I am getting this null pointer exception on the id of the newly created object that I'm putting into  the DB, even though it should be auto generated.  @Entity @Table(name = "REQUEST") @EntityListeners(DatabaseListener.class) @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "requestId", scope = Request.class) public class Request implements Serializable {      private static final long serialVersionUID = 1L;      @Id     @GeneratedValue(strategy = GenerationType.AUTO)     @Column(name = "REQUEST_ID", nullable = false)     private Long requestId;      @Column(name = "REQUEST_NAME")     private String name;      @Column(name = "CATEGORY")     private String category;      @Column(name = "REQUEST_MESSAGE")     private String requestMessage;      @Column(name = "CREATED_ON")     private Date createdOn;      @ManyToOne     private User createdBy;      @ManyToOne     private User assignedTo;      @ManyToOne     private Status status;      @OneToMany(mappedBy = "request", fetch = FetchType.EAGER)     private List<Response> replies;   The endpoint that I am calling...  @RequestMapping(value = "/requests/addRequest", method = RequestMethod.POST) public ResponseEntity<Request> addRequest(@RequestBody final Request request) {     if(request.getStatus() == null){         request.setStatus(request.getAssignedTo() == null ? statusRepo.findByType(StatusType.UNASSIGNED) : statusRepo.findByType(StatusType.ASSIGNED));     }      final Request saved = requestRepo.save(request);      return new ResponseEntity<>(saved, HttpStatus.CREATED); }   I can add more of the other classes if needed as well. But I can not figure out why requestId would be null. I'm calling that enpoint from an app and passing it a Request object, so obviously there it is null but it shouldn't be null when it's going into the DB. Right?   org.springframework.http.converter.HttpMessageNotReadableException: Could not read document: (was java.lang.NullPointerException) (through reference chain: Request["requestId"]); nested exception is com.fasterxml.jackson.databind.JsonMappingException: (was java.lang.NullPointerException) (through reference chain: Request["requestId"]) ... Caused by: com.fasterxml.jackson.databind.JsonMappingException: (was java.lang.NullPointerException) (through reference chain:Request["requestId"]) ... Caused by: java.lang.NullPointerException: null     at com.fasterxml.jackson.annotation.ObjectIdGenerator$IdKey.<init>(ObjectIdGenerator.java:125) ~[jackson-annotations-2.4.0.jar:2.4.0]   Edit: To be clear, I know what a null pointer exception is, I understand that the requestId is null and that is causing a problem, so I don't think this is a duplicate question. What I don't understand is why it is null. It should be auto generated, but as Jason pointed out, I may have to determine the id application side first.

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


I am working in QGIS 3.8.3 in Windows 10,  I have to export all the selected raster layers, is there any way of doing this instead of exporting them one by one?  I only would like to export all of them to the same format, all of them have the same resolution and also the same latitude/longitude size.  I haven't found help page of , and I am totally new to python codes, so I seek any 'easy' to understand or well explained "how to apply the code".

I am trying to export existing raster layers as tif files. I just want to avoid saving each raster manually. this is the code I am trying to use, but it doesn't seem to work:  from PyQt4.QtCore import * from qgis.core import * from qgis.gui import * import os import sys  layers = iface.layerTreeView().selectedLayers() for layer in layers:     output='Desktop'%layer.name%'.tif'   As specified in a comment below, I modified my code to this:  from PyQt4.QtCore import * from qgis.core import * from qgis.gui import * import os import sys   layers = iface.layerTreeView().selectedLayers()  for layer in layers:      file_name = 'D:\tif' + layer.name() + '.tif'      file_writer = QgsRasterFileWriter(file_name)      pipe = QgsRasterPipe()      provider = layer.dataProvider()       if not pipe.set(provider.clone()):         print "Cannot set pipe provider"      continue       file_writer.writeRaster(         pipe,         provider.xSize(),         provider.ySize(),         provider.extent(),         provider.crs())   I can't seem to get it to work and I am not getting any error.  I already selected the layers and it still does not work.


I want to delete all mail from my gmail account. How can I do this all at once? I have over 3000 emails to delete. Any suggestions?

I want to clear my mail from my Gmail account, but don't feel like selecting every message (also the archived ones).  Is there an easy way to delete all my mail from Gmail without having to delete my account?


$$ \sum_{k=0}^{n}{\binom{p}{k}\binom{q}{k}\binom{n+k}{p+q}} = \binom{n}{p}\binom{n}{q} $$  My try was to use multivariable generating functions(I don't know, I just thought). I hope that there is a solution using combinatorical analogy.

Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$   Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.  Could some help me with this, thanks


I'm trying to save the result of the following command to a variable $ peer lifecycle chaincode queryinstalled Installed chaincodes on peer: Package ID: basic_1.0:b558bd2d24b2a1f3404e2055de432aa561b2f08d6e32c67a250279ea76cc9356, Label: basic_1.0 Package ID: basic_1.0:ab07c8d0f7eb00800e99a8b390a8fd6ccc23d3a510bdb443e03dea4f1c456f71, Label: basic_1.0 Package ID: basic_1.0:8df7e007152c44c2253f8349d4d1702b4993ffc42af0cd10a0cec99792f6afc9, Label: basic_1.0 Package ID: basic_1.0:c00fa2d3b1da872d48330e845fc14b178831f5c42f60cfc4205d05bcbcb2fbda, Label: basic_1.0 Package ID: basic_1.0:74050d6a33d6b0ab8865164470c4af5cc8be842fac493f0862982150fdde2669, Label: basic_1.0 Package ID: basic_1.0:bf1f749eca0693dba2b29dacaa05768d867416de64294596bccf3d1fd80e1a9a, Label: basic_1.0 Package ID: basic_1.0:98a3c90d99169dd60a1be85a054e627fd40f160fb6306ee04fcb20f7c2f2b395, Label: basic_1.0  I'm doing PACKAGE_ID_${ORG}=$(peer lifecycle chaincode queryinstalled)  I get the error CC_PACKAGE_ID_1=Installed: command not found  When I do PACKAGE_ID=$(peer lifecycle chaincode queryinstalled) it works fine. Is there a way to do this with PACKAGE_ID_${ORG}?

I want to dynamically create a sequence of strings by manipulate an array of elements and create some arithmetic procedure.  for name in FIRST SECOND THIRD FOURTH FIFTH; do     $name = $(( $6 + 1 ))     $name = "${$name}q;d"     echo "${$name}"; printf "\n" done   The desire outcome would be the below for $6 equals 0.  1q;d 2q;d 3q;d 4q;d 5q;d   But I get this error  reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution   I guess it's something simple. It used to work when I did something like   FIRST=$(( $6 + 1 )) FIRST="${FIRST}q;d"


Any ideas why this image (lower entry) gets its own picture while the other one (upper entry, not visible) does not?      Tried using scale=0.5 instead and it got inline instead. Probably some funky LaTeX thing going on with that it's on the end of the previous page otherwise.

How to influence the position of float environments like figure and table in LaTeX?   This is a general question and should collect useful answers for all users. I hope we can use this as a reference


I'm trying to prove that if two $m\times n$ matrices $A$ and $B$ over a field $F$ have the same row space, then they are row-equivalent.   Intuitively, it makes sense. Indeed, if $\mathrm{row}(A) = \mathrm{row}(B)$, then each row $[a_{i1}, ..., a_{in}]$ of $A$ is generated by rows of $B$ and vice versa, that if, for any $i = 1,...,m$ we have $$[a_{i1},...,a_{in}] = c_{i1}[b_{i_11},...,b_{i_1n}] + ... + c_{it_i}[b_{i_{t_i}1},...,b_{i_{t_i}n}]$$ and $$[b_{i1},...,b_{in}] = d_{i1}[a_{i'_11},...,a_{i'_1n}] + ...  d_{ip_i}[b_{i'_{p_i}1},...,b_{i'_{p_i}n}].$$  This seems similar to the condition an elementary operation being applied to a row, but I can't seem to make this final step to make it rigorous. One idea was to prove by induction is that is $A_k$ is equivalent to $A$, then $A_{k+1}$ is equivalent to $B$, where $(A_k)_{ij}$ is equal to $b_{ij}$ if $i < k$ and to $a_{ij}$ otherwise, but I have a feeling that this is wrong.   Edit: This question has been marked as a duplicate of another question for some reason. However, that question asked how to prove that two row-equivalent matrices have the same row space, while my question asks how to prove the converse.

so I understand row space of a matrix X to mean the subspace of $R^n$ spanned by the row vectors of A  Here is the text that is in question:    Can the rows of B be written as a linear combination of A just because they are row-reduced from A?  Say we have matrix:  $$\begin{bmatrix} 1 & 2 \\ 3 & 8 \end{bmatrix}$$ $$ -> \begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}$$  How is the row-reduced matrix a linear combination of A?


Let $n=2^k$ and $T(2)=1$. Then, if we define $T(n)=\sqrt{n}(T(\sqrt{n}))+n$ for $n>2$, then what would be the exact solution of the recurrence?  The master theorem might not work here, and guessing a solution seems hard. In addition, I suspect $n=2^{2^k}$, for otherwise, we get irrational values. I think $T(n)=O(n\log n)$. But, is a better bound possible? Thanks beforehand.

$$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution.


My moderncv-created resumé compiles without any sort of error: the list of publications regularly appears in its own section, no problems on that side. However, even though I have inserted the urls of my journal papers in my .bib file, publication entries in the resulting pdf document are typed in plain text, devoid of any link. Links work fine in all other sections of my CV. Bibtex handles my bibliography file. A MWE: \PassOptionsToPackage{dvipsnames}{xcolor}   \documentclass[11pt,a4paper,sans]{moderncv}  \moderncvstyle{banking} \moderncvcolor{black}  \usepackage[latin1]{inputenc}   \usepackage[scale=0.8]{geometry} \usepackage[final]{pdfpages} \usepackage{xpatch} \usepackage{url} \setlength{\footskip}{33pt}  \name{I am}{Lingo} \address{Job title}  % I have put my job title in the address section \extrainfo{address}  % and my address as extrainfo  \phone[mobile]{+1(234)567890}                 \email{lingo@lingo.org}  \AtBeginDocument{\hypersetup{     pdftitle={CV},      pdfauthor={Lingo},      colorlinks=true,     linkcolor=gray,     urlcolor=gray }}  \begin{document} \makecvtitle  \section{Education} \cventry{year--year}{Degree}{Institution}{City}{\textit{Grade}}{Description} \cventry{year--year}{Degree}{Institution}{City}{\textit{Grade}}{Description}  \section{Experience} \cventry{year--year}{Job title}{Employer}{City}{}{Description line 1} \cventry{year--year}{Job title}{Employer}{City}{}{Description line 1}  % other sections  \nocite{*} \bibliographystyle{plainyr} \bibliography{publications}   \end{document}       and my publications.bib file has entries such as @article{Lingo2016,     title = {A generic {Title}},     journal = {Yellow Pages},     url = {\url{https://www.google.com}},     year = {2016},     volume = {3},     pages = {251-266},     author = {\textbf{Lingo, I A.} and Doe, J.} }

I'm using bibtex for my bibliography in LaTeX. I have some URL's I need to cite in the paper. How do I add URLs into the .bib file?  A typical section in my .bib file looks like this:  @conference{eigenfacepaper,   title={{Face recognition using eigenfaces}},   author={Turk, M. and Pentland, A.},   booktitle={Proc. IEEE Conf. on Computer Vision and Pattern Recognition},   volume={591},   year={1991} }   I tried some misc sections in bibtex but they don't show up in my document.


