I see several posts about USB-C and TB-3 to TB-2 but are they reversible? As I cannot find a TB-2 to USB-3 device I thought maybe I could connect my MacPro (Late 2013) to a USB-C hub which has USB-3 ports. My primary purpose is to access storage devices not displays.   How can one go from TB-2 to USB-C or USB-3?
Thunderbolt 2 docks have been quite expensive and they will be obsolete once I purchase the new Macbooks with USB-C ports. I currently use a 2014 MacBook Pro with Thunderbolt 2 ports, and I want to futureproof my purchase by acquiring a cheaper USB-C dock now.  The cheapest Thunderbolt 2 dock costs around $200 while USB-C docks cost around $30. Being able to use a USB-C dock with older Thunderbolt 2 Macbook Pros would not only be cheaper but also allows me to use the same USB-C dock if I decided to upgrade to a future Macbook.  For that to work I need a Thunderbolt 2 (male) to USB-C female adaptor, after some searching it doesn't seem to exist. Is there a reason why something like this hasn't been made? Or is the USB-C specification designed such that it is impossible? (hence businesses can still continue to sell expensive Thunderbolt 2 docks...)  I will accept an answer that meets the following requirements:   If there is such an adapter please give a link and some form of evidence that these would have 100% compatibility with USB-C docks. If there is no such adapter please explain:   Is it impossible to produce? If not, why hasn't a vendor created one yet? I've explained that its a better financial investment than buying a TB2 dock... surely there's a market for those still using TB2 MBPs wanting to upgrade in the next couple of years. Not only docks but it would open a gateway to cheaper alternatives to those hideously expensive TB2 peripherals.
PDF editing within Apex/Visualforce Is there a "best" way to automatically restart SQL Server on a regular basis? I read that I could create a batch file to net start, net stop the agent and service in a batch file and schedule that, but I was curious if there was a better way of doing this.  Please nevermind how this is indicative of a larger problem of operation, I've been arguing this for a month and we've ended up here.  Thanks :). Prove $f(S \cup T) = f(S) \cup f(T)$ $f(S \cup T) = f(S) \cup f(T)$  $f(S)$ encompasses all $x$ that is in $S$ $f(T)$ encompasses all $x$ that is in $T$  Thus the domain being the same, both the LHS and RHS map to the same $y$, since the function $f$ is the same for both.  Can you post the solution? Is there something like python's interactive REPL mode, but for Java? Is there something like python's interactive REPL mode, but for Java? So that I can, for example, type InetAddress.getAllByName( localHostName ) in a window, and immediately get results, without all this public static void nightmare() thing? Can you create a strong blockcipher with small blocksize, given a strong blockcipher of conventional blocksize? Suppose I want a strong 20-bit blockcipher.  In other words, I want a function that takes a key (suppose the key is 128 bits), and implements a permutation from 20 bits to 20 bits.  The set of permutations should be close to a randomly-chosen subset of size $2^{128}$ of all $2^{20}!$ permutations on 20 bits.  I don't want you to build a new blockcipher from scratch.  Instead, assume you have a "strong" blockcipher like AES (128-bit blocksize, 128-bit keysize) and use this to build your 20-bit cipher.   Of course, your construction should be practical (i.e., you should be able to run it in both directions with reasonable amounts of time and memory). I was hoping to find a more "mathematical" proof, instead of proving logically $\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$.  I already know the logical Proof:  $${n \choose k}^2 = {n \choose k}{ n \choose n-k}$$  Hence summation can be expressed as:  $$\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \cdots + \binom{n}{n}\binom{n}{0}$$  One can think of it as choosing $n$ people from a group of $2n$  (imagine dividing a group of $2n$ into $2$ groups of $n$ people each. I can get $k$ people from group $1$ and another $n-k$ people from group $2$. We do this from $k = 0$ to $n$. Apply the latest Service Pack and Cumulative Update How can I get the size of an Amazon S3 bucket? How do I deal with a bored or distracted player bringing down the group? My group have been playing various D&amp;D style games for a few years,  I have one player however that often will drift off and start playing on his phone  and several times has actually fallen asleep (with his character having to be "nudged" along by his fellow players).  He insists he's "just tired" and "isn't bored" — and besides, he keeps insisting we all get together for D&amp;D and wants to play.  Unfortunately we're all pretty busy, so if I'm to get the group together at all it's usually late on a Sunday. After talking to the player he insists it's not the game, or the story, he just happens to be tired. But, it's the same excuse every week, and the odd times we've played at different times or situations, the falling asleep has happened.  How do I cope with:   The morale destroying (for me) idea of your players falling asleep through a campaign and story you've put lots of work into. The general tone of the room when this player loses interest. (Occasionally a couple of others zone out into Facebook when they're not "active" because the players have split up / gone unconscious).   