discrete-math

[west hedge]

also the extra (not)subset/element of emptyset things you wrote in each line are unnecessary and/or wrong

[weary tiger]

this is torture

[west hedge]

maybe it will be helpful if I show you a complete proof of this part of the problem
then you can try to write your own proof of the other half of the problem

[weary tiger]

that will be awesome

[west hedge]

here is my half
this is what we're trying to prove: if E and F are both non-empty, then E x F is non-empty
so assume E is non-empty and F is non-empty. then there is an x in E and a y in F
then you have a pair (x,y) in E x F, so E x F is non-empty

remember that this was the contrapositive of if E x F is empty, then E is empty or F is empty, so this is the half of the original problem that is now complete

now you try and write a proof for the other half that looks similar

[weary tiger]

how do we write that e is empty

like E = symbol fi?

[west hedge]

$E = \emptyset$

[vital dew]

Seoin:

[weary tiger]

and we write "if" in the problem?

[west hedge]

where?

P if Q means Q -> P
P iff Q or P if and only if Q means P -> Q and Q -> P
your problem is E x F is empty if and only if E is empty or F is empty

[weary tiger]

""if E and F are both non-empty, then E x F is non-empty"" yea but how do i write these in the problem

[west hedge]

what's wrong with writing it the way you wrote it just now?

[weary tiger]

we write it in steps

oh u are telling me discrete methods in sets is subjective?

[west hedge]

the proof should have a few steps, as you can see from the proof I wrote

[weary tiger]

For example in the course we write steps like these

We dont use english

[west hedge]

I see let hence let then

if you just chop a few words out of what I wrote it will pretty much look like that

Let x in E, y in F. Then (x,y) in E x F.
This is pretty much all I did

[weary tiger]

Is it like that?

[west hedge]

yup exactly

[weary tiger]

thats half way through?

[west hedge]

yes that's half of the problem

[weary tiger]

thats a scary amount of weird subjective logic going in there

[west hedge]

you still need to prove if E is empty or F is empty, then E x F is empty

[weary tiger]

let me try and do that

:D

oh so the "x" to come between ExF has to be "and"?

it doesnt work with or?

[west hedge]

what do you mean?

[weary tiger]

I mean that

[west hedge]

Let x in E or y in F this does not make much sense following E is empty or F is empty

you should use the contrapositive for this part too

the appeal of using the contrapositive is that you get to assume something is non-empty which gives you an element to work with, instead of assuming something is empty and then not having anything to say

[weary tiger]

ah

i had no idea that its allowed to imagine things

the steps are wrong?

excpet the

[west hedge]

yes I don't see any hope for the proof you started to write
I would write the contrapositive of what you're trying to prove and start over

[weary tiger]

all i have to do is make E is not an empty set?

[west hedge]

this is what you're trying to show now if E is empty or F is empty, then E x F is empty
what is the contrapositive of this?

[weary tiger]

E is not empty or F is not empty, then x belongs to E and y belongs to F , then x,y = ExF is not an empty set?

[west hedge]

this is the other half of the problem, which we already completed

[weary tiger]

how can we prove something that has separate values

and we cant combine

[west hedge]

like I said, you need to find the contrapositive of if E is empty or F is empty, then E x F is empty to get started
you said you understood when we did this for the first half of the problem, so you should be able to find it

[weary tiger]

why am i having a hard time figuring out such problems when in calculus i could figure out hard problems on my own

discrete really sucks

its too subjective

@west hedge Can you help me do it and call it a day? thats the only one remaining for me

the doctor should have explained it

u are doing better than his job already

[west hedge]

I would go back and look at the discussion we had about the other half of the problem
follow the same steps, it will be very similar
I can't do both halves for you

is this a true/false question?

do you think it's true?

how come?

10 doesn't divide 2, and 10 doesn't divide 5
but 10 divides 2*5

maybe you can try to do something similar with x and 22 instead of 2 and 5

well you said you don't think it's true
to prove that it's not true, you only need to find one x for which it fails

[weary tiger]

I need to know the 2nd steps to be able to do similar problems, i need to grab the logic of the problem in both ways

[west hedge]

so pick a good x

[weary tiger]

Is this correct?

[west hedge]

yu

yup*

[weary tiger]

Really?

[west hedge]

yep

[weary tiger]

Yayyyy

I couldnt have done it without u

Who knew contraposition proof was so effective in nulifying empty void numbers

But

Its e not empty or f not empty

I wrote it as "and"

Does it matter?

[west hedge]

both are non-empty, in the conclusion of what you wrote in the picture

if you take the contrapositive again, to get back to the original problem statement, this and will become an or

[weary tiger]

oh yeah

U have been such a huge help omg

[fresh flame]

Does the law of excluded middle work in set theory?

