/lit/ - Literature

[Anonymous No.15308300]

I'm 100% convinced I'm smarter than everyone. Objectively smarter.
You have 1 book to prove me wrong. What is it?

[Anonymous No.15308320]

You are rascal Raskolnikov in thinking that you are in possession of a certain superiority that somehow puts you above common decency. When you try and strike down the untermensch you will find that the existence you have made for yourself is not what you want, not what anyone wants

[Anonymous No.15308321]

>>15308300

[Anonymous No.15308329]

>>15308320
>raskolnikov
>is a rascal
>rascal-nikov

[Anonymous No.15308336]

>>15308300
People who are smarter than me have no difficulty solving these two logical problems:
1. There are 6 people in a room. Prove that in this group, there must be either 3 people who all know each other or 3 people who are total strangers to each other.
2. Another puzzle: There is a light bulb and a consciousness switching machine. Two people can enter the machine and switch their consciousness. Whenever two people switch their consciousness, the light bulb switches (on->off and off->on). You have access to as many people as you want. The light bulb starts off being off. Is it possible to using the machine in such a way that in the end, everyone's consciouness is back to where it started but the light bulb is on. If yes, give a method how to do it, and if not, explain why.

If you have an IQ of over 110 you should have no difficulty with them.

[Anonymous No.15308348]

>>15308336
>1. There are 6 people in a room. Prove that in this group, there must be either 3 people who all know each other or 3 people who are total strangers to each other.
>2. Another puzzle: There is a light bulb and a consciousness switching machine. Two people can enter the machine and switch their consciousness. Whenever two people switch their consciousness, the light bulb switches (on->off and off->on). You have access to as many people as you want. The light bulb starts off being off. Is it possible to using the machine in such a way that in the end, everyone's consciouness is back to where it started but the light bulb is on. If yes, give a method how to do it, and if not, explain why.
My IQ is over 150 and I can't solve these

[Anonymous No.15308353]

>>15308348
Then clearly you are lying about your IQ. Those are not hard problems anon.

[Anonymous No.15308370]

>>15308353
No, I'm a genius and you're making up bullshit problems to shit up my thread. What type of schrodinger cat bullshit is problem #1? Complete buffoonery

[Anonymous No.15308386]

>>15308370
Perhaps you need to take more time to think about them? I don't expect you to solve them in 5 minutes. An hour, in my opinion, is plenty, though.

[Anonymous No.15308392]

>>15308370
Also I'm not sure what you mean by "schrodinger cat bullshit" and "complete buffoonery"? Is it that you fail to understand what the problem is asking for? I can clarify the problems, just ask.

[Anonymous No.15308401]

>>15308320
It's not whatr I want and not wht you want but it's what you've made me into

[Anonymous No.15308412]

CTMU

[Anonymous No.15308413]

All it took was two simple problems. All of OP's confidence is gone.

[Anonymous No.15308423]

>>15308300
I find everybody's intellectual intelligence far less superior than that of my own.

[Anonymous No.15308424]

>>15308300
Lacan's Ecrites

Read it and give me a summary/refutation of his ideas without using the words "pseudoscience", "obscurantist garbage" or "word salad"

[Anonymous No.15308430]

>>15308423
Guys, I dont want to think iof that ed gein crossdresser or his loser family. Im gonna cry if I hafve to speak with them.

[Anonymous No.15308433]

>>15308424
>Read it and give me a summary/refutation of his ideas without using the words "pseudoscience"
What would lying about the book prove?

[Anonymous No.15308436]

>>15308423
Literally just thinking of them causes me so much pain. You don't get it.

[Anonymous No.15308442]

>>15308300
>>15308423
You should be able to solve this:
>>15308336

[Anonymous No.15308447]

>>15308442
Norman Bates, if you want to dress like your mommmy tharts fine but go the fuck away please. Go away

[Anonymous No.15308452]

Low IQ people have a hard time believing that it's not B.

[Anonymous No.15308467]

>>15308336
Not sure about the second one but is the first one because they're all satanists?

