Bruteforce GPG passphrase using script
Best way to soft brute-force your own GPG/PGP passphrase?
I try to understand why for a connected Lie group $G$ the first fundamental group $\pi_1(G)$ is abelian, and mainly why the second fundamental group is trivial $\pi_2(G)=0$?  Thanks for anyone who give me references for a 'simple proof' of these results I would appreciate if someone could help me write the mathematical equation for the seasonal ARIMA (2,1,0) x (0,2,2) period 12. I'm a little confused with how to go about this. I would prefer an equation involving $Y_t , e_t, θ$ and $Θ$. How to prove that $E[Y]=E[E[E[Y|X_1, X_2]]]$ ?  PS. I don't see how $E[E(Y|X_{1},X_{2})|X_{1}]=Y[Y|X_{1}]$ and $E[Y]=E[E(Y|X_{1})]$ can be used here. But it feels close. Please help, I'm stuck PPS. Thanks @Deep North. Let's consider the case when X1 is independent of X2 I saw this bug mentioned in image below, where in I was about to mark the question as duplicate..    Has you can see there is no radio buttons to select and go further.. And I am stuck..  Is this a KNOWN bug..  I am using Mozilla Firefox Browser..  Edit:-  As guided by @AzizShaikh, This kind of behaviour is observed when you have already flagged a post. In my opinion, I think the page as I mentioned above should not have come in first place, I mean don't allow the user to enter it doesn't belong here, or it is a duplicate section if his vote has been posted successfully for that section rather than showing go nowhere Im lost Screen ..  Can we do something about this...?? The rules for the monk ability Tongue of the Sun and Moon state:     Starting at 13th level, you learn to touch the ki of other minds so that you understand all spoken languages. Moreover, any creature that can understand a language can understand what you say.   PHB, pp.79   The text says that this ability is something you learn to do, not something that simply happens to you, so it's not necessarily always on. Further, it seems logical to say that this is something you would do actively. That is, you could choose not to touch someone's mind, thereby excluding them from understanding you. You could also choose not to understand someone yourself, just in case they're telling , or using a subliminal trigger.  Assuming you can choose not to touch someone's mind, let's reason further. Suppose you're speaking and don't know that there is someone listening in nearby. Would they understand you, i.e. is the default that you are or aren't touching someone's mind? Similarly, if you are speaking to a large group of people, would you have to focus on all of them at once for them to understand you? Would this take effort? Is there a range?  My questions are:   Can you exclude some characters from understanding you while allowing others to do so? Can you decide not to understand someone who is speaking?   Basically, these boil down to: Can you choose not to touch someone's mind?  Then, assuming that the answer to the above is yes:   Do you have to know someone is listening in order for them to understand you? Is there a limit to the number of people you can affect? A range? How to make ttf-mscorefonts-installer package download fonts, after it says it installed? When I install "Installer for Microsoft TrueType core fonts" (ttf-mscorefonts-installer) from the Software Center on 14.04, it gives a positive response--but after this I get a download error. It seems that the files actually providing the fonts are never retrieved. Is there any way to resolve this? How to Edit PDFs? In Obi-Wan's Order 66 warning transmission why didn't he tell the surviving Jedi about the Sith, including Palpatine's true identity? If you've seen the premiere episode of Rebels, you will have seen Obi-Wan's warning message that he broadcast to Jedi throughout the galaxy in RotS. In this message he speaks cryptically &amp; warns of a "dark shadow" falling &amp; to avoid the temple &amp; await a "new hope".   