I want to show that $f(x)=x^p - x+ a$ is irreducible  in $\mathbb{F}_p [x]$ if $a \neq 0$. I know that if $b$ is a root of $f$, then $b+1$ is also a root of $f$. Can I use this fact to prove that $f$ is irreducible? Any hints or reference are appreciated.

Let $p$ be a prime. How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?   Right now I'm able to prove that it has no roots and that it is separable, but I have not a clue as to how to prove it is irreducible. Any ideas?


I'm not asking how to bevel a shape's edge, I'm asking how to create something like this:  For this, I created a cylinder of default size and scooted it 1m on the Y axis so that it appears as if the cube as a rounded face. However, this leads to Z-fighting between the two separate objects. I would like to know how to alter the cube's face so that it is a single shape with a rounded face. Sorry I don't have the vocabulary to be more clear.

I have this cube which has the basic shape of an escalator but I want to add a horizontal cylinder at the end of the cube like an escalator how can I do that


I have installed the texwork 2017 and I am trying to use biber on it. I am trying to create a bib file to use it as a bibliography database but error keeps appearing us like the screenshot below.  What is wrong? What does it mean by "did you pass the beckend-biber option to Biblatex"? Could anyone please help me how to solve this error?

I've browsed the forums and found a number of posts that have addressed this issue, but none of the solutions seem to work for me. I have the following script that I just copied from the bibtex home page to get familiar with it. Instead of the citation number I get a question mark. I compile using Latex+Bibtex+Latex+Latex+PDFLatex+ViewPDF just as has been previously suggested and the problem persists.   \documentclass[11pt]{article} \usepackage{cite}  \begin{document}  \title{My Article} \author{Nobody Jr.} \date{Today} \maketitle  Blablabla said Nobody ~\cite{Nobody06}.  \bibliography{mybib} \bibliographystyle{plain} \end{document}   My bibliography (Bib.bbl)  @misc{ Nobody06, author = "Nobody Jr", title = "My Article", year = "2006" }    Looking at previous posts one thing that is concerning is that my .bbl looks empty as shown below. Further, I don't have a .blg  \begin{thebibliography}{}  \end{thebibliography}


I have a string: s = "expected exit code none[one] front back rear x y z"  I want to find [one] and capture everything after that including [one].  The output should be my_str = "[one] front back rear x y z"  I tried: import re patt = re.compile(r"[one](.*)") f = re.search(patt, s) f.group(0)  Output: >>> "one] front back rear x y z" Can't figure out how to include the missing [.

I am tired of always trying to guess, if I should escape special characters like '()[]{}|' etc. when using many implementations of regexps.  It is different with, for example, Python, sed, grep, awk, Perl, rename, Apache, find and so on. Is there any rule set which tells when I should, and when I should not, escape special characters? Does it depend on the regexp type, like PCRE, POSIX or extended regexps?


whether a given date lies between other two dates (where dates are manually created using dd-MM-yyyy format)      Date startDate, endDate, dateToCheck; if (dateToCheck.equals(startDate) || dateToCheck.equals(endDate) ||      (dateToCheck.after(startDate) && dateToCheck.before(endDate))

I have a series of ranges with start dates and end dates. I want to check to see if a date is within that range.  Date.before() and Date.after() seem to be a little awkward to use. What I really need is something like this pseudocode:  boolean isWithinRange(Date testDate) {     return testDate >= startDate && testDate <= endDate; }   Not sure if it's relevant, but the dates I'm pulling from the database have timestamps.


I see both sides in everything. What would one call this type of thinking?

A colleague of mine is trying to describe herself as "capable of seeing a situation from multiple perspectives" or "able to look at the big picture from various viewpoints". I feel like there must be a better way to get this point across.  Is there a more concise way to describe this characteristic?


A French character, ç, appears to be breaking my application.  The script harvests data from an external application, then uses post to send this data to our apache/php server to be processed.  For security reasons, anything sent to the server must be accompanied by a hash (md5) of the data, time of sending, user info etcetera to verify the user has permissions to carry out the action.  Now, for whatever reason, actions are getting rejected if there are certain characters in the data string that gets hashed, like the ç.  How can I avoid this? Is there a certain encoding I should be set so they are correctly processed?  A regex that could replace all ç with c, and so on for all non-English letters?  There is also the potential for German and Spanish input to be harvested with this tool so would like to find a catch-all solution!

I'm setting up a new server and want to support UTF-8 fully in my web application. I have tried this in the past on existing servers and always seem to end up having to fall back to ISO-8859-1.  Where exactly do I need to set the encoding/charsets? I'm aware that I need to configure Apache, MySQL, and PHP to do this — is there some standard checklist I can follow, or perhaps troubleshoot where the mismatches occur?  This is for a new Linux server, running MySQL 5, PHP, 5 and Apache 2.


I think it worked two days ago, but since yesterday the result header is "0 jobs near european union". So the keyword is recognized, but apparently the EU has no member states. Search by countries works fine.

Let's say I want to find a company in Ontario, Canada. Ok, I navigate to company pages and type ON, Canada to Where in search:        but there are no companies in there. So why if I type just Canada it finds companies in ontario; and if I type specific area (ON, CA)  it doesn't find anything ?


I hold a Fiji Passport and I have obtained a Schengen Visa for Norway. It is a short stay visa and is for single entry only.  I will be transiting Amsterdam while going to Norway; will my Schengen short stay visa for single entry in Norway be sufficient to guarantee me an air-side transit in Amsterdam?

I am Egyptian citizen holding a single entry visa to Spain , travelling in 1 July but i will have 1 day transit in Swiss airport , so do i need a visa to enter Switzerland ?


I switched from windows 7 to Ubuntu 12.04 LTS a week or so ago on my Toshiba Satellite L755 I believe. It worked fine, but I had Internet problems and internal errors so I thought switching to 12.10 would fix everything, I was wrong.   Long story short, I want to switch back to windows 7 but when I installed over windows 7. I don't have a recovery CD because, can I create a recovery CD on another computer and use that?  Thanks Terry and everyone else who answered!  I was able to restore my computer using a back up from when I had Windows 7 so if all goes well I'll be set.

I have absolutely no experience with Linux, and I desperately need to get my computer back up and running again with Windows.  How do I remove Ubuntu and reinstall Windows?    Editor's note: many of the answers are about removing Ubuntu from dual-boot but keeping Windows (which is a bit complicated), while other answers are about removing Ubuntu from single-boot (which is easy: basically just format the disk while installing Windows). The question as written is ambiguous between dual-boot or single-boot.


I know this question is due to the fact that I don't have my head wrapped around OOP completely.  If I was writing a python script and created 3 methods...  def hello():     print "hello"  def goodbye():     print "goodbye"  def helloandgoodbye():     hello()     goodbye()   I could run it from a single script no problems.  But if I create them in a class I have to use self in order to call them.  class test():     def hello(self):         print "hello"     def goodbye(self):         print "goodbye"     def helloandgoodbye(self):         self.hello()         self.goodbye()   I believe in C# starts internally and works it way out searching for a method before throwing an exception. Why does Python make us use self to explicitly state the location of the method?

When defining a method on a class in Python, it looks something like this:  class MyClass(object):     def __init__(self, x, y):         self.x = x         self.y = y   But in some other languages, such as C#, you have a reference to the object that the method is bound to with the "this" keyword without declaring it as an argument in the method prototype.    Was this an intentional language design decision in Python or are there some implementation details that require the passing of "self" as an argument?


Q. How good is the teacher at that school?   A. They're a good teacher. They just sometimes hit(s)  their students if the students don't pay any attention to what he's saying, and then call(s) for parent-teacher conference.   Are we to use plural verbs with the singular they? It seems to me that with the first one, it should be hit but with the second one it should calls.

another question, I used the following phrase:     Your reader is [...], but they are a busy person.   I have two difficulties here:   In the first phrase, a reader, being singular noun, certainly implies is; In the second phrase, they, being a plural pronoun (regardless that it refers a single person), still requires are, but are implies the plural for busy persons which sounds awkward (a reader ⇆ persons);   Obviously, if I used "...but he is a busy person", everything works fine, but I'm trying to use a gender-agnostic form.  What is the proper number agreement here?


I have a loop that checks if rpm is installed or not and displays output in tabular format, but for the rpm package with long names, the tabular format will not show as expected. Code: for package in ssh yum firewalld policycoreutils-python do         rpm -qa | grep ^$package; exit_stat="$?"         if [[ ${exit_stat} -eq 0 ]]         then                 pkg_output="${pkg_output}\n ${package} \t \t \t | \t INSTALLED"         else                 pkg_output="${pkg_output}\n ${package} \t \t \t | \t NOT INSTALLED"         fi done echo -e $pkg_output  The output will come like:  yum             |   INSTALLED  ssh             |   NOT INSTALLED  firewalld           |   INSTALLED  policycoreutils-python              |   INSTALLED  What I am expecting is:  yum                         |   INSTALLED  ssh                         |   NOT INSTALLED  firewalld                   |   INSTALLED  policycoreutils-python      |   INSTALLED  Is there any way to show the terminal output in the expected format?

