A recent  about &quot;sure&quot; put a question in my head: Wait, why is &quot;sure&quot; pronounced the way it is? The initial consonant is a voiceless postalveolar fricative [ʃ] as opposed to a voiceless alveolar fricative [s]. Another common word example is &quot;sugar&quot;. I have always considered the phonetic oddity of &quot;sugar&quot; easily explainable. It originally came from Sanskrit by way of Arabic and the &quot;sh&quot; sound as a holdover indicating its provenance makes sense. Sugar originated in the Indian subcontinent and for much of European history it had to be imported into Europe by Middle Eastern merchants. Etymology Online:  sugar (n.) late 13c., sugre, from Old French sucre &quot;sugar&quot; (12c.), from Medieval Latin succarum, from Arabic sukkar, from Persian shakar, from Sanskrit sharkara &quot;ground or candied sugar,&quot; originally &quot;grit, gravel&quot; (cognate with Greek kroke &quot;pebble&quot;).  That Etymology Online page has one vague line on its pronunciation change that doesn't tell us much besides unsubstantiated speculation:  The pronunciation shift from s- to sh- is probably from the initial long vowel sound syu- (as in sure).  But what about &quot;sure&quot;? It doesn't seem to have come from either Arabic or Indian languages  sure (adj.) early 13c., &quot;safe against attack, secure,&quot; later &quot;firm, reliable&quot; (c. 1300); &quot;mentally certain, confident&quot; (mid-14c.); &quot;firm, strong, resolute&quot; (c. 1400), from Old French seur, sur &quot;safe, secure; undoubted, dependable, trustworthy&quot; (12c.), from Latin securus &quot;free from care, untroubled, heedless, safe&quot;).  Another word with a similar pronunciation pattern I can think of is &quot;sumac&quot;, but in addition to [ʃ] it can also be pronounced with [s]. But we don't pronounce &quot;super&quot; with [ʃ], nor &quot;suicide&quot;, nor &quot;sue&quot;.
As far as I know, those are the only two.  They should be pronounced Soogher and Soor, shouldn't they?  I looked them up on Dictionary.com, and their etymologies reveal no trace of an SH, except where the listing for sugar had:     Middle English sugre, sucre (noun) &lt; Middle French sucre  &lt; Medieval Latin succārum  &lt; Italian zucchero  &lt; Arabic sukkar;  obscurely akin to Persian shakar, Greek sákcharon   I see an obscure kinship to shakar, but the word morphed so many times since then that the SH disappeared pretty much completely.  And sure is even worse, with no sign of SH:     Middle English sur ( e ) &lt; Middle French sur, Old French seur  &lt; Latin sēcūrus   Why are these the only two like this?
LEFT JOIN vs. LEFT OUTER JOIN in SQL Server What is the difference between LEFT JOIN and LEFT OUTER JOIN? Existence in ZF of a set with countable power set How do I sort a vector of pairs based on the second element of the pair? If I have a vector of pairs: std::vector&lt;std::pair&lt;int, int&gt; &gt; vec;  Is there and easy way to sort the list in increasing order based on the second element of the pair? I know I can write a little function object that will do the work, but is there a way to use existing parts of the STL and std::less to do the work directly? EDIT:  I understand that I can write a separate function or class to pass to the third argument to sort.  The question is whether or not I can build it out of standard stuff.  I'd really something that looks like: std::sort(vec.begin(), vec.end(), std::something_magic&lt;int, int, std::less&gt;()); I have a sight hosted on .ie domain but it is near the NI(Northern Ireland) border and the business has a lot of NI traffic. So when someone searches in NI with certain key words I would like the site to appear in the results, it currently doesn't.   I have seen this post    But was wondering the best way around this. Would it be to create a .co.uk version that redirects to the .ie site? Or would I have to create a separate .co.uk site?  Any help appreciated thanks. I had recently faced a problem:     Solve the Diophantine Equation $x^2 - y! = 2001$.   Solving it was quite easy. You show how $\forall y \ge 6$, $9|y!$ and since $3$ divides the RHS, it must divide the LHS and if $3|x^2 \implies 9|x^2$ and so the LHS is divisible by $9$ and the RHS is not. Contradiction. Hence, the only solution is $(45, 4)$.  That made me wonder, how we can solve the Diophantine Equation $x^2 - y! = 2016$. We cannot apply the same logic here. $2016$ is a multiple of $9$ and it is clear that $3|x$ and $9 \nmid x$. How do I proceed from here? Map not loading when using getLegend and showing error for getView() I was able to run my application on using Legend earlier but now I want to update my Legend when scrolling on the map. So I tried to used this code  . And I got the gist of how he is able to do that but when I am trying the same I keeping getting the error var graphicUrl = wmsSource.getLegendUrl(currentResolution);  I think this is mainly due to the wmsSource not declare. But I tried to initialize  seems wmsSource in TileWMS but does not seems to work. var getExtent = function () {     return map.getView().calculateExtent(); } var getProjectionCode = function () {     