I have an enumerated list, for which I'm using enumitem because I want to resume the enumeration, and one item is rendered in a faded color using xcolor  \usepakage{enumitem, xcolor} % ... \begin{enumerate} \item First. \end{enumerate} other stuff \begin{enumerate}[resume] \item Second. \item \textcolor{gray}{Third.} \item Fourth. \end{enumerate}   I'd like to have the label of "Third." rendered in the same faded color, and because I've seen  I defined  newcommand{\fadeditem}{\item[\stepcounter{enumi}\textcolor{gray}{\arabic{enumi}.}]}   and it works (it's not general, but it works) but now it's not enough... I realize that what I want is a command that takes care also of the item text  What I have now  \fadeditem \textcolor{gray}{Bla bla bla.}   What I would like  \itemcolor{gray} Bla bla bla.   Addendum #1  David Carlisle, in a comment, suggested   \newcommand\itemcolor[1]{\fadeditem{#1}\textcolor{#1}}    and this deals with my task of today, writing optional questions in a list of questions.  An open part of the question is then a generalization of the problem, defining an \itemcolor{somecolor} command that just works in any (in most?) lists...
I am trying to define an alternative \item command that would allow me to use a different color for some of the items. My code has the following form:  \documentclass{article} \usepackage{color}  \definecolor{gray}{rgb}{0.7,0.7,0.7} \newcommand{\grayitem}[1]{{\color{gray} \item #1}}  \begin{document}  \begin{enumerate} \item First item. \grayitem Second item. \item Third item. \end{enumerate}  \end{document}   I was hoping that  \grayitem Second item.   would produce the same result as  {\color{gray} \item Second item.}   but for some reason, these are not equivalent. They produce the following results, respectively:  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;   Why are these two lines of code not equivalent? What am I doing wrong and how can I change the definition of \grayitem to produce the second result?  I have also tried removing a pair of braces, like this  \newcommand{\grayitem}[1]{\color{gray} \item #1}   but then the third item becomes gray as well.  The solution should also work for multi-paragraph items.
How do I bypass restrictions on the length of the PATH variable System Properties      This environment variable is too large. This dialog allows setting values up to 2047 characters long.   I use/have used a lot of software, and my PATH has grown as a result. When I try to add new paths, I get the above message. As a workaround, I'm editing my user path, but prefer not to. Deep in my families history (1600's or there abouts) our family split in two. About three or four generations later two distant cousins got married creating a split / rejoin.                G.G.G.Parents A        +---------+---------------------------+     G.G.Grand Parent                      G.G.Grand Parent        |                                     |     G.Grand Parent                        G.Grand Parent        |                                     |     Grand Parent                          Grand Parent        |                                     | +------+-------+                     +-------+-------+ Aunt       Parent A - - -+- - - Parent B         Uncle                          |                   +------+------+                Child A        Child B                  ...           ...   So in this small example, the family splits in two at G.G.Grandparents and then rejoins later when Parent A and Parent B marry.  About 8 years ago I tried to find a software package which could model this and failed. Has anything happened in the intervening time?  Sorry I'm just asking in SE without doing much research, but I spent a lot of effort last time without any success and I am sure there is someone out there who can point me to an excellent package!    @ColeValleyGirl: I'm looking for a piece of software that is able to print out such a tree/DAG (thanks @ACProctor) in high enough quality to display/frame. All software I tried before didn't seem to be able to display this as they seemed to support mainly variations of Ancestors / Descendants rather than an output of all the people that had been inputed.  @AdrianB38: Essentially, I want to see "everyone". The tree I have was put together by my grandfather, who probably had a lot more data/people, but I've only got the hand written tree which highlights this split. It's essentially a tree of my grandparents ancestors (including this split which is joined at a common ancestor) and descendants (me + my family). I'd like to reproduce this in the computer (there is about 400 years - hundreds of people).  @AdrianB38: Re your point about generations, yes, I have that issue. I think it's only one generation, might be two - i've not looked at the tree in a while. The split between families happened over about 150-200 years iirc.  I can write an SVG generator if I really need to (I'm a Software Engineer) albeit I'm aware such a "loop"/"dependency" is painful to program for, but I was hoping that there was something which could handle this. In the Wikipedia article on positive definite matrices they claim that if $M$ is positive definite, then the determinant of $M$ is bounded by the product of its diagonal entries. How might we show this?  