## Equivalence classes of elements in $X$ vs. equivalence class of $X \times X$

To quote Halmos:

If $$R$$ is an equivalence relation in $$X$$, and if $$x$$ is in $$X$$, the equivalence class of $$x$$ with respect to $$R$$ is the set of all elements $$y$$ in $$X$$ for which $$x R y$$. Examples: if $$R$$ is equality in $$X$$, then each equivalence class is a singleton; if $$R = X \times X$$, then the set $$X$$ itself is the only equivalence class.

~P. R. Halmos, Naive Set Theory (p. 28)

The first one, I think I understand. Each equivalence class is a singleton because each element $$x$$ in $$X$$ is only equal to itself.

The second is confusing me further the more I think about it, perhaps because of the wording. If $$R = X \times X$$, do I still consider it to be ‘in’ $$X$$ or is it ‘in’ the result of $$X \times X$$? If it’s the former, how is that any difference than ‘equality in $$X$$,’ which should yield singletons? If it’s the latter, then surely we’re now dealing with a series of ordered pairs that did not exist in the set $$X$$ beforehand, precluding it from being the equivalence class.

Or, is it that it’s neither of these, and the set $$X$$ used here is being treated like the $$x$$ we are seeking equivalence classes for in his initial definition? This latter definition seems to be the only way I can get my head around how $$X$$ itself ends up being the equivalence class, but also seems like I’m missing something vital in making that assumption.

## Calculating expectation and variance for having rolled 1 and 6 twice out of rolling a die 12 times

First i have calculated the probability to get each possible number {1,2,3,4,5,6} twice from 12 rolls(A).

$$Pr[A]=\frac{\binom{12}{2,2,2,2,2,2}}{6^{12}}.$$

Then there are 2 random variables:

X-number of times that 1 was received.

Y-number of times that 6 was received.

Before calculating $$E(X),Var(X),E(Y),Var(Y)$$ i’m uncertain of how i should calculate the probabilities of X and Y

## ChoiceBox does not populate until it is clicked 3 times

I have the problem, that my ChoiceBox only populates if it is clicked 3 times in a row. I really don’t know why. I also don’t know why it is successfull after exactly three clicks. In a similiar class this code works fine to populate the choiceBox (There are no problems with populating and updating of the ChoiceBox)

public class AuswertungController {  ProjektDAO project = new ProjektDAODBImpl(); MitarbeiterDAO mitarbeiter = new MitarbeiterDAODBImpl(); TaetigkeitDAO taetigkeit = new TaetigkeitDAODBImpl(); ObservableList<Taetigkeit> taetigkeitliste; ObservableList<Mitarbeiter> mitarbeiterliste; ObservableList<Projekt> projektliste; private ObservableList<String> zwischenmitar; private ObservableList<String> zwischenproj;  @FXML private ChoiceBox<String> choiceMitar; 

With this method i populate my choiceMitar choiceBox

public void choiceMitarAktualisieren() {     mitarbeiterliste = FXCollections.observableArrayList(mitarbeiter.getAllMitarbeiter());     zwischenmitar = FXCollections.observableArrayList();      for(Mitarbeiter m : mitarbeiterliste) {         zwischenmitar.add(m.toString());     }      choiceMitar.setItems(zwischenmitar); } 

}

## “A good programmer can be as 10X times more productive than a mediocre one” [closed]

I had read an interview with a great programmer (it is not in English) and in it he said that “a great programmer can be as 10 times as good as a mediocre one” giving reason for why good programmers are very well paid and why programming companies give many facilities for their employees. The idea was that there is a very large demand for good programmers, because of the above reason and that’s why companies pay very much to bring them.

Do you agree with this statement? Do you know any objective facts that could support it?

Edit: The question has nothing to do with experience; if you talk about one great programmer with 1 year experience then s/he should be 10 times more productive than a mediocre programmer with 1 year experience. I agree that from certain experience years onwards, things start to dissipate but that’s not the purpose of the question.

## How can the Same Origin Policy protect the user in times of CORS headers?

How can the Same origin policy protect something if you can bypass it via Cross Origin Resource Sharing headers? And why are those assigned by the server which is tried to access and not by the user?

Where is the risk of a Cross Origin Resource Request?

## Smoothness of $f : S^1 \times S^1 \to T$

Given $$f : S^1 \times S^1 \to T \subset \mathbb{R}^3$$ such that $$(u,v) \mapsto (u_1 (R+ r \ v_1), \ u_2 (R+r \ v_1), \ r \ v_2)$$

where $$S^1$$ is the unit-circle and $$T$$ is the familiar ring torus with $$R > r > 0$$

The book I am studying from mentions that $$f$$ is a diffeomorphism, but I don’t understand its smoothness at all.

I know that both $$S^1 \times S^1$$ and $$T$$ are two-dimensional and if $$f$$ is a diffeomorphism, then $$Df$$ is invertiable, so it must be a $$2\times2$$ matrix. But the map seperates the components of $$u$$ and $$v$$, so I am not sure how the Jacobian looks like. If I consider each components $$u_i$$ and $$v_i$$ seperately, then it’s $$3\times4$$ and is not invertiable.

Any clarification would be great!

## Adding a Binary Number n to Itself 2^{n} times

Say I wanted to add the binary number n = 100100111 to itself 2^{n} times. What would the time complexity of this set of operations be? Would it be O(n2^{n}), or O(log(n2^{n})), or something else? Thank you.

## What is an array that can be iterated circularly an infinite amount of times called?

I have a feeling that the answer is either a “circular linked list” or a finite state machine, but the answer is probably a circular linked list.

## App that blocks sound notifications of SPECIFIED APPS at SPECIFIED TIMES

At the moment I use an app called Nights Keeper that automatically blocks ALL sound notifications for me during the time I sleep.

The problem is I would like to have most sound notifications (calls, SMS messages, E-mails…) automatically blocked during the time I sleep while some other ones would remain intact (so for example I wouldn’t hear if somebody called me or sent an email to me, but I would hear if somebody messaged me on WhatsApp).

Is there an app that can do this?