[Anonymous No.15308480]

>>15308452
C

[Anonymous No.15308482]

>>15308452
Would you stop posting this b8 already? Everyone knows it’s actually B by now.

[Anonymous No.15308496]

>>15308482
It's not bait: it's an actual puzzle. It would make no sense to post this if it was actually B. You are just too low IQ to figure out why it's not B (or perhaps you haven't tried hard enough to figure it out?).

[Anonymous No.15308502]

https://discord.gg/fmeA9SJ

[Anonymous No.15308503]

Puzzles posted ITT: 3
Puzzles solved ITT: 0

[Anonymous No.15308513]

>>15308482
It's not actually B, though. That's just what the mathematically illiterate think.

[Anonymous No.15308522]

>>15308370
You sounds like an extraordinarily stupid person.

[Anonymous No.15308527]

>>15308496
>>15308513
b8 it’s definitely B

[Just some dumb who wandered in(...) No.15308554]

>>15308336
friends:
AB
CD
EF

No 3 people know each other.
Each person has 4 strangers.

Did you misquote the question? Maybe left out an "at least" or something? Or did I go wrong somewhere?

[Just some dumb who wandered in(...) No.15308563]

>>15308503
Please mark my work.
>>15308554

[Anonymous No.15308568]

>>15308554
The question is incredbibly vague and poorly written. It's some pseud's attempt to use Ramsey Theory to flex on brainlets.

[Anonymous No.15308572]

>>15308452
>>15308496
>>15308513
It is bait. It’s not from a “10th grade mathematics exam”, or, if it is, the question was poorly chosen because it is certainly not 10th grade level. The problem is actually from a book written by a mathematician for a scholarly audience.
In my opinion, it is obviously B as B is clearly the furthest point. Even though this might not be mathematically the case, I believe it. My human spirit and rational intuition (which are the inventors of mathematics) tell me that it is clearly B. If mathematics tells us it is not, then it is a problem within the mathematical framework, not my rational human spirit.

[Anonymous No.15308617]

Your statement proves you are a dumbass compared to others on this board. Which is lretty astonishing considering its /lit/

[Anonymous No.15308663]

>>15308554
>>15308563
The group "ACE" are all mutual strangers.
>>15308568
Which part of it is vague? I thought I formulated it quite clearly. Can you point it out?

[Anonymous No.15308665]

>>15308336
The second one is easy so I'll do that first: The answer is No. I solved it like this (don't know remember many maths terms so please pardon the crudeness): For the light to be left on you need an odd number of switches, in order to get everyone back to their bodies you go around in a circle. So I used 7 people, A-G with their consciousnesses being labelled 1-7; they swap with the person next to them clockwise until the 1 reaches back to A. So it looks like this for A: A and G swap, G and F swap, F and E, E and D, D and C, C and B, then B and A. That would take 7 swaps, leaving the light to be on. Then in order to get everyone else back into the bodies, B would swap with C, C D, D E, E F, F G. This would take 5 switches, taking the total to twelve. The amount of switches is amount of people multiples by 2, minus 2. 5 takes 8 switches. 6 takes 10. Maybe I'm wrong but that's what I got.

I don't get the second one as a logic question -- can I do things or can I only speculate? I'll think on it more. Not op

[Anonymous No.15308670]

>>15308554
>Each person has 4 strangers.
So there are 3 people who are strangers to each other -- e.g., A, C and E.

[Anonymous No.15308676]

>>15308572
Lmao.

[Anonymous No.15308684]

>>15308665
You can't be serious.

[Anonymous No.15308689]

>>15308665
>I don't get the second one as a logic question -- can I do things or can I only speculate?
You can answer it however you like, as long as your answer is convincing.
> The second one is easy so I'll do that first: The answer is No
I don't really understand your solution. So are you saying you can do it in a way that leaves the lightbulb on?

[Anonymous No.15308769]

my diary desu

[Anonymous No.15308817]

>>15308568
The formulation is very straightforward desu.