I.e., can I say $x \in A \lor x \in \bar{A}$?

[vital dew]

That guy:

[weary tiger]

how to prove that any tree w/ all leaves removed is still connected?

oh ok

nvm

i can just say its still e = v-1

right

then tree iff e = v-1

[weary tiger]

$\equiv$

[vital dew]

N/𝔄:

[weary tiger]

[sleek swallow]

@fresh flame Sure, set theory makes use of logic as its basis. Certainly, $x \in A$ is a proposition iff x is a set and A is a set. So, you can form an combination of propositions you want, via the standard propositional connectives

[vital dew]

Abhijeet Vats:

[fresh flame]

@sleek swallow thanks, someone answered my question earlier in #discrete-math

Forgot to delete here

[sleek swallow]

Ah I see

[weary tiger]

Hello

I am trying to do the e=f, is it so far correct?

[stray reef]

no

you kind of pulled the "then E = F" out of nowhere

@weary tiger

[weary tiger]

Ah, i wonder what i should do next

Because how do i even expand it

Could i say this instead?

[stray reef]

no you're pulling shit out of nowhere again

you should scratch all the stuff below "assume E, F nonempty"

your goal is to prove E = F. how do you prove that two sets are equal?

[weary tiger]

How do i know?

We never took something like that

I will try one last method

[stray reef]

let x ∈ E, then x ∈ E×F

nope

How do i know?
how do you know what

do you not know that to prove E = F you need to prove E ⊆ F and F ⊆ E

[weary tiger]

No

[stray reef]

what the fuck

this MUST have come up in class

like

how to prove two sets are equal

the defn of set equality

[weary tiger]

He told us to look it up

[stray reef]

what

wait like

have you done any set theory proofs before at all

bc i'm a bit flabbergasted here

[weary tiger]

Thr very basics

[stray reef]

surely "the very basics" would include proving stuff of the form "<this set> = <that set>"

or did they not in your case

[weary tiger]

We only use contradiction and prove by cases

But the idea of two sets equal has not been shown before

In the problems

[stray reef]

and it hasn't been covered in class at all?

not even a mention?

this is mighty weird for sure.

[weary tiger]

Its a homework, all our homeworks are impossible to solve lol

Because its graded he makes it so much harder

[azure lichen]

well, anyway, the usual strategy is to show that each set is a subset of the other, i.e. choose an arbitrary element from one set and show that it’s in the other as well

and then vice versa

[stray reef]

i mentioned it already

i admit, if this is unfamiliar to you, then even me walking you through the proof may be a bit unproductive

but i can try

[weary tiger]

I can have all the time in the world after the deadline in 12 hours lol

This one is the only remains

[stray reef]

okay so would you like me to give you a proof and then have you read it and tell me which parts of it do not make perfect sense to you

[weary tiger]

Yea

That would be perfect

[stray reef]

okay

ORIGINAL STATEMENT:
Let E and F be sets. Then E × F = F × E if and only if E = ∅ or F = ∅ or E = F.

PROOF:
The leftward implication (if E = ∅, F = ∅ or E = F then E×F = F×E) can be verified via considering the three cases:
If E = ∅, then E×F = ∅ = F×E. The same happens if F = ∅.
If E = F then E×F = E×E = F×E.

We prove the rightward implication by proving that if E and F are both nonempty and E×F = F×E, then E=F.
To this end, we set out to prove E ⊆ F and then F ⊆ E.

Let x ∈ E be arbitrary. F is nonempty and as such contains an element f. Therefore, (x,f) ∈ E×F by definition of set product.
Therefore, (x,f) ∈ F×E since E×F = F×E.
Therefore, x ∈ F, again by definition of set product.
Therefore, E ⊆ F, as we have shown that an arbitrary element of E is also an element of F.

Now let y ∈ F be arbitrary. E is nonempty and as such contains an element e. Therefore, (y,e) ∈ F×E by definition of set product.
Therefore, (y,e) ∈ E×F since F×E = E×F.
Therefore, y ∈ E, again by definition of set product.
Therefore, F ⊆ E, as we have shown that an arbitrary element of F is also an element of E.

Therefore, E = F, as we have shown that E ⊆ F and F ⊆ E.

Q.E.D.

this is a bit verbose but i have minimized glossed-over details

@weary tiger i'd like you to read this with utmost care. read each line very, very carefully and make sure you understand exactly what it says. if there is a line that does not make sense, then you should stop reading, ping me and quote the line that didn't make sense and ask me about it.

[weary tiger]

The x,f and x,e, we only took( x,y) in class

I had no idea we had to go this deep to solve this problem

Is the doctor out of his mind?