My script:  date echo -e "${YELLOW}Network check${NC}\n\n"  while read hostname do  ping -c 1 "$hostname" > /dev/null 2>&1 &&  echo -e "Network $hostname : ${GREEN}Online${NC}" || echo -e "${GRAY}Network $hostname${NC} : ${RED}Offline${NC}"  done < list.txt         sleep 30 clear done   Is outputting info like this:    Network 10.x.xx.xxx : Online    Network 10.x.xx.xxx : Offline    Network 10.x.xx.xxx : Offline    Network 10.x.xx.xxx : Offline    Network 10.x.xx.x : Online    Network 139.xxx.x.x : Online    Network 208.xx.xxx.xxx : Online    Network 193.xxx.xxx.x : Online   which I'd like to clean up to get something like this:    Network 10.x.xx.xxx       : Online   Network 10.x.xx.xxx       : Offline    Network 10.x.xx.xxx       : Offline     Network 10.x.xx.x         : Online     Network 139.xxx.x.x       : Online   Network 208.xx.xxx.xxx    : Online     Network 193.xxx.xxx.x     : Online   Network 193.xxx.xxx.xxx   : Offline


One of the rules for targeting spells in the Player's Handbook says that spells must have :  A Clear Path to the Target To target something, you must have a clear path to it, so it can't be behind total cover. If you place an area of effect at a point that you can't see and an obstruction, such as a wall, is between you and that point, the point of origin comes into being on the near side of that obstruction.  And the description for  says:  A target with total cover can't be targeted directly by an attack or a spell, although some spells can reach such a target by including it in an area of effect. A target has total cover if it is completely concealed by an obstacle.  However, the spell  says that the target "gains no benefit from cover". Does this mean sacred flame can target someone that is behind total cover?

We have a PC that is getting very sneaky with his Sacred Flame application. Its fun outside-the-box play but if he carries a hand mirror or uses scrying spells its going to become far easier.  Can a mirror allow use of Sacred Flame around a corner?  Can a PC scry to the other side of a door and use this sight to activate sacred flame?   Rules for Total cover:     A target with total cover can’t be targeted directly by an attack or a spell, although some spells can reach such a target by including it in an area of effect. (PHB 196)   Sacred Flame Text:     Flame-like radiance descends on a creature that you can see within range. The target must succeed on a Dexterity saving throw or take 1d8 radiant damage. The target gains no benefit from cover for this saving throw. (PHB 272)   Sacred flame specifically states "that you can see".   It is also confusing what the "area of effect" might be.


I read some about neutrino oscillations and flavour changing between three types of neutrinos. The question is, what is the significance of that observation? So far, we did not expect them to oscillate and change the flavour. But we do now. Are there any aspects that have changed in physics?

Neutrinos can pass from one family to another (that is, change in flavor) in a process known as . The oscillation between the different families occurs randomly, and the likelihood of change seems to be higher in a material medium than in a vacuum.   Why does the oscillation of neutrinos directly imply that they must have a nonzero mass, since the passage of one flavor to another can only occur in massive particles? And What are the cosmological implications of these oscillations?


Many indefinite integrals cant be solved in terms of elementary functions.e.g.-$$\int \frac{\sin(x)}{x}dx$$ But many hard looking ones are still solvable by weird substitutions and other tricks. Is there a test(like there are tests of solvability/convergence.etc in mathematics) of whether it is solvable or not by which I can see beforehand without wasting hours of time after which I come to know that it is not even solvable?

In the past, I've come across statements  along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations:  addition/subtraction multiplication/division raising to powers and roots trigonometric functions exponential functions logarithmic functions  which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example. What sort of techniques are used to prove statements like this? What is this branch of mathematics called?  Merged with "" by : Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions. I wonder how we can prove that kind of assertion?


Currently I have the following setup with two hostnames and two ips:  cheeze1.mydomain.com    IN  A   123.123.123.111 111.123.123.123.in-addr.arpa    IN  PTR cheeze1.mydomain.com  cheeze2.mydomain.com    IN  A   123.123.123.222 222.123.123.123.in-addr.arpa    IN  PTR cheeze2.mydomain.com   I want to change to having a round-robin dns for a single hostname using the two ips. Are the following PTR records valid? Can a host have multiple IPs and with each having valid reverse dns resolution?  cheezeonly.mydomain.com IN  A   123.123.123.111 cheezeonly.mydomain.com IN  A   123.123.123.222 111.123.123.123.in-addr.arpa    IN  PTR cheezeonly.mydomain.com 222.123.123.123.in-addr.arpa    IN  PTR cheezeonly.mydomain.com   EDIT: Please note that this question isn't asking the same as suggested in comments.   I'm going to answer this one myself, you guys are no help marking it as duplicate and not reading the question properly. Hopefully this answer will help others.  ANSWER:::  Yes, it is valid, and yes it will pass a FcRDNS check.  From the "Pro DNS and BIND" book:   An alternate approach is to define multiple A records with the same mail server name: ; zone file fragment         IN  MX  10  mail.example.com. .... mail    IN  A       192.168.0.4         IN  A       192.168.0.5         IN  A       192.168.0.6   In this case the load-balancing effect is under the control of BIND and the rrset-order record. In order to avoid problems if the receiving mail system does reverse look-up as a spam check then the PTR records for 192.168.0.4, 192.168.0.5, 192.168.0.6 above must all define (resolve to) mail.example.com.  And from RFC1912: Common DNS Operational and Configuration Errors   "For every IP address, there should be a matching PTR record in the in-addr.arpa domain. If a host is multi-homed, (more than one IP address) make sure that all IP addresses have a corresponding PTR record (not just the first one). Failure to have matching PTR and A records can cause loss of Internet services similar to not being registered in the DNS at all. " –   Therefor this configuration is valid and passes FcRDNS checks.

I   that using multiple PTR records in a DNS configuration is not recommended.   However, the reasons are often vague, or not so obvious, naming:    "it can cause problems",  "can trigger bugs in programs expecting a single answer" : it's the software's problem then, isn't it?! "can make DNS answer packet too large" : isn't this fixed with ?   Are these good reasons? Do you know of any other (good) reasons? All this kinda looks like a "legacy fear"...


I'm working in a pawnshop and when people attempt to sell or pawn an iDevice, I used to check its serial number or IMEI in icloud.com/activationlock. Now, without that tool, I'm concerned about activated phones that bypassed the activation lock with third party solutions.  Once a co-worked bought an iPhone 5 with iOS 7 apparently without iCloud account, when she tried to reset the phone, the activation lock asked for the owner's Apple ID, then I found the Apple's official tool to check phones IMEI and since, it has been very useful.  First, I want to know why Apple got rid of that page? And there is another alternative to check the activation lock status of an iDevice?

I have an used iPhone 6s that is pin locked and I wanted to check whether it was locked to iCloud or not.  I opened  to find out that the service is no longer available.    Is there any reason Apple decided to disable this tool? Now it will be harder to tell if an used iPhone that you want to buy is stolen.


I have a Lenovo y40.   intel i7 8gb memory 250 ssd hardrive AMD Radeon R9 M275 grapchis card windows 8.1 upgraded to windows 10 (did a fresh install)   I've tried to install Ubuntu 14.04 from USB, setup on pendrive and I also tried rufus. I've read similar questions and have tried all the answers. turned of secure boot, tried legacy mode in bios, etc ... followed different guides to the T. I haven't disabled SRT, but there seems to be no option for it in the Lenovo BIOS.  Here is what happens: I boot up the usb,  I have the option to try or install. I've tried both with the same results. Ubuntu loading screen comes on, and then after 30 seconds or so, it gives me this error: (initramfs) Unable to find a medium containing a live file system.  I've tried it on another pc and it all works fine.

I am trying to boot Ubuntu on my computer.  When I boot Ubuntu, it boots to a black screen.  How can I fix this? Table of Contents:


If function $f(z)$ is analytic and one valued in $|z-a|<R$ , prove that for $0<r<R$  $$f'(a)=\frac{1}{r\pi}\int_{0}^{2\pi}P(\theta)e^{-i\theta}d\theta$$  where $P(\theta)$ is real part of $f(a+re^{i\theta})$  EDIT : As pointed out by @Kavi Ram Murthy the question has been answered correctly here :   Since the answer has not been accepted I will copy the answer and mark this as closed (with due credits to @Kavi Ram Murthy)  My approach :  We know that $$f'(a)=\oint\frac{f(z)}{(z-a)^2}dz$$ subsituting $z=a+re^{i\theta}$ and simplifying i get   $$f'(a)=\frac{1}{2r\pi}\int_{0}^{2\pi}f(a+re^{i\theta})e^{-i\theta}d\theta$$  I am struck here I dont know how to procced further and prove first equation is equal to last equation . Any hints or sugestions ? (Also I may be be totally wrong in approaching the solution please let me know if I am

If the function $f(z)$ is analytic and one valued in $|z-a|<R$, How do we prove that $0<r<R$, $f^{'}(a)$ = $\frac{1}{{\pi}r}\int P(\theta)e^{-i\theta}d\theta$, where $P(\theta)$ is the real part of $f(a+re^{i\theta})$. I am guessing Cauchy's integral formula can be used but that has lead me to $f^{'}(a)$ = $\frac{1}{2{\pi}r}\int f(a+ re^{i\theta})e^{-i\theta}d\theta$. How to proceed from here.


A coin is tossed three times. The probability of zero heads is 1/8 and the probability of zero tails is 1/8. What is the probability that there is at least one head and at least one tail?  So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  We would use the complement rule and Multiplication Rule?   P(at least one head) = 1 - P(no heads) = 1 - 1/8= 7/8 P(at least one tail) = 1 - P(no tails) = 1 - 1/8= 7/8  would I multiply the two values?

A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used?


How can I find the solution of the distributional equation      $$ \frac{d^2u}{dx^2}-xu = 0 $$   And prove that there are at least two linear independent solutions? I've tried to to with the Fourier Transform but it is not possible to find the solution that diverges as $x$ tends to infinity, because the Fourier Transform is not defined for such functions. (they are so called Airy functions). I've tried also to solve it by testing the equation with an appropriate test function, but I'm stuck.   This is the method used. Once I've found Ai(x), I cannot prove that there are two independent solutions.

How can I prove that the Airy equation $$ \frac{d^2u}{dx^2}-xu = 0 $$  has at least two linear independent solutions?  Once I've found it how can I prove the existence of two independent DISTRIBUTIONAL solutions


I would like to know how many reputation points I need to perform each possible task in Stack Overflow. Is there a publicly available page on Stack Overflow from where I can get this information?

Am I so myopic as to not see the obvious link somewhere? I came looking for it immediately after seeing King Atwood's post but couldn't find it.  It seems appropriate to just put the list on the "reputation" tab of user profile, or at least a prominent link.


This is the error coming. I have shut down windows properly and yet this error comes.  Error mounting /dev/sda7 at /media/harshit/Movies and songs:  Command-line `mount -t "ntfs" -o "uhelper=udisks2,nodev,nosuid,uid=1000,gid=1000,dmask=0077,fmask=0177" "/dev/sda7" "/media/harshit/Movies and songs"' exited with non-zero exit status 14: The disk contains an unclean file system (0, 0). Metadata kept in Windows cache, refused to mount. Failed to mount '/dev/sda7': Operation not permitted The NTFS partition is in an unsafe state. Please resume and shutdown Windows fully (no hibernation or fast restarting), or mount the volume read-only with the 'ro' mount option.

