I'm trying to remember how to do polygons from lines, I remember it was a very simple procedure. have a layer of lines intersecting or touching, use the tool and voila, qgis took the lines layer  and created a layer of polygons  with the spaces between the lines. very simple tool with Polygonize here   However today I try, I can not get a good result. For this exercise I made a layer of lines with different shapes as you can see below.    A= it is a single line that does not close (in the table object id 0)  B= a square with lines in, that joined to form a single object (in table object id 1)  C= a triangle made of different lines that are touching, for this I used the snapping tool (in the table objects id 2,3,4)  This is the result of: vectorial-> Geometry Tools -> lines to polygon    As observed the tool close the object A to forms a good polygon (table object id 0), object B is formed with errors and under a single polygon (table object id 1) and the C object (the triangle) is completely ignored  This is the result of: processing-> vectorial geometry -> lines to polygons    As observed the tool close the object A to forms a good polygon (table object id 0), object B is formed with errors and under a single polygon (table object id 1) and the C object (the triangle) is completely ignored  This is the result of processing-> vectorial geometry -> poligonize    the poligonize tool gives a different result, only a general polygon shape without taking into account the interior lines of the B object and the table is filled with the area and perimeter of this.  As you can see the results are not what I expected, as I recall it was very easy to make and very good results were obtained and usually had to make small adjustments to the polygons.  and I read this answer from 2014:     with a very similar problem where the possibility that the tool is BROKEN , however I dont know if this will be the case or I doing something wrong with the tool  QGIS: 2.14.3  Installer: standalone QGIS-OSGeo4W-2.14.3-1-Setup-x86_64 OS: Windows 10 64bits (1115 10586.318)
In QGIS, I want to make polygons from a set of intersecting lines. Looking at the screenshot here, I want chop the purple line (it's a closed line, not a polygon) where it crosses the green and blue lines. The result hopefully would be a polygon where I have hatched in red. I have quite a few to do.    I have tried the QGIS tool previously called Polygonizer (see ), but now called Polygonize. This is a tool that, given a layer containing a set of crossing lines arranged essentially like a # symbol, should create a polygon from the enclosed space in the middle.  (In QGIS 2.01, Polygonize is in the top menu bar: go to Processing -> Commander, then type "Polygonize" on the command line, select "Processing algorithm: Polygonize" from the list that appears, and press Enter.)   At the momemt Polygonize doesn't work properly. Is there another way to do what I want to do?   For example, wouldn't it be great if the QGIS 'Split Feature' command was able to work like MapInfo's 'Polyline Split', where you select the polygons to split, then select an already-drawn polyline to split by, and voila. But Split Feature seems not to work that way.    UPDATE: Detail of problems with running Polygonize.  I have 64-bit QGIS 2.01 installed to Windows 7 from the installer package, QGIS-OSGeo4W-2.0.1-3-Setup-x86_64.exe.  The routine processed an output from my shapefile, but did not create a polygon from the region enclosed by intersections between the blue, green and purple lines. (Note: in the actual input file, all these lines are on the same layer; the screenshot just colours them differently for illustrative purposes). What I got instead was a simple conversion of the purple outline to a polygon, as though the program completely ignored the blue and green lines.  When I node-edited the closed purple line, to open it so that I had a set of open crossing lines like in the # symbol, and then ran Polygonize again, nothing happened. No crash, but no visible polygon in the output file.    When I used Nick Hopton's own  data, I got a crash; see screenshot:
