i saw this code on the internet but i don't know what that starred expression in front of map function do.  here's the code  print(*map(input().find,map(chr,range(97,123))))   i was thinking about args but it doesn't apply for this code because this is map function right? does it relate to map function or print function? or both?  also i tried running this code without that star in front map function like this  print(map(input().find,map(chr,range(97,123))))   but the output was this  &lt;map object at 0x7f42409d50f0&gt;
In the following method definitions, what does the * and ** do for param2?  def foo(param1, *param2): def bar(param1, **param2):
I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference? Can you read my proof and tell me if it's correct? Thanks.  Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon &gt; 0$, $k,j$ large enough, $$ \varepsilon &gt; \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:  Let $\varepsilon &gt; 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$  for $k$ large enough. Ok, so PHP isn't the best language to be dealing with arbitrarily large integers in, considering that it only natively supports 32-bit signed integers. What I'm trying to do though is create a class that could represent an arbitrarily large binary number and be able to perform simple arithmetic operations on two of them (add/subtract/multiply/divide).  My target is dealing with 128-bit integers.  There's a couple of approaches I'm looking at, and problems I see with them. Any input or commentary on what you would choose and how you might go about it would be greatly appreciated.  Approach #1: Create a 128-bit integer class that stores its integer internally as four 32-bit integers. The only problem with this approach is that I'm not sure how to go about handling overflow/underflow issues when manipulating individual chunks of the two operands.  Approach #2: Use the bcmath extension, as this looks like something it was designed to tackle. My only worry in taking this approach is the scale setting of the bcmath extension, because there can't be any rounding errors in my 128-bit integers; they must be precise. I'm also worried about being able to eventually convert the result of the bcmath functions into a binary string (which I'll later need to shove into some mcrypt encryption functions).  Approach #3: Store the numbers as binary strings (probably LSB first). Theoretically I should be able to store integers of any arbitrary size this way. All I would have to do is write the four basic arithmetic functions to perform add/sub/mult/div on two binary strings and produce a binary string result. This is exactly the format I need to hand over to mcrypt as well, so that's an added plus. This is the approach I think has the most promise at the moment, but the one sticking point I've got is that PHP doesn't offer me any way to manipulate the individual bits (that I know of). I believe I'd have to break it up into byte-sized chunks (no pun intended), at which point my questions about handling overflow/underflow from Approach #1 apply. Minimum file permissions to rename file I put a 64-bit ELF file in my unprivileged user's home directory. I wonder which permissions do I need to have only possiblity to rename this file being logged in as unprivileged user. From what I understand I only need execute permission for user to whom file belongs to, set for parent's directory. Is that right?  So if path to my file is /home/Simon_dir/zip and I wrote chmod 0100 /home/Simon_dir and file permissions are set to 0000, should I have rights to rename the zip file being logged in as unprivileged user? This was posted on facebook and people are saying it is incorrect, it should be: "...as you and I"   Which is correct? Names of higher-order derivatives Specific derivatives have specific names. First order is often called tangency/velocity, second order is curvature/acceleration. I've also come across words like Jerk, Yank, Jounce, Jolt, Surge and Lurch for 3rd and 4th order derivatives. Is there a widely agreed list of names for these? How many orders have specific names?  In this case, I'm dealing with NURBs curves, so the "tangency" and "curvature" related words are to be preferred over "velocity" and "acceleration" words. Would fewer Jedi be more powerful? Should we sign-then-encrypt, or encrypt-then-sign? When to use "I" and "me" in an answer A problem regarding to Schur's lemma Let $\rho: G \to GL_n(\mathbb{C})$ be an irreducible representation, and $g\in Z(G)$. Show $\rho(g)$ is a scalar multiple of the identity matrix $I$.   