Ideas: The sum of the diagonal entries of $M$ is the trace, which is also the sum of the eigenvalues. The product of the eigenvalues is the determinant. All the diagonal entries of $A$ must be real and strictly positive. Also we have $|m_{ij}|\leq \sqrt{m_{ii}m_{jj}}\leq \frac{m_{ii}+m_{jj}}{2}$. So far nothing is working...  We might try to write the Gram matrix $m_{ij} = \langle v_i, v_j \rangle$, and then $\prod_i m_{ii} = \prod_i \|v_i\|^2$. Now we'd need to show that the determinant, which is a polynomial in the entries of $v_i$, is less than $\prod_i \|v_i\|^2$ somehow. How to do this manually, not installing any tools? Blown out blue/red light making photos look out of focus Ubuntu MATE freeze on logout/shutdown/reboot when running on Intel graphics card Filter questions with no accepted answer I have some questions with answers, but none of the answers are acceptable (just wrong).  I'm sure there are a lot more people with the same problem.  Can you add a button to filter questions with answers but no accepted answer?  This way, people will see those questions and maybe try to offer a new perspective. How is consensus determined on Meta sites? On our main sites &quot;the best answer&quot; is  as the superposition of the highest voted answer and the accepted answer.  Truth be told, the system is rather insistent on allowing multiple &quot;right answers&quot; and that's one of its strengths: you can be the judge of who you believe. But that doesn't always translate well on the per-site metas, which serve as a sort of policy-creation body.  People who are active on meta help decide how policies are enforced on the main site.  Moderators who have ♦-powers tend to take mandates on meta very seriously.  , accepted answers to meta-questions are no different than any other answer.  Does that leave just votes to determine consensus? Closely related: Not everyone participates on meta-sites.  It can be hard to tell if enough people have looked and voted on questions for real consensus to be declared.  Size of the site determines how many people could be voting on questions.  What counts as a quorum? Finally, sometimes people change their minds about a meta-question, but, , it's not really possible to change one's mind without a pointless edit to someone else's answer.  When should meta-questions be revisited?  On graduated sites, ♦ moderators are elected, which makes them representatives of the active participants, but on beta sites, we are appointed.  It's doubly pressing for beta sites to determine &quot;the will of the people&quot; since part of the point of a beta is to get a working policy framework in place before electing permanent moderators. I'm interested in figuring out how sites already answer these questions and what might be done network-wide to make discovering consensus a little bit easier.  N.B.: Obviously this assumes that consensus matters.  It does to me when I'm making my decisions and I suspect it matters to others.  But that isn't to say that it always makes a difference.  I'm really not interested in answers that hope to use this question as a grindstone for all the axes stored up against certain people who have made unpopular decisions and have the power to make those decisions stick.  Feel free to ask your own question(s).  (Thank you.) better Curve/Text to Mesh generation? The triangulation process of Text to Mesh or Curve to Mesh generates some nasty topology.   I get why, but maybe it could be done a bit nicer? Since text is flat, have ngons instead of hundreds of triangles you'll need to modify anyway?  Another issue is when you then want to use the text as a Volume for particles, which for some reason behave weird when you have unusual topology even though they should be using the volume, not vertex or face positions. How do I change a close vote to a duplicate? Voted to close a question based on "lack of minimal understanding".  Later realized the question was a duplicate.  How to I change vote to mark as duplicate instead? Is there a realistic way of implementing a multi-threaded model in PHP whether truly, or just simulating it. Some time back it was suggested that you could force the operating system to load another instance of the PHP executable and handle other simultaneous processes.  The problem with this is that when the PHP code finished executing the PHP instance remains in memory because there is no way to kill it from within PHP.  So if you are simulating several threads you can imagine whats going to happen.  So I am still looking for a way multi-threading can be done or simulated effectively from within PHP. Any ideas? What books, journals, and/or electronic resources have you found most valuable for expanding your knowledge in the GIS field, and why? Eclipse freeze accessing project -> properties -> android tab Dot Product Intuition I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ).      For example, suppose I have vector $\mathbf{a} = (a_1,a_2)$ and vector $\mathbf{b}=(b_1,b_2)$.  What's the physical or geometrical meaning that   $$a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos(\theta)\;?