@masterthesis doesn't work for bibtex citation I'm using bibtex and I would like to know how to cite a @masterthesis. I've exported Mendeley's citations and it has only the @phdthesis style, which actually works. Is there any way to make work with @masterthesis instead? My bibliography at the end of the paper gets wrong.  I'm using abntcite.sty.  Here goes the code:  @masterthesis{Filho2016Automatic,     author = {Silva{ }Filho, P. F. F.},     institution = {Dissertação (Mestrado) - ITA},     pages = 159,     school = {Dissertação (Mestrado) - ITA},     title = {Automatic Landmark Recognition in aerial images for the autonomous navigation system of Unmanned Aerial Vehicles},     year = 2016 }   Thanks!!
Institute Name in @masterthesis Citations I need to cite master thesis and the lecture notes into a document. But the problem with @masterthesis and @unpublished citations is they don't print an institution name in the pdflatex output of bibliography.   `\documentclass{article}   \usepackage{filecontents}  \usepackage[ backend=biber, style=numeric, sortlocale=de_DE, natbib=true, url=true,  doi=true, eprint=false ]{biblatex} \addbibresource{bibliography.bib}  \begin{filecontents}{bibliography.bib} @masterthesis{bhgt, title={Bike Model}, author={Fris}, year={2016}, school={University A}, address={XXX} } @unpublished{Map, title = {Electrical Engineering Book}, author={Koch}, howpublished = {Lecture Notes}, Institution= {University B}, year = {2015} } } \end{filecontents}  \begin{document}  Citation Test:  \cite{bhgt} and \cite{Map} \printbibliography  \end{document}`   I get this bibliography output:         [1] Fris. Bike Model. XXX, 2016.   [2] Koch. “Electrical Engineering Book”. Lecture Notes. 2015.       Any suggestions, how we can also print an institute name?
Why is the Petersen graph no Cayley graph? Using proof by contradiction vs proof of the contrapositive How to properly ignore exceptions Strong Induction Proof: Fibonacci number even if and only if 3 divides index The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? I have a DEM that I would like to smooth or generalize to remove topographic extremes (chop off peaks and fill-in valleys). Ideally, I would also like to have control over the radius or level of "blurriness". In the end, I will need a set of rasters that range from slightly blurry to really blurry. (Theoretically, the blurriest would be a constant raster of the arithmetic mean of all values).  Are there any tools or methods I can use (based on Esri, GDAL, GRASS)?   Do I need to home bake my own  routine?   Could I use a low-pass filter (e.g. ArcGIS's ), and if so, would I need to run it a bunch of times to get the effect of a large radius? Implicit conversion issue in a ternary condition My linear algebra professor proved that      Diagonalizable matrices with complex values are dense in set of $n \times n$ complex matrices.   He defined a metric (I believe) that was somehow related to the usual metric on $\mathbb{R}^{n^2}$.  Then somehow proved that diagonalizable matrices were dense because for any matrix $A$ if $\det(A - \lambda I) = 0$ on an open subset, then $\det(A - \lambda I)$ was the zero polynomial.  I Googled around a bit and found some stuff talking about the Zariski topology, and I am not sure this is what I want. Most of what I have found on this topology is much more general than what he was doing.  Does anyone know any resources that would explain how to go about proving this statement?  Edit: I found out how to prove this the way my professor did.  We define the norm of a matrix by $$ |A| = \max\{|Ax| : |x| = 1 \}.$$ Now we have a distance $d(A, B) &lt; \epsilon$. You can prove that if $(A - B)_{ij} &lt; \epsilon/n$, then $|A - B| &lt; \epsilon$.  Let $p(x)$ be the characteristic polynomial of $A$, an $n \times n$ matrix.  $$p(x) = \prod_1^n (x - x_i) = x^n + (-1)\sigma_1 x^{n - 1} + \cdots + (-1)^n\sigma_n$$ where $\sigma_1, \ldots, \sigma_n$ are the elementary symmetric polynomials of $x_1, \ldots, x_n$ which are the eigenvalues.  The discriminant of the characteristic polynomial is a symmetric polynomial, therefore it can be written in terms of the elementary symmetric polynomials, which in turn can be written in terms of the entries of the matrix.  But since the discriminant is a polynomial, it only has finitely many roots. Therefore for $\epsilon &gt; 0$, by changing the entries of the matrix less than $\epsilon / n$ we can find a new matrix $B$ such that $|B - A| &lt; \epsilon$ and the discriminant is not zero.   The discriminant being not zero means $B$ has distinct eigenvalues, thus has a basis of eigenvectors. Therefore, it is diagonalizable.  Thus the set of diagonalizable matrices is dense in the set of matrices with respect to that metric. How to run Visual Studio post-build events for debug build only How can I limit my post-build events to running only for one type of build?  I'm using the events to copy DLL files to a local IIS virtual directory, but I don't want this happening on the build server in release mode. Can we implement a dead time on answers for high reputation user Summary: High reputation users on SO come swooping in the second a simple question is posted, and get the first answer in before I've even opened my Python interpreter to test my idea out.  This actually is a problem because it prevents use lower users who aren't as good  at programming from getting any answers in.  This makes it very hard to build up reputation.  Thus, I propose a short dead time for high reputation users where their answers do not appear.  The Problem: This isn't just reputation.  By answering questions and having discussions with OP, we all get better at our language.  