>>15308348
We will call the necessary groups 3-groups.
Consider one of those people. He has 5 binary relations (knows that person or not), meaning that he has at least 3 of the same relation. Let's call those relations R1, R2, and R3.
- If there's at least one relation between R1, R2, or R3, the original person, and the R's which have the relation form a 3-group.
-If there are no such relations, then R1, R2, R3 form a 3-group.

[Anonymous No.15308822]

>>15308336
>1. There are 6 people in a room. Prove that in this group, there must be either 3 people who all know each other or 3 people who are total strangers to each other.

I'm guessing the "all" in "all know each other" is the key point here. Otherwise in a group of people [a,b,c,d,e,f], 'a knowing b' and 'b knowing c' already gives you 3 people (a,b,c) who know someone in the group. But they don't "all know each other". [As in 'a knows b', 'b knows c', 'c knows a']
If this is what you meant, one can notice the more general statement "Either X people all know each other or 6-X people don't know each other" is true because by definition if you want to limit the people who know each other to X, you must necessarily demand that the other 6-X don't know each other.
Your statement is just the case where X = 3.

[Anonymous No.15308853]

>>15308452
its (0.75,0.75,2)

[Anonymous No.15308890]

>>15308572
Based

[Anonymous No.15308909]

Why do pseuds always use stupid maths questions to prove they’re smart? Why not show your ability to think critically, or post some of your published work? Unless you don’t have any, of course...

[Anonymous No.15308912]

>>15308300

A Book of statistics

But you got me interested. What are you are studying and working on right now, OP. How does it feel to be that smart?

[Anonymous No.15308916]

>>15308321
BAAAASED

[Anonymous No.15308920]

>>15308300
Only an idiot thinks they are objectively smarter than everyone, qed

[Anonymous No.15308924]

>>15308300
must feel good. i hate feeling like a complete and utter retard. i got trolled so bad yesterday i didn't know what hit me. i took what was said at facevalue like an imbecile. im still reeling over it. i cant believe im so stupid

[Anonymous No.15308974]

>>15308909
Because it's a simple, quick way to test intelligence.

[Anonymous No.15308990]

>>15308974
It’s not though. Intelligence isn’t something that can be tested in a ‘simple’ and ‘quick’ way.

[Anonymous No.15309009]

>>15308990
It's just a riddle lmao. A quick and fun way to see if OP is full of shit. A genius would have no difficulties solving those.
You sound angry. Is it because you couldn't solve them? That's ok, most people of average intelligence wouldn't solve it quickly either.

[Anonymous No.15309029]

>>15308336
Come on, anons. I know Finland is supposed to be better with this, but is this not basic high school math where you're from?


1)
Let's call the people in the room A, B, C, D, E and F. Let's then say that everyone on the same level knows each other. Any one of them can of course be put on multiple rows. For example, if A knows B and C but no one else knows anyone else, we can write that as:

AB
AC
D
E
F

Now, if we have a three people in a horizontal row, for example DEF, then three people will all know each other. Similarly, three in a vertical row means three people who don't know eachother.

Our job is therefore to show that these six people cannot be order so that there would be no row of three. Of course, the largest set without a row of three wouldn't have space for all six. It would look like a 2x2 square:

AB
CD

Therefore, with six people, there will always be a row of three. And thus, this is proven.


2)
No. I cannot be bothered to solve and write this one out because I'm on my phone, but I'd assume it will always take an event amount of switches to get where you started. Doesn't look like a very difficult problem if you had pen, paper and ten minutes.

[Anonymous No.15309030]

>>15308572
>It’s not from a “10th grade mathematics exam”, or, if it is, the question was poorly chosen because it is certainly not 10th grade level.
you're right, it's actually lower than 10th grade level. either way - nice cope

[Anonymous No.15309070]

How do I get good at solving riddles?

Is there a starting riddles program? What are some books that'll make me good at solving riddles?

[Anonymous No.15309084]

>>15308300
your diary desu

[Anonymous No.15309104]

>>15309070
>How do I get good at solving riddles?
be born with good genes. if you aren't naturally good at math and logical reasoning, you must instead choose yourself an edgy cringe ideology to conceal your midwit brain. perhaps Marxism will be more your speed?

