I find going to /node as being very useful to pinpoint exactly what content non-authorized users can see. But for two reasons I don't want anyone to access "/node".   I don't want that to end up in Google Search Results. I don't want prying eyes to go through each and every content in case something should accidentally not have been made public.   So my question is: what is the most correct way to block /node? Or should I keep it open and rather let robots.txt avoid that path?  UPDATE  To make matters worse, it seems I have set everywhere in my permissions that certain content types must not be viewable by certain roles. However, those pages are still showing up under /node ?
Drupal out of the box comes with a page (which is the default homepage) that is assigned to the node path.  How do I avoid users reach it, and redirect them to the actual homepage of the site?
iPhone SDK: what is the difference between loadView and viewDidLoad? When working with views and view controllers in an iPhone app, can anyone explain the difference between loadView and viewDidLoad?    My personal context, is that I build all my views from code, I do not and will not use Interface Builder, should that make any difference.  I've found that often when I add init code to loadView, I end up with an infinite stack trace, so I typically do all my child-view building in viewDidLoad...but it's really unclear to me when each gets executed, and what is the more appropriate place to put init code.  What would be perfect, is a simple diagram of the initialization calls.  Thanks! I've been working with ArcGis for a while now and was wondering if there was a way to change positions of parameters inside the toolbox for the user. The only time I have seen this done in arc toolbox is with the Raster clip tool with the Rectangle parameter where the 4 parameters are aligned in a circle in the toolbox.     I don't know if what I want is possible since I haven't seen anyone else do it. Confusion in symbol of subset. If every element of set A ia an element of set B then we say that A is subset of B. A is said to be proper subset if A does not equal B.  My question is - whether the symbol we use to show subset is  same as that for proper subset? I have seen that A is subset of A itself is shown by the same symbol as that for the subset while when we write improper subset, we usually write equal to sign with symbol of subset. So, please help me to know this. Procedural texture for longitudinal lines How to isolate the text from a bitmap logo? I have been given a logo from a client that they want me to 'extract' or 'cut' the text out of and use just the text with a transparent background to fit with the design ethos of their marketing.   I'm using Photoshop CS6 and have tried using channels, magic wand and eraser but I can't get clean edges on the text.  I'm pretty new to Photoshop and any help appreciated. Where can I find stock or custom ROMs for my Android device? There are several reasons to be looking for a ROM:   Upgrading or downgrading your stock ROM Looking for firmware from a different region/language Flashing a custom ROM for different features, removing bloat, etc.   But where can one find ROMs in the first place?  What sites are likely to have the ROM I need?  ; please add to the wiki answer instead of posting separate answers.  See also: I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.  In the proof of the theorem, they wrote:     $$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$  ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$  [...]   I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?  Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so? Prevent malicious bots from posting spam I remember  due to misuse and I wonder if bots have a part of it. If the bot is POSTing something to my site what are ways I can combat it? I was thinking of setting some cookies and having the cookies changed via JavaScript + timestamp and sign (so yesterday's cookies can't be used today and next week).  I'm sure most people/bots would just use another site instead of enabling JavaScript in their bot.  What else can I do? I'm thinking daily POST limit and a honeypot for generic bots who just randomly post spam. Moving all files to a new computer, my concern is links How long can you say "the late so and so"? When you refer to the deceased, you say "the late so and so." How long can you say that? Is JFK referred to as the late John F. Kennedy? How about Abraham Lincoln? Here is my candidate for  the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.  $***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$  It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x &gt; -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 &lt; x &lt; 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx &lt; 0$, $1+mx &lt; 0$ for all $m \ge n$ so certainly $(1+x)^m &gt; 1+mx$.)  This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$  (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.  Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.  We will prove that $a_n$ is an increasing sequence  and $b_n$ is an decreasing sequence.  Since $a_n &lt; b_n$, this implies,  for any positive integers $n$ and $m$ with $m &lt; n$  that $a_m &lt; a_n &lt; b_n &lt; b_m$.  For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.  Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI,  $((n+2)/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $(1+1/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $a_{n+1} &gt; a_n$.  For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.  Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI,  $(n/(n+1))^{n+1} &gt; (1-1/n)^n$  or $(1+1/n)^{n+1} &lt; (1+1/(n-1))^n$,  or $b_n &gt; b_{n+1}$.  Since $b_n-a_n  = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 &lt; 4$, $b_n-a_n &lt; 4/n$, so $b_n$ and $a_n$ converge to a common limit.  These proofs do not seem to be really elementary,  since they use the AGMI.  However, they use a special form of the AGMI,  where all but one of the values are the same,  and this will now be shown to be implied by BI,  and thus be truly elementary.   Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.  The AGMI for these values is  $(((n-1)u+v)/n)^n \ge u^{n-1}v$  with equality if and only if $u = v$.  We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$  is the same as  $(u+(v-u)/n)^n \ge u^n(v/u)$.  Dividing by $u^n$,  this is equivalent to  $(1+(v/u-1)/n)^n \ge v/u$.  By BI,  since $(v/u-1)/n &gt; -1/n &gt; -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$  with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI. Unfortunately MyApp has stopped. How can I solve this? Least upper bound not an element in the cut In his proof "Any bounded nonempty subset $\mathbb{R}$ has a least upper bound in $\mathbb{R}$", Enderton showed that the least upper bound is just $\bigcup A$. Since by definition $x\subseteq \bigcup A\quad \&amp; \quad x\in A$ . In the case $x=\bigcup A\quad \&amp; \quad x\in A$, $x$ becomes the largest element in $A$, but this cannot be true since the Dedekind Cut is a subset of $\mathbb{Q}$ that must have no largest element. How do you explain that? I have seen quoted several places on the internet a rule that one can apply bonuses and penalties in the most beneficial order.  An example I come across frequently (as a druid) is the combination of Natural Bond and . Natural Bond increases your effective druid level for determining the stats of animal companions but cannot increase it beyond your character level. Choosing an alternative animal companion lowers your effective druid level for determining the same stats. The "most beneficial order" rule is used to justify applying the level penalty before the natural bond bonus in order to get around the character level cap on natural bond.  Does this rule actually exist? If so, where does this rule originate? I would like to be able to cite a reliable source for my GM. Apply SP3 on SQL Server 2012 AlwaysOn AAGs bash function cannot change the value of a local variable when called by $() The following code works differently depending on which way Parse_Short_Switches() is called.  After calling the function 'Parse_Short_Switches' by using result=$(Parse_Short_Switches), the value of $error_flag is NOT set to -1 as I would expect.  After calling the function 'Parse_Short_Switches' by simply using Parse_Short_Switches, the the value of $error_flag is set to -1 as I would expect.  Any help would be greatly appreciated.  Thanks  #!/bin/bash  function Parse_Short_Switches() { echo "error_flag inside of the function= $error_flag" 1&gt;&amp;2 error_flag="-1" echo "blah blah ..." }  # --- MAIN --- error_flag="999" echo "error_flag= $error_flag" #result=$(Parse_Short_Switches) Parse_Short_Switches echo "error_flag= $error_flag" How can I calculate this limit: $\lim\limits_{x\to 0} x\left\lfloor\frac{1}{x}\right\rfloor$? Calculate the following limit. $$ \lim_{x\to 0} x\left\lfloor\frac{1}{x}\right\rfloor $$ Where $\left\lfloor x \right\rfloor$ represents greatest integer function or floor function, i.e greatest integer less than or equal to $x$.  Thanks. I got a flat/puncture on my bike tube, how can I fix it myself without going to the nearest bike shop? What kind of equipment would I need? How do the bonus perks from the Advanced Warfare Center work? Proper way to exit iPhone application?