Given that it's an "all or nothing" situation with the group (there are only 3 players and they won't play unless all 3 are around), how do I best deal with it? While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:  Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not? Why are there hexagonal holes in my donut? I'm currently making a donut using blender gurus tutorial. I have no idea why, but there are holes in my icing. Does anyone know what could be causing it? Thanks for any help. How is dark energy consistent with conservation of mass and energy? I initially thought that dark energy must in some way violate conservation of mass and energy since the component of the energy density of space that comes from dark energy is constant, and space is expanding. Therefore, as space expands the energy in the universe that comes from dark energy would increase. I presumed the source of this energy was not coming from the conversion of other types of energy to dark energy, so it must violate conservation.  I decided to Google this and came upon this article:    It says that dark energy does NOT violate conservation and quotes Carroll, Press, and Turner (1992):  "…the patch does negative work on its surroundings, because it has negative pressure. Assuming the patch expands adiabatically [i.e. without loss or gain of heat], one may equate this negative work to the increase of mass/energy of the patch. One thereby recovers the correct equation of state for dark energy: P = – ρ c2. So the mathematics is consistent."  Is there a way to explain this in layman's terms? (The blog attempted to do this, but it was very unclear to me.) More specifically, can you explain where my initial train of thought described above fails when I erroneously concluded that dark energy violates conservation?  Thank you. How do you make clothes move along with model how do you make clothes move along with your model? Do you rig the model first and then parent it to the body of the character then click cloth simulation? I am a little bit confused with the process. Thanks for the help. Find the determinant of the following matrix Prove that $d(n)\leq 2\sqrt{n}$ In Waclaw Sierpinski's book Elementary Theory of Numbers on page 168 there is the following exercise:  "Exercises. 1. Prove that for natural numbers $n$ we have $d(n) \leq 2\sqrt{n}$," where $d(n)$ is the number of divisors of n.  As a hint right below is given: "The proof follows from the fact that of two complementary divisors of a natural number $n$ one is always not greater than $\sqrt{n}$.  I understand the hint but I don't know how it can be used to prove $d(n)\leq 2\sqrt{n}$.  Complementary divisors are pairs of divisors that when multiplied gives the number that is to be divided. For example the number $120$ has the complementary divisors: \begin{align*} &amp; 1, 120 \\ &amp; 2, 60 \\ &amp; 3, 40 \\ &amp; 4, 30 \\ &amp; 5, 24 \\ &amp; 6, 20 \\ &amp; 8, 15 \\ &amp; 10, 12 \\ \end{align*} How do you prove that $d(n) \leq 2\sqrt{n}$? How to politely negotiate PhD funding offer? I have been fortunate enough to get accepted with 5 years guaranteed funding to my 2nd choice graduate school. While the guarantee is great, the funding is only for 9/months out of the year. Would it be rude to ask them to guarantee summer funding to me for the first year until I can write a grant to get on a RAship? I know I should be more appreciative, since not everyone gets funding, but just an increase of about ~3k a year would make living there so much better and I do not want to accept an offer at a lesser institution just because of money.  What would be the most polite way of doing this?  I want to make it clear that I am not trying to get more money for the sake of getting more money, moreover that I am just not comfortable with so little money and such a high living cost in that area.  Thank you for your time. How Do I get into the castle walls? I already have the pink and red leather gloves, the armor, the crown, and the monkey's staff (and they're all enchanted). Regardless of how many health potions I use, I always die. Help! How can I go about determining the number of unique simple paths within an undirected graph? Either for a certain length, or a range of acceptable lengths.  Recall that a simple path is a path with no cycles, so I'm talking about counting the number of paths with no cycle. how to find $|GL_2(F_p)|$ I need to find $|GL_2(F_p)|$, I am very very uncomfortable in counting, so please help here I proceed,  $\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}$ be such an invertible matrix, so we need and sufficient that $ad\neq bc$, so if i chose all non zero $a,b,c$ arbitrarily and chose $d\neq a^{-1}bc$ and form matrices then the cardinality   will be $(q-1)^3\times (q-2)$, well now if one of $a,b,c,d$ be zero then other three can be anything nonzero so in this case cardinality will be $4(q-1)^3$, so finally if two entries are $0$ then they must be opposite to each other i.e diagonally, but other two must be nonzero, so cardinality in that case $2(q-1)^2$, so total number is just sum of three cases. Can I prevent others from editing my question? edited my question,  where I specifically said that our software sending back stacktrace to one of our servers was a user-known public feature.  He then modified the title of my post, along with the entire content, making it sound like I snuck in the feature and that our application was phoning home without our users knowing.    I specifically wrote in the original post that the feature was not sneaky. He completely changed the meaning of my post. I find this to be rather unacceptable.  A few questions:   Is it possible to prevent editing of my posts? How can I retaliate? Can I file a complaint?   Can people with  do anything they want on Stack Overflow?