i have added too many subdivisions and do not know how to reverse these. i need help with this    this would help understand the issue
I am wondering if there is a modifier or tool that will essentially undo a subsurf or subdivide command.
edges with thickness and color Is there a way to vary the thickness and color of lines created by just edges and vertices? I realize that a face can be defined and then color and material applied, but is there any way to apply a color to the edges? Now that the , I've been seeing tag-only suggested edits from 1500-rep users.  For example, one had the summary of 'burninating [tagThatWasRemoved]'.  It's very tempting to reject these as Too Minor, but I know that these users used to have a Retag privilege (and are obviously trying to participate in a burnination.)  What should we do for tag-only edits from high(er) rep users? I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$,  we have $S+T\geq 0$, and $ST\geq 0$.  $T\geq 0$ means $(Tx,x)\geq 0$.    To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$. But for $ST$ i like some help.   There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:   $T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.   Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.   Is there a more straightforward way? I know that both can mean "inside" but what I don't have clear is whether both mean the same when talking about time. For example:   The party is in two days = The party is within two days ??   According to this link  within means a limit that can be exceeded, so I'm wondering too if say "the party is within two days max" is correct or redundant. Unwanted Lines when rendering using GPU and OpenCL (RX 580) Package natbib Error: Bibliography not compatible with author-year citations I have a basic tex file and a bibliography which is exported from Mendeley  However, when I want to cite in the author-year style, I get the error:   Package natbib Error: Bibliography not compatible with author-year citations. (natbib) Press &lt;return&gt; to continue in numerical citation style.  Any clue what's wrong?  Minimum working example:  \documentclass[letterpaper, 10pt]{article}  \usepackage[authoryear]{natbib} \usepackage{lipsum}  \begin{document}  \lipsum[1-4] \nocite{*}  \bibliographystyle{apalike} \bibliography{articles}  \end{document}    A snippet of my .bib library:  @article{shortcode, author = {Lastname, Firstname}, file = {:/path/to/file.pdf:pdf}, journal = {Journal name}, number = {xx}, pages = {x--xx}, title = {Title}, volume = {xx}, year = {xxxx}     } Travelling from India to canada via Dubai and London. Do I need a transit visa I am travelling to Calgary from Delhi. I have an Air India flight from Delhi to Dubai then British Airways flight from Dubai to London and Air Canada flight from London to Calgary. My query is "Do I require a transit visa." Plz help as I am travelling in four days "Can not setup lock file:/tmp/g2ipmsg.lock errno : -11 (Resource temporarily unavailable)" I installed IP Messenger (GNOME 2) from Ubuntu software center but its not working. It's always showing an error message on Ubuntu 12.04.  FileName:ipmsg.c Function:create_lock_file Line:157 Can not setup lock file:/tmp/g2ipmsg.lock errno : -11 (Resource temporarily unavailable) To make your opinion known on MSE you need both up and downvotes. Given at the moment you need 125 rep to downvote on this site, the voting patterns will be skewed positive because not everyone is active on the site.   This is to be expected for regular Stack Exchange sites as you are expected to be involved in the community before you are deemed knowledgeable enough to be part of the community voice. However, MSE is different because you are voting for the entirety of the Stack Exchange network and you shouldn't need to be active on MSE itself to have a voice on changes to the entire network.  This is especially important for questions like  because it will effect every network to do with coding at all. This was brought up in .  I would suggest reducing the rep required to downvote on MSE down to 100 (or 101) rep. This is the amount you get for the site association bonus. This is given once you achieve 200 rep on any network site and shows that you are active in at least one community. I think this would be the perfect use of the association bonus in showing that you are a member of the Stack Exchange community. Why use a ReentrantLock if one can use synchronized(this)? What is the result of $ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $ and why? What is the result of the next limit:   $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$  Why (theorem)? How to disable the "Get Windows 10" icon shown in the notification area (tray)? This icon showed up in my taskbar notification area today and I cannot seem to get rid of it:    Clicking on it displays the following screen:    So how do I disable or remove the "Get Windows 10" icon? I went to the doctor's If you provide 0 as the dayValue in Date.setFullYear you get the last day of the previous month:  d = new Date(); d.setFullYear(2008, 11, 0); //  Sun Nov 30 2008   There is reference to this behaviour at . Is this a reliable cross-browser feature or should I look at alternative methods? What are the differences between Generics in C# and Java... and Templates in C++? Prove that the graph is connected I was wondering if someone can help me understand how prove that this graph is connected.  Given a graph with n vertices, prove that if the degree of each vertex is at least $(n − 1)/2$ then the graph is connected.  So far I know that about connected graphs:  An undirected graph is called connected if there is a path between every pair of distinct vertices of the graph  The distance between two vertices in a graph is the length of the shortest path between them.  The diameter of a graph is the distance between the two vertices that are farthest apart. Solve $f(x)f(2x^2) = f(2x^3+x)$ The full question is to solve $f(x)f(2x^2) = f(2x^3+x)$, prove there are at most one solution per degree of $f(x)$ and then find all the solutions. (Here $f(x)$ is assumed to be polynomial.)  So far I have proven that $(x^2+1)^n$ works for even degree polynomials, and I'm pretty near certain there are no odd solutions.  Is the following a valid argument for there only being at most 1 solution per degree? Let $K$ be the degree of $f(x)$ then $f(x)f(2x^2) = f(2x^3+x)$ produces $3k+1$ linearly independent equations with only $k+1$ independent variables. Therefore for any degree $K$ polynomial there can be at most 1 solution.  Finally, is there any elegant way to prove that no odd solutions exist? Ubuntu files worthless on Windows? How many combinations can I make? let $n \gt 1$ be an integer, and consider $n$ people; $P_1, P_2,..., P_n$ let $A_n$ be the number of ways these $n$ people can be divided into groups, such that each group have either one or two people  determine $A_1, A_2, A_3$  So I have  $A_1 = 1$ way $A_2 = 4$ way $A_3 = 12$ way  But I am not sure if this is right... Can anyone confirm?  edit:  My way of getting this is  $A_1 = 1 $ because {P1} only has 1 way to sort  $A_2 = 4$ because {P1,P2} = {P1,P2},{P2,P1},{P1P2},{P2P1}  etc. Changing a CSS rule-set from Javascript