In the images I'm trying to add a loop cut vertically through the rectangles, however, it only allows me to add a horizontal loop cut.  The mesh is only a plane and I am using the mirror modifier.  I'm not sure if the mirror modifier is somehow messing it up or if it's something else.  Also, the mesh is part of a larger mesh that I have hidden, not sure if that helps or not either.  I've already tried removing any duplicate vertices and such but to no avail. Can someone tell me what I'm doing wrong?
Alright so I've seen some postings about loop cuts and done some research. I am very new to blender and modeling overall. I am trying to loop cut this concrete barrier. Now I'm following a tutorial and kind of did my own thing in some areas. I cleaned it up following some instructions in another answer here on stack exchange.  I can't for the life of me figure out why I can loop cut around the whole bottom:   ...but not the whole middle:   I see this marked as answered by another question. That question refers to ngons and inside faces. Now once again I'm totally new to this. I looked on the inside. I don't believe I have inside faces or any ngons by the terms I've read online in my research.  Here is a wireframe picture and the inside view:     Maybe I'm missing something the answers on other posts are saying. If so and you know what it is I'd be very appreciative of what it is.
Optimal step size in gradient descent Okay, this should be super easy to do but I just can't find the way.  I have a background image with a solid colour layer on top of it, and I would like to add text on top of everything in such a way that the filling of the text is the image in the background. I have tried selecting the text shape and deleting that selection shape in the solid colour layer, but it just looks wrong.   Can anyone help me with this? Many thanks in advance. When writing letter to an academic professor (not necessarily from student to professor), what is the appropriate way to address his/her title?  I have seen in letters using  Dear Prof. X even when the addressee is not a full professor  Dear Dr. X though, knowing that the addressee is a full professor  Dear Mr/Mrs X though, knowing that s/he has an academic title In my  I was printing a double using cout that got rounded when I wasn't expecting it.  How can I make cout print a double using full precision? How to resume baking fluid simulation? It happens sometimes that after finishing bake (e.g., for fluid simulation) we wish we could continue the simulation for two more seconds :(    How to use the cached bake to continue the simulation?  We played with the settings, but with no success. How can I compile this successfully? I have to use algpseudocode package to improve my paper look like more professional   But I can't compile successfully.  \usepackage{amsmath} \usepackage{algorithm} \usepackage[noend]{algpseudocode}  \makeatletter \def\BState{\State\hskip-\ALG@thistlm} \makeatother  \begin{document} \begin{algorithm} \caption{AIP}\label{euclid} \begin{algorithmic}[1] \Procedure{MyProcedure}{} \State $\textit{locationAgent} \gets \text{position of }\textit{{agent}}$ \State $\textit{locationMD} \gets \text{position of }\textit{{MD}}$ \If {\textit{distance(eNB,$agent_i$)}}&gt;$\delta$ \BState \emph{loop}: \State $j \gets \textit{patlen}$ \If {$\textit{string}(i) = \textit{path}(j)$}\Return TURE \State $j \gets j-1$. \State $i \gets i-1$. \State \textbf{goto} \emph{loop}. \State \textbf{close}; \EndIf \State $i \gets i+\max(\textit{delta}_1(\textit{string}(i)),\textit{delta}_2(j))$. \State \textbf{goto} loop \EndProcedure English proverb for when a solution comes too late In Flemish we have a saying "Vijgen na pasen". Translated: "figs after Easter". It means a solution comes too late to be of any use.  What is the English equivalent for this?  Some googling gives me "Closing the barn door when the cow has bolted", but the explanation seems to point more towards the wrong solution for a problem. Are proofs by contradiction really logical? Let's say that I prove statement $A$ by showing that the negation of $A$ leads to a contradiction.   My question is this: How does one go from "so there's a contradiction if we don't have $A$" to concluding that "we have $A$"?  That, to me, seems the exact opposite of logical. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true". Android Studio shortcuts are not working I use Android Studio in Ubuntu and everything was working fine. Since two weeks that shortcuts stopped working, Ctrl+C, Ctrl+V, Ctrl+Z. I tried to change Keymap in Settings but without success.  What could be the possible problem? So I know the basic gist is that fusion power's main issue is sustaining the fusion. I also know that there are two methods. The Torus method and the laser method. The torus magnetically contains plasma and heats it with radiation and accelerates the plasma around to make strong enough collisions that protons fuse. The laser method uses 192 lasers and focuses it on tiny frozen hydrogen pellets and aims to initiate fusion each time pellets are dropped.  The though struck me when we could sorta combine the two designs together. The torus doesn't have to worry about making fusion happen at a specific location but it has issues in that the plasma is unevenly heated and leaks. On the other hand, the laser design is extremely complicated in the level of precision needed and would have to repeat this for every pellet. This lead me to think to make something precise and contained at the same time.  