Whenever I boot Ubuntu, I get a message that it cannot mount my windows partition, and I can choose to either wait, skip or manually mount.  When I try to enter my Windows partition through Nautilus I get a message saying that this partition is hibernated and that I need to enter the file system and properly close it, something I have done with no problem so I don't know why this happens.    Here's my partition table, if any more data is needed please let me know.     Device Boot      Start         End      Blocks   Id  System /dev/sda1            2048    20000767     9999360   83  Linux /dev/sda2        20002814   478001151   228999169    5  Extended /dev/sda3   *   478001152   622532607    72265728    7  HPFS/NTFS/exFAT /dev/sda4       622532608   625141759     1304576   82  Linux swap / Solaris /dev/sda5        20002816   478001151   228999168   83  Linux


This question stems from curiosity rather than a need to solve some particular problem. Given the two snippets of code below, why does the first produce an unbounded local error while the second does not? They appear to have the same critical sequence of operations in my mind, but there must be some under-the-hood magic at work that I am missing.      x=5      def fun1():       x=x+2                # or x+=2       print x      def fun2():       z=x+2                       print z   I understand that to make the first function work, the global function must be passed. I just want to know what the interpreter does to make those two statements much more different than they seem. Thanks.   This question is different from similar questions in that it asks for the underlying cause of this behavior and not a solution for mitigating it.

How can I create or use a global variable in a function?  If I create a global variable in one function, how can I use that global variable in another function? Do I need to store the global variable in a local variable of the function which needs its access?


Yesterday I just installed the Ubuntu 12.04 for the first time, the problem is that I can't play and of the MP3 files that I have.    I get this error:       Required plugin could not be found Python (v2.7) requires to install plugins to play media files of the following type: MPEG-1 Layer 3 (MP3) decoder   Please help me so I can play the files. Thanks

How can I install the most used plugins like flash player and mp3 in Ubuntu?


I'm trying to install the latest version of Git.  The ubuntu installser apt-get install versions 1.9.1, while the latest version is 2.9.1.  The same thing happens with Docker.  Apt-get installs only version 1.11.2 while the latest version of Docker is version 1.12.  I want to install the latest version of Docker, that is version 1.12 to work with the latest features.  Is there a way of specifying to install the latest version?

Why are packages in the official Ubuntu repositories older than the latest (upstream) versions from Debian Sid, PPAs, the authors, etc.?


My question is about my home station which doesn't turn off, and return the following message:    Wait-for-state stop/waiting *Stopping rsync daemon rsync  *Speech-dispatcher disabled: edit /etc/default/speech_dispatcher *asking all remaining processes to terminate...

Hope you can help me, 'cause this problem really made me sick already.  Hardware: Asus X555SJ (Dualcore Intel celeron 2166MHz, GeForce GT920 1GB, RAM 4GB) Systems: Windows 10 LTSB x64, Ubuntu 16.04.1 i386.  Here's the problem: when i try to reboot ubuntu, i see the ubuntu logo, then the screen gets black, and that's all that happens. The notebook still works and never turns off. When i'm shutting down ubuntu, i see the ubuntu logo on screen, but the screen doesn't fade out, i can observe the logo for hours and nothing happens either. In both cases I have to poweroff the notebook with a button. I've read couple of articles, tried to sudo shutdown, sudo reboot, but it didn't help. The problem still remains. Any ideas? Some people suggest to do sudo apt-get update, sudo apt-get upgrade, tried that too. Failed. All the updates/upgrades install correctly but it doesn't solve the problem. Any help will be useful!  ps. sorry if it's hard to read, may be mistakes, i'm not a native englishspeaker and writer =)


With Star Wars Episode 7 being released this year, this obviously throws a lot of the novel plots out of the window, especially post Return of the Jedi.   Which EU novels are considered Canon?

For all types of derivative works, namely:   Movies Novels Comic books Other written works (including RPG materials) Cartoons Video games Derivative movies (such as the infamous Ewok movies and even more infamous )   According to what rules does something become "canon" and who is responsible for the evaluation of a published work?


I am looking for a good introductory book about general relativity at undergraduate level.

What are some good books for learning general relativity?


I am a UK state pensioner living in canada with Dual citizenship. will the NHS cover me during my 3 week trip to the UK starting 22 June? I will travel with my Canadian passport

I want to know if I am entitled to medical assistance in the UK while visiting for 3 weeks in June. I am travelling on my Canadian passport since my UK one is outdated.


I recently bought the nikon 50mm 1.8 af-s lens. I bought a uv filter for it. It was a "multi-coated" filter, but when I put it on, the first thing I noticed was that it causes reflections. I returned it. Now I'm wondering what is a good UV filter for this lens?

What should I consider when shopping for a lens filter?    I see filters that range from $15 to $115 and more.  What are the differences between the $15 and the $115 filter?    I'm asking in the generic sense but if you need specifics, I'm looking to buy a UV filter and a polarizing filter for a Panasonic GF1 w/ 20mm pancake lens.  EDIT: I've read some of the other posts and some refer to multi-coating but doesn't explain what is this and why it matters.  What I've gathered from reading related posts is that price is generally the only indication of quality in the case of choosing a filter.  Is this right?


Spivak, Calculus, prologue, page 6 says:     Moreover, $1\ne 0$. (The assertion may seem a strange fact to list, but   we have to list it, because there is no way it could possibly be   proved on the basis of the other properties listed - these properties   would all hold if there were only one number, namely, $0$).   I did not understand what would go wrong without that assertion. $a*1 = 1*a = a$ couldn't hold true if $1 = 0$.

Consider the set $F$ consisting of the single element $I$.  Define addition and multiplication such that $I+I=I$ and $I \times I=I$ .  This ring satisfies the field axioms:   Closure under addition.  If $x, y \in F$, then $x = y = I$, so $x + y = I + I = I \in F$. Closure under multiplication.  $x \times y = I \times I = I \in F$ Existence of additive identity.  $\forall x \in F$ (i.e., for $x=I$), $x + I = x$, so $I$ is the additive identity. Existence of mulitiplicative identity. $\forall x \in F, x \times I = x$, so $I$ is the multiplicative identity. Additive inverse. $\forall x \in F, \exists y = I \in F: x + y = I$ Multiplicative inverse. $\forall x \in F, \exists y = I \in F: x \times y = I$.  However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth. Commutativity of addition.  $\forall x, y \in F, x + y = I = y + x$ Commutativity of multiplication.  $\forall x, y \in F, x \times y = I = y \times x$ Associativity of addition.  $\forall x, y, z \in F$, $(x + y) + z = I + I = I$ and $x + (y + z) = I + I = I$, so $(x + y) + z = x + (y + z)$ Associativity of multiplication.  $\forall x, y, z \in F$, $(x \times y) \times z = I \times I = I$ and $x \times (y \times z) = I \times I = I$, so $(x \times y) \times z = x \times (y \times z)$ Distributivity of multiplication over addition. $I \times (I + I) = I$ and $I \times I + I \times I = I$, so $\forall x,y,z \in F, x(y+z) = xy+xz$   Based on the above, $\{I\}$ seems to qualify as a field.  If $I$ is assumed to be a real number, then the unique solution of $I + I = I$ and $I \times I = I$ is, of course, $I = 0$.  So, is {0} a field, or is there generally considered to be an additional field axiom which would exclude it?  Specifically, is it required for the multiplicative identity to be distinct from the additive identity?


Please help!  The search / address bar in Safari normally autocompletes my typing. So for example, I can type em and it will autocomplete for me. It has randomly stopped functioning and is now forcing me to type a complete web address before it will take me to a site.  Secondly, I can normally type a word and press return and it will fire up google (my preferred search engine) with a list of suggestions but it has just stopped.  I don't remember changing anything - any suggestions its incredibly frustrating!  thanks in advance

All of a sudden, Safari on my Mac has started misbehaving. I can't select the text in the address bar, and so I can't go to any new websites. When it redraws, it goes all weird:    I've tried deleting the cache, clearing all cookies, disabling all plugins and rebooting. These things didn't help. I've had to switch to Chrome for the time being. Can anyone suggest some other possible remedies?


P(A) $\subset$ P(B) implies A $\subset $ B proof or disproof.  I have a strange feeling this is false but I do not know. Something to do with P(A) $\subset$ P(B) seems strange since P(B) is itself a powerself with P(A) being a subset.

Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement?


I want to automate multiple rsync commands like below:   Execute rsync1 Display output of rsync1 in the terminal Ask confirmation if want to continue rsync2 Execute rsync2   Can someone help?

Sometimes I need to ask the user for yes/no to confirm something.  Usually I use something like this:  # Yes/no dialog. The first argument is the message that the user will see. # If the user enters n/N, send exit 1. check_yes_no(){     while true; do         read -p "$1" yn         if [ "$yn" = "" ]; then             yn='Y'         fi         case "$yn" in             [Yy] )                 break;;             [Nn] )                 echo "Aborting..."                 exit 1;;             * )                 echo "Please answer y or n for yes or no.";;         esac     done; }   Is there a better way to do it? Is this utility maybe already in my /bin folder?


Hello everyone I'm building a license plate detection model in Tensorflow. I built a function that chooses a license plate at random from a collection of ~5000 plates and puts it in a random place in on a random background and saves the coordinates. At first I thought to generate about 40K images this way and train the network on with the generated data. But wouldn't it be a good idea to just continiously keep generating new data to feed to the network and basically eliminate any chance of it getting overfitted?

I'm training a convolutional neural network by adding small perturbations (like rotation and shifting) to the images each time it gets a batch data.  I think a better way of doing this could be to generate a bigger data set by oversampling with these transformations before training, however I don't have much RAM to store a bigger data set, so I decided to transform them on the run. But it means that now I'm supplying a different data set in each epoch.   I think it should make sense, and the training, in fact, converges to a better performance, but it watches a zig-zaggy road, which is understandable I guess. So am I safe to assume that I'm not making a mistake? Also is there a better way for this, i.e. training a generalizable classifier under a small data set and limited computational resources?  Thanks for any help !


Let's assume a rogue assassin level 5 with the 5thE Assassinate. When an enemy hasn’t taken a turn yet, you get advantage and, if they are surprised, any hit is critical.   I am wondering if you roll a natural 20 in this situation where any hit is considered critical already, will it be a "double critical" or give you any type of bonus for rolling 20? I am assuming no but wanted to check!

Is the 5e Rogue Assassin's Assassinate feature affected by rolling a natural 20?  My fellow players and DMs have different opinions.    The Rogue Assassin's Assassinate feature on page 97 of the PHB says:     You have advantage on attacks rolls against any creature that hasn't   taken a turn in the combat yet.  In addition, any hit you score   against a creature that is surprised is a critical hit.   During that attack roll on the surprised creature, what if you roll a natural 20, which counts as a critical hit in 5e?  Do these two instances of a critical hit stack?  If so, would it be double the amount of dice rolled, would it be tripled, or even worse quadrupled?   Maybe you can help end the debate amongst my group.


