Mathematics - 2018-01-07 (page 2 of 4)

Monte

11:00

I am writing a program to check 4 integer values form a rectangle or not..

That is what the problem is asking me in the link given above

But simply comparing the equal sides doesn't seem to do the trick

Leaky Nun

did you send me the wrong question?

@Mr.Xcoder hi

Monte

codechef.com/JAN18/problems/RECTANGL

Mr. Xcoder

@LeakyNun Hi

Monte

Yeah I sent the wrong link

This is the correct link codechef.com/JAN18/problems/RECTANGL

Leaky Nun

@Monte it's can form

Monte

11:02

It <b> must </b> form bro

Leaky Nun

> Determine if there's a rectangle such that the lengths of its sides are a, b, c and d (in any order).

not:

> Determine if any quadrilateral with side lengths a, b, c and d must be a rectangle

Mr. Xcoder

> Source Limit: 50000 Bytes

Lol that's not very golfy

Leaky Nun

@Mr.Xcoder lol, I can do it under 20 bytes with Pyth

Mr. Xcoder

Ya? Well, I might give it a go

Monte

Yeah. the problem here is that I my algorithm is only comparing two equal sides as given in the test cases

My algorithm passes the test cases

but when I submit the problem it gives wrong answer

Mr. Xcoder

11:04

Oh you just have to check if they are equal 2-by-2?

Monte

So surely comparing the equal sides is not the only thing

Mr. Xcoder

Well you might want to test if they are all positive

Leaky Nun

@Monte maybe you forgot order

@Mr.Xcoder 10 bytes in Pyth (spoiler)

Mr. Xcoder

@LeakyNun I don't wanna see the spoiler, but do you check if they are positive or you don't bother to?

Leaky Nun

the latter

Mr. Xcoder

11:06

Oh that's going to be easy

Leaky Nun

Constraints
1 ≤ T ≤ 1,000
1 ≤ a, b, c, d ≤ 10,000

Monte

@Mr.Xcoder what do you mean by equal 2-by-2?

and numbers are positive

Balarka Sen

I have reached my chemistry queue for the day. Now I'll do math and no power in the universe can stop me

Monte

as pointed out by @LeakyNun check the constraints

Balarka Sen

No more aldehydes; Gromov is my animu babe now

Leaky Nun

11:06

1 hour ago, by Leaky Nun

Compute the limit: $$\lim_{x \to \infty} x \left( 1 + \frac 1 x \right)^x - ex$$

@BalarkaSen @Mr.Xcoder

Balarka Sen

@LeakyNun Why would I want to compute it?

Leaky Nun

because it's fun

Balarka Sen

To you perhaps

Mr. Xcoder

@Monte If you take one of them... it is equal to either 1 or 3 other values

@LeakyNun ooh, interesting. I doubt I have the necessary knowledge though, but I'll try

Leaky Nun

@BalarkaSen well I do have a calculus exam tomorrow

Balarka Sen

11:09

That's not an answer to my question!

Mr. Xcoder

Why would I want to compute it

:P

Monte

@LeakyNun @Mr.Xcoder any idea on how to determine the rectangle problem?

Mr. Xcoder

Yup.

Leaky Nun

@Monte we already said it a thousand times

check if there are two pairs of equal numbers

Monte

Would you mind submitting your solution there?

Because I tried this approach and it gives WRONG ANSWER

Mr. Xcoder

11:11

We did it in a language that isn't there >_<

@LeakyNun Yours is invalid.

Try [3, 3, 3, 3]

Leaky Nun

ah, lol

@Monte wait

Monte

@Mr.Xcoder which language are you talking about?