How many connected graphs over V vertices and E edges? Is there a way to calculate the number of simple connected graphs possible over given edges and vertices?   Eg: 3 vertices and 2 edges will have 3 connected graphs But 3 vertices and 3 edges will have 1 connected graph  Then 4 edges and 3 will have 4 connected graphs  Till such values...it is easy to see its  V choose E  But what about when the number of vertices are less than number of edges...how to calculate then?  I am not able to visualize that  Can it be a variation of the Stars and Bars problem  Like...number of ways 7 edges(balls) can be connected to (kept in) 5 vertices(bags) such that no vertex(bag) is isolated (is empty)  Here maybe we might have to consider the number of edges twice as each edge needs two vertices... Virtualbox Kernel driver not installed Integral vanishes on all intervals implies the function is a.e. zero I am having trouble with the following problem:      $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:   $$\int_{[0,a]}f\,dm=0.$$   Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).    I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.  My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.  Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.  Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions? Is there an option to output a manual page in a different language? I don't want to change the computer language completely, but only specific manual pages. For example  $ man -English man Man is a manual program … $ man -Russian man Инструцтия для Unix, BSD и Linux. … How do I print a double value with full precision using cout? How to install (old) packages for Ubuntu 9.04? Based on some excellent feedback by Mark here (), today I was able to use the vmware converter to clone my production server for a staging server.  However the nic won't come up no matter what I do. I attempted to inistall vmware tools, as I suspect that the fact that it is not installed might prevent the nic from working. (I have the nic set as a vmxnet3 card in the vm settings). The install failed because there were several dependencies missing as well as the Linux headers. Given that Ubuntu 9.04 has been EOL'd, the packages I need to install to get the vmware tools to install are no longer available. I doubt the ubuntu 9.04 install CD has the packages on it.   What are my options? I'd rather not upgrade the version of Ubuntu yet, as the point of the vm right now is to maintain parity with the production server. Might I have better luck resetting the driver to use vmxnet2 instead of the vmxnet3?  Thanks in advance!  Chris Existence of irreducible polynomials over finite field Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?  One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely  $ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$  where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.  This trick doesn't work for $n &lt; |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.  I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence. How should we treat questions that are, or have seeds of, evil or malicious purposes and actions, such as writing spambots and malware?    Should they be answered like regular "everybody keeps their pants on" questions? Or should there be some other type of response to this bad behaviour lot? Time-Stream Crossing and Forgetfulness In The Day of the Doctor, it's explained that the Doctor doesn't remember having interacted with himself, and if he interacts with a future version of himself, the younger version forgets.  But in Time Crash, 10th solves the problem of the TARDISes occupying the same space because he remembers himself doing it from when he was 5th.  If this is true, 5th wouldn't have wondered whether or not he was going to regenerate when he did.  Does the TARDIS occupying the same space with itself re-instate those memories? Or is it just possible that I'm asking a ridiculous question that will never have an answer because Doctor Who.  :D cmd has wildcard bug? Files in folder:  a.j ab.jp abc.jpg abcd.jpeg   dir command results:  Command       Result (Files shown) ----------    -------------------- dir ?.*       a.j dir ??.*      a.j, ab.jp dir ???.*     a.j, ab.jp, abc.jpg dir ????.*    a.j, ab.jp, abc.jpg, abcd.jpeg   So we can see single ? it means "0 or 1 letter".  Now more dir command results:  Command       Result (Files shown) ----------    -------------------- dir *.?       a.j dir *.??      a.j, ab.jp dir *.???     a.j, ab.jp, abc.jpg, abcd.jpeg --&gt; What is this?! dir *.????    