Which one is correct? Leaning towards first one but I am unsure because a more basic version would be "How good is it?"
I want to know which sentence is correct. If i say     Guess, what it is?   or     Guess, what is it?   which one is correct?
How to lock compiled Java classes to prevent decompilation? How do I lock compiled Java classes to prevent decompilation?  I know this must be very well discussed topic on the Internet, but I could not come to any conclusion after referring them.  Many people do suggest obfuscator, but they just do renaming of classes, methods, and fields with tough-to-remember character sequences but what about sensitive constant values?  For example, you have developed the encryption and decryption component based on a password based encryption technique. Now in this case, any average Java person can use  to decompile the class file and easily retrieve the password value (defined as constant) as well as  and in turn can decrypt the data by writing small independent program!  Or should such sensitive components be built in native code (for example, VC++) and call them via ? In-universe, a lightsaber is basically a controlled, fixed-length plasma stream that can be used as a sword. As such, the blade wouldn't have any mass (or infinitesimal mass if you want to get technical about E=mc2).  Anyone familiar with fencing or swordplay knows that the weight of the handle and blade, and thus the overall balance of the weapon, is critically important while fighting. A sword that is too blade-heavy is unwieldy; it is difficult to start, stop, and change the path of the blade as it's swung, which is vital to outmaneuvering your opponent in a duel. Conversely, a sword with too light a blade (equivalently, too heavy a hilt) is too easy to swing around. it's hard to "feel" where the blade is, and thus to make the minute changes needed to block a blow or land one. It can also be just as exhausting, in a different way, to wield such a sword; because most of the mass is behind your hand, it takes muscle power to point the blade forward instead of upward, while a balanced weapon would point directly forward from the user's hands. You also get no mechanical advantage while swinging it.  Obviously, out-of-universe the props were full "swords" that were then rotoscoped to paint the beam blade over the white wooden (later plastic) dowel that protruded from the saber. This gives the actual prop used by the actors a more realistic weight and balance, which can be seen in the duels (Darth Maul has to heave a bit to get that double-saber staff moving, and you can see the effort behind a full baseball swing of the sword by Luke or Vader in the original trilogy).  Was there any attempt in-universe to explain the weight and balance that an active lightsaber seemed to possess? PhD in Europe without scholarship from non-European homeland I earned my bachelors in Computer Engineering from my home country. I am now enrolled in a Physics masters program in the same university. Next year I plan to find a different university in Europe in which to enroll in a Computer Engineering PhD program.  However I doubt that I'll receive any scholarships from my homeland's government when studying abroad. Can I survive if I start to live out of my own my pocket? Is that enough to live in Europe just for a half year. Are there any available funds that can supplement my income such as a TA/RA's salary?  My main question is:   How to survive if I want to study PhD in Europe without any scholarship from my home country? Which country might provide sufficient support for a foreigner to study at graduate level?   I am interested to Germany and Finland but any European country would be acceptable! Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$ I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$   $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.      I think the first step should look something like this:      $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$   From here I tried to proceed with two cases. 1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$ 2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$  My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.  Any help is appreciated. Thx in advance. Cherio Woltan How to remove the lines which appear on file B from another file A? I'm fairly certain this has happened to all of us at some point: You go into a room to get something, but once you're there, you can't remember what you intended to get. It seems like a specific, common phenomenon like this must have a name, but I can't think of it. Does anyone know a name (either formal or slang) for this?  I don't want a general term like forgetfulness or absent-mindedness, which could apply to all sorts of forgetting, or a term for not being able to remember a certain word. I'm looking for a word or phrase that describes this one specific situation of forgetting what you need to get as soon as you go to get it. Proof: $\|X\|_2\leq \sqrt{n}\|X\|_\infty$ (Euclidean Norm)  Proof that $\|X\|_2\leq \sqrt{n}\|X\|_\infty$, if $x \in \mathbb{R^n}$.   With that, proof $\|X\|_\infty \to 0 \implies \|X\|_2\to 0$  I'm stuck:  $\|X\|_2= \sqrt{x^2_1+x^2_2+...+x^2_n}\leq\sqrt{n}\sqrt{x^2_1+x^2_2+...+x^2_n}$  I don't know how to use the fact: $\|X\|_\infty=\max(|x_1|,|x_2|,...,|x_n|)$ Is there any well-ordered uncountable set of real numbers under the original ordering? I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $&lt;_\mathbb R$?   Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set. Is leaving airport at a stop before the final destination illegal? There are tricks such as  or  which involve booking a ticket composed of several connecting flights, then leaving the airport at a stop, possibly even "hopping off" a plane (intermediate landing). Rationale: Sometimes these tickets are cheaper than a ticket to one of the stops. So one adds an additional leg and discards it.  