calculation of Transitive Closure

This question is not from homework, but rather as preparation for the test:

The calculation of place-in is a calculation in which the algorithm does not need space beyond the output size (beyond beyond a fixed space complexity). Show how, given a graph (E, V = G), the transitive closure can be calculated in-place in the calculation of G * = (V, E *)

I tried to represent the graph using a adjacency matrix. what is the next step?

Prove that (if $R$ is set-like, then its transitive closure is set-like) implies the axiom of replacement

This is exercise I.9.6 from Kunen’s Set Theory (2011). $ R$ is set-like on the class $ A$ iff $ y \in A$ implies $ \{ x \in A: xRy \}$ is a set. I am not so sure about this, but it is implied that this can be done without using the axiom of choice.

The obvious choice of $ R$ is $ xRy$ iff $ x= F(y)$ , however what I am most unsure of is how to pick the class such that it cointains all of $ A$ and all of $ F(A)$ . I think the simple $ A \cup F(A)$ doesn’t work because for an $ x \in A$ , $ F(F(x))$ may not be a set and hence $ R$ wouldn’t be set-like. This reasoning seems to apply to any set $ B$ , since $ A \cup F(A)$ must be cointained in $ B$ in order for the transitive closure to include $ F(A)$ .

As an specific example, consider the set $ \mathbb{N}$ of natural numbers and the formula $ \phi(x,y)$ $ y$ is $ \mathbb{N}$ if $ x$ is a natural number and, $ y$ is $ \{w: w=w\}$ if $ x$ is $ \mathbb{N}$ “. Clearly while $ R$ is set-like in $ \mathbb{N}$ , it is not set-like in $ \mathbb{N} \cup F(\mathbb{N})$ , and in general I can’t modify $ \phi(x,y)$ as to exclude the $ x$ s in $ F(A)$ , as they may be part of $ A$ , too.

So, how can we prove it?

How to include external javascript files that rely on jquery but did not wrap in closure

Hello I am including an external javascript file using

drupal_add_js('https://wwww.sample.com/js/sample.js', 'external'); 

But this file is using jquery but didnt wrap it with

(function($  ){   //some code here })(jQuery); 

I am getting this error

Uncaught TypeError: Cannot read property ‘ajax’ of undefined

Thanks in advance.

Proving non regularity by closure properties

I have a question about a specific proof in my book.

Show that L = {wv∈ {a, b}*| |w|=|v| and w!=v} is not regular.

They proceed to proof this with proving the closure properties don’t hold, but my question here is that why does the following statement hold?

K = L ∪ {w ∈ {a, b}*| |w| is odd}

Wouldn’t one be able to take the union of any (also regular) L’s with {w ∈ {a, b}*| |w| is odd} and then state that L is non regular?

How to fix “object of type ‘closure’ is not subsettable”?

I’m studing on the book. when only reapting what wrriten, i got that kind of eror”Error in coef[2] : object of type ‘closure’ is not subsettable”. if u guys let me know, i’ll really appreciate it.

text(4, 27, labels=paste0(“y=”, round(sr$ coef[[1]], 4), “(+)”, round(sr&coef[[2]], 4), “*X”), col=2, cex=1.2)

Closure implementation

I have custom cell of tableview which is used as header as well as normal cell.

In custom cell I have this properties

public var btnReplyTappedClousre:((CommentCell) -> (Void))? public var btnBulbTappedClousre:((CommentCell) -> (Void))? public var btnOtherReplyTapped:((CommentCell) -> (Void))? 

And Call it with

@IBAction func btnReplyTapped(_ sender: Any) {     self.btnReplyTappedClousre?(self) }  //--------------------------------------------------------------------------------  @IBAction func btnBulbTapped(_ sender: Any) {     self.btnBulbTappedClousre?(self) }  //--------------------------------------------------------------------------------  @IBAction func btnViewPreviousReplyTapped(_ sender: Any) {     self.btnOtherReplyTapped?(self) } 

In view Controller i have implemented that closure

func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {     let cell =  tableView.dequeueReusableCell(withIdentifier: StoryBoard.Cells.CommentCell, for: indexPath) as! CommentCell     cell.isForHeaderCell = false     cell.btnReplyTappedClousre = {[weak self] (cell) in         self?.btnViewAllReplyTapped(cell)     }     cell.btnBulbTappedClousre = {[weak self] (cell) in         self?.btnBulbTapped(cell)     }     return cell }  func tableView(_ tableView: UITableView, viewForHeaderInSection section: Int) -> UIView? {     let cell =  tableView.dequeueReusableCell(withIdentifier: StoryBoard.Cells.CommentCell) as! CommentCell     cell.isForHeaderCell = true      cell.btnReplyTappedClousre = {[weak self] (cell) in         self?.btnViewAllReplyTapped(cell)     }     cell.btnBulbTappedClousre = {[weak self] (cell) in         self?.btnBulbTapped(cell)     }      cell.btnReplyTappedClousre = {[weak self] (cell) in         self?.btnReplyTapped(cell)     }       return cell  } 

Is there any better way to do it ? Or can i improve it ?

Finding the interior and closure of set $A \subset \mathbb{R}^2$

I am trying to determine the interior and closure of the following set $ $ A := \{ (x,y) \in \mathbb{R}^2 : |x| \geq |y|\} $ $ and would like to vertify my answer

For every $ (x,y) \in A$ we can find an $ \epsilon$ -neighbourhood that is completely contained in $ A$ unless $ |x| = |y|$ . If we let $ a \in A$ such that $ a = (x,x)$ and $ \epsilon > 0$ then $ b = (x-\epsilon, x+\epsilon) \in B_{\epsilon}(a)$ and $ b\notin A$ which means $ A^{°} := \{ (x,y) \in A : |x| \neq |y| \}$

The set $ A$ is closed, which means $ A = \overline{A}$ . I tried to show that $ \mathbb{R}^2 \setminus A$ is open, but I am having trouble finding an $ \epsilon$ such that $ $ c \in \mathbb{R}^2 \setminus A \ \text{and} \ d = (x,y) \in B_{\epsilon}(c) \Longrightarrow |x| < |y| \ \text{and} \ d \in \mathbb{R}^2 \setminus A $ $

Note: we are using the Euclidean metric on $ \mathbb{R}^2$