Mr. Xcoder

Pyth

Monte

Dude the site supports python

Balarka Sen

I have forgotten how to define the second fundamental form

rip

Monte

11:13

also PYTH

Mr. Xcoder

> PYTH, PYTH 3.5

@LeakyNun

Lol what is that

skullpatrol

pir

Monte

@Mr.Xcoder Try submitting your solution too over ther

Mr. Xcoder

Wait, not done yet. Gotta fix a bug

Balarka Sen

rippity rippity rip rip [starts beatboxing]

Monte

11:15

wabalalbadubdub

Mr. Xcoder

@LeakyNun 8 bytes in Pyth (spoiler)

Leaky Nun

@Mr.Xcoder lol

Balarka Sen

i see we have a man of high IQ with us

Mr. Xcoder

Hope it's valid.

skullpatrol

::starts break dancing::

Balarka Sen

11:16

break dancing stopped being hip in the 90's

Mr. Xcoder

Thankfully

Leaky Nun

@Mr.Xcoder fails for [1, 2, 3, 4]

Mr. Xcoder

>_>

skullpatrol

ya, too many broken necks

Monte

@LeakyNun and @Mr.Xcoder try submitting your solutions here codechef.com/JAN18/problems/RECTANGL

Mr. Xcoder

11:17

Yeah how could I be that dumb

@Monte We understood. Give us some time please

Monte

haha

skullpatrol

patience

Monte

Guys make sure we are not checking a parallelogram in our algorithms

Leaky Nun

@Monte ugh

a parallelogram and a rectangle have exactly the same lengths

and my answer passed

Monte

yes dude

Leaky Nun

11:18

import java.util.Scanner;
import java.util.Arrays;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for(int i=0;i<T;i++){
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			int d = sc.nextInt();
			boolean x = (a==b && c==d) || (a==c && b==d) || (a==d && b==c);
			System.out.println(x ? "YES" : "NO");
		}
	}
}

codechef.com/viewsolution/16839753

Mr. Xcoder

@LeakyNun 9: iI2hlM.gk (Try it online!)

Now I hope it is valid

Monte

GOD DAMN.. Tried the same with C++.. Getting Wrong answer

Leaky Nun

@Monte post your code

Monte

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
int t;
cin>>t;
while(t--)
{
int a[4];
for(int i=0;i<4;i++)
cin>>a[i];

sort(a,a+4);

if(a[0] == a[1] && a[2] == a[3])
{
cout<<"YES"<<endl;
}
else
{
bool flag = true;
for(int i=0;i<2;i+=2)
{
if(a[i] != a[i+1])
{
flag = false;
break;
}
}
if(flag)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
}
return 0;
}

Leaky Nun

press ctrl+K before sending it

Monte

11:20

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int a[4];
        for(int i=0;i<4;i++)
            cin>>a[i];

        sort(a,a+4);

        if(a[0] == a[1] && a[2] == a[3])
        {
            cout<<"YES"<<endl;
        }
        else
        {
            bool flag = true;
            for(int i=0;i<2;i+=2)
            {
                if(a[i] != a[i+1])
                {
                    flag = false;
                    break;

Mr. Xcoder

Ugh... Ugh..

Monte

here basically I am sorting such that my comparisons are reduced

But doing the same thing

Balarka Sen

why are these coders invading the math chat

Gabriel Romon

wrong room ?

Monte

@BalarkaSen Sorry brother.. Actually it was a math problem..

Balarka Sen

11:21

go back to PPCG

Mr. Xcoder

IT is related to math (we are checking whether there exist rectangles with the given set of sides)

Balarka Sen

go back to the opium dens

Mr. Xcoder

@BalarkaSen ಠ_ಠ

Monte

@LeakyNun any comments bro?

Leaky Nun

@Monte do you know how to spell "patience"?

Monte

11:23

hahahah sorry once again XD

Leaky Nun

idk what you did, but this passed:

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    int t;
    cin>>t;
    int a[4];
    while(t--)
    {
        for(int i=0;i<4;i++)
            cin>>a[i];

        sort(a,a+4);

        if(a[0] == a[1] && a[2] == a[3])
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
    return 0;
}

Mr. Xcoder

I did it in C

Leaky Nun

your program outputs nothing for 1 1 1 2 because your extraneous checking in the else part contains a break because a[0] == a[2] (both are 1)

@Monte

@Mr.Xcoder ` || (a == b && a == c && a == d)` is not needed

Mr. Xcoder

Actually

Ninja'd

Much, much nicer in C

Leaky Nun

sure

Monte

11:27

@LeakyNun Yeah got my mistake.. Thank you

Mr. Xcoder

Where do you submit? Not the link, where on the page?

