I was rendering some scenes and the little white dots just don't go away. I find they are primarily there when i have some light shining on glass or even just a normal diffuse. Lots of people say different things. BlenderGuru says put the 'Indirect Clamp' up, but that doesn't seem to work.   Thanks!
I often use Cycles' built in Glass BSDF shader to achieve realistic refraction effects, but in many cases, it seems that Cycles' glass caustics creates an unfortunately large number of fireflies (pixels that are abnormally bright relative to their surroundings). In some models of mine, even when I render for 3000 or more samples/pixel, my final image is still covered in fireflies. While I can improve this in many cases by increasing the total amount of light available in the scene, this loses a lot of the potency of the refractive effects I'm looking for, especially in scenes with low light levels.  So my question is, how can I reduce the amount of fireflies while using Cycles' caustics without sacrificing the quality of the refractive effects?  Note: I am aware that Cycles' caustics will perform much better once Cycles has full support for bidirectional path-tracing, but it looks unlikely that that will become , unfortunately.
To use Windows and Ubuntu on one computer, which should I install first? Why are we limited to six questions per day? Proof of "well-founded iff there is no sequence $x_n$ s.t. $(x_{n+1}, x_n) \in R$" I tried to do the following exercise from a book I'm reading, can you tell me if my solution is correct? Thanks!  Theorem 2: Let $R$ be a binary relation on a set $X$. Then $R$ is strictly well-founded iff there is no sequence $(x_n)_{n \in \mathbb N}$ of elements of $X$ such that $(x_{n+1}, x_n) \in R$ for all $n \in \mathbb N$.  Exercise:  (i) Formulate an analogue of Theorem 2 for  well-founded relations.   (ii) Prove Theorem 2.    (i) Theorem 2': Let $R$ be a binary relation on a set $X$. Then $R$ is well-founded iff there is no sequence $(x_n)_{n \in \mathbb N}$ of elements of $X$ such that $(x_{n+1}, x_n) \in R$ and $x_n \neq x_{n+1}$ for all $n \in \mathbb N$.  (ii)   $\implies$: Assume there is a sequence $x_n$ with $(x_{n+1},x_n) \in R$ for all $n \in \mathbb N$. Then the set $Y = \{x_n \mid n \in \mathbb N\}$ and hence $R$ is not well-founded since for every element $x_n$ there is $x_{n+1} \in Y$ with $(x_{n+1},x_n) \in R$.  $\Longleftarrow$: Let $R$ be strictly well-founded. Assume we have a sequence $x_n$ with $(x_{n+1},x_n) \in R$ for all $n \in \mathbb N$. Then by the same argument as above $R$ is not well-founded, hence there cannot be such a sequence.   I must be missing something but I can't seem to figure out what. Inspired by :  In my answer, I had stated that Patronuses appear to not be physical beings because they have no effect on anything other than Dementors.     They don't cause leaves to stir in the movies. They passed straight through the tent. They don't appear to be bound by normal laws pf physics and can travel great distances almost immediately (Shacklebolt was presumably in London when his Patronus issued the warning). They can apparently walk across water without sinking (as they did in The Prisoner of Azkaban). They can jump, walk, and float in mid-air (as seen in the Room of Requirement in the movie version of The Order of the Phoenix). They make no footprints in lightly powdered snow (The Deathly Hallows Chapter 19) and make no sound of crunching snow as Harry does (further down the same page in Chapter 19).   I stated incorrectly that the only place in canon where there is an exception is with Dementors, where the Patronus physically fends them off with its horns.  A Dementor is clearly a physical being, if it prised Dudley's mouth open and (in the movie anyway) physically slid open the train door.)   Based on comments and further thought, I can think of at least two other examples when a Patronus interacts physically with a physical being.   As pointed out by Slytherincess, a Patronus can fend off a Lethifold (Living Shroud). When training, the Patonus took form against a Boggart, which had taken the form of a Dementor.   All three of these things have, AFAIK, only three things in common (but I may be off).   They are evil They are magical They are, indeed, physical beings.   So the question is, why can a Patronus interact with all three of these, but seems to have no physical effect on everyday things? How to simulate traceroute using ping? I am learning computer networks and looking for a way to simulate traceroute using the ping command. I think it has something to do with ttl, but I don't know how to develop that further. I have inserted an image in my LaTeX document and the image always displays with the file path of the image directly above it.  How can I prevent this file path from displaying?  I am inserting the image like this:  \begin{figure} \centering     \includegraphics{C:/Users/Name/Pictures/image1.png} \caption{a caption} \label{fig:reference} \end{figure}   Here is a working example of document that produces this behaviour:  \documentclass{article}  \usepackage{graphicx} \begin{document} \begin{figure} \includegraphics{C:/Users/Name/Pictures/image 1.png}  \end{figure}  \end{document}   The image was inserted using the TeXnicCenter Insert>Picture function.  When I generate the pdf document, directly above any image (I have inserted multiple images using this technique) the file path displays above the image, in the case above it would show: "file\path\of\image", how can I prevent this from displaying? I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$. a) There's a radially symmetric charge density $\rho(r)$ centered around the origin. Determine the dipolemoment of that charge density. b) Let $\rho(r)$ be an arbitrary charge density now. Under what circumstances does the dipole moment of the displaced charge density  $\rho '(\vec{r}) = \rho (\vec{r}-\vec{b})$ differ from the one not displaced at all.  