Hello I am trying to understand how Python's pass by reference works.  I have an example:  &gt;&gt;&gt;a = 1 &gt;&gt;&gt;b = 1 &gt;&gt;&gt;id(a);id(b) 140522779858088 140522779858088   This makes perfect sense since a and b are both referencing the same value that they would have the identity. What I dont quite understand is how this example:  &gt;&gt;&gt;a = 4.4 &gt;&gt;&gt;b = 1.0+3.4 &gt;&gt;&gt;id(a);id(b) 140522778796184 140522778796136   Is different from this example:  &gt;&gt;&gt;a = 2 &gt;&gt;&gt;b = 2 + 0 &gt;&gt;&gt;id(a);id(b) 140522779858064 140522779858064   Is it because in the 3rd example the 0 int object is being viewed as "None" by the interpreter and is not being recognized as needing a different identity from the object which variable "a" is referencing(2)? Whereas in the 2nd example "b" is adding two different int objects and the interpreter is allocating memory for both of those objects to be added, which gives variable "a", a different identity from variable "b"?
Why does the following behave unexpectedly in Python?  &gt;&gt;&gt; a = 256 &gt;&gt;&gt; b = 256 &gt;&gt;&gt; a is b True           # This is an expected result &gt;&gt;&gt; a = 257 &gt;&gt;&gt; b = 257 &gt;&gt;&gt; a is b False          # What happened here? Why is this False? &gt;&gt;&gt; 257 is 257 True           # Yet the literal numbers compare properly   I am using Python 2.5.2. Trying some different versions of Python, it appears that Python 2.3.3 shows the above behaviour between 99 and 100.  Based on the above, I can hypothesize that Python is internally implemented such that "small" integers are stored in a different way than larger integers and the is operator can tell the difference. Why the leaky abstraction? What is a better way of comparing two arbitrary objects to see whether they are the same when I don't know in advance whether they are numbers or not?
I clicked on  to a comment in my email and it prompted a list of apps to open link. All browsers (Chrome, Firefox and etc) are in the list, but not SE app.  Most apps that can open site specific links get prompted for that app to open  the link, but not Stack Exchange links for some reason.    App Version: 1.0.95 Device Manufacturer: Asus Device Model: ASUS_Z010D   OS Version: 6.0.1 (WW_Phone-13.8.26.76-20161230) Disable cache for some images I have a big pot full of chicken necks and backs on the stove which I need to separate into (a) stock and (b) everything else.  I'm wondering the best way to go about doing this. I've previously laid cheesecloth into a fine-mesh strainer and poured it through that, but the cheesecloth always seems to kind of clog up, making straining 4 litres of stock a very lengthy process. I also tried a coffee filter laid inside a strainer and got more-or-less the same results.   Alternatively, I ditched the cheesecloth and just used the fine-mesh strainer by itself, but, unsurprisingly, this method leaves some bits I'd rather not have.   I'm wondering: is there a better way? Or should I stick with the cheesecloth and work on being more patient? Why do Windows 10 foreground apps mysteriously launch as background processes? Black screen on boot after AMDGPU-Pro install (16.04) I have been having some screen freezing issues on my new Ubuntu 16.04 system, and have been in the troubleshooting process for this issue. I thought it was a graphics driver problem, as I am running an AMD R9 380x card which was supposed to be fully supported by the open-source AMDGPU driver preinstalled with 16.04, but after looking around online, many people seem to have issues with. I thought the answer would be to install the recently released AMDGPU-Pro driver , following the instructions from . However, when I reboot I am getting a black screen. Is there an easy way to revert my changes back? My machine was semi-functional with the freezing issue but now with no video at all is useless to me. How to proceed? Normally Star Trek seems to based on some sort of logic, but the time travel in "Star Trek: Enterprise" seems to have no logic at all, or  am I just missing it?  Example: If they go back in time to change something it takes time until it has any effect in the future? I can't see any reason for this. How do I turn off the arrow for desktop icons? Setting-up SSH-tunnel for VPN? I was wondering, why there are passport checks at the check-in counter, Immigration and even at the gate?    For example, I was at Istanbul flying to either UK or US. There's a security check at the check-in counter asking a bunch of questions about my immigration status - even more than the Immigration officer at the border. Also, at the gate, there's also Security officer checking my immigration status and passport asking the same questions.    I thought that would be the responsibility of an Immigration officer rather than some third party company?   The thing I'm curious the most is they don't even know which countries this passport can go to with or without visa. They have to go back to the counter and check with the system which takes a long time. I was wondering why they hire these kind of companies to slow down the process? Here $n$ belongs to natural numbers. Firstly, I proved the relation by putting $n = 1$ . Then, taking $$|\sin(mx)| \leq m|\sin x|$$ true, I had to prove $$|\sin(m + 1)x| \leq (m + 1)|\sin x|$$  Now, here I got stuck. How to prove it?? Please help. (c) Let $\mathcal{L}^*\subseteq\mathcal{L}^{\text{FOPC}}$ be a first order language of predicate calculus consisting of three unary function symbols $f,g,h$.  (i) Write down an $\mathcal{L}^*$-sentence $\phi$ such if $\mathcal{A}$ is an $\mathcal{L}^*$-structure with $\mathcal{A}\vDash\phi$ then the restriction of $h_\mathcal{A}$ to the range of $f_\mathcal{A}$ gives a one-to-one function onto the range of $g_\mathcal{A}$.  (ii) Write down an $\mathcal{L}^*$-sentence $\psi$ such that if $\mathcal{A}$ is an $\mathcal{L}^*$-structure with $\mathcal{A}\vDash\psi$ and the domain $A$ of $\mathcal{A}$ is finite then the cardinality $|A|$ is a square.    Part (i) is easy enough, but I'm having trouble with part (ii). I assume that you can use part (i). My initial idea was to use $h$ to ensure $|Range(f_A)| =|Range(g_A)|$ but I'm not sure where to go from there. Thanks. I have this problem as part of a school assignment:  NOTE: In case it's hard to read in the image above, here's the directory tree: $ tree . `-- sample_dir1     `-- sample_dir         |-- admin         |-- cambridge         |   |-- cafeteria         |   |-- library         |   `-- security         |       |-- annex         |       |-- building         |       `-- parking         |-- faculty         |-- history.exe         |-- markham         |   |-- annex         |   |-- building1         |   `-- parking         |-- oxford         |   |-- outline.doc         |   |-- programming         |   |   `-- report.pdf         |   `-- security         `-- stenton             |-- gen_ed             `-- lib_arts                 |-- english.txt                 `-- match.doc  15 directories, 11 files  I tried doing the command: $ cp ../cambridge/security/parking ./parking2  ...but it isn't working. The question in the image above states that the current directory is stenton, and that we need to make a copy of the file named parking (from the security directory), and to name the new file parking2, and place it in the current directory, stenton, using relative pathnames. What am I doing wrong? My cp command seems to be correct. What is the easiest way to handle huge numbers in C? I need to store values in the Area 1000^900, or in more human readable form 10^2700. Does anybody know of an easy way to do that? Any help would really be appreciated! Find the radius of the inscribed circle of a right triangle. Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.  This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any help?  EDIT: Here is a picture of the problem given a polynomial $p(x,y)$ from $\mathbb Z[x,y]$. I want to substitute $x=1$ leaving $y$ as it is. The command Value(p,[1,y]) does not work. Can you give me a hint?  After that it gets a little more complicated. Let  $$p(x,y)=x^6+y^6$$ After substituting $x=1$ we have $p(1,y)=1+y^6$. Given a natural number $j$. I want to find all expressions $y^j$ in $p$ and substitute these $y^j$ by $1$ and the others by $0$. In this case (and for $j=6$) we would get $1+1=2$. If $$p(x,y)=x^4y^2+x^2y^6$$ we should get $0+1=1$. Is there any way to do this?  Thanks for reading! edit: Well, it turns out I used the function Value wrong. The right code for my problem would be Value(p,[x,y],[1,y]). So my first question is solved. Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.  Can we more generally conclude that $T$ is compact?  Assume $p&lt;\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology. On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$. EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true. In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition   Let $h: Y \to X$ be a covering. For any path $\gamma : [0,1] \to X$ and any $y \in h^{-1}(x) (x = \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to Y$ of $\gamma$ with $\hat{\gamma}(0) = y$. Furthermore, for any homotopy $\gamma_t (t \in [0,1])$ of $\gamma$ with $\gamma_t = \gamma(0)$ and $\gamma_t(1) = \gamma(1)$, there exists a unique lift of $\hat{\gamma_t}$ such that $\hat{\gamma_t}$ is the homotopy of $\hat{\gamma}$ with $\hat{\gamma_t}(0) = \hat{\gamma}(0)$ and $\hat{\gamma_t}(1) = \hat{\gamma}(1)$.  The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$."   I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different $x_1, x_2$ and their fibers $h^{-1}(x_1), h^{-1}(x_2)$ such that $y_1 \in h^{-1}(x_1)$ and $y_2 \in h^{-1}(x_2)$. Take two paths $\gamma_1, \gamma_2$ such that $\gamma_1(0) = x_1$ and $\gamma_2(0) = x_2$. Then we get two lifts $\hat{\gamma_1}, \hat{\gamma_2}$ with $\hat{\gamma_1}(0) = y_1$ and $\hat{\gamma_2}(0) = y_2$. Now, since our covering space is connected we can continuously deform $\hat{\gamma_1}$ into $\hat{\gamma_2}$ and conclude that every elements in $h^{-1}(x_1)$ is also in $h^{-1}(x_2)$ and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated. Pass extra parameters to an event handler? Is there a "reasonable" way to get a sense of "how many" measurable subsets there are of $\mathbb{R}$? Computational complexity of Fibonacci Sequence I understand Big-O notation, but I don't know how to calculate it for many functions. In particular, I've been trying to figure out the computational complexity of the naive version of the Fibonacci sequence:  int Fibonacci(int n) {     if (n &lt;= 1)         return n;     else         return Fibonacci(n - 1) + Fibonacci(n - 2); }   What is the computational complexity of the Fibonacci sequence and how is it calculated?