that the following conditions hold:   $\forall a \in A : \exists b \in B : (a,b) \in f$ $\forall a \in A : \forall b, b' \in B : (a,b), (a,b') \in f \rightarrow b=b'$   Is there a purely set-theoretical definition of the function evaluation $f(a)$ for every $a \in A$? I get as far as defining  $f(a) = \{ b \in B : (a,b) \in f \}$ but that defines rather the set of values of a multi-valued function. LWC URL reading parameters I am trying to authorize another sandbox/production to create connection. When I first time authorize the connected app, it is returning the Auhorization Code in the URL, however when it gets redirected to LWC app, code is disappeared because the parameter is 'code' not like 'c__code'. So, pagereference state is unable to read the param value. It was working till the previous release, with Spring release it started behaving differently.  Original URL: not working    I tested with adding 'c__' to the param manually and it worked fine.  Modified URL: Working    is there a way we can rename the redirected URL params dynamically?  'code' to 'c__code' Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$     Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)?    The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension. What is truecm? In examples found around the web, some people specify lengths such as width=5truecm with \includegraphics or the geometry package. How is this different from regular 5cm? "I left smoking", "I quit smoking", "I gave up smoking", "I stopped smoking" are these same? For various such sentences in many cases I get confused, it seems different for different cases. I want to know if someone left doing something for permanantly then what should be used? Also give some examples for different cases. Suppose... "I left my bag" means "I forgot my bag". but If I want to say the same for permanentaly "I left bringing my bag". Where should we use "stop" and "quit" then? What is the meaning of equation of a circle or equation of a cube? We have an equation of a Circle $$x^2 + y^2 = 1$$ and equation of a Cube $${||x-y|+|x+y|-2z|+||x-y|+|x+y|+2z|=1}$$  When I substitute x,y in first equation or x,y,z in the second equation then the equations are said to be satisfied when LHS and RHS are equal.  I have a quick question. If the first equation or second equation get satisfied, because LHS becomes equal to RHS, then those values of x and y of first equation are said to be lying on the circumference of the circle and not inside or outside of the circle and x,y,z of the second equation are said to be lying on the surface of the cube and not on the inside and not on the circle of the cube. Correct? Comparing and contrasting, p-values, significance levels and type I error I was wondering if anybody could give a concise rundown as to the definitions and uses of p-values, significance level and type I error.  I understand that p-values are defined as "the probability of obtaining a test statistic at least as extreme as the one we actually observed", while a significance level is just an arbitrary cutoff value to gauge if the p-value is significant or not. Type I error is the error of rejected a null hypothesis that was true. However, I am unsure regarding the difference between significance level and the type I error, are they not the same concept?  For example, assume a very simple experiment where I flip a coin 1000 times and count the number of times it lands on 'heads'. My null hypothesis, H0, is that heads = 500 (unbiased coin). I then set my significance level at alpha = 0.05.  I flip the coin 1000 times and then I calculate the p-value, if the p-value is > 0.05 then I fail to reject the null hypothesis and if the p-value is &lt; 0.05 then I reject the null hypothesis.  Now if I did this experiment repeatedly, each time calculating the p-value and either rejecting or failing to reject the null hypothesis and keeping a count of how many I rejected/failed to reject, then I would end up rejecting 5% of null hypotheses which were in actuality true, is that correct? This is the definition of type I error. Therefore, the significance level in Fisher significance testing is essentially the type I error from Neyman-Pearson hypothesis testing if you performed repeated experiments.  Now as for p-values, if I had gotten a p-value of 0.06 from my last experiment and I did multiple experiments and counted all the ones that I got a p-value from 0 to 0.06, then would I also not have a 6% chance of rejecting a true null hypothesis? What is the appropriate public key encryption for secure coin flipping? I quote Bobby's question here since I encountered the same one...    Consider the following protocol for two parties to flip a fair coin.   Trusted party $T$ publishes her public key $K_{pub}$ $A$ chooses a random bit $b_A$, encrypts it under $K_{pub}$, and announces the ciphertext $c_A$ to $B$ and $T$ $B$ does the same and announces a ciphertext $c_B$ (where $c_B \neq c_A$) $T$ decrypts $c_A$ and $c_B$, and announces the results. Both parties $XOR$ the results to obtain the random value $b_A\oplus b_B$.   It was shown that  was not appropriate for the encryption scheme.  seems to fail to work also. Since $c_A=m^e \mod N$. Then if $m=1$, $c=1$. If $B$ receives $A$'s cipher he will know the exact value of $m$.   So what should the appropriate encryption scheme be? Improving Review Queues - Design overview II: Changing review bans and other new features If random variables are drawn from an identical distribution, why doesn't this guarantee they are independent? Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only heat weighting: failed to find solution for one or more bones Determining coordinates of a SHP file i have a SHP file of a building converted from a DXF file. This SHP file is unprojected and have no specific coordinates (origin 0,0). I find WGS84 coordinates (lat, long) of this building from Google Maps. My aim is to replace these coordinates to the SHP file on QGIS.  Could you explain the process step by step?  (I read lots of similar questions, about affine transformation. But i don't know the paramaters.)    What i ask is how to make this transformation in QGIS (step by step)?   SHP file coordinates (not projected)  Scale: 1:142.670.713  X1,Y1: 16.70824, 253.74534  X2,Y2: 1.03521,-6.34918  X3,Y3: 244.00248,-7.49189  Target WGS84 coordinates   X1,Y1: 41.047773, 28.896241  X2,Y2: 41.045452, 28.895929   X3,Y3: 41.045436, 28.899039    I need a second vector layer with coordinates. I have just a SHP file of a building converted from a DXF file. That is not projected. I know where this building is, and can find the WGS84 coordinates from Google Maps. My aim is add WGS84 coordinates to this SHP file and export to replace the building on Google Maps.  My Questions are:  1) I think my SHP file has measurements in meters and unprojected. I have WGS84 lat long coordinates from Google Maps. How can i project this SHP file, for being ready to transformation? Right Click -> Set Layer CRS -> Choose WGS84 EPSG: 4326. Is this enough?  2) qgsAffine plugin just need transformation matrix paramaters. But the answer of @Jochen Schwarze is a bit complex for me. Is there a online calculator? (Just give the source XYs and target XYs and calculate the parameters.)    I think, I have a projection problem before affine transformation. I have a DXF file of a building from AutoCAD. I saved this DXF file as a SHP file in QGIS. But this SHP file is converted from a plain DXF file and is not referenced/projected, it is drawn from coordinates 0,0 in AutoCAD in meters. And the drawing has meters, WGS84 has decimal degrees.  When I set Layer CRS as WGS84 in QGIS, the coordinates don't change.  How can I project this SHP file as WGS84 in QGIS? If I take a screenshot, often the next action is to paint on it. Sometimes I paint a circle around something, sometimes I blank parts to anonymize them.  Below is an example of the window after the screenshot was taken:    It would be great if I could directly go to "edit image".  BTW, I use Pinta, not Gimp.  I guess this is impossible.  Off-topic but nevertheless relevant:  I know the terms "Wayland", "Xorg", "Gnome".  But my grand mother is a user, not a geek. She does not know these terms. And it would be great, if she could use ubuntu without learning these terms.