I recently bought a SanDisk 128GB USB flash drive.  And after formatting the USB flash drive in exFAT format, I copied a folder whose capacity is around 10GB. There are lots of small files in it, so it took some time.  However, when I see in the Windows Explorer after copying the folder, it says that around 43GB of the storage is occupied and now only 70GB of the storage is free to use.  What is happening and how should I deal with it? Is my USB flash drive physically broken?  It is still weird because when I copied a single file with 7 GB capacity, it showed the remaining capacity correctly at around 110 GB available..

I recently bought a 750 GB 2.5" hard drive and put it in an external enclosure.  It has , but I use Windows 7, so I assumed I would be safe.   I decided to try exFAT, because I was enticed by the reduced overhead For "Allocation unit size" I selected "Default allocation size" Tried with "Quick Format" un-selected first, but it was taking too long, so I stopped and started over with "Quick Format" selected After drive was formatted, looking at the format options again shows that the "Allocation unit size" is 256 KB   The drive works fine, but I had a question about the difference between "Size" and "Size on disk."  On my 120 GB NTFS formatted internal hard drive, the difference is only 0.16%.  However, on my much larger, exFAT formatted external drive, the difference is 17.38%.  At that rate, the most I will fit on this 750 GB drive is 637 GB (with 0 free space).  What format options should I use to maximize usable disk space?  What impact will my current settings have on performance?  What impact will your recommended settings have on performance?  Format settings    Internal hard drive    External hard drive (with big space difference!)


I noted a discussion on groups being abelian under a certain restriction on powers of elements, e.g. . Maybe this result (probably not too well-known) concludes it all.  Let $m$ and $n$ be coprime natural numbers. Assume that $G$ is a group such that $m$-th powers commute and $n$-th powers commute (that is for all $g, h$ $\in$ $G$: $g^mh^m=h^mg^m$ and $g^nh^n=h^ng^n$). Then $G$ is abelian.

Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that $$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$  Then how prove $G$ is an abelian group?  Some context: Some of these commutation relations often imply that $G$ is abelian, for example , or . This looks like another example of this phenomenon, but the same techniques do not apply.


I know that Elastic-net regularization is the combination of L1 and L2 regularization.  My question is what are the limitations of Elastic-net regularization?    My question is not related to Elastic-net regularization compared to LASSO or Ridge. Just, I would like to know the possible flaws of Elastic-net regularization.

What are the disadvantages of using elastic net in comparison to lasso.  I know that the elastic net is able to select groups of variables when they are highly correlated.    It doesn't have the problem of selecting more than $n$ predictors when $p \gg  n$. Whereas lasso saturates when $p \gg n$. When there are highly correlated predictors lasso tends to just pick one predictor out of the group.  When $n \gg p$ and the predictors are correlated, the prediction performance of lasso is smaller than that of ridge.    All these disadvantages of lasso are overcome by the elasic net.   What I don't understand is when should the lasso be then used? Is there any reason to use it when the elastic net performs better than lasso? What are the disadvantages of using elastic net in some cases? In which cases would the lasso be a better choice?


The Celsius unit is arbitrarily defined, based on the boiling and freezing point of water. Is it a coincidence, then, that the SI unit of temperature Kelvin, which is used in all natural equations, has the same length as the Celsius unit?

How on earth is it possible that the difference between two temperatures in Celsius and Kelvin is exactly the same. Given the historical definition of Celsius, I find it hard to believe that this is pure coincidence.


In Dune, the Atreides family disregarded any evidence of Dr. Yueh being a traitor since he had been certified by Imperial Suk conditioning. This conditioning apparently ensured loyalty to the Emperor and/or their assigned master, although no details are given regarding how it works. The novel repeatedly mentions that Suk conditioning is unbreakable, with Yueh's inner monologue hinting at some sort of mental triggers or hypnotism.  However, as we see in the novel, all it takes for the Harkonnens to break Yueh is to      torture his wife in a pain amplifier.   Given the seeming ease with which this "conditioning" was broken, why is it considered such an infallible stamp of trust?  Note: I am not trying to find the specifics of how the conditioning is done. What I'm trying to ask is why such conditioning is so highly regarded. If torturing a loved one is all it takes to break an initiate of the program - and that seems to be the case for Yueh - it would be useless.

I have read the first five Dune novels and after the first I do not recall any further mention of Imperial conditioning. In the novels written since Frank Herbert's death, were any further references or details given about what this process actually involved and why it was considered virtually impossible to remove?


It says that I've accepted 99% of the answers and for the life of me, I can't find the question to which I've not accepted the answer. I've gone to my profile and gone through the list of my questions with no luck.   How to find questions where I've not accepted an answer?

I have asked many questions on SO. There are about 30 tabs I have to scroll through to find question for which I have not yet accepted an answer.  Some questions of mine are long-term projects. They may require years to be answered satisfactorily.  In contrast, some questions have shown bugs in Vim and OS X, such that I cannot solve them alone. An example is my question   In short, please add a feature to SO which allows me see my questions without accepted answers effectively.


I want to prove at any group of order $4$ is isomorphic to either $\Bbb Z_4$ or $\Bbb Z^*_8$. I know that these two groups are not isomorphic to each other because they have different order, but I cannot go further to prove at any group of order 4 should be isomorphic to either of them. Help would be greatly appreciated.  Thanks!

I've figured out that if I know $G$ is not cyclic, then it for any $a \in G, o(a) \neq 4$ (or the order of any element in group $G$ is not 4).  I know ahead of time that the elements in the group ($\forall x \in G$) must have order $o(x)=k$ where $0 < k \leq 4$ where $k \in \mathbb{Z}$, so $k = 1, 2, 3, 4$.  If $G$ not cyclic, then we know $k \neq 4$ so we have $k = 1, 2, 3$ left.  I know how to show that $k \neq 3$ but I am not sure if I am supposed to do it for all possible $k$ because obviously if our order was higher (let's say $n$) then it would get messy. It ends up being in this case that $k = 2$ is good and $k = 1$ is trivial. I have looked up online and it says using Lagrange's Thm we know it has to be $k = 1, 2$ since $3 \nmid 4$, but I cannot use that theorem as we have not have learned it in class. How can I show that $k = 1$ or $2$ in another way?  However working from using the result that if we have $x \in G$ and $H = \langle a \rangle$ and knowing that $\left\vert{H}\right\vert=1$ or $2$, then $x^2=1$ for either case. Then I am done right?  What is another, cleaner, better way of trying to answer the question in the title?  I apologize if my formatting is poor since I am new at LaTeX and extraordinarily bad at algebra it seems. Thank you in advance.


Can I use ChaCha with four rounds as a good non-cryptographic PRNG with different streams if I use 64 bit integers instead of the standard 32 bit integers? I need a bigger state and this seems the easiest way to do it.

The fastest known attack against the  requires 2137 simple operations against Salsa20/7, or 2244 against Salsa20/8. The 8-round version encrypts data at 1.88 cycles-per-byte on a Core2Duo, which is already extremely fast. As there's no real key schedule, cutting this down to two rounds should bring the speed up to 0.47 cycles-per-byte.    is based on the Salsa core, but improves diffusion per round (quite significantly, as an attack taking 232 operations on Salsa20/6 instead takes 2127.5 operations on ChaCha6). The speed on the same CPU is equivalent. This is as fast as  (0.48 cycles-per-byte on Kaby Lake). Even worse for XorShift128+, the lowest bit is an LFSR, which may not be ideal for certain use cases such as Monte Carlo simulations.  This leads me to wonder, in situations where cryptographic security is not needed, could two-round ChaCha be used in place of another, fast PRNG? Would the output be indistinguishable from random data, sans cryptographic attacks?


I'm supposed to show that the set of real numbers, $R$, is an incomplete metric space with metric $d(x,y)=|\text{tan}^{-1}(x)-\text{tan}^{-1}(y)|$.  My issue: What we usually do in such problems is that we find a cauchy sequence which doesn't converge in same metric space. Now since all of the terms of the sequence $(x_n)_{n\geq1}$ are supposed to come from real numbers, shouldn't the limit itself be in real numbers too? Also, we were taught a theorem in class "Every Cauchy sequence of real numbers converges" which means that the limit will exist too. Then why would it be incomplete? Thanks!

I have to show that the real numbers equipped with the metric $ d (x,y) = | \arctan(x) - \arctan(y)| $ is an incomplete metric space.  Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.  Can anybody help me with this.   Thanks for helping me.


I'm running a command through a program that needs to create an OpenVPN account. After putting in this line to start making the account:  cd /etc/openvpn/easy-rsa; . ./vars; ./build-key username   I need to have it put in no input 10 times [like pressing enter,] and then press y twice. It needs to all be in this one line. Is this possible? Thanks in advance.  EDIT:  root@suffice-vpn:~# cd /etc/openvpn/easy-rsa; . ./vars; <<<$'\n\n\n\n\n\n\n\n\n\nyy' ./build-key username NOTE: If you run ./clean-all, I will be doing a rm -rf on /etc/openvpn/easy-rsa/keys Generating a 2048 bit RSA private key ..............+++ .....................+++ writing new private key to 'username.key' ----- You are about to be asked to enter information that will be incorporated into your certificate request. What you are about to enter is what is called a Distinguished Name or a DN. There are quite a few fields but you can leave some blank For some fields there will be a default value, If you enter '.', the field will be left blank. ----- Country Name (2 letter code) [US]:State or Province Name (full name) [NY]:Locality Name (eg, city) [Merrick]:Organization Name (eg, company) [IceWare]:Organizational Unit Name (eg, section) [IceWare]:Common Name (eg, your name or your server's hostname) [username]:Name [server]:Email Address [iceware@programmer.net]: Please enter the following 'extra' attributes to be sent with your certificate request A challenge password []:An optional company name []:Using configuration from /etc/openvpn/easy-rsa/openssl-1.0.0.cnf Check that the request matches the signature Signature ok The Subject's Distinguished Name is as follows countryName           :PRINTABLE:'US' stateOrProvinceName   :PRINTABLE:'NY' localityName          :PRINTABLE:'Merrick' organizationName      :PRINTABLE:'IceWare' organizationalUnitName:PRINTABLE:'IceWare' commonName            :PRINTABLE:'username' name                  :PRINTABLE:'server' emailAddress          :IA5STRING:'iceware@programmer.net' Certificate is to be certified until Jan 14 16:59:03 2026 GMT (3650 days) Sign the certificate? [y/n]:CERTIFICATE WILL NOT BE CERTIFIED: I/O error root@suffice-vpn:/etc/openvpn/easy-rsa#   As you can see, it goes through the process filling in the blanks [keeping all that information that I want to remain the same] but then I get this I/O error.