a.j, ab.jp, abc.jpg, abcd.jpeg   In 3rd command why ??? is showing jpeg? Can you explain? Is this bug in cmd? How to pass Unicode characters as JSP/Servlet request.getParameter? Is this study stating red algae reduces cold virus by 92% believable? I was using this spray meant to hasten recovery from the common cold. , sponsored by the pharmaceutical company that produces the spray, stated it reduced virus by 92%, and so shortens the duration of the common cold.  Meanwhile a doctor told me that these kinds of things don't work.  Does someone have some perspective on this? Is this a believable study? Was it well designed, and has it been duplicated?   If so, why wouldn't the pharmaceutical produced a certified and expensive drug instead of a no-prescription drug? Dual booting Windows and Ubuntu My laptop has 500gb hard disk in which for 250gb I installed Windows xp on C drive(primary partition). I want to install Ubuntu 12.04 in the remaing hard drive with dual boot option. When I am going to install Ubuntu it is showing something like  /boot /boot/   Can anybody help me in this regard that where and how to install Ubuntu? Particle effect showing only on one side of mirrored object I'm trying to create a fur particle effect on sonic the hedgehog, and I'm using weight paints to keep it in the areas I want.   Problem: I can only get the hair effect to show up on certain random areas on model, and only on one side.  I'm using a mirror effect to keep model symmetrical, but hair only shows on one side of the mirror.  I'm new used Blender, so I don't know what all the settings for the particle effects do.  Whats going on and how do I fix it? Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence Let $f_n:[a.b]\rightarrow \mathbb{R}$ be sequence of $L-$Lipschitz functions, that is: $$\forall x,y\in[a,b]: |f_n(x)-f_n(y)|\leq L|x-y|$$   Suppose $f_n \rightarrow f$ pointwise, prove $f_n \rightrightarrows f$   I have all the parts of the puzzle for the proof, and I'm trying to put them all together, I'm using  in my answer.  I would appreciate is you could correct my proof, and if you have an alternative proof, I would be more then happy to see it.  My proof:   Let $\epsilon&gt;0.$  $f_n$ are uniformly continuous on $[a,b]:$  $\tag{1} \exists \delta&gt;0\ \forall x,y\in[a,b]: |f_n(x)-f_n(y)|&lt;\frac{\epsilon}{3}$  $f$ is also $L-$Lipschitz:  $\tag{2} \forall x,y\in[a,b]:|f(x)-f(y)|&lt;L|x-y|=\frac{\epsilon}{3}$  Let us set a partition of $[a,b]$ such as Stephen Montgomery-Smith suggests:     Pick points $x_1,\dots,x_m \in [a,b]$ which are distance   $\frac{\epsilon}{3}$ from each other.        For each $1 \le i \le m$, find a number $N_i$ so that for all $n \ge  N_i$ we have $|f_n(x_i)-f(x_i)| \le \epsilon/3$.   Let $N = \max_{1 \le i \le m} N_i$      Now given any $x \in [a,b]$, pick $1 \le i \le m$ such that $|x-x_i| &lt;\frac{\epsilon}{3L}:  |f_n(x)-f(x)|&lt;\frac{\epsilon}{3} \tag{3}$   $$\begin{align}|f_n(y)-f(y)| &amp;=|f_n(y)-f_n(x)+f_n(x)-f(x)+f(x)-f(y)| \\ &amp;\leq |f_n(y)-f_n(x)| + |f_n(x)-f(x)|+|f(x)-f(y)| \\ &amp;&lt; \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \\\end{align}$$ Best way to save data on the iPhone I am writing an iPhone application, and need to save the state of my application (5K or so).  My main worry is data persisting across upgrades. Some of the applications I use clearly got this wrong, and I would prefer not to! Is there a more accurate way to create a Javascript timer than setTimeout? Something that has always bugged me is how unpredictable the setTimeout() method in Javascript is.  In my experience, the timer is horribly inaccurate in a lot of situations. By inaccurate, I mean the actual delay time seems to vary by 250-500ms more or less. Although this isn't a huge amount of time, when using it to hide/show UI elements the time can be visibly noticeable.  Are there any tricks that can be done to ensure that setTimeout() performs accurately (without resorting to an external API) or is this a lost cause? How can we prove that for every pair $N \in \mathbb{N}$, and natural number $\beta\in [2, \infty)$ there exists a unique set of integers $x_i \in [0, \beta -1]$, $k\in [0,\infty)$ such that:  $$N = \sum_{i=0}^{k}\beta^ix_{i}, \quad $$  It's a question if every positive integer has its unique representation in every base $\beta$. Is the validity of above statement somehow related to Horner's rule? I bought two pieces of land. One from Morthal, and one from Falkreath. Morthal was the first plot of land I purchased. I only finished half of it. I bought the plot of land in falkreath because I liked the nature better. 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