Teaching is one of the best ways of learning, and a more well versed population is absolutely good for SO.  (If anyone wants to argue that, go home)  While its nice to have folks who can quickly answer nearly all questions in a jiffy, its a little frustrating for those of use who simply aren't as fast, but still want to contribute.  The result is that many people do not answer questions because they feel like they just can't get an edge in.  SO should aim to incentivize as many answers by as many users as possible.  The most answers from the most users is the most beneficial situation for the site and the community. Once we accept this idea, it is clear that something needs to be done to address the issue at hand    The Solution: Luckily, this is an easy problem to solve; we use a dead time.  High reputation users can post answers whenever they like, but they won't appear until after the dead time is completed.  The trick is balance, a dead time which is too long will harm the OP as the question will remain unanswered, however, a dead time which is too short won't actually fix the problem.  There definitely needs to be a discussion about what "high reputation" means and what dead time to use, but it should be independent of whether y'all agree that this is a problem.  So, what do you think?  EDIT: Wow, -18, I leave for lunch for 30 minutes and this is what happens!  I guess y'all think this is a bad idea, but let me try to clarify a few things on which I was unclear earlier.  This is not about reputation.  This is not about getting the first answer just for the sake of getting to say "in before..."  This is about learning.  By getting a shot at answering questions before the high reputation users, people learn. Think of it as doing an example problem in class before the teacher does it. You get a chance to try it yourself, test your knowledge, and if you fail, then the teacher is right behind you to show the correct solution.  This is what I'm trying to accomplish. However, seeing the solution, and then trying to answer the question, is not very helpful. This is how I view the status quo.    In response to Mr. Pieters' answer and comments: You're absolutely right, this is not about quantity, its about quality.  The fastest answer isn't always the best.  But asking the high reputation users to take a step back for a few minutes is a means of getting there.    Hopefully this clears things up a bit. shortcut for commands I use following commands very often, and they're too long.  sudo /etc/init.d/kerio-kvc start sudo /etc/init.d/kerio-kvc stop   how can I create a shortcut command for them? something like sudo kerio start or sudo kerio stop Merging Raster Datasets of Different Resolutions using ArcGIS Desktop? I am using ArcGIS Desktop.  I have two DEMs in .tif format and I am looking to merge the two. The main problem with merging these are that they are different resolutions (one is 10m and one is ~20m). Also, another problem with merging the two is that they overlap and I am not sure of the most effective way of dealing with that overlap. Obviously I want to keep the values of the better resolution.   So what are the best methods here?   Is there a python script available that works best with merging rasters? I would like to automate this with a script if possible. Which process of resampling is best for DEM data? Bipolar, Cubic... ? What is the best way to merge overlapping DEMs? "Temporarily unreachable" when using fetch as Googlebot from Webmasters Tools The page I'm trying to fetch is public.  The server logs show no request.  Google reports that any page on my site is "Temporarily unreachable". It won't even fetch the verification file, a static HTML file that was successfully used to verify with the webmaster tool application itself.   My site is running on the .  What is causing this? Can the "Sacrifice (this card):" activated ability be placed back on the stack? This is a broad question so for simplicity I will use  and .  Scenario 1: Common exchange- creating a baseline.   Player A uses Naturalize on Seal of Fire  Player B Sacrifices SoF in response. SoF is highest on the stack and goes off first.  This leaves Naturalize without a valid target.   Scenario 2: Complicated exchange- Is this legal?   Player B Sacrifices SoF. In response Player A targets it with Naturalize. SoF is on the stack and it's ability has not resolved. Player B (as above) sacrifices SoF in response to the Naturalize.    Can Seal of Fire be placed back on the stack by re-activating its ability, potentially with a new target? What should I do when my neural network doesn't learn? I'm training a neural network but the training loss doesn't decrease. How can I fix this?  I'm not asking about overfitting or regularization. I'm asking about how to solve the problem where my network's performance doesn't improve on the training set.    This question is intentionally general so that other questions about how to train a neural network can be closed as a duplicate of this one, with the attitude that "if you give a man a fish you feed him for a day, but if you teach a man to fish, you can feed him for the rest of his life." See this Meta thread for a discussion:   If your neural network does not generalize well, see: Read Gif Images using WebView android How to make a Rigid Body fall off a conveyor belt? I've created conveyor belt and and two objects with rigid body physics. But the first object won't fall and second object falls backwards.  How can I fix it? Extract URLs from text in PHP Why high voltage transmission lines? Is there any scrappable junk I should hold on to that could otherwise be used for other purposes? "Me and my wife" or "my wife and me"