Heres the thread I'm bouncing off of:   So I'm trying to run this command: cd /etc/openvpn/easy-rsa; . ./vars; ./build-key username  However, after running that command, I need to input nothing [just press enter] 10 times, and press y twice. The user on the last thread suggested I use except, but I couldn't get it working. Here are the questions it asks:  root@suffice-vpn:/etc/openvpn/easy-rsa# ./build-key usernamtest Generating a 2048 bit RSA private key ..................+++ ........+++ writing new private key to 'usernamtest.key' ----- You are about to be asked to enter information that will be incorporated into your certificate request. What you are about to enter is what is called a Distinguished Name or a DN. There are quite a few fields but you can leave some blank For some fields there will be a default value, If you enter '.', the field will be left blank. ----- Country Name (2 letter code) [US]: State or Province Name (full name) [NY]: Locality Name (eg, city) [Merrick]: Organization Name (eg, company) [IceWare]: Organizational Unit Name (eg, section) [IceWare]: Common Name (eg, your name or your server's hostname) [usernamtest]: Name [server]: Email Address [iceware@programmer.net]:  Please enter the following 'extra' attributes to be sent with your certificate request A challenge password []: An optional company name []: Using configuration from /etc/openvpn/easy-rsa/openssl-1.0.0.cnf Check that the request matches the signature Signature ok The Subject's Distinguished Name is as follows countryName           :PRINTABLE:'US' stateOrProvinceName   :PRINTABLE:'NY' localityName          :PRINTABLE:'Merrick' organizationName      :PRINTABLE:'IceWare' organizationalUnitName:PRINTABLE:'IceWare' commonName            :PRINTABLE:'usernamtest' name                  :PRINTABLE:'server' emailAddress          :IA5STRING:'iceware@programmer.net' Certificate is to be certified until Jan 14 18:27:48 2026 GMT (3650 days) Sign the certificate? [y/n]:y   1 out of 1 certificate requests certified, commit? [y/n]y   As you can see, I press enter 8 times initially [to keep the default input] and then press enter another 2 times to skip the optional commands. Then, I need to press y twice. How can I automate this? I need to have this all happen in one command. Any suggestions? Thanks in advance.


When I build my server image, I write the following to a temporary file;  * * * * * php /var/www/artisan schedule:run   and then load it with  crontab < /tmp/cron   After deployment of the image, I run cron -f and I can see that the process is indeed running, but the cronjob is not being initiated.  If I run crontab -e, add a space, save it, run crontab -e again and remove the space, the cronjob working just fine without reload of cron.  I've tried reloading the cron on build with /etc/init.d/cron reload, but this does not solve the issue.

This is a  about using cron & crontab.   You have been directed here because the community is fairly sure that the answer to your question can be found below. If your question is not answered below then the answers will help you gather information that will help the community help you. This information should be edited into your original question.  The answer for 'Why is my crontab not working, and how can I troubleshoot it?' can be seen below.  This addresses the cron system with the crontab highlighted.


As an avid reviewer (but also internet user at large), I typically have a tab open for a site's review queue, but it's mildly annoying to have to constantly go over to that tab and constantly F5 to check if there are any new review items.  Ideally, I'd like to have the /review page auto-refresh counts, and display the new values live as they come in. For example, a new Suggested Edit is available, the number for Suggested Edits should instantly jump up.  In the same vein, if I'm not on the Review tab, a little notification counter, like on Chat, should be present, counting up for each new review item added since I've last visited the tab:    Again, like chat, this number should dismiss when I visit the tab and check on what's going on. Perhaps, if at all possible, a desktop notification could also come up on a new Review item, provided I've enabled that feature.  While this isn't a big (or really, important) change, it would still be a nice quality-of-life improvement and let frequent reviewers respond more quickly to things happening in the queues.

I would like to see the review page refresh automatically if I have done some reviews and get the message that there are currently no more reviews to perform.  Refreshing the whole page is, in my opinion, a little too much overhead. The review itself seems to work with AJAX. So, a small refresh button to trigger would be nice too, if the real-time updates are too complex.


I am having trouble with converting .htaccess files to nginx. I have 3 .htaccess files. The first .htaccess file is located in the document root and is as follows:  Options +FollowSymLinks RewriteEngine On RewriteRule ^img-(.*)\.html img.php?id=$1 [L] RewriteRule ^slide-(.*)\.html slider.php?id=$1 [L] RewriteRule ^page-(.*)\.html page.php?name=$1 [L] RewriteRule ^contact\.html$ contact.php [QSA,L,NC]   The second .htaccess file is located in a folder called upload:  RewriteEngine On RewriteCond %{HTTP_REFERER} !^http://(.+\.)?foo\.com/ [NC] RewriteCond %{HTTP_REFERER} !^$ RewriteRule .*\.(jpe?g|gif|bmp|png)$ nohotlink.gif [L] <Files ~ "\.(php|sql|php3|php4|phtml|pl|py|jsp|asp|htm|shtml|sh|cgi)$">   order allow,deny   deny from all </Files>   And the third and final .htaccess file is found in a subdirectory of the upload folder named "small":  RewriteEngine Off   Now I have created a folder in /etc/nginx called includes and made 3 seperate .access files with these rewrite rules:  For the first .htaccess file located in the document root directory I have created an file in /etc/nginx/includes called root.access. In this file I have:  # nginx configuration   location /img {  rewrite ^/img-(.*)\.html /img.php?id=$1 break;  }  location /slide {      rewrite ^/slide-(.*)\.html /slider.php?id=$1 break;  }  location /page {      rewrite ^/page-(.*)\.html /page.php?name=$1 break;  }  location /contact {      rewrite ^/contact\.html$ /contact.php break;  }   For the second file located in the "upload" folder I have crated a file in /etc/nginx/includes called upload.access. This file contains the following:  # nginx configuration   location /upload {  if ($http_referer !~ "^http://(.+\.)?foo\.com/"){  rewrite .*\.(jpe?g|gif|bmp|png)$ /nohotlink.gif break;  }  }  location ~ \.(php|sql|php3|php4|phtml|pl|py|jsp|asp|htm|shtml|sh|cgi)$ {  deny all;  }   And for the third and final file I have created a file in /etc/nginx/includes names small.access. The contents of this file are as follows:  # nginx configuration   location /upload/small {  }  In the server block configuration file I have:   location / {               try_files $uri $uri/ /index.php;               include /etc/nginx/includes/root.access;      }   location /upload {               include /etc/nginx/includes/upload.access;      }   location /upload/small {               include /etc/nginx/includes/small.access;      }   Now with this configuration when trying the site I receive 403 errors. Nginx error log reports:  [error] 18156#0: *7 access forbidden by rule, client: 111.**.**.**, server: foo.com, request: "POST /upload.php HTTP/1.1", host: "foo.com", referrer: "http://foo.com/"   Now under apache everything works no problems. But I can not see why I am getting 403 errors. I am also worried that the rewrite rules as I have stated them, beyond the 403 error will not function properly at all. Can anyone help me with this? I can not see what is wrong here.


I'm trying to install offline g++ software on Ubuntu 14.04:  $ sudo dpkg -i g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb  [sudo] password for ecil:  dpkg: error processing archive g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb (--install):  cannot access archive: No such file or directory Errors were encountered while processing:  g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb ecil@ECIL-HP:~$ sudo dpkg -r g++-4.4 (Reading database ... 164836 files and directories currently installed.) Removing g++-4.4 (4.4.3-4ubuntu5.1) ... Processing triggers for man-db (2.6.7.1-1) ... ecil@ECIL-HP:~$ sudo dpkg -i g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb  ecil@ECIL-HP:~$ sudo dpkg -i g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb  dpkg: error processing archive g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb (--install):  cannot access archive: No such file or directory Errors were encountered while processing:  g++-4.4.4_4.4.3-4ubuntu5.1_i386.deb ecil@ECIL-HP:~$ sudo dpkg -r g++-4.4 dpkg: warning: ignoring request to remove g++-4.4 which isn't installed ecil@ECIL-HP:~$ sudo dpkg -r g++-4.4 dpkg: warning: ignoring request to remove g++-4.4 which isn't installed ecil@ECIL-HP:~$ sudo dpkg -r g++-4.4 dpkg: warning: ignoring request to remove g++-4.4 which isn't installed ecil@ECIL-HP:~$ cd Desktop/ ecil@ECIL-HP:~/Desktop$ sudo dpkg -i g++-3.3_3.3.5-13_i386.deb  Selecting previously unselected package g++-3.3. (Reading database ... 164820 files and directories currently installed.) Preparing to unpack g++-3.3_3.3.5-13_i386.deb ... Unpacking g++-3.3 (1:3.3.5-13) ... dpkg: dependency problems prevent configuration of g++-3.3:  g++-3.3 depends on gcc-3.3-base (>= 1:3.3.5-13); however:   Package gcc-3.3-base is not installed.  g++-3.3 depends on gcc-3.3 (>= 1:3.3.5-13); however:   Package gcc-3.3 is not installed.  g++-3.3 depends on gcc-3.3 (<< 1:3.3.6); however:   Package gcc-3.3 is not installed.  g++-3.3 depends on libstdc++5-3.3-dev (>= 1:3.3.5-13); however:   Package libstdc++5-3.3-dev is not installed.  dpkg: error processing package g++-3.3 (--install):  dependency problems - leaving unconfigured Processing triggers for man-db (2.6.7.1-1) ... Errors were encountered while processing:  g++-3.3

I have a friend who has got a computer that is not connected to the Internet. Is there any way to install software offline easily?


Trying to install Ubuntu Server 20.04.01 - the Ethernet driver is not detected. The motherboard does not have onboard Wi-Fi. I have a USB Wi-Fi dongle available but not sure how to mount etc. and if its possible to use. Is there a way to get this machine going, either via the Wi-Fi dongle or potentially installing the onboard Ethernet driver via a USB drive after downloading from another machine? (Realtek RTL8125)

So I did a fresh install of Ubuntu 20.04 on a computer with the new RTL8125B Realtek network card and I cannot get the network to work. I was very suprised since this is supposed to be the new feature of the 5.4 kernel (I have 5.4.0-26-generic installed). Does anyone have any idea on how to get it working? Thank you. $ lspci -knn | grep Eth -A3 06:00.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8125 2.5GbE Controller [10ec:8125] (rev 04) Subsystem: ASUSTeK Computer Inc. RTL8125 2.5GbE Controller [1043:87d7] Kernel modules: r8169 07:00.0 VGA compatible controller [0300]: Advanced Micro Devices, Inc. [AMD/ATI] Ellesmere [Radeon RX 470/480/570/570X/580/580X/590] [1002:67df] (rev e7)  $ dmesg | grep -i r8169 [    0.911236] r8169 0000:06:00.0: unknown chip XID 641


Consider the following code:   0.1 + 0.2 == 0.3  ->  false   0.1 + 0.2         ->  0.30000000000000004   Why do these inaccuracies happen?