Oh perhaps I need an account

Leaky Nun

you need to create an account

sniped

Monte

@LeakyNun best of luck for your exams tomorrow bro

Leaky Nun

thanks

Mr. Xcoder

Yes ^^

Balarka Sen

11:31

Good luck; even though you don't need it

Mr. Xcoder

Yeah I'm sure your grade will be 200%

> Execution Time: 0.00 ಠ_ಠ

Monte

Not sure if he'll be giving the exam or taking the exam

Leaky Nun

@Monte take

Balarka Sen

I need to crack open the pages of DIFFERENTIAL GEOMETRY: A GEOMETRIC APPROACH

Mr. Xcoder

@LeakyNun You're having an exam tomorrow and I'm still in my winter holiday :)

Leaky Nun

11:34

I see

Mr. Xcoder

Lol

Balarka Sen

I think my rapper name is going to be A$AP Gau$$

Mr. Xcoder

@LeakyNun $-\frac e 2$

Leaky Nun

@Mr.Xcoder how?

Mr. Xcoder

According to Wolfram Alpha :P

Leaky Nun

11:39

very funny

skullpatrol

his highness A$AP Gau$$

:P

Balarka Sen

kingly adjectives don't go with rapper names

Monte

Leaky Nun already sounds like a rapper's name

Balarka Sen

it doesn't sound very thuggish

also a little creepy

skullpatrol

sounds like a nun that leaks

Monte

11:41

hahahah

philmcole

Hi guys. I'm unsure if my proof that the set of cluster points is closed is any good. Didn't want to post a new question on this, maybe you can take a look?

Let $A \subseteq \mathbb R$ be a set and $H$ the set of all cluster points of $A$.

Let $x \in \mathbb R \setminus H$. Since $x$ is no cluster point there exists $\varepsilon \gt 0$, such that for all $a \in A$ we have $|a-x| \ge 2\varepsilon$.

For a cluster point $x_0 \in H$ there exists $a \in A$, such that $|a-x_0| \lt \varepsilon$. By the reverse triangle inequality

$$|x-x_0| = |x-a+a-x_0| \ge ||x-a| - |a-x_0|| \ge 2\varepsilon - \varepsilon = \varepsilon$$

Therefore $(x-\varepsilon,x+\varepsilon) \nsubseteq H$. But then $(x-\varepsilon,x+\varepsilon) \subseteq \mathbb R \setminus H$…

Leaky Nun

I'm sure there's a topological proof @philmcole

philmcole

Sure I wanted to keep it simple and used just $\mathbb R$.

Leaky Nun

don't you think using metric is not keeping it simple?

the "simple" proof is to use the definitions of those terms alone, in point-set topology, without assuming that the topology is equipped with a metric

Balarka Sen

that's controversial opinion

Leaky Nun

11:45

unless you follow crap like Rudin who defines metric before point-set topology

Balarka Sen

analysis I would say the simple proof is to use epsilon delta

philmcole

For the more advanced mathmaticians it's sure a better thing to prove everything right away in the most abstract sense, but I'll stick with $\mathbb R$ for learning.

Balarka Sen

@LeakyNun metric should be defined before point-set topology. 90% of the time people work with metrizable spaces

Leaky Nun

> Let $x \in \mathbb R \setminus H$. Since $x$ is no cluster point there exists $\varepsilon \gt 0$, such that for all $a \in A$ we have $|a-x| \ge 2\varepsilon$.

@philmcole this line smells a little bit fishy

philmcole

@LeakyNun Why?

Leaky Nun

11:48

how do you define a cluster point?

(we may have different definitions)

skullpatrol

\o @MatheinBoulomenos

MatheinBoulomenos

o/ @skullpatrol

philmcole

@LeakyNun $x$ is a cluster point of $A$ if for all $\varepsilon \gt 0$ there exists $a \in A$, such that $0 \lt |a-x| \lt \varepsilon$

Leaky Nun

@philmcole exactly

so that line I just quoted is false

philmcole

because of the $2 \varepsilon$?

Leaky Nun

11:49

no

because $a$ may be $x$

philmcole

Ah I see.

