I have a few domains with Dreamhost where I have set a load of records using their web based domain name system, and I've like to move them to another provider that lets me enter info directly as a text file for their name server, bind 9 to use. (If you're interested, I'm moving them to Gandi.net).  Previously when I used a cpanel based system to do something similar, there was a tool that let me simply enter a domain name, and any available domains were automagically entered into a system, saving me typing it myself (and bringing down sites with silly typos in the process).  What open source tool can I use to query a domain for all the relevant subdomains and records and list them in a format like a zone file, that I can use with other name servers?
I want to see what nodes exist under a certain domain from within the domain. Short of querying every valid IP address within that domain, is there a way to get all records/map the network?
I'm using Microsoft Word 2010 and I worked in a word-2003 compatible document (.doc). After hours of working, I selected a half of text and did Ctrl+X to copy it into another document. Surprising, after closing the current document and opening a new one and pressing Ctrl+V, nothing happened. After closing original document, ms word asked me if I want to save the clipboard buffer and I pressed No, as I didn't knew that MS Word has it's own clipboard.  Is there a way to recover that text that was "cutten" from original document and never "pasted" to the new one? The PC was not rebooted so i guess all my temp files are here. May $y=e^x$ be satisfied with both $x$ and $y$ be positive integers?  I think it is not possible as $e$ ,a transcendental number, when multiplied by itself would never result in rational number.  Am I right? I cant change to faces or edges view! Where can we find frequently updated details about travel/entry restrictions due to Coronavirus (nCoV-19)? Due to the outbreak of nCoV-19 (Novel Coronavirus) many Countries have imposed several entry restrictions. But these are quite confusing for a lot of people. Some countries allow transit through parts of China (like Shanghai) but some restrict transit even through Hong Kong/Macau. Philippines restricted passengers coming from Taiwan too, but I think that that was later changed.  Where can one find a list of such restrictions for all the countries, ideally a source that is updated frequently "GPT detected. Please create a BIOS-Boot partition" while using boot-repair Since I updated Windows to 8.1 my Grub menu disappeared. I ran Ubuntu using Live USB, I used Boot-Repair but something went wrong. If I click on Reccomended repair I get this error:     GPT detected. Please create a BIOS-Boot partition (>1MB, unformatted filesystem, bios_grub flag). This can be performed via tools such as Gparted. Then try again. Alternatively, you can retry after activating the [Separate /boot/efi partition:] option.   So I tried to set my sda9 (I think that it's the Ubuntu system partition, since it's ext4) with the bios_grub flag and then to run Boot-Repair again. It didn't work.  I don't know what to do. My Grub was working fine (using UEFI) before I updated Windows.  I removed the bios_grub flag. Then I used fsck and it didn't find any error. So I tried again with Boot-Repair but it always gives me the "GPT detected" error. My boot is on UEFI mode, Secure Boot is "off".  When I boot my PC with SecureBoot OFF, I can reach the GRUB menu. But it's empty (the only voice is "System Setup").  I mounted my Ubuntu system disk:  sudo mount /dev/sda9 /mnt   and the other things:  sudo mount --bind /dev /mnt/dev sudo mount --bind /proc /mnt/proc sudo mount --bind /sys /mnt/sys   I logged on the system with chroot:  sudo chroot /mnt   I re-installed grub:  grub-install /dev/sda update-grub2   But it still doesn't work. Is logic subjective? If logic is constructed from axioms, and axioms are depended on observation which in term could be subjective, does this means that logic could be limited to our observation, and not really absolute and fundamental? Parsing DateTime on Localized Systems A toplogical space $X$ is said to be second countable   if there exists a countable basis for the topology.      $X$ is separable if there exist a countable dense subset.   Show that a second countable space always is separable.      Give an example of a space that is separable but not    second countable   The first part was easy and i managed to solve it on my own  Assume that $X$ is second countable and let $B_1, B_2, B_3...$ be a countable basis. Construct the set $A$ by taking one element $x_n$ out of every $B_n$. Then $A$ becomes dense because every open set $U$ contain a $B_n$ which in turn contain the element $x_n$ in $A$.    Again, like in the question i posted yesterday, i had trouble to come up with an example but our professor gave an example during the last lecture but i dont fully understand it:  "For an example of a separable but not second countable space let $X$ be the real line provided with the cofinite topology, i.e. where a set is open if its either empty or if its complement is finite.   Then every infinite set is dense and hence $X$ is separable.  But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  We asked him to explain it in more detail but his english is poor and nobody understood anything...:)       Def: A subset $A$ of a topological space $X$ is dense in $X$ if for any point $x$ in $X$, any neighborhood of $x$ contains at least one point from $A$.   $1.$ "Then every infinite set is dense"  Take $A$ to be one of these infinite sets. If $x \in A \subset X$ then it trivially fulfills the definition. But if we take a $x \not\in A$ why is it not possible to find a neighborhood in these finite sets that does not contain an element in $A$?  