Let $A = \{ 1,2,3,4 \}$ Let $F$ be a set of all functions from $A \to A$.  Let $S$ be a relation defined by : $\forall f,g \in F$ $fSg \iff f(i) = g(i)$ for some $i \in A$  Let $h: A \to A$ be the function $h(x) = 1 $ for all $x \in A$.   How many 1:1 functions $g \in F$ are there so that $gSh$ ?  Is it $4*3*2*1$ $($total $1:1$ functions)$ - 3*2*1 (1:1$ that don't connect to $1)$ the answer?
Let $A = \{ 1,2,3,4 \}$ Let $F$ be a set of all functions from $A \to A$.  Let $S$ be a relation defined by : $\forall f,g \in F$ $fSg \iff f(i) = g(i)$ for some $i \in A$  Let $h: A \to A$ be the function $h(x) = 1 $ for all $x \in A$.   How many functions $g \in F$ are there so that $gSh$ ?  My solution : $gSh$ means that $g(i) = h(i)$ for some $i \in A$. So $g(i) = 1$.  So number 1 always needs to connect to some x - 4 choices.  Then we have $3$ left-over numbers that can connect to $4$ numbers each.  So solution is $4 \cdot 4 \cdot 4 \cdot 4$. Is this correct at all? Thanks in advance !! :)
Following , this is how we correctly make the r sound:   Let the throat vibrate since it is voiced. Round the lip a little. Raise the tip of the tongue towards the hard bump behind the upper front teeth,  but do not let the tip of tongue touch the bump. This also makes the center of the tongue go down. The back part of the tongue is raised because you pull it into a tight ball Fold and raise the sides of the tongue so they touch the upper side-teeth.   If we follow the above requirement, our tongue should have an s shape when we pronounce the r sound as depicted in the following sketch:    Is it possible to raise the tip while raising the back and lowering the center of the tongue when making an r sound?   How do native speakers make an r?  P.S. My tongue is very short anyway! What happens in a circuit, when the wire and the battery are superconducting. And shorted I am new to QGIS. 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We simulate this by generating a sample from the posterior distribution (e.g., using a Metropolis-Hasting algorithm to generate values, and accept them if they are above a certain threshold of probability to belong to the posterior distribution). Once we have generated this sample, we use it to approximate the posterior distribution, and things like its mean.   But, I feel like I must be misunderstanding something. It sounds like we have a posterior distribution and then sample from it, and then use that sample as an approximation of the posterior distribution. But if we have the posterior distribution to begin with why do we need to sample from it to approximate it? Finding the nth non square free number We all know what  numbers are. I was more interested in checking an efficient way to find nth non square free number. How do we explain the behavior of \rput, \psGetNodeCenter and saveNodeCoors? Introduction  I am writing a tutorial on a short introduction to PSTricks for newbies. I got obstacles to explain how  a node A (for example), \rput, (!\psGetNodeCenter{A} A.x A.y) and (!N-A.x N-A.y) work by design.  I have read the pst-node documentation as shown below.      And the following is taken from page 35-36:    The manual says:   The transformation matrix will be reset by \psGetNodeCenter. (!N-A.x N-A.y) is not affected by \rput.   Even though I read it many times, I still get confused. The statements are too short.  Problem  Please consider the following figure for the remaining discussion.     Case 1  A node A is defined outside \rput and a dot is translated by \rput.   with (!\psGetNodeCenter{A} A.x A.y)  \pnode(2,2){A} \rput(1,1){\psdots(!\psGetNodeCenter{A} A.x A.y)}  with (!N-A.x N-A.y)  \pnode(2,2){A} \rput(1,1){\psdots(!N-A.x N-A.y)}    The results are different. \rput only affects (!N-A.x N-A.y). Case 2  A node A is defined inside \rput and a dot is translated by \rput.   with (!\psGetNodeCenter{A} A.x A.y)  \rput(1,1){\pnode(2,2){A}\psdots(!\psGetNodeCenter{A} A.x A.y)}  with (!N-A.x N-A.y)  \rput(1,1){\pnode(2,2){A}\psdots(!N-A.x N-A.y)}    The results are identical. \rput affects both. Case 3  A node A is defined inside \rput and a dot is defined outside \rput.   with (!\psGetNodeCenter{A} A.x A.y)   \rput(1,1){\pnode(2,2){A}}  \psdots(!\psGetNodeCenter{A} A.x A.y)  with (!N-A.x N-A.y)   \rput(1,1){\pnode(2,2){A}}  \psdots(!N-A.x N-A.y)    The results are different. \rput only affects (!\psGetNodeCenter{A} A.x A.y).   The MWE is given as follows.  \documentclass[margin=12pt]{standalone} \usepackage{pst-node} \addtopsstyle{gridstyle}{gridlabels=5pt} \psset{saveNodeCoors,linecolor=red} \everypsbox{\color{blue}} \begin{document}  \begin{psmatrix} \begin{pspicture}[showgrid=bottom](3,3)     \rput(1,1){Case 1-A}     \pnode(2,2){A}     \rput(1,1){\psdots(!\psGetNodeCenter{A} A.x A.y)} \end{pspicture} &amp; \begin{pspicture}[showgrid=bottom](3,3)     \rput(1,1){Case 1-B}     \pnode(2,2){A}     \rput(1,1){\psdots(!N-A.x N-A.y)} \end{pspicture} \\ % case 2 \begin{pspicture}[showgrid=bottom](3,3)     \rput(1,1){Case 2-A}     \rput(1,1){\pnode(2,2){A}\psdots(!\psGetNodeCenter{A} A.x A.y)} \end{pspicture} &amp; \begin{pspicture}[showgrid=bottom](3,3)     \rput(1,1){Case 2-B}     \rput(1,1){\pnode(2,2){A}\psdots(!N-A.x N-A.y)} \end{pspicture} \\ % case 3 \begin{pspicture}[showgrid=bottom](3,3)     \rput(1,1){Case 3-A}     \rput(1,1){\pnode(2,2){A}}     \psdots(!