Mr. Xcoder

How to you evaluate limits to infinity in general?

Balarka Sen

with gusto

Leaky Nun

@Mr.Xcoder I'm not aware of an algorithm

such an algorithm would be able to solve the halting problem, so it cannot exist

Balarka Sen

oh shaddup

Leaky Nun

11:51

@BalarkaSen I'm not kidding

Manolis Lyviakis

if rudin is crap which book you believe is good?

Balarka Sen

you and your halting problem

Leaky Nun

you can encode the halting problem in real equality

so real equality is not decidable

whatever decidable means

Balarka Sen

i'll halt your problems up

:roasted:

skullpatrol

rotfl

Mr. Xcoder

11:53

@LeakyNun But, what do you think of first when you attempt to evaluate limits to infinity?

skullpatrol

Large numbers

Leaky Nun

@Mr.Xcoder there are different tricks for different questions

Mr. Xcoder

I understand

Leaky Nun

conjugates

taylor expansion

l'hopital

plotting graphs :P

skullpatrol

^

Mr. Xcoder

11:54

Yeah plotting the graph definitely shows it's $\approx -\frac e 2$

Balarka Sen

@LeakyNun my response to your "Rudin is crap" comment

Leaky Nun

@BalarkaSen lol

philmcole

@LeakyNun I changed two parts and excluded $\{a\}$. Did I solve it?

Let $A \subseteq \mathbb R$ be a set and $H$ the set of all cluster points of $A$.

Let $x \in \mathbb R \setminus H$. Since $x$ is no cluster point there exists $\varepsilon \gt 0$, such that for all $a \in A \setminus \{x\}$ we have $|a-x| \ge 2\varepsilon$.

For a cluster point $x_0 \in H$ there exists $a \in A$, such that $|a-x_0| \lt \varepsilon$. By the reverse triangle inequality

$$|x-x_0| = |x-a+a-x_0| \ge ||x-a| - |a-x_0|| \ge 2\varepsilon - \varepsilon = \varepsilon$$

Therefore $(x-\varepsilon,x+\varepsilon) \setminus \{x\} \nsubseteq H$. But since also $x \notin H$ we get $(x-\…

Leaky Nun

@philmcole there is a slight error

$(x-\varepsilon,x+\varepsilon) \setminus \{x\} \nsubseteq H$ does not imply $(x-\varepsilon,x+\varepsilon) \subseteq \mathbb R \setminus H$

philmcole

Why?

Leaky Nun

12:01

well, $\{1, 2\} \nsubseteq \{2, 3\}$ but also $\{1, 2\} \nsubseteq \{1, 4\}$, where $\Bbb R = \{1, 2, 3, 4\}$ and $H = \{2, 3\}$

it's an extraneous and erraneous statement

MatheinBoulomenos

but we know that $x\in \Bbb R \setminus H$ by assumption

Leaky Nun

@MatheinBoulomenos that isn't the point

his overall logic is correct, but he mistakenly inserted that wrong statement

philmcole

Can you tell me what the fix is so I can understand it?

Leaky Nun

you just need to delete that statement

MatheinBoulomenos

you want to say $(x-\varepsilon,x+\varepsilon) \cap H = \emptyset$ instead of $(x-\varepsilon,x+\varepsilon) \setminus \{x\} \nsubseteq H$

Leaky Nun

12:05

$\varnothing$ \varnothing

$\emptyset$ \emptyset looks very ugly

philmcole

@MatheinBoulomenos Thanks

MatheinBoulomenos

I'll just use $\{x \mid x \in x\}$ instead

2

Leaky Nun

@MatheinBoulomenos lol

Secret

12:48

lol wtf?

desmos.com/calculator/34jgyhcwfi

The terms of the limit is not separable, but how does one knew that without drawing the graph?

[Random]

Yet another quasi finite proposal

(see rambles)

Dattier

$f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

Hi,

2/ Determinate $n\in\mathbb N^*$ with $n^{n+1}+1$ is prime.

3/$u_0=3$, $u_{n+1}=u_n^2+1$. Is it true that : $\exists n \in \mathbb N, v_n \mod (2^{89}-1)=0$ ? Why ?