Here were my ideas so far: a) Just thinking about the situation it has to be zero, right? I mean, since there's no real dipole. But how do I show that mathematically? I was thinking of just going like this (it may be wrong): Let the charge density be $\rho (r)=kr$, then we can get the charge q by integrating: $$q=4\pi \int_0^R kr\cdot r^2dr=\pi k R^4$$ I'm looking at the charge distribution as a spherical electron cloud with radius $R$. Then, since $p=qd$ and $d$ is zero because there are no two different charges the dipole moment is zero. Is that sufficient as an answer? b) I don't know how to approach this one. My guess is that its dipole moment is also zero because we're only looking at a displacement here. Anyone got any idea? I would appreciate any advice on this. I know this issue has been addressed before with different software, but how do I do it with Shapely?  I have a polygon with a very small gap on the interior of the shape:    Here is a close up:    The polygon is valid so I can't use the buffer trick, and simplifying the polygon also doesn't work. I've been attempting to learn C in my spare time, and other languages (C#, Java, etc.) have the same concept (and often the same operators) ...   What I'm wondering is, at a core level, what does bit-shifting (&lt;&lt;, &gt;&gt;, &gt;&gt;&gt;) do, what problems can it help solve, and what gotchas lurk around the bend? In other words, an absolute beginner's guide to bit shifting in all its goodness. "Use self" doesn't work? I'm currently working with drivers in Blender 2.81 and want to use self but it doesn't work.   For example drive the rotation by the location, I'm adding a driver to the rotation, check the Use Self box and write self.location.x  Blender return the message: ERROR Invalid Python expression.  Have I missed something, is it a bug?  edit: Prove that if $f$ is a real continuous function such that $|f|\le 1$ then $|\int_{|z|=1} f(z)dz| \le 4$ I've encountered the following problem in complex analysis:  Prove that if $f$ is a real continuous function such that $|f|\le 1$ then $|\int_{|z|=1} f(z)dz| \le 4$.  I've decomposed $f$ to it's real and imaginary parts and got that it's enough to prove that:  $(\int_0^{2\pi}cos(\theta)g(\theta)d\theta)^2 +(\int_0^{2\pi}sin(\theta)g(\theta)d\theta)^2 \le 16$   Where $g(\theta)=f(e^{i\theta})$.   But now I'm stuck. Any ideas?  Thanks I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.  Provided:  $$a+b+c=0$$  Find the value of:  $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$  Like I'm not sure where to start, and the provided clue doesn't even make sense.  There's no way I can think of to factor this big polynomial into a form like $a\times p+b\times q+c\times r=s$ where $p,q,r,s\in\mathbb{Z}$.  Thanks! Correct plural form of a noun preceded by "zero" When using zero as a quantifier, is it correct to use the singular form on the object of the quantifier, or the plural form?  It sounds confusing when I put it that way, but what I mean is: Which is correct?   Your password expires in 0 days. Your password expires in 0 day.   Essentially I suppose I'm asking, does "singular" mean "one" or "the opposite of more than one", as zero is not "plural" in the traditional "more than one" sense?  I'm pretty sure "days" sounds correct, but I can't be sure. Calculating latitude and longitude of points using QGIS How do I easily calculate the lat/long of points using the QGIS Field Calculator? I saw a previous related question (), but it deals with a more advanced approach. In ArcGIS Desktop this is similar to the Calculate Geometry function; I just can't find where do this in QGIS. Control Android with broken screen from PC My rooted Galaxy Nexus has a broken screen. How can I use my PC (ie, via ADB) to control the phone?  androidscreencast would be the ideal solution, but its keyboard/mouse control feature does not work in Jelly Bean. Perhaps someone knows why and we can patch the code? (see: )  Installing a VNC server is another, but for the love of me I can't get it to work. Eg, using Droid VNC Server, I install the apk, and am supposed to execute /data/data/org.onaips.vnc/files/androidvncserver but there is no files directory inside org.onaips.vnc (instructions from here: ). VMLite is another VNC server that can be started with a special Windows utility (and doesn't need root!) but it's not OSS so I can't get the apk in order to install (Google Play is not pushing apps to my phone prob because I disabled some service--let me know if I can force Google Play to update itself from ADB or install the app). fastdroid-vnc is a binary (not apk) and it runs fine via shell, but something is broken and it shows nothing.  What are my options? Let's make this a community wiki, to help anyone who's broken their screen! How to get less spacing in math mode LaTeX commands like this:  Here again, four different simulations are run on networks of size  $C^{G} = 25 \times\ 25 = 625$, $C^{G} = 50 \times\ 50 = 2500$, $C^{G} = 250 \times\ 250 = 62500$, and $C^{G} = 500 \times\ 500 = 250000$...   When compiled generate the following:    The spacing in math mode really irritates me. There's too much white space around the equal sign, and the multiplication sign has more right padding than left padding. Is there any way to reduce this spacing? If a superconductor has zero resistance, does it have infinite amperage? Let $x*y=\gcd(x,y)$ on $D_n=\{x\in \mathbb N:x |n\}$. I have to prove that $(D_n,*)$ is a commutative monoid and I have a problem when finding the identity element.  $x*e=x$  So $gcd(x,e)=x \to e=xk,k\in \mathbb N^*$ and $e \leq n$  A good $e$ is $0$ but it's not in $D_n$ Why was this very low quality flag declined? Question (since closed and deleted): . I know about the  of VLQ flags, and I can't fathom why anyone would think that question wouldn't be a candidate for removal.  Question text, for posterity:     Hey I want a Cyrus beck line algorithm program in c or c++. I have   searched a lot but hadn't got anything over Web..!!      If anyone of you have Program please give it to me and help me out.      Thank you in advance :)   I find  not compelling because it requires me to accept the premise that this question was not "unlikely to be salvageable".