@stray reef thank you a million tons!

[stray reef]

uh

f and e are just

letters

[weary tiger]

So its like

X,y ,z and t

[stray reef]

i could've given my variables other names

[weary tiger]

Like we need 4 unknown variables?

[stray reef]

not really

i could've said "now let x ∈ F be arbitrary" and continued the last 1/3 of the proof that way

the only names i have no control over are E and F

[weary tiger]

In the proof is there a way you can make it only x,y?

[stray reef]

i just chose to use different names to help alleviate potential confusion

i mean does it matter what names your vars are tho

as long as you know what's what, you can name things whatever you want

[weary tiger]

Ah

[stray reef]

and you don't give the same name to two different things ofc

i mean i can explain some of my naming choices from an informal standpoint if you want

or a stylistic standpoint ig

[weary tiger]

Yea its crazy to think our doctor was annoyed because a student choose a different variable point

I think it helps him correct the exams faster

But thats just how it is in college life

Lol

Maths is not my major, its CS and i am confusrd how these statements or bool are going to help me improve my programming skills

[stray reef]

i mean this particular problem won't help you any more than a single gym session would help you get buff

this is the mathematical equivalent of a workout

a brain workout if you will

it helps you stretch your brain to like. think mathematically

[weary tiger]

I have lumosity for that lol

[stray reef]

nah but proofs tho. it helps you get into that mindset

where you can for example

prove that your program is working correctly

[weary tiger]

It makes me think in black and white

Thats what it felt like

[stray reef]

or prove that your algorithm has such and such complexity

there are some things in math that are black and white

[weary tiger]

The subjectivity of this course is what bothered me

Its just too subjective

[stray reef]

what about it is too subjective

[weary tiger]

For example if 1+1=3, then 5+7=12

Thats clearly false because the patterns dont match

However the logic rule in other patterns are different

For example 2+2=5, then 3+4= 670123

Thats false because its too far apart

[stray reef]

For example if 1+1=3, then 5+7=12
Thats clearly false because the patterns dont match

actually no

"if 1+1=3, then 5+7=12" is a true implication

as is "if 1+1=3, then 5+7=0"

as is "if pigs can fly, then i'm a millionaire"

[weary tiger]

Thats not how i thought about it

[stray reef]

"if [false statement], then [blah blah blah]" is true always

[weary tiger]

I thought about it like a cross multiplication thing

[stray reef]

well ig that's something to work on

bending your old mindset

[weary tiger]

So the mindset is related to the course material hmm

Yea thats the thing that bothered me

Thank u a ton again!

[faint narwhal]

I mean, this is the way you should be thinking about if statements in CS too

[weary tiger]

Yeah but i can bend it to however i want

[faint narwhal]

uh what?

[weary tiger]

@faint narwhal Like i can shape the loop to however i need. It doesnt have to be a bool (yes or no)

[faint narwhal]

but just the concept of false implies true being a true statement is important

[weary tiger]

Yes, if has only one mindset, and that is condition

Unlike sets which has a certain way to solve

[faint narwhal]

What are you even trying to say

When presented with a programming problem, it won't be about one single thing, and there will be many right ways to solve it

[obsidian kraken]

@weary tiger do you understand that a computer program will run from top to bottom, line after line, and that a statement like if(x>2) implies that x is currently holding one definitive number value?

[faint narwhal]

that's really not what I was talking about

I was talking about the "false implies true" is true

[weary tiger]

I want to end the conversation here its getting pointless

Programming is better than discrete methods

If u ask me discrete math ruins my logic

[dusk laurel]

how would i go about proving it using a bijection?

[obtuse lance]

by constructing the bijection explicitly

[dusk laurel]

what does that mean?

[obtuse lance]

just make the bijection

[dusk laurel]

yea idk how to choose f

[obtuse lance]

ok instead of those two sets, can you split the integers up into two infinite sets?

what's the simplest way you can think of to split integers up into two infinite sets

[dusk laurel]

yes even and odds

[obtuse lance]

ok good

so now associate an even number to every one in the first set

and an odd number to every one in the second set

and you're done

[dusk laurel]

mmm

i dont get that part

how would i show a one to one correspondence even the evens and the first set?

like should there be a rule to map the evens to the first set?

[obtuse lance]

well the first set is a map from n to 1/n

so you need to take your even numbers which are of the form 2k and just divide by 2

similarly for the odds subtract 1 and divide by 2

[dusk laurel]

yea i got it

youve been of great help

thanks a lot

[weary tiger]

what was it

whats the function

im curious

[obtuse lance]

you're welcome

[dusk laurel]

i have another one tho

if you can still heklp

idk how to choose the cases for indction

unless im totally wrong and can't use induction at all

specifically for the last case where m>0 and n>0

i proved it for m=1, n= 1 but i dont know how to use it prove for all m,n in N x N

[weary tiger]

its induction

[dusk laurel]

how do i choose the cases?