I see that particle colliders are able to direct two beams of protons and have them collide at a specific spot with a very precise energy. Couldn't we tune the energy of the two beams of protons to the energy required for them to fuse? We have the ability to smash them into bits, surely we have the ability to have them fuse. (I'm thinking about the type of collider that circles two beams in opposite directions)  It would be at much lower energies than normal colliders and would be very precise and it would be possible to fuse at a specific location that has greater leeway because for protons that missed collision, they'd just circle around again! Thus protons would efficiently be used and very little would be wasted. There wouldn't be problems of plasma leakage because we are focusing them in a thin tight beam.   It seems that this idea has girth, or I feel this way at least, can someone back me up by offering some calculations on how to calculate the efficiency? How would I go about calculating the two circling beams of protons and at what specific velocity would be needed? etc. I want to know the correct form:     He and I will go there   or   He and me will go there.   Which is correct? I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically."  I've finally sat down to work through this, and -  Okay, yes, it is easy to see that the "canonical isomorphism" from $V$ to $V^{**}$ is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor.  Also, I see that there is no way that the functor $V\mapsto V^*$ could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to:     Is contravariance the whole problem?   To elaborate:   I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism $V\rightarrow V^*$ would feel more like a theorem as opposed to an artifact of the inapplicability of a definition.  Then I tried to create a definition of a natural transformation from a covariant functor $F:\mathscr{A}\rightarrow\mathscr{B}$ to a contravariant functor $G:\mathscr{A}\rightarrow\mathscr{B}$. It seems to me that this definition should be that all objects $A\in\mathscr{A}$ get a morphism $m_A:F(A)\rightarrow G(A)$ such that for all morphisms $f:A\rightarrow A'$ of $\mathscr{A}$, the following diagram (in $\mathscr{B}$) commutes:  $$\require{AMScd}\begin{CD} F(A) @&gt;m_A&gt;&gt; G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @&gt;&gt;m_{A'}&gt; G(A') \end{CD}$$  This is much more stringent a demand on the $m_A$ than the typical definition of a natural transformation.  Indeed, it is asking that $m_A=G(f)\circ m_{A'}\circ F(f)$, regardless of how $f$ or $A'$ may vary. Taking $\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k$, $F$ the identity functor and $G$ the dualizing functor, it is clear that this definition can never be satisfied unless $m_V$ is the zero map for all $V\in\text{f.d.Vec}_k$ (because take $f$ to be the zero map). In particular, it cannot be satisfied if $m_V$ is required to be an isomorphism.     Is this the right way to understand (categorically) why there is no natural isomorphism $V\rightarrow V^*$?   As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors?  Note: I have read a number of math.SE answers regarding why $V^*$ is not naturally isomorphic to $V$. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. ( was closest.) Hence my question here. Parametric Form for a General Parabola It is well known that a parametric form of the parabola $y^2=4ax$ is $(at^2, 2at)$.      What are possible parametric forms of the general parabola   $$(Ax+Cy)^2+Dx+Ey+F=0$$   ? Overriding public virtual functions with private functions in C++ How to get rid of navigation symbols in beamer? Counting nodes in a tree in Java First of all, I swear this is not homework, it's a question I was asked in an interview. I think I made a mess of it (though I did realise the solution requires recursion). Here is the question:  Implement the count() method which returns the number of nodes in a tree. If a node doesn't have either a left or right child, the relevant getXXChild() method will return null  class Tree {    Tree getRightChild() {     // Assume this is already implemented   }    Tree getLeftChild() {     // Assume this is already implemented   }    int count() {     // Implement me   } }   My reason for asking the question is simply curious to see the correct solution, and thereby measure how bad mine was.  Cheers, Tony What does !function ($) { $(function(){ }) }(window.jQuery) do? Closures in a for loop Extra Backslash Required in First Appended Newline in sed Suppose $f$ is a continuous, strictly increasing function defined on a closed interval $[a,b]$ such that $f^{-1}$ is the inverse function of $f$. Prove that,   $$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$   A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.  However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with $f$ continuous, I cannot apply change of variables technique.  Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.  The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given $f$ is differentiable. However, this seems to hold for non-differentiable functions also.  I would really want some help. Pictorial proofs and/or area arguments are invalid.