I'm sure this is a FAQ, but I can't find any clear answer. From what I understand, IPv6 addresses should be allocated in /64 blocks. Many articles are warning against using anything less, which might break some protocol expectations.  Does this mean that in IPv6 a /64 block should be treated as a single IPv4 address?  Can someone give me examples of practical problems I'd be facing if I went ahead and built a local network inside the /64 block?
This is a  about IPv6 Subnetting.      Related:               I know a lot about , and as I prepare to (deploy|work on) an IPv6 network I need to know how much of this knowledge is transferable and what I still need to learn. IPv6 seems at first glance to be much more complex than IPv4. So I would like to know:   IPv6 is 128 bits, so why is /64 the smallest recommended subnet for hosts? Related to this:   Why is it recommended to use /127 for point to point links between routers, and why was it recommended against in the past? Should I change existing router links to use /127? Why would virtual machines be provisioned with less than a /64 worth of addresses? Are there other situations in which I would use a subnet smaller than /64?  Can I map directly from IPv4 subnets to IPv6 subnets? For instance, does an IPv4 /24 correspond directly to an IPv6 /56 or /120? My interfaces have several IPv6 addresses. Must the subnet be the same for all of them? Why do I sometimes see a % rather than a / in an IPv6 address and what does it mean? Am I wasting too many subnets? Aren't we just going to run out again? In what other major ways is IPv6 subnetting different from IPv4 subnetting?
How to disable "time travel" entirely? Putting a matrix inside of caption gives \caption has an extra } I've just begun to work learn Linear Algebra on my own through Hoffman and Kunze's book and the first problem set already has a question that I can't solve:  Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.  I can't seem to figure out how to prove this without resorting to case work where you account for the cases where one of the coefficients are zero and when both are.  Is there an elegant way to prove in general that when two systems of linear equations have the same solutions, they are equivalent? The converse is obvious enough though.  Definition of equivalence from the text :     Let us say that two systems of linear equations are equivalent if each equation in each system is a linear combination of the equations   in the other system. What are the differences between Generics in C# and Java... and Templates in C++? I mostly use Java and generics are relatively new. I keep reading that Java made the wrong decision or that .NET has better implementations etc. etc.  So, what are the main differences between C++, C#, Java in generics? Pros/cons of each? How Special Relativity causes magnetism center align equation in ‎\documentclass[11pt,twoside,fleqn]{book}‎ If $I$ is a non-zero proper ideal of a commutative integral domain, is it possible for $I$ to be its own square? tcolorbox manual tip box, how to clone? Wildcard not working in subprocess call using shlex How should I handle possible revenge downvotes? I don't like such kind of questions on Meta, so I will try to make it a good question and describe all with a details.  Today I down-voted  added a comment with explanations why I down-voted it, and flag it as non constructive (my flag was marked as helpful). An hour later  was downvoted, without any comment left about the reason. Yes, that wasn't my best answer, but in my opinion it answers the question. Well, even if somebody considered my answer incorrect he could add the correct answer, or up-vote the competitive ones. I checked the  who made competitve answer, and he didn't get any rep today for that answer.   It was my second down-vote on Stack Overflow, so I was a little bit down, after that. I think who could considered my answer so completely wrong, and that I thought that maybe it was revenge down-vote. So I checked  of the user I down-voted today, and it was clear that he down-voted some answers today. After I checked that that user cast only 19 downvotes in 8 months, so chances were very big that this was revenge down-voting.  I came to this meta, and searched what I should do in this case. I found that most questions about down-votes, were closed as duplicates. After I found , and found that a team member of Stack Exchange suggested to cast a flag for moderator attention, and explain the situation there.  I did that, and create next flag:     I think after I down-voted this question (and a comment about this), the OP down-voted me in revenge; it's clear from his today's reputation and from timing of my down-vote of his question, and time when one of my answers was downvoted. I cannot be 100% sure about that, but if somebody can check this, and point him that it's not the good way to do.   As you can see, I didn't ask for returning any reputation, or banning that user, I just wanted, somebody to point him, that he did wrong. And after I get this respons on my flag: "declined - flags should only be used to make moderators aware of content that requires their intervention." So in the end I also lost 10 points of the flag weight.  