4/ $v_0=3$, $v_{n+1}=v_n^2-2$. Is it true that : $\exists n \in \mathbb N, u_n \mod (2^{89}-1)=0$ ? Why ?

5/Determine $F_{2 ^{2^{2018}}} \text { mod } (2^{89} -1)$ of the Fibonacci sequence (remember that $2^{89} -1$ is a prime integer).

Vrouvrou

13:24

Hello, please i have a function $f(t)=\int_0^{t}\dfrac{\Phi^{-1}(\tau)}{\tau^{1+\frac1n}}d\tau$ $f$ we have that $f$ is strictly increasing, continuous and satisfies $ \int_1^{\infty}\dfrac{\Phi^{-1}(\tau)}{\tau^{1+\frac1n}}d\tau=+\infty.$ why $f$ is surjective

BAYMAX

13:35

Any help on matlab?

Leaky Nun

14:03

in other news:

in Rambles, 4 mins ago, by Secret

|S| = Holy holy holy holy holy ....

Secret

14:21

Imagine a set that failed to destroy itself. That... if consistent, would be something strictly larger than a proper class in a vague sense (as obviously no bijection can be written)

And thus you met God himself, holyholyholyholyholhyolholyohloylhoylohly..

static

Leaky Nun

is it true that $\displaystyle \left\lceil \frac 2 {2^{1/n}-1} \right\rceil = \left\lfloor \frac {2n} {\ln 2} \right\rfloor$ for all $n \in \Bbb N$?

Alessandro Codenotti

What's the point of asking a question you know the answer to and that can be answered only through numerical computation or reading the appropriate entry of the OEIS?

3

Manolis Lyviakis

15:48

FInd the Error of this Proof : $S^2 $ is simply connected: Proof. Let a loop on $x_0$ of the sphere and take a point $p$ in the $ S^2 $ also not in the loop. Now $S^2 \setminus p$ is homeomorphic with the plane and tha plane is simply connected so the loop is homotopic to a point hence $S^2$ is simply connected

Alessandro Codenotti

\setminus

Manolis Lyviakis

:)

i think there might be a problem if the point is on the interior of the loop or something

Alessandro Codenotti

Balarka gave me this same exercise an year or so ago so I know where the problem is, but I won't tell you directly :P So which steps are surely correct in this "proof"?

Manolis Lyviakis

haha really?

Alessandro Codenotti

Yep, I don't know if it is in some standard book :P Where did you find it?

Manolis Lyviakis

15:51

my proff gave it to me

dont know where he got it

DarkRunner

Hi folks; I'm having trouble working through a Number Theory question. If anyone could help that would be greatly appreciated

"Show that one more than the product of any pair of twin primes is a perfect square"

What I've gotten up to is

Manolis Lyviakis

soo i was thinking if the point is in the interior of the loop the way i will project the sphere to the plane will kinda "break" the loop? or it has nothing to do with that? @AlessandroCodenotti

because i always thought of projecting the sphere to the plane like opening up an onion or something

Alessandro Codenotti

@ManolisLyviakis That'd be a problem indeed, but can another point which is not problematic always be found?

Manolis Lyviakis

ohhh

Alessandro Codenotti

Hi @Mathei

DarkRunner

15:55

Manolis Lyviakis

i guess no since i can always have a loop homotopic to the given loop that has this point on the inside? @AlessandroCodenotti ?

DarkRunner

I'm stuck at this point

Question: "Show that one more than the product of any pair of twin primes is a perfect square"

Alessandro Codenotti

but the loop is fixed here, while the point has to be picked

@Dark hint: what's the difference of $2$ consecutive squares?

DarkRunner

@AlessandroCodenotti An odd number, right!

Manolis Lyviakis

i cant see the problem if the point is on the outside

DarkRunner

15:58

wow; thanks

Alessandro Codenotti

Ok, so $S^2\setminus p$ is definitely homeomorphic to $\Bbb R^2$ and $\Bbb R^2$ is definitely simply connected so I agree that this looks right if the point is not in the image of the loop

MatheinBoulomenos

Hi @Alessandro

Manolis Lyviakis

or in the inside

Transcript for

$Mathematics$ Mathematics