spring jsf facelets and rest - i want them all but have mapping problems
Why does Spring MVC respond with a 404 and report "No mapping found for HTTP request with URI [...] in DispatcherServlet"?
What does Fermi level in the band gap mean? What does it mean that the  for some semiconductors lie in the band gap?  Is Fermi level definition different from what is know as usual?  We define the Fermi level as the highest level of energy which is occupied by electrons.  Then what is missing here? Quick directory navigation in the bash shell Imagine that a PC gets reduced to 0 hit points during a battle. The battle continues for a couple more rounds and they roll their death saves, one success and one failure (not sure if this part of the question is relevant).  Then the rest of the party defeats the last enemy and battle is over. Now what?    Does the downed PC keep rolling death saves until they strike 3 one way or the other? Does everyone else repeatedly roll for Medicine to stabalise the dying PC? Or (as we've been ruling it) they are just assumed to become stabilised because combat is over (I'm fairly certain this probably isn't RAW or RAI though).   I've looked through the PHB and I've only managed to find descriptions of being stabilised during combat, not after it.  Ideally I'd like a rule quoted (or a tweet or something) to point to where such a thing is outlined in the rules, if it is written anywhere. How to force Wine into acting like 32-bit Windows on 64-bit Ubuntu? Shapefile projections not matching up in QGIS? To start off with, I'm very new to QGIS. I'm attempting to place a shapefile with locations for 3 substations on a map of Peniche, Portugal. The locations for the substations are given in latitude and longitude coordinates.   When I start off by importing the substation shapefile, I make sure that both the layer CRS and project CRS are set to EPSG:32629, which gives each substation point the correct coordinates, however the scale is at 1:1, which is not correct. As soon as I then add a raster map of the region or a shapefile of Portugal using the same CRS the position of the substations also becomes incorrect.   I'm sure there's a simple solution I'm not getting, but I've been trying for a while and have had no success. Here's a dropbox link to the files Why do 'get' and 'have' work similarly in      I got/had my car repaired.    but differently – that is are not complemented in the same way although they still mean the same – in      I got someone to repair my car.    and      I had someone repair my car. Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$  but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say). Which is correct, "neither is" or "neither are"? Bob: "Can I set the font color? Can I customize the text?"  Frank: "Neither of these options is available. Sorry!"    Is "neither is" always correct or should one use "neither are" in some cases and what are the exact rules?  of this-is-why-stackexchange-exists :P A lady and a monster A famous problem:  A lady is in the center of the circular lake and a monster is on the boundary of the lake. The speed of the monster is $v_m$, and the speed of the swimming lady is $v_l$. The goal of the lady is to come to the ground without meeting the monster, and the goal of the monster is to meet the lady.  Under some conditions on $v_m,v_l$ the lady can always win. What if these conditions are not satisfied?  Edited: the monster cannot swim. If the conditions are not satisfied, then monster can always perform a strategy such that the lady will not escape the lake. On the other hand this strategy is not desirable for both of them because they do not reach their goals. As there was mentioned, this deals with undecidability of the problem. On the other hand, if you imagine yourself to be this lady/monster, you can be interested in the strategy which is not optimal. What is it? If there are such strategies in the game theory?  Edited2: My question is more general in fact. If we have a game with one parameter $v$ when two players $P_1, P_2$ are enemies and if $v&gt;0$ then for any strategy of $P_2$ the player $P_1$ wins. If $v\leq 0$ then for any strategy of $P_2$ there is a strategy of $P_1$ such that $P_2$ does not win and vice versa. I am interested in this case. From the mathematical point of view as I have understood the problem is undecidable since there is no an ultimate strategy neither for $P_1$ nor for $P_2$. But we are solving somehow these problem IRL. Imagine that you are a lady in this game - then you would like to win anyway even while knowing that your strategy can be covered by the strategy of the monster. On the other hand, the monster knows that if he will cover all strategies of the lady she will never reach the shore and he will never catch her. I mean they have to develop some non-optimal strategies. I hope now it's more clear. If a bob attached to a string is swinging in vertical circular motion, we say that the bob leaves the circular path when the tension in the string becomes zero, but even when tension becomes zero, there's always a component of the bob's weight towards the center of the circle. Why then does the bob leave the circular trajectory? I tried to get the number of available brew packages by running  brew search | wc -l   but I don't know how accurate this really is. What do I need to take into account to get a more correct number?  PS: macports lists 19,889 packages online Let $f$ be a bounded measurable function on a set of finite measure $E$.  For a measurable subset $A$ of $E$, show that $\int_A f = \int_E f \cdot \chi_A$     Lemma: Let $f$ be a bounded measurable function on a set of finite measure $E$.  Suppose $A$ and $B$ are disjoint measurable subsets of $E$.  Then $\int_{A \cup B} f = \int_A f + \int_B f$.   My first approach was to view the problem like this utilizing the above lemma:  $$\int_E  f \cdot \chi_A = \int_{E\setminus A} f \cdot X_A + \int_A f \cdot \chi_A $$  I'm not really sure how the result follows from the above. If $$(1+ \cos A) (1+ \cos B) (1+\cos C)= y = (1- \cos A) (1-\cos B) (1-\cos C)$$ then prove that $$y = \pm \sin A \sin B \sin C$$ I have a dataset in csv file format that I'm loading into pgfplotstable. I can't change the format of the original csv, and being financial data it has a column of item names and two or three explanatory data points at the top of the table. e.g.  \documentclass{article} \usepackage{array,datatool,pgfplotstable}   \begin{filecontents}{sample.csv}     item,2008,2009,2010,2011      ,Actual,Actual,Forecast,Forecast      ,GBP,GBP,GBP,GBP      ,('000),('000),('000),('000) rounded     Area 1 OP, 1000000,1500000,1750000,250000     Area 2 OP, 400000,500000, 450000,-50000     Area 51 OP, 300000,375000,390000,15000     No P, 1250000, 1000000, 950000, 50000     Residuals, 800000, 80000, 90000, 10000   \end{filecontents}  \pgfplotstableread[col sep=comma]{sample.csv}\mystuff   I'm stumped at the next stage which is to typeset \mystuff so I have comma separators and can add a row that sums the 'Area' columns. At the moment to display the table I have to do something like  \begin{document}  \pgfplotstableread[col sep=comma,header=false]{sample.csv}\mystuff  \pgfplotstabletypeset[   columns/0/.style={string type},   columns/1/.style={string type},   columns/2/.style={string type},   columns/3/.style={string type},   columns/4/.style={string type}]\mystuff  \end{document}   How do I tag the top three rows as string and the rest of the last four columns as integer type so I can format and transform them? I often have to write up reports based on the analysis of some data. I use R to analyse the data and export tables, figures, and text. This is then included into a LaTeX document either using input or Sweave ().  However, when I collaborate with others, I sometimes need to provide a document in Open Office / MS Word format.  Question:  Thus, assume the simplest scenario   I have a LaTeX document with text, tables, and figures I need to export this reliably into Open Office or MS Word format: this includes mathematical formulas, table formatting, and quality figures) I don't need to go back from MS Word to LaTeX   What is a reliable, efficient, and preferably free process?  Initial Thoughts:   Export to HTML,  LaTeX - LyX - ODF Use odfWeave  that deals with some specific issues associated with LaTeX to Open Office conversion    I was hoping that there is an expert out there who has worked out a good system already. Why is the change in Hamiltonian for an active infinitesimal canonical transformation defined the way it is? I'm trying to understand infinitesimal canonical transformations and conservation theorems (section 9.6 Goldstein ed3). My specific problem is with understanding eq 9.104,         $\partial H = H(B) - K(A^{'}) $, where $\partial H$ represents the change in the hamiltonian under an infinitesimal canonical transformation (from the active view point ofc). The issue only arises for generating functions that explicitly depend on time.  Here's what I've thought of so far. I understand what $K$ is from the passive viewpoint (or so I think), its the hamiltonian that would furnish Hamilton's equations for the transformed variables $Q=q+ \delta q$ and $P = p + \delta p$. In the specific case of the generating function being a time dependent hamiltonian, $K = H + \delta t \frac{\partial H}{\partial t} = H + \delta t\dot H$, which makes sense because once you transform the variables to what they'll be in time $\delta t$, you'll also need to add on a term that comes from the Hamiltonian being at a later time $t+ \delta t$.   In the book, this is explained as a transformation depending on time changes the hamiltonian, so $H(A) \rightarrow K(A^{'}) $. Which I sorta get. But why not for $H(B)$? And what about the case when the Hamiltonian is time dependent and it is the generating function? Is $H(B)$ at a later time then?   Note: $A$ represents the original variables, $B$ is the transformed variables in the active sense, so in the same phase space as $A$, and $A^{'}$ is the same point as A but in the transformed phase space (passive). I know that for continuous variable $P[X=x]=0$.   But i can't visualize that if $P[X=x]=0$, there is infinite number of possible $x$'s. And also why do their probabilities get infinitely small ? Is there a proof from the first principle that the Lagrangian L = T - V? How to Compact geodatabases in multiple directories using ArcPy? I am trying to write a script to walk through all folders in my workspace and find and compact all geodatabases in my workspace.  I feel like I'm on the right track but I am stuck.  import acrpy, os from arcpy import env  arcpy.env.workspace "C:"  for (path, dirs, files) in os.walk(arcpy.env.workspace)     for file in files         if ".gdb" in file:         workspaces = arcpy.ListWorkspaces("*", "FileGDB")             for workspace in workspaces:             arcpy.Compact_management(workspace)   I am using ArcGIS 10.2.2 for Desktop. When abstract logic circuits get translated to the electrical equivilant, how are conversions from 0 to 1 handled? I've searched around and there's lots of good information out there, but I'm a bit of a n00b when it comes to the precise meanings of words like current, voltage, amplify, resistor, diode etc. So what at first presents itself as a simple explanation quickly turns into a wikipedia tabfest of ever increasing complexity. With that in mind please forgive my probably inaccurate use of these terms in this post.  The basic problem for me is that gates on a logic circuit appear to be creating power from nothing, which is impossible. For example imagine two NOT gates connected in sequence: To my uninitiated mind what seems to be happening is that the first NOT gate cuts the power to the second NOT gate, which responds by magically conjuring new power out of thin air and sending it on it's way. The only explanation I can come up with feels a bit redundant and absurd (That every single gate on the diagram is implied to be directly connected to a power source)  The closest I've come to an answer is something to do with changing the "signal" from "high" to "low" but again, I'm not familiar enough with the language. It still suffers from the same problem, if you've got a low signal how is it that a NOT gate can turn it into a high one without directly drawing on a power source?  To wrap up and present the question in all it's rookie glory: How do logic gates physically convert 0 to 1 without violating the law of conservation of energy?