[Anonymous No.15309116]

>>15309029
lmao you're low IQ

[Anonymous No.15309128]

>>15309104
>if you aren't naturally good at math and logical reasoning
I'm studying math, but I want to get better at these recreational puzzles. Any book progression you guys recommend?

Thanks for the edgy response

[Anonymous No.15309154]

>>15308909
Massive retard detected.

[Anonymous No.15309161]

>>15308990
Sounds like you're embarrassed to be an egregious moron who can't solve a simple brainteaser.

[Anonymous No.15309207]

>>15308370
lol OP is retarded

[Anonymous No.15309212]

>>15309161
>>15309104

Jesus Christ this is a nasty place ahaha

[Anonymous No.15309274]

>>15308336
>There is a light bulb and a consciousness switching machine. Two people can enter the machine and switch their consciousness. Whenever two people switch their consciousness, the light bulb switches (on->off and off->on). You have access to as many people as you want. The light bulb starts off being off. Is it possible to using the machine in such a way that in the end, everyone's consciouness is back to where it started but the light bulb is on. If yes, give a method how to do it, and if not, explain why.
Xx = Person X with consciousness of X.

You need an even number of switches to get OFF to ON, but an odd number of switches to get a person's consciousness back to themselves. So there is no number of switches that satisfies both [can't be an odd number and an event number at once]

Did I do it?

[Anonymous No.15309277]

wittgenstein makes me feel v dumb

[Anonymous No.15309290]

>>15309274
Sorry, other way around, change odd/even in my post.

[Anonymous No.15309296]

>>15309274
>You need an even number of switches to get OFF to ON
You mean an odd number of switches?
>but an odd number of switches to get a person's consciousness back to themselves
Why is that true?
For one person that's generally not true. A switches with B, B with C, A with C. Then A has his consciousness back but an odd number of switches has occured.

[Anonymous No.15309304]

>>15309296
But you created unmatched people, and the problem states no one is unmatched in the end.

[Anonymous No.15309332]

>>15309304
Yeah, so the question is why you need an even number of switches so that no one is unmatched in the end?

[Anonymous No.15309438]

>>15308300
Explain a brainlet what this book is about

[Anonymous No.15309465]

>>15308336
the second one is easier. We will have 3 people, a,b,c. and the light is l.
for starters we have:
a=1;b=1;c=1;l=0;
then
a=0;b=0;c=1;l=1;
a=1;b=0;c=0;l=0;
a=1;b=1;c=1;l=1;

now for the first one, i will try a different approach:
with 6 people we can get 20 subgroups of 3. 6C3=20.
alright let x be the number of subgroups in which all people know eachother, y be the number of sungroups in which everyone is a complete stranger to everyone and z be the whats lefts.
x+y+z=20
know to prove what we want, that there must be atleast x=1 or y=1, that means that z cant be 20, right?
~(P OR Q)= ~P AND ~Q, so x and y are zero.
if everybody knows eachother, then x=20. we can also get y=20. in none of those cases z=20. but those are the extremes cases.
lets examine what is z made of. z is then made up of 1 one guy who knows the other two, and they are strangers to each other or two people who know eachother and 1 stranger. this might be a bit hard to follow i guess.
lets start with the first case, lets say that A knows B and C, but B and C dont know eachother. so its only enough to A and B/C to know one other person in common to make up a group of 3 people who all know eachother, for z to be smaller than 20.now things get more interesting
rememberm A knows B and C, and they dont know each other. Lets add some other person D
lets say that both A and B/C dont know D, then z gets smaller than 20. again. but what if A knows D but B doesnt? then C is the key. if C knows D, then we get a subgroup of A,C,D that all know eachother. If C doesnt, then we get a subgroup B,C,D with are all strangers.
The same logic can be applied i hope i dont need to go on further. so z must be less than 20, therefore x or y need to be atleast 1, which proves the question.

im also the anon who said (0.75, 0.75,2), where is my nobel?