I have an ellipse that is defined by center, width and height. The axes of the ellipse parallel to the x and y. I want to find the largest rectangle that completely fits inside this ellipse. Is there an easy way to do this?  And sorry if my terminology is a bit off... it's been a long time since I last tackled with geometry

The actual problem reads:      Find the area of the largest rectangle that can be inscribed in the ellipse    $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$   I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to find the derivative I don't think I'm doing it right.


How do we find the chisq value for a two tailed test? For left tail we use qchisq(x,df,lower.tail="True") For right tail we use qchisq(x,df,lower.tail="False") What about two tailed?

A sporting goods manufacturer claims that the variance of string tensions for any decent tennis racquet should be about 9 pounds. The string tensions of 18 randomly selected tennis racquets produced a variance of 8.13 pounds. Find the p-value to test the manufacturer's claim (assume the population is normally distributed).  I tried attempting the question as follows: ${H_o}: {\sigma ^2} = 9$ ${H_1}: {\sigma ^2} \ne 9$ We are also given the following: $n=18$ $s^2 = 8.13$ Test Statistics: $\chi^2 = {{(n-1)S^2}\over{\sigma^2}}$ Calculating Test Statistics under $H_o$ gives us: $\chi^2 = {{(18-1)8.13}\over 9} = 15.35667$  For calculating the p-value I proceeded as follows: $p-value = 2\min\{P[\chi^2>15.35667],\ P[\chi^2<15.35667]\}\ =2\times 0.430192 = 0.860384$  I am not certain if my method for calculating the p-value is correct or not. Can someone please tell if the above process is valid?  Thanks in advance!


The  starts with a default scan range of 192.168.1.1 to 192.168.1.254 (presumably this changes dependant what C level IP the gateway is set to.   Is there a technical reason the scanning range isnt set to 192.168.1.0 to 192.168.1.254 (note the 0 not 1 at the beginning and 255 not 254 at the end) ?  Is the x.x.x.0 and x.x.x.255 address not commonly used for some reason ?

It was specified somewhere (presumably in an RFC) that Class A addresses range from 1-126 in the first octet, and that class E addresses range from 240-254, but that leaves out two potential networks on either side, namely a 0 and a 255 network.  I believe that addresses starting with these IPs are invalid. Does this mean that they wouldn't work on a private network, and is there a reason for them being invalid?


I have just installed Ubuntu 16.04 LTS and the installation went smoothly but I can't seem to connect to WiFi, the button to select networks is greyed and it shows WiFi Network disconnected. I have looked through the other threads and none seem to be working...  My laptop uses Realtek RTL8723BE 802.11 bgn Wi-Fi Adapter

I have installed Ubuntu 15.04 on my laptop with RTL8723BE Wi-fi card. But it is always disconnecting from network. I have tried echo "options rtl8723be fwlps=N ips=N" | sudo tee /etc/modprobe.d/rtl8723be.conf but that didn't helped. What can I do to prevent the wifi from disconnecting? If I upgrade to kernel 4.X, will it help?  ~$ lspci -knn | grep Net -A2  09:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8723BE PCIe Wireless Network Adapter [10ec:b723]     Subsystem: Hewlett-Packard Company Device [103c:2231]     Kernel driver in use: rtl8723be   ~$ rfkill list 0: hci0: Bluetooth     Soft blocked: yes     Hard blocked: no 1: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no   ~$ ifconfig && iwconfig && route -n && ping -c 1 google.com eth0      Link encap:Ethernet  HWaddr 38:63:bb:cd:4a:7e             UP BROADCAST MULTICAST  MTU:1500  Metric:1           RX packets:3 errors:0 dropped:0 overruns:0 frame:0           TX packets:53 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000            RX bytes:752 (752.0 B)  TX bytes:8445 (8.4 KB)  lo        Link encap:Local Loopback             inet addr:127.0.0.1  Mask:255.0.0.0           inet6 addr: ::1/128 Scope:Host           UP LOOPBACK RUNNING  MTU:65536  Metric:1           RX packets:3870 errors:0 dropped:0 overruns:0 frame:0           TX packets:3870 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0            RX bytes:314613 (314.6 KB)  TX bytes:314613 (314.6 KB)  wlan0     Link encap:Ethernet  HWaddr c0:38:96:6d:c4:83             inet addr:192.168.1.205  Bcast:192.168.1.255  Mask:255.255.255.0           inet6 addr: fe80::c238:96ff:fe6d:c483/64 Scope:Link           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:15240 errors:0 dropped:0 overruns:0 frame:0           TX packets:14627 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000            RX bytes:14410285 (14.4 MB)  TX bytes:2192744 (2.1 MB)  eth0      no wireless extensions.  wlan0     IEEE 802.11bgn  ESSID:"DIR-615"             Mode:Managed  Frequency:2.437 GHz  Access Point: 00:90:4C:08:00:0D              Bit Rate=150 Mb/s   Tx-Power=20 dBm              Retry short limit:7   RTS thr=2347 B   Fragment thr:off           Power Management:off           Link Quality=70/70  Signal level=-22 dBm             Rx invalid nwid:0  Rx invalid crypt:0  Rx invalid frag:0           Tx excessive retries:0  Invalid misc:186   Missed beacon:0  lo        no wireless extensions.  Kernel IP routing table Destination     Gateway         Genmask         Flags Metric Ref    Use Iface 0.0.0.0         192.168.1.1     0.0.0.0         UG    400    0        0 wlan0 169.254.0.0     0.0.0.0         255.255.0.0     U     1000   0        0 wlan0 192.168.1.0     0.0.0.0         255.255.255.0   U     0      0        0 wlan0 PING google.com.Dlink (172.26.136.19) 56(84) bytes of data. 64 bytes from 19.136.26.172.in-addr.arpa (172.26.136.19): icmp_seq=1 ttl=249 time=102 ms  --- google.com.Dlink ping statistics --- 1 packets transmitted, 1 received, 0% packet loss, time 0ms rtt min/avg/max/mdev = 102.547/102.547/102.547/0.000 ms


I recently installed ncurses and from then on I have been getting similar errors while trying to install new libraries  You might want to run 'apt-get -f install' to correct these: The following packages have unmet dependencies:  lib64ncurses5-dev:i386 : Depends: lib64c-dev:i386   After I run apt-get -f install I hit another wall  The following NEW packages will be installed:   libc6-dev-amd64:i386 0 upgraded, 1 newly installed, 0 to remove and 548 not upgraded. 6 not fully installed or removed. Need to get 0 B/1,429 kB of archives. After this operation, 9,590 kB of additional disk space will be used. Do you want to continue? [Y/n] y (Reading database ... 451168 files and directories currently installed.) Preparing to unpack .../libc6-dev-amd64_2.19-0ubuntu6.6_i386.deb ... Unpacking libc6-dev-amd64 (2.19-0ubuntu6.6) ... dpkg: error processing archive /var/cache/apt/archives/libc6-dev-amd64_2.19-0ubuntu6.6_i386.deb (--unpack):  trying to overwrite '/usr/include/gnu', which is also in package libc6-dev-i386 2.19-0ubuntu6.6 Errors were encountered while processing:  /var/cache/apt/archives/libc6-dev-amd64_2.19-0ubuntu6.6_i386.deb E: Sub-process /usr/bin/dpkg returned an error code (1)

Occasionally, when I'm installing stuff, I get an error like the following:  Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  package1 : Depends: package2 (>= 1.8) but 1.7.5-1ubuntu1 is to be installed E: Unable to correct problems, you have held broken packages.   How can I resolve this?


I have 700 Unique values “Communities” and each one of this 700 unique communities has a unique reference. For example the community “Abc” has a reference “A001”, Now I have two fields in my attribute table the first one is CommunityName and the second is REF_Number this two fields should be filled. The question now is there any method that allow me fill the field “REF_Number” based on my section form the domain of the communityName field to limit the errors and to avoid filling the data two times. For example: If I select the community Abc form the domain list then the value of the field REF_Number should be updated automatically to A001.

Is there a way to create attribute subdomains in ArcGIS with clauses?  For example, if I'm editing a line feature class of pipes, I create a domain for the pipe type to be either plastic or metal. I want to setup another domain field "Rusted" with a yes/no value. When editing the data in ArcMap, if the pipe is metal, I should get a yes/no coded value option on the "Rusted" field. If it is plastic, I want the "Rusted" field to calculate N/A or perhaps give me other options.


This is the error!:  Overfull \hbox (12.10773pt too wide) in paragraph at lines 13--19  I don´t understand it!  It has apareces when i try to compilate with the next document:  \documentclass[11pt]{book} \usepackage[utf8]{inputenc} \usepackage[spanish,mexico]{babel} \usepackage[T1]{fontenc} %\usepackage{amsmath} %\usepackage{amsfonts} %\usepackage{amssymb} %\usepackage{graphicx} \usepackage{apacite}  \title{Manual de estrategias psicopedagógicas en el manejo de grupo desde la perspectiva sociocultural} \author{Itzel Viridiana Gutiérrez Ruiz\\Mayra Yurieth Hernández Robledo} \date{} \begin{document} \frontmatter \maketitle %\tableofcontents %\mainmatter \include{introduccion} \backmatter \bibliographystyle{apacite} \bibliography{bibliografia} \end{document}

I often see output from TeX with the warning overfull hbox, badness 10000. What does this message mean?


Given any triangle in the plane, associate to each side an outward pointing normal vector of the same length as the side. Show that the sum of these three vectors is always 0.

I'm able to prove that, for a triangle $PQR$,  $$ PQ+ QR+ RP = 0. $$ I am not sure how to show that for  $A \perp PQ$, $B\perp QR$, and $C \perp RP$, where $A$ has the same length as $PQ$, $B$ has the same length as $QR$, and $C$ has the same length as $RP$, we have  $$A+B+C=0.$$  I've made a sketch of the triangle and perpendicular vectors, but I'm not sure how to proceed.


I've been using texstudio for a while on ubuntu 16.04, and one day the toolbar graphics became very big and very ugly  They used to be smaller. I'm not sure but I think it has somtheing to do with a package I installed "libqt5scripttools".   Now I can't figure out how to fix this problem.  Any kind of help is appreciated. Thanks.

Using TeXstudio, the icons and menus are almost invisible on my 4k display (3840 x 2160 pixels, 28").  Is it possible to adjust the graphics so that TeXstudio could be usable on my 4k display?