[weary tiger]

base case is 0

[dusk laurel]

yeaaa

[weary tiger]

prove m+1 and n+1

just like any other induction problem

[dusk laurel]

to proe the last one?

where m and n are greater than 1

oh crap nvm i got it

thank you

[obtuse lance]

what you have it backwards

this is a problem which doesn't seem like a typical combination problem

and this is a cool trick to turn it into one

that's how you should be appreciating it lol

[lyric pumice]

@neon thorn Using the standard reasoning for choosing subsets, you can think of generating the multisets as follows. You have 4 different objects of a type that represents different kinds of pastries and you have 6 different objects (duplicators/replicators) of a type that represents how the pastries are multiplied. A multiset is created by choosing 4 of these 10 different objects. The type objects appear in an order, and picking the i'th replicator produces whatever object appears on the i'th pick. So, suppose we construct the multiset AAABBBC from the example that you have provided. Pick A, B, and C. Let them appear in alphabetical order. Pick the 1st and 2nd replicators to produce AAA. Pick the 4th and 5th replicators to produce BBB. Then we are done.

[lyric pumice]

@neon thorn You are considering two different kinds of problems.

One deals with an application of order, and the other does not.

Indeed, these problems invoke underlying abstractions with respect to sets and ordering.

[sleek swallow]

Using a bijection? No, you establish that you cannot form a bijection.

[faint narwhal]

read it again

[lyric pumice]

@neon thorn Traditionally, combinatorics is often unintuitive because the organization of fundamental principles is not explicit.

[faint narwhal]

its not a bijection from A to P(A)

[sleek swallow]

g(A) is the image set of A under the function g

[lyric pumice]

@iron crescent Each element of A is mapped to a subset consisting of only that element.

[sleek swallow]

So, you can identify elements in P(A) that don't have a pre-image. For example, consider two element subsets. So, g: A -> P(A) is not surjective and, hence, cannot be bijective.

g(A) = {g(1),g(2),g(3)}

P(A) = {emptyset, {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}}

[faint narwhal]

$g(A) = {g(a) \mid a \in A}$

[vital dew]

Zopherus:

[faint narwhal]

so what would g(A) be when A is {1,2,3}?

[sleek swallow]

Your proof

Has a definition for g

[faint narwhal]

and what is that?

[sleek swallow]

g(x) = {x}

Use that

Also, g(A) isn't g(1),g(2),g(3). It's g(A) = {g(1),g(2),g(3)}. This is a set.

No!

g(1) is the image of the element 1 from the set {1,2,3}

[lyric pumice]

@neon thorn Those methods are the explanations.

[faint narwhal]

Just because its visual doesn't mean its not rigorous

you can rigorously give a bijection between the number of ways to do the encoder/divider and the number of ways to do your combination w/ repetition

and this bijection will tell you that the two numbers are the same

[weary tiger]

There are n items on the circle. Show that number of ways of choosing k of them in a way that none two are next to each other is equal $\frac{n}{k} \binom{n-k-1}{k-1}$

[vital dew]

Godel:

[weary tiger]

I know induction probably works pretty well, but anyone sees the combinatorial proof for that?

I thought it may have to do with cutting the circle somehwere and putting items in line and multiplying by n to get all possible lines, but not sure (I think so because of the n in the fraction)

[faint narwhal]

Because the questions are just different, in combinations with repetitions, you have to distribute all your items, but in this, you're only choosing 3 of them to distribute

And you can only give one book to each shelf space

[lyric pumice]

@neon thorn If you are assuming that the books are distinct, then the correct number is 3^4=81.

[lyric pumice]

In that case, you are dealing with permutations of multisets.

[smoky heath]

https://gyazo.com/92ad4bc1d045aafad529ad6872603d13 for this question, how can i map everything in the domain to the codomain if n and m are different?

[faint narwhal]

why do you thin that causes a problem?

[smoky heath]

the sizes of the sets would be 2^n, and the hint is telling me to map everything from the domain to everything in the codomain but they are different sizes?

[faint narwhal]

Can you find a function from {1,2} to {1}?

[smoky heath]

would it be to the power of 0

[faint narwhal]

what?

[smoky heath]

like 1^0 goes to 1 and 2^0 goes to 1... idk if im dumb

[faint narwhal]

I'm asking a question that's not related to the question you're working on

Can you find a function from {1,2} to {1}?

[smoky heath]

uh yes?

[faint narwhal]

And the function is?

[smoky heath]

n^0?

[faint narwhal]

What does that mean?