I think I have it, here is my solution:  Since $\rho(g) \in Hom_G(\mathbb{C}, \mathbb{C})$ and $\rho$ is irreducible, consider a nonzero eigenvalue of $\rho(g)$, say $\lambda$, we have $\rho(g) -\lambda $ is a zero map by Schur's lemma, as the map contains a non-trivial kernel (i.e. the eigenvectors associated to $\lambda$ are in the kernel).  But I didn't use the condition that  $g\in Z(G)$, so are there something wrong with my solution? How does background blur (bokeh) relate to sensor size? This is a somewhat theoretical question.  Suppose I first take a photo of a subject using a full frame sensor DSLR, with a given lens (say 50mm prime lens at f/3.5).  Now suppose I exchange the camera for an APS-C sensor DSLR (with a 1.6× crop factor). I keep the same lens (same focal length, same aperture) and I step back a few meters in order to maintain the field of view (at least keep the same magnification of the subject). I now take a second photo.  Clearly depth of field will have increased between the two photos. But what about background blur (for example, trees at infinity)? Will I have the same amount of background blur, or will that have changed?  I have read somewhere that background blur depends on the physical aperture size. In this case the physical aperture (physical focal length divided by f/stop) stays the same. But should this number be taken in relation to the sensor size? In which case with the smaller APS-C sensor, the physical aperture will be relatively bigger, which would mean more background blur. This would be rather counter-intuitive as we usually consider it to be more difficult to get background blur on an APS-C camera.  Please give the reasoning behind the answer. I would answer the question myself using this  but I can't manage to get it to run on my computer. I used to be able to go to Accounts and Import → Add another email address you own and add an email account that's only a forwarding address. That is, it's not a mailbox and I don't have SMTP access to it.  Gmail would send me a confirmation email to that address, I'd confirm it's mine, and done. Now it only gives me the option to add SMTP credentials.  Did they stop supporting the previous method? Is there any other way I can get that option back? I know there a lot of posts about ram but I'm still confused about it. I'm building a new system and because it was cheaper, I bought a 6GB kit from Corsair. So I'm wondering if I would be better off just using 4GB of it in a dual channel configuration or just add all 6GB (3 DIMMS) in a single channel configuration.   Thanks Can I apply for a Schengen visa of the first port of entry or does it need to be the country of the longest duration? I'm planning to travel to the Netherlands, Spain and Bulgaria. My longest stay will probably be in Bulgaria. Now, Bulgaria doesn't issue a Schengen visa but allows people with a Schengen visa to visit. Which visa should I be applying for? Bulgarian Visa AND Schengen visa or just the Schengen visa for Netherlands or Spain? $(\Bbb R,+)$ is a topological group. Is there any other group structure on $\Bbb R$ such that it is still a topological group and this group is not isomorphic to $(\Bbb R,+)$ ?  Refer to  for the general problem. In that problem I have conjectured that there is no such group however I couldn't prove it. Sarrus Rule with LaTeX I stand up and I look at two parallel railroad tracks. I find that converge away from me. Why? Can someone explain me why parallel lines seem to converge, please? The bottom bracket of my bike was completely stripped to the point that when you push the left side crank arm, the right crank moves and you can see the bottom bracket lock move out the frame. The bike shop tried putting Teflon tape but it's temporary. It's a 26' xc hard tail. For all I know both sides of the bottom bracket(square tapered) are stripped clean hence the Teflon tape put by the previous owner. Can I fix the problem by changing into a hallow tech cranks because it works like a thread less bb, or should I just buy a thread less square tapered bb? Any suggestion on what to do? Does the equivalence $$\neg\neg (P \rightarrow Q) \leftrightarrow (\neg\neg P \rightarrow \neg\neg Q)$$ hold in propositional intuitionistic logic? In propositional classical logic the equivalence holds obviously since $P \leftrightarrow \neg\neg P$. However, the similar-looking implication $\neg\neg (P \vee Q) \rightarrow (\neg\neg P \vee \neg\neg Q)$ is . The validity of the backward implication comes up in  as well. Are data members allocated in the same memory space as their objects in C++?