I was in a room and one of the characters there whose name was in italics, didn't like the discussion another was engaged in. The person whose name was in italics, then wrote some incorrect information (probably trolling saying something blatantly wrong) (I'll include it here since he appeared here, introduced himself, and provided the transcript) "THERE IS NO REFERENCES OR POINTERS IN JAVA GET OVER IT" Vogel612 (the person whose name was in italics) (in the Java chat room) And he tried to imply that any discussion of references in java was off topic, because java didn't have references.  (a rather absurd claim). The fact that he was wrong (references do exist in java) was pointed out to him,  at which point he suggested that he had some power, as indicated by his name being in italics, and that he shouldn't be messed with, and he declared that the topic should not be discussed further. I went to see who owns the room, but I saw no list. Though I see clicking on names of people whose names are in italics, lists that individual as an owner of the room. So i'm curious. Do these owners have the power to ban somebody from a room? If so, then for how long? Do they have the power to mute somebody? Interestingly the person whose name was in italics, then started telling other people whose names were also in italics that the discussion was finished. Is it possible that one person whose name is in italics has more power than another. And is it possible to see what the ranking is? vogel has brought up the chat transcript itself, link here

Chat is an offshoot of the main site, and using it is  during their time here.  Chat is our place for:        Real-time collaboration   Meeting fellow members of the community   Less structured, casual (but still roughly on-topic) conversation      Our chat system exists to serve the main site; users of each site can gather and commune with each other.  When in chat, we expect all users to abide by our  policy, which means that we expect community members to treat each other with respect.    While it is a place where the community can go to interact on a more casual level, chat still needs to be moderated by the community and there are tools in place to help with the moderation. This raises a few questions, that are addressed here:


Under a for loop , i am constructing div this way , adding dynamic content whereever applicable .

Popcorn

                 <form>                     <input type="checkbox" class="checkboxclas" name="checkbox-mini-0" id="checkbox-mini-0" data-mini="true">                     Small Pack 150g                 </form>

                                          12

                     Rs: 150/-

   Actual code   for (var k = 0; k < responseinner[i].type.length; k++)  { activateUiHTML.append('

' + obj.name + '

<form><input type="checkbox" class="checkboxclas" name="checkbox-mini-0" id="' + random_number + '" data-mini="true">' + responseinner[i].type[k] + '</form>

1

Rs: ' + responseinner[i].price[k] + '/-

'); }   When i registered with the checkbox event  this way , i was getting multiple values on click of checkbox . (I mean the text present for other checkboxes also present in for loop )  $(document).on("click", ".checkboxclas", function (e) { if($(this).is(':checked')) {       var labelAttr = $('label').text(); alert(labelAttr); } }   How to return only that particular text of checkbox ??

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


I am using Ubuntu 14.04 LTS. I want to upgarde to Ubuntu 16.04 via the software updater application. I have lots of data in my laptop. Also I am having useful softwares related with my work.  Now I wish to upgrade my system without losing any data and software. Ca n I do it?. or else what is the procedure to upgrade Ubuntu without losing old data.

Question is about Ubuntu release process. Are migrations from n to n+1 releases expected?  Assuming we have server with installed Ubuntu 14.04 LTS. Will it be smoothly upgradable to Ubuntu 16.04 LTS via apt-get upgrade ?  Assumptions:   scope is software therefore we assume that given hardware is supported on both releases (upgrading from and to) that's why, always before upgrade you are asked to make backup to be able to revert in case of failure


The problem: Running any game in full screen causes the screen to turn on and off. It goes completely black for 1-2 seconds, then turns back on. This will keep happening at uneven intervals. Sometimes it will stay on for a minute or two, sometimes it will stay on for just a few seconds. Everything else continues to work while it's off - sound work, mouse and keyboard input works, etc. While the screen is on there are no visible issues - fps is steady, no video artifacts, etc.   The weird part: This only happens if I leave the computer running overnight. A restart fixes it every time and I've never experienced the issue within the same day that I've restarted or turned it on. There are also no issues with anything that's not full screen, exiting the game stops the behavior. It also doesn't happen if playing in windowed mode.  I've checked the event log and I don't see anything being logged when this happens. This is a desktop running windows 10 with a GeForce GTX 1070 in it that's connected via display port to a single monitor. I don't have any sort of sleep settings set up where the monitor, hard drives, or any peripherals turn off after X minutes, at least everywhere I've checked.  What are some troubleshooting steps I could take to try and narrow this down?

Currently, when I boot into Windows, one monitor stays black (undetected), and the other can only display 800x600 resolution.  When I look at Device Manager > Display Adapters > NVIDIA GeForce GTX 970 > General > Device Status, it says "Windows has stopped this device because it has reported problems. (Code 43)".  Details > Problem Code says "00002b". Details > Status says:  01802400 DN_HAS_PROBLEM DN_DISABLEABLE DN_NT_ENUMERATOR DN_NT_DRIVER   When I then click into the Events tab, it says "Device PCI\VEN_10DE&DEV_13C2&SUBSYS_29763842&REV_A1\4&25438c51&0&0008 requires further installation."  I have a EVGA Nvidia GeForce GTX 970 graphics card with the latest drivers (376.19 from 12/1/2016).  My Windows 10 Pro has the latest updates and is at Version 1607, OS Build 14393.576.  My ASRock Z170 Pro4 motherboard BIOS is also updated to the latest version (7.00 from 10/4/2016).  I have dual Acer monitors with 1920x1080 resolution.  I downloaded Ubuntu 16.04.1 LTS onto a bootable USB drive and booted into the Ubuntu operating system instead of Windows. Both monitors immediately worked at full 1920x1080 resolution. So the hardware seems functional! My remaining challenge is to figure out why my video card has stopped working within Windows.  What I Have Tried   I've tried using Display Driver Uninstaller from Guru3d in Safe Mode to delete all drivers and try to install fresh.   I've tried many different versions of the driver dating back more than 12 months. I've tried reseating the card. I've tried moving the card to a different slot. EVGA's phone support told me to reinstall windows (Windows > Reset this PC > Keep my files). Unfortunately, I followed their advice (and now don't have any of my programs), and I still get Code 43 with the EVGA Nvidia card.   What I Have NOT Tried   I would love to try installing this card onto a different Windows 10 computer to see what happens, but I don't have access to any other computers. If you live northeast of Atlanta, let me know. ;-)   I'd appreciate any other ideas you have!.


This came up during a game this weekend where the rogue player claimed that, since she was taking damage from it and she could see where the "attack" was coming from, she could reduce that damage with this feature.  I allowed it at the time, for the pace of the game. The Rogue feature Uncanny Dodge states:  Starting at 5th level, when an attacker that you can see hits you with an attack, you can use your reaction to halve the attack's damage against you.  In my mind, this excludes the Magic Missile spell, since nowhere in the spell description does it state the word "attack".  You create three glowing darts of magical force. Each dart hits a creature of your choice that you can see within range.  In the future, I would have to say "no", based on my reading of the ability and the spell. Looking for confirmation/counter-arguments.

I was looking over the class feature of the 'fiend' patron warlock in 5e. I'm a bit confused on what actions the player needs to take before they can use the 'Hurl Through Hell' ability on an enemy. The PHB states:     Starting at 14th level, when you hit a creature with an attack, you can use this feature to instantly transport the target through the lower planes.   I'm not clear on what an 'attack' means in this context. Does it have to be a physical attack like a sword or crossbow? Do magical attacks like Eldritch Blast that require an attack roll count? What about spells with no attack roll?


His  and  give some details on his housing situation and how he helped out a couple surviving Jedi, but is there any story or series of stories that better explain the hardships of his exile? Like how he kept sheltered after his lifeboat fell apart and then eventually built the house we see if the films, or how he figured out which plants/creatures were safe to eat and which were poisonous. Or even how he specifically dealt with the inner-conflict (and possible guilt) of the downfall of the Jedi Order. What about after Luke left to fight Vader? Did business simply return to usual for him?  There is  and subsequent answers which I first considered mine to be a dupe of, but they all seem silly and appear to not fully address the question in favor of nailing the perfect Yoda-ese. Plus it's over a year old and something new might have come out since then.  I'd prefer anything that explains it from Yoda's POV.  Please link any shops that sell whatever (if any) materials on this subject, as this is something that I would actually like to read about first hand.

Leader of the Jedi Council, Yoda was. For centuries, a Jedi.   To Dagobah he went. For twenty years, lived there did he. Bored, he might be.   Do what, did he?


How to move the polygons of an object into other location by bmesh frame by frame? I tried below way but failed. Is it possible to generate polygon movement animation effects like in Animation Nodes but by python? bpy.ops.mesh.primitive_monkey_add(size=2, location = (0,0,0)) obj = bpy.context.object obj.name='myobj' me = obj.data  ico0 = np.array([v.co for v in me.vertices]) # get ico vertices  #find out Zvalue zmin, zmax = ico0[:,2].min(), ico0[:,2].max() zc = np.linspace(zmin, zmax, n_frames) #print(zmin, zmax,zc) #return ico = np.zeros_like(ico0)

I can only move the vertices of the monkey to (2,4,6) but not by polygons and effects look strange. What is the right way to calculate the polygons by the vertices according to their Z value? bpy.ops.mesh.primitive_monkey_add(size=2, location = (0,0,0)) obj = bpy.context.object obj.name='myobj' me = obj.data  ico0 = np.array([v.co for v in me.vertices]) # get ico vertices  #find out Zvalue zmin, zmax = ico0[:,2].min(), ico0[:,2].max() zc = np.linspace(zmin, zmax, n_frames)  icop0 = np.array([p.index for p in me.polygons]) # get ico polygons zpmin, zpmax = icop0[0], icop0[-1] zpc = np.linspace(zpmin, zpmax, n_frames) ico = np.zeros_like(ico0) data = [] for i in range(n_frames):     ico=ico0[np.where(ico0[:,2]<=zc[i])]+[(2,4,6)]     data.append(ico.copy()) ## ------- part 2  ---set up frame change event handler, def my_handler(scene):     i_frame = scene.frame_current     if not (0 <= i_frame < len(data)):         return      obj = bpy.data.objects['myobj'] # be explicit.     me = obj.data     for (vert, co) in zip(me.vertices, data[i_frame]):         vert.co = co     me.update() #clear the post frame handler bpy.app.handlers.frame_change_pre.clear() bpy.app.handlers.frame_change_post.clear()  bpy.app.handlers.frame_change_pre.append(my_handler)


If a planet emerges from multiple colliding pieces, that causes its material heat up.  Later, an existing planet is continuously heated by radioactive decay, tidal forces and other effects.  But are planets like the Earth hot because they are   still hot from that time and did not cool, or heated later with only minor initial heat, or even both in significant fractions?

How does the Earth's internal heat exist? Is it available from its creation as collision heat? Or is there any other source feeding the heat to the core?