[smoky heath]

you map different sizes to each other

[faint narwhal]

Do you understand what a function is?

[smoky heath]

i think?

[faint narwhal]

Okay, then what is your function from {1,2} to {1}?

[smoky heath]

f: x^0 -> f(x)?

like hwat

[faint narwhal]

That makes zero notational sense

I feel like you mean to write that f(n) = n^0

or in other words, f(n) = 1

[smoky heath]

yeah

ok

[faint narwhal]

Anyways, this is a function from sets of different size

so n and m being different doesn't matter

[smoky heath]

what if n < m

so u map a smaller set to a bigger set

[quasi igloo]

how can you map everything in the domain to the codomain if the size of domain > size of codomain

[faint narwhal]

Can you find a function from {1} to {1,2}?

[smoky heath]

no?

[faint narwhal]

Why not?

[smoky heath]

cause u cant have something in ur domain go to more than one thing in codomain

[faint narwhal]

Sure that's right

But why does that stop the mapping

f(1) = 1

from being a function?

@quasi igloo Because you can have multiple things from your domain go to the same thing of your codomain

[smoky heath]

ohh

ok thanks

[weary tiger]

Can anyone help me with my HW? I’m failing this class and I need some help badly

[reef thistle]

Please don't crosspost

#help-7｜zen1thxyz for anyone interested

[weary tiger]

<@&286206848099549185>

[quasi igloo]

@weary tiger converse just means the other way, so instead of y implies x or whatever u have its x implies y

[umbral solstice]

Really cant figure out how to solve (87+65) *43 (mod 21) on two different ways.

[deft patrol]

anyone good and comfortable with discrete maths?

[near prawn]

[deft patrol]

peux tu m'evoyer un msg prv

[near prawn]

post it here, cuz I gotta go soon

[lilac pivot]

this seems like a #probability-statistics

question

I can help you there

[vocal solar]

What is discrete math?

[pale epoch]

math that deals with discrete (as opposed to continuous) structures

[river reef]

like SETS

[pale epoch]

literally what

[gleaming zephyr]

@pale epoch can you tell me what does dsicrete structures mean

i was always curious about this

and what are continuous structures

i know a few examples like graphs or trees but like idk what they are

[pale epoch]

discrete structures are finite or countable

like graphs we mostly care about the ones with finite amount of vertices

[gleaming zephyr]

finite in what way

finite in order for example?

the only structure ik ever is gropus

[pale epoch]

yeah

[gleaming zephyr]

col

[pale epoch]

a graph is a structure as well

just a set with some additional information

but the graphs we mostly care about are finite sets

[weary tiger]

Graph theory?

Or do you mean for functions?

[pale epoch]

as in graph theory

[gleaming zephyr]

cool

ty

[fiery pivot]

Hadwiger and Nelson would like a word

[weary tiger]

Disregard the "e"

I'm trying to find the characteristic equation of the homogeneous linear recurrence relation above

^ This is the form we are given for the characteristic equation

^ This is the answer we are given

Why is it 8r^2, shouldn't it be 8r^3 ?

Because k is 4, and we have c1*r^(k-1)

Oh

Is it because 8*a(n-2) is actually c2*r^(k-2)

[weary tiger]

What is this question even asking?

[sour arrow]

No

[stray reef]

everyone's dead

coronavirus has taken over the world

[quasi tiger]

hey i need help with context free grammar , "language theory" , i have solutions but not sure if they're correct !

oh , am having trouble with these types of exercices , is there a method to follow , as i've said the trial and error isn't taking me far in this !

okay can i post my solutions here so you can check if they're correct ?

i know 😅

it's giving an extra number at the end

Oh so it's not récursive

Oh... Crap. 😮 , should i add another non term variable

Sorry , i will

[quasi tiger]

for this, is : a* U b* a valid solution

?

[quasi tiger]

i dont understand, is it correct ?

[weary tiger]

What are some good resources on recurrence relations

I'm learning about characteristic equations

Homogenous, non-homogenous

Idk what this represents in real life

But I can apply the steps to solve

[lyric pumice]

@weary tiger Hello. I have found a nice combinatorial proof.

[lyric pumice]

@weary tiger I just realized that I have assessed the problem before and there is a second, simpler combinatorial proof.

[weary tiger]

Ok mind showing?