After describing all situation in details, I want to know were exactly I was wrong here. And what should I do next time something like this happens? How does equality $\mathfrak{m}^\mathfrak{m} = 2^\mathfrak{m}$ in $\sf{ZF}$ relate to the axiom of choice? Usually, the equality $\mathfrak{m}^\mathfrak{m} = 2^\mathfrak{m}$ for infinite cardinal $\mathfrak{m}$ is proved like that: $$ 2^\mathfrak{m} \leqslant \mathfrak{m}^\mathfrak{m} \leqslant (2^\mathfrak{m})^\mathfrak{m} = 2^{\mathfrak{m} \cdot \mathfrak{m}} = 2^\mathfrak{m} $$ The last step relies much on the axiom of choice, because the fact $\mathfrak{m}\cdot\mathfrak{m} = \mathfrak{m}$ does. Moreover,  states that in $\sf{ZF}$ this fact is equivalent to $\sf{AC}$.  Note, by a cardinal I mean an equipollence class of an arbitrary set, not necessarily admitting a well-ordering.  That has motivated my question: what can we say about relationship of the initial equality and the axiom of choice in pure $\sf{ZF}$?    To avoid question about definition of cardinals without $\sf{AC}$, we may consider the following statement instead:     For any infinite set $X$ exists a bijection $\; F \, \colon X^X \rightarrow 2^X$   where $X^X$ is the set of all functions from $X$ to $X$, $\; 2^X = \mathcal P(X)$ is the power set, and an infinite set is such that cannot be bijectively mapped onto a finite ordinal. Capitalization of explanation of abbreviations Limit of $\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} }$ Print escaped representation of a str The Internet is awash with criticisms (or is that praise) of how Star Wars Episode VII: The Force Awakens borrows, steals, homages, references, archetype-uses or parallels from Star Wars Episode IV: A New Hope.  What are those similarities? Let $P$ be a probability measure on $(\Omega, \mathscr{F})$, If $a,b,\in\mathscr{F}$ and $A\subset B$, then $P(B)-P(A)=P(B\setminus A)\ge 0$.  So far I have:  $A\subset B \implies B\setminus A = B\cap A^c = \emptyset$ as $A$ and $B^c$ are disjoint. Since $A\cup (B\setminus A)=A\cup (B\cap A^c)=(A\cup B)\cap (A\cup A^c)=B\cap\Omega = B$  Then for $A,B\in \mathscr{F}\implies P(A\cup (B\setminus A))=P(A)+P(B\setminus A) = P(B)$ by countable additivity.  Rearranging, $P(B)-P(A)=P(B\setminus A) \ge 0$. Projecting a point onto a convex set decreases its distance from each point of that set For a convex set $K$, $x\in \mathbb{R}^d$ and $y=\Pi_K(x) $ (the projection of $x$ onto $K$), we want to show that for any $z\in K$ it holds that: $\|x-z\| \ge \|y-z\|$.  $$ $$  I'm not sure if the following proof is valid:  $$ y=\Pi_K(x) = \arg \min_{y\in K} ||y-x||  \Rightarrow \forall z\in K \langle y-x,y-z\rangle =0 $$ $$ ||y-x||^2_2+||y-z||^2_2=||x-z||^2_2  \quad (Pythagoras)$$  $$ \Rightarrow ||y-z||^2_2 \le ||x-z||^2_2 $$ $$ \Rightarrow||y-z|| \le ||x-z|| $$ I'm trying to crack a cypher that I believe is Vigenère encrypted and I'm currently stuck.  I calculated the key length by finding repeated sequences in the cypher and calculating the the common factors.  Word    distance factors agzd    16       2,4,8,16        zna     92       2,4,23,46,92 zdc     54       2,3,6,9,18,27,54 csio    44       2,4,11,22,44 omy     46       2,23,46 zab     5        5   I assume the key-length is 4 as these are the once that are common factors, but from here on trying to map the correct caesar shiffre for each position I'm stuck.  Can somebody confirm that my key-length calculation has no errors which would explain why I can't solve the cypher.  Here is the cypher:  jD 2GSA 7HNA ZNA w.m. V7KXA 36HBM MKE5 /H SGP9 +R GKKJ 2I A4OTC KDBCUR. 19+R 8Y TKJ JG4 IGO6 6U4T OJ hH/BIK. 196 qFKT+9 BNJK ZD6+Q AGZD. kJ'R EaOP6 4NCR ZK 36 / AGZD6C/HOIE2D. x0ZNAC2S8IO8DI HB xX8D4D 0XK O66M 0Y MN62S 8TZABBD2Za8BI. lBJ HA+DF 6UU/ 2J /FOZDC6S8I OJ kDC8G OO 2BLCYZ O66M 0Y G 925F4 UL J2JHCTGH FHH3K I'd like to typeset documents using the '' of page layout. Good examples are on pages 40-53 of . The  package is helpful for this, but two additional features would be useful:   Automatic flow of text across (arbitrarily placed) columns / textpos blocks. Vertical line spacing enforced to be an integer multiple of a base line spacing, even when interrupted (for example) by an equation of arbitrary height.   Any ideas as to how either or both of these could be accomplished? Can I use SPI for asynchronous serial output? I need a second serial async output channel on the MSP430G2553 and would like to use the SPI USCI to do this.  Is the USCI SPI baud rate stable enough to do this at 300 baud, 2400 baud or 9600 baud? I am currently using the 8 MHz clock.  Is there any other option that can save me from bit-banging?  My current thinking for 300 bd Tx is as follows:   set prescaler to 26666 LSB first load byte into Txbuf load 0x7F into Txbuf to generate stop and start bits repeat 2-3 send final stop bits 0xFF   It might also be possible to send ten bits of data on other members of the MSP430 family that contains a SPI.  Thanks for your responses.