return map.getView().getProjection().getCode(); }  var updateLegend = function() {     var currentResolution = map.getView().getResolution();     var graphicUrl = wmsSource.getLegendUrl(currentResolution);        const crs = `&amp;CRS=${getProjectionCode()}`;     const bbox = `&amp;BBOX=${getExtent().join(',')}`;     const height = getHeight(getExtent());     const width = getWidth(getExtent());     const heightAndWidth = `&amp;srcwidth=${height}&amp;srcheight=${width}` ;     const legenOptions = &quot;&amp;legend_options=countMatched:true;fontAntiAliasing:true;hideEmptyRules:true;forceLabels:on&quot; ;        var img = document.getElementById('legend');     img.src = `${graphicUrl}${crs}${bbox}${heightAndWidth}${legenOptions}`; }; //Updated Legend End   //View var myview = new ol.View({     center : [8717490.201868,2788422.791843],     zoom : 4 })  var firstSel = document.getElementById(&quot;City&quot;); var secSel = document.getElementById(&quot;Sector&quot;);  //var LayerArray = [baseLayer, firstLayer]  //Map var map = new ol.Map({       target : 'mymap',       view: myview,       layers: [         new ol.layer.Tile({             source : new ol.source.OSM({                 attributions: 'Surveyor Allpication'             })         }),         new ol.layer.Tile({             source : new ol.source.TileWMS({                url: 'http://localhost:8089/geoserver/India/wms',                params: {'LAYERS': 'India:India', 'TILED': true,'TRANSPARENT':true},                ratio: 1,                serverType: 'geoserver',                transition: 0,             })         })       ] }) updateLegend();   This here is my JS &lt;div class=&quot;innerMap&quot;&gt;     &lt;div class=&quot;map&quot; id=&quot;mymap&quot;&gt;&lt;/div&gt;         Legend:     &lt;div&gt;&lt;img id=&quot;legend&quot;/&gt;&lt;/div&gt;     &lt;!--    &lt;div&gt;&lt;img src=&quot;http://localhost:8089/geoserver/wms?REQUEST=GetLegendGraphic&amp;VERSION=1.0.0&amp;FORMAT=image/png&amp;WIDTH=20&amp;HEIGHT=20&amp;LAYER=India%3AIndia&amp;RULE=Agra&quot; /&gt;%Name&lt;/div&gt; --&gt; &lt;!--    &lt;div&gt;&lt;img src=&quot;http://localhost:8089/geoserver/India/wms?SERVICE=WMS&amp;VERSION=1.1.1&amp;REQUEST=GetLegendGraphic&amp;FORMAT=image/png&amp;WIDTH=20&amp;HEIGHT=20&amp;LAYER=India%3AIndia&amp;SRS=EPSG%3A4326&amp;STYLES=&amp;SRCWIDTH=742&amp;SRCHEIGHT=768&amp;BBOX=75.4541015625%2C14.21630859375%2C91.7578125%2C31.09130859375&amp;LEGEND_OPTIONS=countMatched:true;fontAntiAliasing:true;hideEmptyRules:true&quot; /&gt; &lt;/div&gt;  --&gt;     &lt;/div&gt;  This is my HTML. How do I make this work? Image uploads from the editor strip preceding line breaks Recently, using the image uploader in the post editor has starting stripping preceding line breaks; placing the image directly after any previous text instead of at the current cursor position. This means that the inserted image is placed inline as part of the previous line of text rather than in a new paragraph as it did previously...  Before inserting the image (note the cursor position):        After inserting the image:        This also happens after the Markdown for a horizontal rule, which destroys the rule and gives us a bunch of hyphens followed by an image instead:        Seems the behaviour is a bit inconsistent... Sometimes I get the behaviour shown above and sometimes it leaves a single newline, I'm not sure when or why but single line breaks are ignored in Markdown so the rendered result is the same regardless.  Obviously it's useful to be able to have inline images, but I'm pretty certain most images aren't intended to (and shouldn't) be. Multivariable limit - perhaps a trickier problem I am stuck on. I am trying to solve the following limit:  $\lim_{(x,y) \to (0,0)} \frac{x^4y^4}{(x^2 + y^4)^3}$  (This is a more challenging problem from Folland Calculus, it seems).  I am pretty sure this limit does not exist (however, this is just a guess, and I am not 100% sure.)  I am trying to approach via the paths y = mx and x = my but... they don't seem to work. I have tried simpler cases such as x = 0 and y = 0.   Any help appreciate. Change Folder Permissions Reproject GeoJSON from EPSG:3857 to EPSG:4326 I am using this  to convert a GeoJSON point from EPSG:3857 to EPSG:4326.   My command fails. What am I doing wrong?   $ echo '{"type":"Point","coordinates":[-13720479.997764934,5688254.733325758]}' | reproject --from=EPSG:3857 --to=EPSG:4326   /usr/local/lib/node_modules/reproject/cli.js:98             throw new Error("Could not find definition for CRS \"" + crsName + "\".");                   ^ Error: Could not find definition for CRS "EPSG:3857".     at lookupCrs (/usr/local/lib/node_modules/reproject/cli.js:98:10)     at Object.