[short wyvern]

So i had a task where i needed to prove an existence of a particular type of a set. On math exchange i was proposed to look at a given set

but there is a consition if $!X \in !S$, then $!X \sim \mathbb{R}$

[vital dew]

DoubleKnot:

[short wyvern]

and i am not sure if this kind of set actually fits it

Because as far as i understand, we take just {x} as an element and it would belong to S, yet its cardinality wouldnt be equinumerous to $\mathbb{R}$

[vital dew]

DoubleKnot:

[weary tiger]

Can someone help with counting elements in a set problems

[pale epoch]

you could try just asking

[weary tiger]

"There are 350 who signed up for MAT210 and MAT265. Also, 120 students are signed up for MAT210 and 275 students are signed up for MAT265."

a) How many students are signed up for MAT210 AND MAT265 => 350 ?
b) How many are signed up for MAT210 and NOT MAT265 => 120 ?
c) How many are signed up for MAT265 and NOT MAT210 => 275 ?

I'm skeptical if this is good because the answers are literally in the sentence

[azure lichen]

okay I read it about 5 times but the only way it actually makes sense to me is that the first number is supposed to be “total number of students who are in at least one of those two classes”

the use of “and” in that sentence is super confusing though

but basically in the second sentence, that 120 tells you how many students are in MAT210, but you don’t know whether those students are also in the other course

(not right away at least, you can obviously calculate the numbers)

@weary tiger

[weary tiger]

I'm translating from french

"Il y a 350 etudiants inscrits aux cours MAT 210 et MAT 265"

There are 350 students enrolled in MAT 210 and MAT 265

@azure lichen

and then the a) is "a) Combien d’etudiants sont inscrits a la fois aux cours de MAT 210 et MAT 265 ?"

In english: "How many students are enrolled in both MAT 210 and MAT 265"

[south knoll]

My guess if I may is that it’s (120 + 275) - 350 = 45 students for the first question
To simplify things :
Let’s say there are 9 students in total, 6 students in group A & 5 in group B
The students that are in group A & B would be 2

@weary tiger

For the rest of the questions take for example the 120 that are in MAT210 & eliminate the number of students that are also in MAT265 which would be 45
So you’ll have : 120 - 45 = 75

Hope this helps 🙂

[weary tiger]

@south knoll Thanks a ton

So we got 350 students total

45 are in both mat210 and mat265

75 are in mat210

230 are in mat265

[south knoll]

I believe you’re correct
& you’re welcome 😁

[open narwhal]

how does the definition of congruence turn a = b mod n and c = d mod n turn into n | (a-b) and n| (c-d) or am i just doing this wrong

ignore that. had a brain death moment

[open narwhal]

I need to do a proof by contradiction. where the main statement is There is no smallest positive real number.
The Negation should be: There exists a smallest positive real number.
But what do i do next?

[wary nexus]

@open narwhal I’ve given you a hint in the question channel you asked in

[stray reef]

let's say one person in particular is named Bob

consider that Bob has 3 enemies or 3 friends within the group

then consider the relationships among these three

[stray reef]

"consider a particular person in the group named bob. since bob has a total of 5 connections within the group, he must have at least 3 enemies or at least 3 friends, as he cannot have less than 3 of both.
suppose he has 3 friends; consider the relationships between those three. if any two of them are friends, then they form a friendship triangle with bob. if all three are enemies to each other, they form an enemy triangle.
the case where bob has 3 enemies is entirely analogous, but with the words friend and enemy interchanged."

no

well

okay what the FUCK is up with your placement of "either"

if he has 4 friends then pick any three and apply the same argument

if he has 4 friends then pick any three and apply the same argument

what

no you're overthinking it

pick any three of his friends

and apply my argument to them

my argument considers the connections i've drawn with dotted lines

if any of them is a friendship, you've got a friendship triangle with bob

if none of them are friendships then you have an enemy triangle

do you reject the fact that either there IS a friendship among the dotted relationships or there ISN'T a friendship among the dotted relationships

or is it the fact that i arbitrarily chose three of bob's four friends that is scaring you

there's something you're misunderstanding

any pair of people in this goddamn group of 6 are either friends or enemies

if they are not friends then they are enemies

if they are not enemies then they are friends

how hard is this to understand

if your lines represent friendships

yes, my use of "dotted" was a mistake and i edited it out

if your lines represent friendships

then considering the three people you marked as bob's friends

do you or do you not agree that either there IS a friendship between two of these three people or there ISN'T

i didn't hear a "yes, i do agree" nor a "no, i do not agree"

ok great

so

IN THE CASE that there ISN'T a friendship between any two of these three people

the three people are all ENEMIES to each other.

do you or do you not agree with this

i didn't hear a "yes, i do agree" nor a "no, i do not agree"

ok great

so as you just saw

IN THIS CASE, there is an ENEMY TRIANGLE.

does it matter

you aren't asked to show that there exists one and only one enemy triangle

there's a difference between existence and uniqueness

this didn't occur to you?

there need not be one

you are not asked to show that there always exists an enemy triangle AND a friend triangle SIMULTANEOUSLY.