edit: its obvious if it is only open neighborhoods, open sets in the topology, but that is not specified in the definition.  $2.$ I am not fully understanding this either  "But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  Is it possible to express this in a simpler way? I want to find all files which contain a specific string of text. The grep command works, but I don't know how to use it for every directory (I can only do it for my current directory). I tried reading man grep, but it didn't yield any help. Question mark for sentence which starts with a question but doesn't end with one I'm basically looking at sentences which start with a question, and end with an explanation to it. But there may be more instances of a similar structure. Sample sentence :      Can you book a room in advance, because otherwise we'd be in trouble   Should this sentence end with a period or a question mark?  PS: I read the apparently similar question  but the sentence structure there is different from what I'm looking for.  Edit: I just discovered that this question is a duplicate. A resolved question already  exists:   But i do like the explanations here better. Prove that the set of palindromes are not regular languages using the pumping lemma. Firstly I pick a language $xyz$ where $x = \epsilon$, $y = (abb)^{k}$, $z = (bba)^{k}$ where $|y| \ge$ the number of states in the automaton representing my language. Then $xyz = (abb)^k(bba)^k$ is a palindrome.  Now I split $y$ into $uvw$, where $v \ne \epsilon$ contains a state visited more  than once. If the language is regular then $xuv^iwz$ is within the language $\forall i \ge 0$.  I choose $i = 2$, then $uv^2w = (abb)^n$ where $n &gt; k$.  Then $xuv^2wz = (abb)^n(bba)^k$ where $n &gt; k$, which is not a palindrome.   Firstly, is my reasoning for this particular proof correct?  I'm unsure about how to properly apply the pumping lemma to this problem. And secondly the question asks me to state a proof for palindromes in general, whereas I only attempted to prove it for a single case. Is there an example I can choose which proves the conjecture for all palindromes? I assume not, seeing as I can't think of how that could be represented using regular expressions. Replacement for alliums? Prove that  $ n &lt; 2^{n}  $ for all natural numbers $n$.  I tried this with induction: Inequality clearly holds when $n=1$. Supposing that when $n=k$, $k&lt;2^{k}$. Considering $k+1 &lt;2^{k}+1$, but where do I go from here?  Any other methods maybe? How do we prove that  $\operatorname{rank}(A) = \operatorname{rank}(AA^T) = \operatorname{rank}(A^TA)$ ?  Is it always true? Induction on a sum The left hand side has terms involving  $\binom{n}{m}= \dfrac{n!}{(n-m)!m!}$    $$1+\dfrac{1}{2}\binom{n}{1} +\frac{1}{3}\binom{n}{2}+...........+\frac{1}{n+1}\binom{n}{n} = \dfrac{2^{n+1}-1}{n+1}$$  I've done induction and proved $P(0)$ holds and also $P(1),P(2)$ holds (Just in case)  Now I've found $P(K+1)$ so the sum on the left is going to equal $\dfrac{2^{n+1}-1}{n+1}$ + $\dfrac{1}{n+2}$  and the right hand side is going to equal $\dfrac{2^{n+2}-1}{n+2}$  My problem is coming with the algebra though. I can't get those sides to equal. I can't get rid of the $n+1$ in the denominator.   Anyone have any ideas for me? They'd be much appreciated!  P.S. Someone gave me a hint and said that you can use the binomial theorem to solve this. What is the best word describing the nature of expiring in a short amount of time, connoting "losing public attention"? Is there an adjective that describes the nature of expiring, connoting a sense of losing public attention, in a very short amount of time? For example,  In a fast-paced era, news articles are ____. During the 1990s, compared with traditional technologies, computer  technologies are ___; for example, a new generation of CPU technologies usually comes out every one to two years.  (I thought about &quot;perishable,&quot; but it does not seem connoting &quot;losing public attention.&quot;) I have seen many softwares such as Update Manager and Synaptic Package Manager, they wait if some other program is using the /var/lib/dpkg/lock and is locked. How can we do this through the Terminal? I saw apt-get's manual but didn't find anything useful. Previously, someone posted his/her proof of "every neighborhood is open" . I am focusing on the same statement, but it seems that my approach is different from his/hers. Could someone comment on my own proof? Here we go:  Let $X$ be a metric space and $p\in X$. Fix $r&gt;0$. Define a neighborhood around $p$ as $N_r (p)=\{q\in X| d(p, q)&lt;r\}$. Now, take $w = \frac{r}{2}$. Then, $N_w (p)\subset N_r(p)$. In other words, every point of $N_r(p)$ is interior. Therefore, $N_r (p)$ is open. Since $r$ is arbitrary, this proves every neighborhood is open.  Update:  Under my definition of "neighborhood," can I prove in this way instead: For all $m\in N_r(p)$, take $w=\frac{r−d(m,p)}{2}$. Then, $N_w(m)\subset N_r(p)$.   Update2  From the hints given in comments, I solve it:  Assume $m\in N_r(p)$ and $q\in N_w(m)$. Take $w=r-d(m, p)$. Then, $d(p, q)\leq d(p, m) + d(m, q)&lt; d(p, m) + r-d(m,p) =r$.   Thanks to @JoséCarlosSantos and @fleablood ! Is it true that all numbers of the form $\sqrt[b_1]{a_1}+\sqrt[b_2]{a_2}+\cdots+\sqrt[b_k]{a_k}$ are not integers, where $a_i,b_i$ are positive integers such that none of the $k$ terms is an integer? I think this must be a theorem somewhere, probably a fundamental theorem in number theory, but not sure what it would be called. What is a good strategy to deal with lots of engineers turtling on the other team? Now that the Engineer update has come, there will be lots of Engineers building up everywhere.  How should this best be handled?