&lt;anonymous&gt; (/usr/local/lib/node_modules/reproject/cli.js:25:1)     at Module._compile (module.js:456:26)     at Object.Module._extensions..js (module.js:474:10)     at Module.load (module.js:356:32)     at Function.Module._load (module.js:312:12)     at Function.Module.runMain (module.js:497:10)     at startup (node.js:119:16)     at node.js:901:3 How to convert Strings to and from UTF8 byte arrays in Java In Java, I have a String and I want to encode it as a byte array (in UTF8, or some other encoding). Alternately, I have a byte array (in some known encoding) and I want to convert it into a Java String. How do I do these conversions? iptables drop all incoming ICMP requests except from one IP Currently, I have something like:  iptables -A INPUT -p ICMP --icmp-type 8 -j DROP iptables -A INPUT -s x.x.x.x -p ICMP --icmp-type 8 -j ACCEPT   However, when I run the second command, it looks as if iptables just stops.  I have to break out of it to get back to terminal.  Perhaps I am doing it all wrong, but some insight would be helpful. How to use a decimal range() step value? Is there a way to step between 0 and 1 by 0.1?   I thought I could do it like the following, but it failed:  for i in range(0, 1, 0.1):     print i   Instead, it says that the step argument cannot be zero, which I did not expect. How to typeset the symbol “^” (caret/circumflex/hat) I need to display the symbol '^'  d &lt;- dist(fascores, method = "euclidean")^2   How do I do that? How to understand the combination formula? I'm reading an algorithm book. In which it mentioned:     Imagine (once again) you have n people lined up to see a movie, but   there are only k places left in  the theater. How many possible   subsets of size k could possibly get in? That’s exactly $C(n,k)$, of   course,  and the metaphor may do some work for us here. We already   know that we have $n!$ possible orderings of  the entire line. What if   we just count all these possibilities and let in the first $k$? The only   problem then is  that we’ve counted the subsets too many times. A   certain group of k friends could stand at the head of  the line in a   lot of the permutations; in fact, we could allow these friends to   stand in any of their $k!$  possible permutations, and the remainder of   the line could stand in any of their $(n–k)!$ possible  permutations   without affecting who’s getting in. Aaaand this gives us the answer!       $$C(n,k)= \frac{n!}{k!(n −k)!}$$      This formula just counts all possible permutations of the line (n!),   and divides by the number of  times we count each “winning subset,” as   explained.   I'm not quite understand why the reminder can in any $(n-k)!$ permutation, if he stands at the end of the queue, won't he let all these people in?  And why I should use $n!$ to divide by $k!(n-k)!$ ? Didn't see that consequence from the metaphor.  If you have better ideas to explain this formula, you can use them of course. No cycle containing edges $e$ and $g$ implies there is a vertex $u$ so that every path sharing one end with $e$ and another with $g$ contains $u$ There is a proof in my textbook for the following claim, which doesn't make a whole lot of sense to me.  My annotations are in bold.  Could someone perhaps elaborate on what's going on?  Claim.  If there does not exist a cycle containing edges $e$ and $g$ then there exists a vertex $u \in V (G)$ such that every path in $G$ sharing one end with e and another with $g$ contains $u$.  Proof: The claim trivially holds if $e$ or $g$ is a loop OK, so we assume that neither is. Let $P$ with vertex set $v_1, v_2, . . . , v_k,$ in order, be a path with $e$ joining $v_1$ to $v_2$ and $g$ joining $v_{k−1}$ and $v_k$. Let $f_i \in E(P_i)$ be the edge with ends $v_i$ and $v_{i+1}$. Let $j$ be chosen minimum so that no cycle in $G$ contains $e$ and $f_j$ We can do this because we know that at least the edge $g$ will not create a cycle by assumption, right?. We will show that $u = v_j$ satisfies the claim.  Suppose not. Let $C$ be a cycle containing $e$ and $f_{j−1}$ What if $f_j = e$?  then what cycle? a single vertex? and let $P′$ be a path from an end of $e$ to an end of $f$ Not sure what the book meant by $f$ here, any guesses? avoiding $u$. Choose a subpath $Q$ of $P′$ with one end in $V (C)$ and another in ${v_{j+1}, v_{j+2}, . . . , v_k}$ as short as possible. Then $C \cup Q \cup P$ contains a cycle containing both $e$ and $f_j$, a contradiction. (The last statement requires some case checking.) This ending seems a bit abrupt and non-obvious to me  Thanks for the help Is there a way to export my artwork as PDF so that it has a transparent background? Something like exporting objects only... whenever I try to save my .ai as pdf, it always comes with a full white background. If I try to save as eps, I get a chunky white background (where it has white rectangles behind the artwork only where the artwork exists, so there are transparent chunks).  If I export as png, I get an image like what I'm looking for. Is there any way to do that in a vector format? Geometric meaning of the determinant of a matrix What is the geometric meaning of the determinant of a matrix? I know that "The determinant of a matrix represents the area of ​​a rectangle." Perhaps this phrase is imprecise, but I would like to know something more, please.  Thank you very much. What would be the symbol for 'multiplication on division'? The symbol I would like to obtain ought to be like the symbol '\pm'. What antivirus programs are available for Ubuntu?  We previously used Symantec Endpoint Protection but it does not work in Ubuntu.