all you're asked to show is that at least one of these is guaranteed to eixst

that's math in general for you

details MATTER.

here why don't i draw them for you

this isn't paint, soap

[loud copper]

😔

[stray reef]

@iron crescent

does that answer your question

each column consists of 3 points each of which is painted one of 2 colors

therefore the number of possible column colorings is 2 * 2 * 2

or 2^3

or 8

since there are 9 columns but only 8 possible patterns for a column to have

by the pigeonhole principle there must be two cols that are identical

context?

uh

this looks like it's supposed to be capital i

not a stick

identity function

$I_A$ is the identity function on $A$

[vital dew]

Ann:

[stray reef]

i.e. $I_A : A \to A$ is defined by $I_A(x) = x$

[vital dew]

Ann:

[stray reef]

uh

...

.........

.......................................

i didn't even mention any function named f

it is a bijection from A to A but this is not a complete description of what it is

[craggy gale]

you can show that gcd(b, d) divides gcd(a, b) and that gcd(a, b) divides gcd(b, d)

[glossy relic]

hey there ! i have an alphabet sigma with two elements and i've been asked how many elements there are in sigma to the power of 0, for me that's an empty set since there should be no elements but i saw a video on youtube where that person followed a certain formula (2 to the power of the power asked ) so following the logic of that person , the answer for my question should be 2 to the power of 0 which is 1. Does that mean that the empty set itself is considered an element?

[stray reef]

Σ^0 consists of the empty string and nothing else

but the empty string is an element

[glossy relic]

oh and are the empty set and the empty string two different things?

like i know the empty set is when your set has no elements in it right?

[stray reef]

the empty set is the set that doesn't have anything in it

[glossy relic]

and that would an empty string be?

is it just something you write to symbolize null elements?

[stray reef]

the empty string is the empty string

[glossy relic]

okay , thank you ann

[weary tiger]

no

how do you know that there isn't a greater common divisor

if the greatest common divisor is 1, that means a-b and a+b are coprime

but lets take a=7 and b=5

5+7=12

7-5=2

gcd(12,2) = 2

well, you can see that i picked two odd numbers in my counterexample

maybe check the parity?

also according to this theorem, if a-b is divisible by any number greater than 2, a+b cannot be divisible by that number

like if a-b is divisible by 3, a+b is not divisible by 3

@iron crescent do you see a possible solution

ok

so lets say that

$a-b=0\ \equiv {mod}n$

[vital dew]

sanath1237:

[weary tiger]

ughh

latex

ok then

lets say a-b is divisible by a number

any number

the difference between a+b and a-b is 2b

a-b=2b?

how

heres what i was thinking

lets say x=a-b

then we're essentially finding gcd(x,x+2b)

ok so lets say that gcd(x,x+2b) = 3

that means that both x and x+2b is divisible by 3

this means that 2b is divisible by 3

which means that b is divisible by 3

right?

but if x is divisible by 3, it means a-b is divisible by 3

and since b is divisible by 3, a has to be divisible by 3

meaning that a and b and not coprime

you can say this for any odd number above 2

but replace the 3 with any odd number

so we've proven that gcd(x,x+2b) cannot be an odd number above 2

what about an even number above 2?

lets pick 4

if gcd(x,x+2b) = 4, then both x and x+2b are divisible by 4

this means that 2b is divisible by 4

which means b is divisble by 2

im saying that you can generalize what i said for all odd numbers

ok so lets say that gcd(x,x+2b) = n, and n is an odd number above 2
that means that both x and x+2b is divisible by n
this means that 2b is divisible by n
which means that b is divisible by n
but if x is divisible by n, it means a-b is divisible by n
and since b is divisible by n, a has to be divisible by n
meaning that a and b and not coprime

i just copied the proof for 3 but replaced all the 3's with n

and it still holds true

im just making it easier to understand

ok back to what i was saying

if a-b is divisible by 4, it is also divisible by 2

and if it is divisible by 2, and b is divisible by 2, a has to be divisible by 2

if both a and b are divisible by 2, it means that a and b are not co prime

once again we can generalize this proof for all even numbers above 2

coprime

also i imagine theres a simpler way to do this

oh wait

what about this

we said the difference between a+b and a-b is 2b

but the addition is 2a

the difference and addition between 2 numbers must divide the gcd of those two numbers

so 2a and 2b divide gcd(a-b, a+b)

which means gcd(2a,2b) divides gcd(a-b,a+b)

gcd(2a,2b) = 2gcd(a,b) = 2

2 divides gcd(a-b,a+b)

so the only posiblities are 1 and 2

sorry for jumping around so much

[weary tiger]

@south knoll

the prof sent an email saying the correction is: "There are 350 students enrolled to mat210 or mat265"

does that change what we did ?