$$  Why is multiplying $|\mathbf{b}|$ times $|\mathbf{a}|$ in direction of $\mathbf{b}$ the same as multiplying the first and second components of $\mathbf{a}$ and $\mathbf{b}$ and summing ?       I know this relationship comes out when we use the law of cosines to prove, but even then i cant get a intuition in this relationship.  This image clarifies my doubt:    Thanks How to check if Node connected to another node input in Python I am doing a process to some switch links in materials. I need to know if something is ever linked or not before to process. I can't find out how to :/ The script works well if something is linked only.  import bpy for m in bpy.context.object.material_slots:     mat = m.material     nodes = mat.node_tree.nodes      # get node input (socket)     alphasocket = mat.node_tree.nodes.get("Principled BSDF").inputs[18]     # get node link that connects into the socket     lnk = (link.from_socket for link in mat.node_tree.links if link.to_socket == alphasocket)      #add emission node     emi= nodes.new('ShaderNodeEmission')     emi.name= "emibak"     outepute= mat.node_tree.nodes.get("Material Output").inputs[0]      #make new link destination     #here 's where I cant find out how to check     if lnk != None:         print(mat.name + " alphanod found")        # mat.node_tree.links.new(lnk, emi.inputs[0])     else :         print(mat.name + " alphanod missing")      mat.node_tree.links.new(emi.outputs[0], outepute)     here is another way I am searching...  import bpy  for m in bpy.context.object.material_slots:      mat = m.material     nodes = mat.node_tree.nodes          print("**** Material " + mat.name+ " ****")      for mat_node in mat.node_tree.nodes:         if mat_node.type == 'BSDF_PRINCIPLED':             print("From " + mat_node.name )                         alphasocket = mat_node.inputs[18]             #print(alphasocket)             node_in = mat_node.inputs             #print(node_in)                    for n in alphasocket.links:                                print("Alpha connected to " + n.from_node.name)   This show me what is connected to the alpha channel but if nothing is connected I want to connect a new node. How to disconnect selected mesh geometry? I don't know how to disconnect a face from the rest of the mesh in edit mode.  This isn't like pressing P to separate the selection into a completely different object. I just want to select a face and disconnect/rip it away from the rest of the mesh but keep it as a part of the original object.  One way to do it would be to duplicate the face with ShiftD and then delete the original face. But I suspect there is a "native" command for disconnecting selected faces and I would like to learn it. For every twice differentiable function $f : \bf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statements are TRUE? For every twice differentiable function $f : \mathbf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statement(s) is (are) TRUE ?  (A) There exist $r, s\in \bf R$, where $r &lt; s$, such that $f$ is one-one on the open interval $(r, s)$  (B) There exists $x_0 \in (–4, 0)$ such that $|f'(x_0)| &lt; 1$   (C) $\lim\limits_{x\to\infty}f(x) = 1$   (D) There exists $\alpha\in(–4, 4)$ such that $f(\alpha) + f''(\alpha) = 0$ and $f '(\alpha)\ne0$  My attempt is as follows:-  $A)$ If function is continuous, then definitely in some part of the interval it will be increasing or decreasing hence will be one-one on the open interval $(r,s)$   $f(x)$ cannot be a constant function as in that case $f'(x)=0$ which means $f(0)=\sqrt{85}$, but it doesn't belong to co-domain of $f$  B) By mean value theorem we can say that there exists some $c\in(-4,0)$ for which $4f'(c)=f(0)-f(-4)$  Suppose if $f(0)=2$ and $f(-4)=-2$, then we can only say that there exists $c$ such that $f'(c)=1$. It will not be necessary that there exists a $c$ for which $|f'(c)|&lt;1$  C) $\lim\limits_{x\to\infty}f(x) = 1$, this is not at all necessary.  D) If function is not constant in $(-4,4)$, then we can safely say that at some point $\alpha$, $f(\alpha)\ne0$  But how to know $f(\alpha)+f''(\alpha)=0$  Actual answer is $A,B,D$ but I am able to ascertain only option $A)$. Please help me in this. I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power.  Here's the theorem (with proof) and two applications:    (Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.  Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 &gt; 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$.   Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 &gt; 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$    Here is one application (taken from ):  Claim: $[0,1]$ contains uncountably many elements.  Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in (0,1)} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable.     And here is another one (taken from ):  Claim: The linear space of all polynomials in one variable is not a Banach space in any norm.  Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem." Illogical GTK theme behaviour on Ubuntu 18.04 Strange behaviour of 'sudo chmod -rwx': Does not remove group write permissions