[Anonymous No.15309485]

>>15309438
Yes

no

maybe

[Anonymous No.15309507]

>>15309465
pardon me for bad english, hope its alright :)

[Anonymous No.15309529]

>>15309465
>im also the anon who said (0.75, 0.75,2), where is my nobel?
You are correct about that, so congratulations on solving it!
For the second problem I think you misunderstood it. What does
>a=0;b=0;c=1;l=1;
even mean?

Having a hard time reading your proposed solution to the first problem, so perhaps later.

[Anonymous No.15309534]

>>15309465
clever sophistry, but no. I take it that when 2 variables change their assigned boolean values then that's supposed to mean they swap their conciousnesses. if so, then your solution to 2nd challenge is clearly wrong as even though the variables a,b,c have the same boolean values assigned to them in the final step they nonetheless have conciousnesses other than ones they started with. in the first step A and B swap theirs so A has the conciousness of B and B that of A. in the 2nd step C and A switch so A has now conciousness of C and C has conciousness of B. in the last step B and C switch therefore B now has the conciousness of B and C has the conciousness of A which is clearly not the result with wanted.

[Anonymous No.15309541]

>>15308336
obviously all 6 people know each other because they're in the same room

[Anonymous No.15309549]

>>15309541
Have you ever been to a party?

[Anonymous No.15309556]

>>15309541
HOLY BASED, mathematicians and their fake and gay problems eternally BTFO

[Anonymous No.15309573]

>>15309549
Where do you think we are?

[Anonymous No.15309659]

>>15309534
Ah i thought switching meant turning on/off like a light bulb. my bad. when i have time i will do the correct answer then.

[Anonymous No.15309675]

>>15309541
>>15309573
BASED AND WITTY PILLED

[Anonymous No.15309676]

>>15309659
Like switching your consciousness on/off? LOL. How did it make sense in your mind? I'm really interested in hearing this.

[Anonymous No.15309680]

>>15309676
I mean switching consciousness is already sci fi enough so why not?

[Anonymous No.15309688]

>>15309680
I just wanted to know that your post made me laugh. What do you think would happen when your consciousness switches off?

[Anonymous No.15309697]

>>15309688
That's when daddy puts the cream on me.

[Anonymous No.15309705]

>>15309688
no problem fren :)
solving puzzles and making people laugh is always a comfy time.
the machine would then be a punching glove that knocks you out!

[Anonymous No.15309714]

>>15309207
are you fucking surprised?

[Anonymous No.15310382]

>>15308684
>>15308689
The question asked if it's possible to have to lightbulb be on while everyone is back in their own bodies. I think it's not possible because the methods I used all resulted in an even number, but I'm no mathematician so there's no need to be rude.

[Anonymous No.15310400]

>>15310382
If your answer is "No", then you have to provide a proof. Giving an example only works if you are answering "Yes".

[Anonymous No.15310570]

>>15310400
I haven't done maths since I was 15, so I'm sorry I can't elaborate more. I was also busy with a barbeque with my family so I didn't have much time to spend on it. But I find it strange that no a single person, many of whom are probably STEM students or at least contain some erudition for arithmetic, have not solved what's been deceptively disguised as a simple logic puzzle. Here's my 5 minutes of working I did on MSpaint if you want a layman to laugh at.

[Anonymous No.15310641]

>>15310400
Yeah im also starting to think that its impossible. our number of chances needs to be an for the light bulb to be on.
So if we have the person A with the mind 1, being its own, and B2, C3 and so on for any number of persons we start get this permutation (pay attention only to the numbers and their positions):
{1,2,3,4,..,n}
and we want to finish in the same place
{1,2,3,4,...n}
well the permutation is the same, so it seems that we did zero moves, if we only look from start to finish. but there is this theorem that says that even if a chain of switches might have different number of switches,all of the possible paths must have the same parity. Therefore, as zero is ussually classified as even, all possible paths to the start will need an even number of switches, thus being impossible by the end to have the light turn on.
even if you disagree that zero isnt nor even nor odd you will find that there is a way with even number of steps, independent of n, so all of them are. the parity is invariant.

[Anonymous No.15310661]

>>15310641
needs to be an odd number*