Symmetric difference Puns

So I've been stuck on this for hours and had to look at solutions. This is the solution I've been using:

function sym(args) {
  let argumentsArray = [...arguments]
  function symDiff(arr1, arr2){
    let result = []
    arr1.forEach(a=>{
      if (arr2.indexOf(a)<0 && result.indexOf(a)<0){
        result.push(a)
      }
    })
    arr2.forEach(a=>{
      if (arr1.indexOf(a)<0 && result.indexOf(a)<0){
        result.push(a)
      }
    })
    return result
  }
  return argumentsArray.reduce(symDiff)
}

sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])

The result to this is [ 1, 4, 5 ]

From what I understand, it:

1. Turns the arguments into a 2d array so it's easier to work with

2. Creates a callback function called symDiff, where it takes 2 arrays, looping through each one, and if the other array doesn't contain the element AND if it doesn't already exist in the result array, it pushes it into the result array

3. Applies the callback function on the reduce method used on the 2d array of arguments, the reduce method goes through each array of arguments until it returns the result

I have 2 questions here:

How is 5 in result? Doesn't the symDiff function only push elements that don't exist in the other array? If arr1 is [1, 1, 2, 5] and arr2 is [2, 2, 3, 5], isn't arr2.indexOf(5) going to be 3, and same for arr1.indexOf(5), which means it shouldn't be pushed into result?

And my other question is, I noticed if I declare the result variable outside of the symDiff function, it gives a different result, it becomes [ 1, 3, 4, 5 ] instead. How does that work?

This would help me out a lot. Any tips are welcome!

Kindest regards, BHTA!

EDIT right away: this is not homework, the actual questions are not at all answered by this. I just cant understand it from the resources I find on internet. Im merely asking for an easier explanation of the concepts :)

Lately a trick I've been using on 2fort is to go in the front door and then sticky jump up the little hole in the grate to get to the upper level. This allows you to avoid the courtyard, which is where like 80% of 2fort sentries end up getting placed.

This has caused me to notice a difference: in the Blu fort, the upper area here is enclosed, with a roof. But on the red side, it's wide open, up to the sky. This means that when I enter the red side, I fly way up and end up taking fall damage, while on Blu I usually just bounce off the roof and stand right on the grate. If I took damage on my way across the bridge, I can easily die entering the red fort, while on the Blu fort I know I'm probably okay (and the health pickup at the top of the spiral helps top me off until I can get to the next health, right outside the Intel room).

This is a big difference in the two halves, and I would have thought the game would be made to be perfectly balanced. Are there other examples of big differences between the two sides of a map that initially seem functionally identical?

Hi All,

I am using the Wilcoxon signed-rank test to pairwise compare data. This test assumes an approximately symmetric distribution of differences. Right now, I am verifying this by making a boxplot of the differences between the data and looking at the shape of the boxplot.

Is there maybe a more clear way to test this? Like a statistical test that investigates the distribution of differnces? I thought that maybe using the Levene's test to check equal variances between 2 groups could be an indication, but this is something different I think?

Thanks!

Hey guys, I plan on going through each of the algorithms under the coding prep's algorithm section. Here is the first one: https://youtu.be/XG3O5xJgMA0

Enjoy!

In my class i’m working with sets. I defined

• via sub & via and | via or But how do you define symmetric difference (^)? I can’t find it

I've been looking at this one for hours, but can't seem to figure out where I'm going wrong The arrays do get passed in correctly, their values are:

this->pInteger {15,16,17,18,19,20,21,22,23,24}

That.pInteger {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}

By the time the program finishes, IntSet3's values are {15,16,17,18,19} Which is the opposite of what I want, the values I want it to return are {1,2,3,4,5,6,7,8,9,10,11,12,13,14,20,21,22,23,24}

Here's the function: https://www.pastebucket.com/564484

My thinking is that since match2 is initialized as false, then it should only flag as true when the numbers do line up, and NOT add the value to the array. Where am I going wrong on this one?

Is this the same thing? Are one of these evens or one of these odd? If that makes sense? I'm not understanding the difference between the two and wonder if it is the same process to solve.

So I get that it omits the intersections. What throws me off is the venn diagrams for >3 events also include some of the overlaps and I don’t understand why.

Can someone explain what it means in a real-world scenario without using the word intersection? Everywhere I look explains it as omitting the intersection but I need a non-jargon explanation.

Is it the same as “the probability that exactly one of these events occur” (but not more than one, hence the intersection being omitted)?

If so why is the middle overlaps included (for >3 events)? I can only find examples using discrete sets which I can understand. But what if the venn’s represent events with probabilities? Like 10% of people choose ABC, while only 40% choose A, 30% choose B, 20% choose C, etc. Why would you include the people that select all 3 in the centre of the venn?

Hi guys,

I have to tell you guys that the long blocky selection issue has been solved by one of the Krita developer that worked on selection tool a long time ago, and thus that means symmetric difference selection can technically be pushed soon.

There's one or two thing that needs to be done on my end - Define the shortcuts, and add layer-based selection shortcut as Dmitry has added those.

What should the shortcut be for symmetric difference selection tool, and layer-based symmetric difference selection be?

Image of the new selection tool (Note the 5th selection tool)

• https://i.imgur.com/HfQsOxo.png

Brief info of the new tool - Symmetric Difference invert the selected area which means that any area that intersect are not included as selection, and any area that don't intersect are included.