\psGetNodeCenter{A} A.x A.y) \end{pspicture} &amp; \begin{pspicture}[showgrid=bottom](3,3)     \rput(1,1){Case 3-B}     \rput(1,1){\pnode(2,2){A}}     \psdots(!N-A.x N-A.y) \end{pspicture} \end{psmatrix} \end{document}   Question  How do I have to explain their behaviors to the newbies as the statements in the documentation seem to be unclear?  Note: As (!\psGetNodeCenter{A} A.x A.y) is identical to (A), (A) will not be discussed here.     If you are a teacher, how do you explain it to your   students? Finding an Inverse Function and Composition of Functions? Under what conditions does loot despawn? I have found that when my bags get full and I need to vendor loot I can sometimes leave loot on the ground and return to it.  Other times it seems to disappear almost instantly once I leave the area.  Is the loot despawn time based, based on my distance from the dropped loot, or some other factor(s)? To make a JavaScript class with a public method I'd do something like:  function Restaurant() {}  Restaurant.prototype.buy_food = function(){    // something here }  Restaurant.prototype.use_restroom = function(){    // something here }   That way users of my class can:  var restaurant = new Restaurant(); restaurant.buy_food(); restaurant.use_restroom();   How do I create a private method that can be called by the buy_food and use_restroom methods but not externally by users of the class?  In other words, I want my method implementation to be able to do:  Restaurant.prototype.use_restroom = function() {    this.private_stuff(); }   But this shouldn't work:  var r = new Restaurant(); r.private_stuff();   How do I define private_stuff as a private method so both of these hold true?  I've read  a few times but it doesn't seem like "private" methods can be called by public methods and "privileged" methods can be called externally. Is Feyerabend confusing discovery and justification when he criticizes the scientific method? I am reading Feyerabend's "Against Method", where he uses Copernicus's (and Galileo's confirmation) discovery of the fact that the Earth orbits around the Sun and other examples to show that irrational approaches can lead to legitimate scientific progress. He then claims that no unified epistemic principle or principles can be used to define science, and that science is best characterized by anarchy.   Is he confusing the way scientific truth is discovered with the way scientific truth is epistemically justified?   Even if it is true that Copernicus, Galileo and others used irrational methods to arrive at valid scientific discoveries, why does that affect science's special epistemic status compared to other fields of inquiry? jQuery Validation plugin: disable validation for specified submit buttons I have a form with multiple fields that I'm validating (some with methods added for custom validation) with Jörn Zaeffere's excellent jQuery Validation plugin. How do you circumvent validation with specified submit controls (in other words, fire validation with some submit inputs, but do not fire validation with others)? This would be similar to ValidationGroups with standard ASP.NET validator controls.  My situation:  It's with ASP.NET WebForms, but you can ignore that if you wish. However, I am using the validation more as a "recommendation": in other words, when the form is submitted, validation fires but instead of a "required" message displaying, a "recommendation" shows that says something along the line of "you missed the following fields.... do you wish to proceed anyways?" At that point in the error container there's another submit button now visible that can be pressed which would ignore the validation and submit anyways. How to circumvent the forms .validate() for this button control and still post?  The Buy and Sell a House sample at  allows for this in order to hit the previous links, but it does so through creating custom methods and adding it to the validator. I would prefer to not have to create custom methods duplicating functionality already in the validation plugin.  The following is a shortened version of the immediately applicable script that I've got right now:  var container = $("#&lt;%= Form.ClientID %&gt; div.validationSuggestion");  $('#&lt;%= Form.ClientID %&gt;').validate({               errorContainer: container,     errorLabelContainer: $("ul",container),     rules: {         &lt;%= YesNo.UniqueID %&gt;: { required: true },         &lt;%= ShortText.UniqueID %&gt;: { required: true } // etc.      },     messages: {         &lt;%= YesNo.UniqueID %&gt;: 'A message.',         &lt;%= ShortText.UniqueID %&gt;: 'Another message.' // etc.     },     highlight: function(element, errorClass) {         $(element).addClass(errorClass);         $(element.form).find("label[for=" + element.id + "]").addClass(errorClass);         $(element.form).find("label[for=" + element.id + "]").removeClass("valid");     },     unhighlight: function(element, errorClass) {         $(element).removeClass(errorClass);         $(element.form).find("label[for=" + element.id + "]").removeClass(errorClass);         $(element.form).find("label[for=" + element.id + "]").addClass("valid");     },     wrapper: 'li' }); How does the Schengen 90/180 rule work? 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