Wouldn't it have been useful to inform the Jedi that their mortal enemy the Sith had returned? And specifically that Palpatine himself was a Sith? You'd assume that the majority of the Jedi who were not in the loop of the Council would be unaware of the existence of the Sith. Yoda mentions a few times that they should keep it under wraps. I think now would've been the time to reveal the truth. The more info the surviving Jedi have the better. Who knows if the likes of Kanan knows anything about the Sith's return! I find Obi-Wan's omission of this information bizarre. How can we disable triggers, Validation Rules and workflow rules in salesforce org. [on off - switch] How can we disable triggers, Validation Rules and workflow rules in salesforce org.  This becomes really needed and important when we are doing data migration(data imports) in salesforce.  The way I handled this is create a custom label and include it in workflow and validation rule logic.   example - Custom label 'Disable workflow' = True then do not fire workflows.            if its False then fire workflow  But wanted to understand if there is any more efficient way to handle these scenarios. $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes See the related question . This is the second part of question 4-C in Milnor and Stasheff's book on characteristic classes. In the solution to the first part, we rely on the fact that having a field of tangent $1$-planes allows us to split the tangent bundle into a direct sum. It doesn't seem like this should be true for a field of tangent $2$-planes (otherwise the question wouldn't just be asking about $P^4$ and $P^6$), so I'm not sure exactly where to start.  edit: actually, since $P^n$ can be given a Riemannian metric, any subbundle should be a Whitney summand; this doesn't answer the question as to why they only ask for $P^4$ and $P^6$ though, so I'm still confused. Let $a$ and $b$ be positive reals. Show that $$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$ Prove Borel Measurable Set A with the following property has measure 0. This question is exercise 4.10 of Richard F. Bass's Real Analysis for Graduate Students, 2nd edition.  Let $\epsilon \in (0,1)$, let $m$ be Lebesgue measure, and suppose $A$ is Borel Measurable subset of $\mathbb R$. Prove that if $$ m(A\cap I)\leq (1-\epsilon)m(I)$$  for every interval $I$, then $m(A)=0$.  Collection of my thoughts  1. Try to prove by contradiction, suppose $m(A)=a&gt;0$ Borel Measurable Set satisfies the condition given. A property of any Borel Measurable Set on $\mathbb R$ is $A$ contains a closed set $F$ and is contained in open set $O$, and the measure of $F$ and $O$ can be arbitrarily close to $a$. 2. Obviously $A$ contains no interval, but it does not lead to contradiction, because we can find an example of $m(A)&gt;0$ that contains no interval(fat Cantor Set(remove middle 1/4 each time) has measure $1/2$ and contains no interval).  2'. For fat Cantor Set $C$ and given $\epsilon&gt;0$ how could we construct an interval I satisfies $m(C \cap I)&gt;(1-\epsilon)m(I) ?$   Maybe try to write Borel measurable set as countable union/intersection of intervals.   I'm stucked here and will appreciate any help, some examples/special case will be great too. How to create a pattern or tiles from rotated elements What is the difference between Interest Rate and APY on savings accounts, CDs, etc?  Is there a simple formula to convert the two? The Pauli group for $n$-qubits is defined as $G_n=\{I,X,Y,Z \}^{\otimes n}$, that is as the group containing all the possible tensor products between $n$ Pauli matrices. It is clear that the Pauli matrices form a basis for the $2\times 2$ complex matrix vector spaces, that is $\mathbb{C}^{2\times 2}$. Apart from it, from the definition of the tensor product, it is known that the $n$-qubit Pauli group will form a basis for the tensor product space $(\mathbb{C}^{2\times 2})^{\otimes n}$.  I am wondering if the Pauli group in $n$-qubits forms a basis for the complex vector space where the elements of this tensor product space act, that is $\mathbb{C}^{2^n\times 2^n}$. Summarizing, the question would be, is $(\mathbb{C}^{2\times 2})^{\otimes n}=\mathbb{C}^{2^n\times 2^n}$ true?  