More info here - https://en.wikipedia.org/wiki/Symmetric_difference

Phabricator Discussion - https://phabricator.kde.org/D15085

Hey Pls i need some help to do the logical approach for showing that:

(A+B = A inter B ) implies A=B={ }

Thanks in advance

So, I was supposed to implement symmetric difference for two sets of numbers of, respectively, n and m size. I was supposed to assume that every set is valid (doesn't have duplicates, have only integers etc.), the sets can range in size (min. 0, max. 10^7) and have numbers no bigger than 10^9. The result should be a single line consisting of three numbers separeted by spaces (in that order: number of elements of the resulting set, it's smallest element and it's largest element, both -1 if the resulting set is empty). The catch is I can use only primitive data types and primitive, one dimensional arrays. Here is my code:

http://pastie.org/private/bzp22ayvg851bq5efiqaq

Now, here's the fun part. I'm evaluated by automated platform. This code works for first five of it's 16 test cases. The next nine report wrong answer and final two say that I took too long (reported run time is more than 1.01 seconds). And that's all the information I have. No details whatsoever. Me and my friends have been trying to figure out what's wrong witht that code for the past week. We haven't come up with any possible reasons, so at this point I'm desperate.

Also, I should mention, while this is the most recent version of that implementation, it isn't the most succesful one. This one scored the most points:

http://paste.ofcode.org/38T7H2eh8QsxDUPHc7h8FvQ

It passed succesfully eight checks and took too long in the rest.

Prove: A△(A△B)=B

Knowing that A∆B = (A − B)∪(B − A)

I have to prove the above without using its associativity. All kinds of help are appreciated since I'm stuck on this one.

Hi, I need to prove the problem stated in the title.
I do know that the problem basically simplifies to proving that AΔ(BΔC)=(AΔB)ΔC.

Would it be sufficient to show that there exists an element x within AΔ(BΔC) that is also (AΔB)ΔC and also show the converse, that there exists an element within (AΔB)ΔC that is also in AΔ(BΔC)? I don't really feel like expanding out the definition of the symmetric difference but I will if that's the only reasonable way.

Thank you!

One blocker bug remaining on my end, but it's at the point where the full functionality can be explored. There's a issue of needing a custom shortcut to assign symmetric difference selection using opaque information from layers, but Dmitry had worked on that end without the new symmetric difference selection feature, and since I have no idea about how to go on about that, I'll probably leave that up to him. So, technically two blocker bug, but one on my end. I think it should be cntrl+shift+space to activate selection using opaque on layers, but feel free to oppose this design.

In any case, here's a video of the symmetric difference selection patch. - https://streamable.com/i0k6w

And the differentials - https://phabricator.kde.org/D15085

Hello, i have a problem i am trying to solve. Basically i has two sets of emails. One email per line and i have two files a.txt and b.txt. Is there a way to maybe use a built in java method to get the Symmetric set difference from these two text files? So far what i did was take each email and add it to two different hash sets, s1 and s2. Now i need to compare these two hash sets and basically find the emails that are missing in a from b.txt. Is there a java method to do this? or anything that someone can think of to help me come up with a solution. I appreciate any help, thanks. Here is what i have so far.

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.HashSet;
import java.util.Set;

public class test {

public static void main(String[] args) throws FileNotFoundException,IOException {
	File f1 = new File("a.txt");// OUTFILE
	File f2 = new File("b.txt");// INPUT

	Set<String> s1 = new HashSet<String>();

	try (BufferedReader br = new BufferedReader(new FileReader(f1))) {
		String line;
		while ((line = br.readLine()) != null) {
			s1.add(line);
		}
	}
	
	Set<String> s2 = new HashSet<String>();
	
	try (BufferedReader br2 = new BufferedReader(new FileReader(f2))) {
		String line;
		while ((line = br2.readLine()) != null) {
			s2.add(line);
		}
	}
	
	System.out.println("" + s1.containsAll(s2)); 


}

}

Let A and B be sets. Prove: If A(+)B subset A, Then B subset A [(+)= symmetric difference and I'll use C's for subset e= element of]

Claim: If (AUB)-(AnB) C A, Then B C A Suppose x e (AUB)-(AnB) Claim: x e A SInce x e(AUB)-(AnB) x e A or x e B, but x not e of (AnB)

It seems to me that many people think one way to determine whether a situation is discriminatory by taking that situation, inverting the roles and seeing if it appears problematic.

But I think that is a detrimental oversimplification, as it completely occludes the associated context.

For instance, if a black man disguises as a white character, many people will get riled up against them because "if it were a white man doing a blackface everyone would think it's racist". But that's forgetting completely that for decades, blackface was used to severely mock black people.

Another example I can take is that of my own university. Women make up 25% of the students and in this male dominated environment, they're too often victim of sexual harassment and assault. To help with that, my university has hired someone to oversee gender equality on the campus. It would have been perfectly sexist to require a man for that job. But requiring a woman is not only okay but necessary as female victims of sexual assault are typically reluctant to open up to a man.

So yeah, sometimes situations are problematic when the roles are reversed, but that's because those roles aren't symmetrical, and we need to stop pretending that they are.

The question asks to show that

P(A XOR B) = P(A) + P(B) - 2P(A ∩ B)

directly using the axioms of probability.

I have no clue how to even start this proof.