Some people say that when using such tricks, one enters a legal gray area. But I wonder:  Could leaving the airport at a stop really be illegal, i.e. violating an airline's or a travel agency's terms?  Also there are suggestions that one should not request an upgrade, because then the trick may be discovered by the airline. Here I wonder:  How can they know that one will throw away the last leg? How do I test if all minecraft players on a server are in a radius? This has got me. I need a command that tells me if all players on a server are in a radius of a command block. If one of them isn't in the radius, the command will be false. Any ideas? Thanks! The code below compiles, but has different behavior for the char type than for the int types.  In particular     cout &lt;&lt; getIsTrue&lt; isX&lt;int8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;uint8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;char&gt;::ikIsX  &gt;() &lt;&lt; endl;   result in 3 instantiations of templates for three types: int8, uint8 and char. What gives?  The same is not true for ints: int and uint32 which result in the same template instantiation, and signed int another.  The reason seems to be that C++ sees char, signed char and unsigned char as three different types. Whereas int is the same as a signed int. Is this right or am I missing something?  #include &lt;iostream&gt;  using namespace std;  typedef   signed char       int8; typedef unsigned char      uint8; typedef   signed short      int16; typedef unsigned short     uint16; typedef   signed int        int32; typedef unsigned int       uint32; typedef   signed long long  int64; typedef unsigned long long uint64;  struct TrueType {}; struct FalseType {};  template &lt;typename T&gt; struct isX {    typedef typename T::ikIsX ikIsX; };   // This  int==int32 is ambiguous //template &lt;&gt;            struct isX&lt;int  &gt;    { typedef FalseType ikIsX; };  // Fails template &lt;&gt;            struct isX&lt;int32  &gt;  { typedef FalseType ikIsX; }; template &lt;&gt;            struct isX&lt;uint32 &gt;  { typedef FalseType ikIsX; };   // Whay isn't this ambiguous? char==int8 template &lt;&gt;            struct isX&lt;char  &gt;  { typedef FalseType ikIsX; }; template &lt;&gt;            struct isX&lt;int8  &gt;  { typedef FalseType ikIsX; }; template &lt;&gt;            struct isX&lt;uint8 &gt;  { typedef FalseType ikIsX; };   template &lt;typename T&gt; bool getIsTrue(); template &lt;&gt;           bool getIsTrue&lt;TrueType&gt;() { return true; } template &lt;&gt;           bool getIsTrue&lt;FalseType&gt;() { return false; }  int main(int, char **t ) {    cout &lt;&lt; sizeof(int8) &lt;&lt; endl;  // 1    cout &lt;&lt; sizeof(uint8) &lt;&lt; endl; // 1    cout &lt;&lt; sizeof(char) &lt;&lt; endl;  // 1     cout &lt;&lt; getIsTrue&lt; isX&lt;int8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;uint8&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;char&gt;::ikIsX  &gt;() &lt;&lt; endl;     cout &lt;&lt; getIsTrue&lt; isX&lt;int32&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;uint32&gt;::ikIsX  &gt;() &lt;&lt; endl;    cout &lt;&lt; getIsTrue&lt; isX&lt;int&gt;::ikIsX  &gt;() &lt;&lt; endl;  }   I'm using g++ 4.something If $f(0)=0$ and $f'$ is increasing, then $\frac{f(x)}{x}$ is increasing. Is AncestryDNA match of 1,655 cM a half sibling? Shared DNA: 1,655 cM across 46 segments.   Can someone that shows this as one of my matches on AncestryDNA be a half sibling? Why do the measurements of this object seem erroneous? I just found out about the Mesh Display : Length feature in the "N" panel (thanks to ) that allows one to measure the length of a face or vertice.  But my first encounter with this tool gives me complex feelings of love and hate, see how, as I was trying to measure the aspect ratio of this TV screen, I am getting WIDTH = 20.33 and HEIGHT = 37.57.  These values seem awfully wrong, since my betrusted eyes tell me the height should be shorter than the width.    I have tried measuring the single vertices as well, to identical results. Motivated by :   Which finite groups are the group of units of some ring? Which finite groups are the group of units of some finite ring? Which finite abelian groups are the group of units of some commutative ring? Which finite abelian groups are the group of units of some finite commutative ring?   It  that even for fields this is not simple to answer.  I expect the general question (Which groups are the group of units of some ring?, with no finiteness hypotheses) is even harder. When do I need to invoke \phantomsection? SGS2 screen gets more and more dark blue, possibly after overheated charging Windows 8 OEM Reinstalling I know this question is a repetition but I didn't find any solution of my problem from any of the previous questions. Recently I formatted my Laptop HDD and now I want to re install windows 8 I've a Windows 8 OEM license but the problem is I don't have a disc , I downloaded en_windows_8_x64_dvd_915440 from msdn , but when I boot the windows it shows the error that the current version is not suitable with my serial key, so where I can find an OEM Windows 8 and also I've a retail windows 8 disc so if I install that windows 8 in my pc and is there any way to activate it using my OEM key How can I detect if my apple device supports Bluetooth Low Energy $\DeclareMathOperator{\Char}{char}$$\DeclareMathOperator{\Aut}{Aut}$ Let $G$ be a finite group, $H$ a subgroup, such that $H \Char G$. If $\phi \in \Aut(H)$ is there an automorphism $\widehat{\phi} \in \Aut(G)$ such that $\widehat{\phi}|_H$ = $\phi$?  In words, if $\phi$ is an automorphism of $H$, is there necessarily an automorphism $\widehat{\phi}$ of $G$ where $\widehat{\phi}$ maps the elements in $H$ to $H$ in exactly the same way?  I am thinking this is false (consider the non-finite example of $\bar{Q}$ and $\mathbb{C}$), but would like a counter example (for the finite case).