I have been trying to prove it using arguments about the dimensions of both spaces, but I have not been able to get anything yet. Drawing a 3D cylinder, and a circle shape on the x---y plane I'm trying to draw a cylinder on the three dimensional space, and I also want its projection on the x--y plane. I found the first answer  to be very useful. I just copy-paste it to my document, and modify it as follows (the modifications are only inside the \begin{tikzpicture} tags):   \documentclass[parskip]{scrartcl} \usepackage[margin=15mm]{geometry} \usepackage{tikz}  \pgfmathsetmacro{\xdeg}{30} \pgfmathsetmacro{\xx}{cos(\xdeg)} \pgfmathsetmacro{\xy}{sin(\xdeg)}  \pgfmathsetmacro{\ydeg}{120} \pgfmathsetmacro{\yx}{cos(\ydeg)} \pgfmathsetmacro{\yy}{sin(\ydeg)}  \pgfmathsetmacro{\zdeg}{80} \pgfmathsetmacro{\zx}{cos(\zdeg)} \pgfmathsetmacro{\zy}{sin(\zdeg)}  \newcommand{\tdcyl}[5]{% origin x, origin y, origin z, radius, height     \path (1,0,0);     \pgfgetlastxy{\cylxx}{\cylxy}     \path (0,1,0);     \pgfgetlastxy{\cylyx}{\cylyy}     \path (0,0,1);     \pgfgetlastxy{\cylzx}{\cylzy}     \pgfmathsetmacro{\cylt}{(\cylzy * \cylyx - \cylzx * \cylyy)/ (\cylzy * \cylxx - \cylzx * \cylxy)}     \pgfmathsetmacro{\ang}{atan(\cylt)}     \pgfmathsetmacro{\ct}{1/sqrt(1 + (\cylt)^2)}     \pgfmathsetmacro{\st}{\cylt * \ct}     \filldraw[fill=white] (#4*\ct+#1,#4*\st+#2,#3) -- ++(0,0,#5) arc[start angle=\ang,delta angle=-180,radius=#4] -- ++(0,0,-#5) arc[start angle=\ang+180,delta angle=180,radius=#4];     \filldraw[fill=white] (#1,#2,#3+#5) circle[radius=#4]; }       \begin{tikzpicture}[x={(\xx*1cm,\xy*1cm)},y={(\yx*1cm,\yy*1cm)},z={(\zx*1cm,\zy*1cm)}]          \tdcyl{-3}{-3}{0}{1}{3} % x y z   r h         \tdcyl{0}{0}{0}{1}{3} % x y z   r h           \draw (-3,0,0) -- (3,0,0) node[circle,fill=white] {x};         \draw (0,-3,0) -- (0,3,0) node[circle] {y};         \draw (0,0,-3) -- (0,0,3) node[circle] {z};     \end{tikzpicture}     \end{document}   I added another cylinder and tweaked the positions to fit what I need. This produces the following:      Which is almost what I want, except that I want to see the projection of that cylinder on the x--y plane, but unfortunately, the top cylinder hides the full circle. When I disable the top cylinder:      \begin{tikzpicture}[x={(\xx*1cm,\xy*1cm)},y={(\yx*1cm,\yy*1cm)},z={(\zx*1cm,\zy*1cm)}]          \tdcyl{-3}{-3}{0}{1}{3} % x y z   r h            \draw (-3,0,0) -- (3,0,0) node[circle,fill=white] {x};         \draw (0,-3,0) -- (0,3,0) node[circle] {y};         \draw (0,0,-3) -- (0,0,3) node[circle] {z};     \end{tikzpicture}   Now the projection is shown again:   But the top cylinder is missing...  How can I make the top cylinder transparent, or, alternatively, and even better; how do I create one long cylinder and emphasize its projection on the x--y plane?   BTW, what determines the position of the labels x,y,z in the above code? I want to reposition them so they won't overlap with the axis. Thanks in advanced! Is it possible to remove a user's password using the net command?  I tried "net user &lt;name&gt; *" and just hitting return twice, but this does not work as expected. Is there any other way to do this? (I'd gladly take non-net commands, as long as they are built-in in XP SP3).  Edit: Seems, I did not make my intention clear. I don't just want to change the password, but remove it completely (so that the user may log in without having to supply a password).  I'm familiar with the "net"-reference and have searched online for several hours, but could not find a hint on how to do this. On the one hand, I'm aware that  in the US do not have a valid government ID. On the other hand, traveling within the US requires you to show a piece of ID wherever you go - even when taking a Greyhound.   So let's say you're an American who has a large number of cash in their pocket but no ID whatsoever. Would you be able to get from New York to San Francisco or from Seattle to Miami? If so, what are your options? Even though I do have plenty of identification documents, I find it enjoyable to be able to travel anonymously. How to schedule a task to run when shutting down windows Proof that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq a+b+c$ I've tried getting everything on the left side and transforming it into something squared so that I can prove it's bigger or equals to 0 but I've been unsuccessful.