[azure lichen]

that means what I said originally is correct (didn’t read the rest of the convo)

[weary tiger]

Is not h|(a-b) what you have to show?

[weary tiger]

You assumed h|a-b and the conclusion follows easily, though is that not what you’re trying to show? Unless this has been show by some other theorem in your book/course

[tacit marlin]

Can somebody explain why they are multiplying the first equation by 2?

[weary tiger]

@iron crescent just show that the implication if h divides a and h divides b then h divides a-b

Should be doable with the definition of divisibility

Because that is the crux of the argument

[sleek swallow]

@tacit marlin They want to solve the two equations in 2 variables simultaneously

[wispy osprey]

Hi. I'm learning enumerative combinatorics, and would appreciate help with the following problem.
I understand how there are 3^n ways to choose two subsets two subsets A and B of {1, 2, ...n} so that A ⊆ B.
I understand its proof based on 3 choices per each possible element from [n].
But I struggle to come up with a similar proof to the problem of number of ways to choose two subsets A and B of {1, 2, ...n} so that A ⊆ B, and |B| is even.

[stray reef]

hmm

[lyric pumice]

@wispy osprey You can count by constructing subsets so that a sum occurs over the possible cardinalities of B.

The solution should be easy to manage.

Make sure to properly formulate the upper bound of the summation.

[wispy osprey]

@lyric pumice thanks, I have done that, and that technique works (with floot(n/2) as upper bound), but I was wondering if there was a proof similar to the one with the choices

also with the summation you still need additional steps and proofs to get its closed form of (3^n + (-1)^n) / 2

[glossy relic]

hey , can a dfa have two same outputs (let's say two outgoing edges with the element 1)

[pale epoch]

no

transition is uniquely determined by current state and input symbol

[azure lichen]

if you have two same outputs then by definition it’s no longer a deterministic fa, but a nondeterministic one

[glossy relic]

then how am i supposed to make the b) if i have one one and zero zero (zero is an even number)

[azure lichen]

you’re supposed to count the number of times the element 0 shows up

it doesn’t matter that 0 is an even number

you can just like, have a state that counts whether 0 has shown up an even or odd number of times and switch between the states every time you read a 0

you can make a DFA with four states total that solves b

[stray reef]

am i correct that a DFA has like. an "accept" node

[azure lichen]

each node either is accepting or rejecting

[stray reef]

er

wait

[azure lichen]

and when it halts you just check where you are

[stray reef]

ah

er

so when the string ends

[azure lichen]

if you’re on an accepting node you accept, otherwise you reject

[stray reef]

right

[azure lichen]

(if you know you’re gonna accept before reading the whole string you just enter an unconditional loop on an accepting state)

[glossy relic]

are you sure that four states should be enough? i don't need the actual answer since i'd like to do it myself but the number of states would be a good help

and when i talked about zero being even , i meant that the element 0 appearing 0 times exists in that set too

[pale epoch]

4 cases are enough

there are 4 cases of 0 and 1 appearing an odd/even amount of times

[glossy relic]

by cases you mean states right? like q1 q2 q3 and q4

[azure lichen]

the empty string has both elements appearing an even number of times. when you read a 0, you now have 1 appearing an even number of times and 0 an odd number of times. and so on

[azure lichen]

…they just said they didn’t want the solution ann ffs

[pale epoch]

i mean if you observe a string and check if 1 appears an odd/even amount of times and 0 appears an odd/even amount of times

there are 4 total cases

and those will correspond to the states of the DFA

[glossy relic]

dw i'm not looking at anns solution even though i appreciate the help ^^

can a state get multiple same inputs?

[stray reef]

FUCK

sorry.

i should've spoilered this

bc i wanted to check my own sol

i guess?

[glossy relic]

it's okay

my current dfa totally does not have 4 states so ours won't be similar at all either way

[stray reef]

the "length 3 or more" thing

[pale epoch]

tbh i think it will be way harder if you have more than 4 states

[stray reef]

you can annotate states in any way you want right

[pale epoch]

but a state can have multiple inputs, sure

[glossy relic]

thank god

wait

maybe i didnt say it

but multiple inputs of the same element?

does that work

[pale epoch]

states are just a finite set, Ann

[stray reef]

a state can have as many arrows leading into it as you want

i think

[glossy relic]

because you guys told me multiple outputs of the same element from one state is not possible

[pale epoch]

also your solution didnt mark accepting states

[glossy relic]

so im not sure about inputs

[stray reef]

yeah it didn't

i think i should

[pale epoch]

nor start state

[stray reef]

but ||it's obvious by the annotations that L0 is the start state and L3 is the accept state||

[pale epoch]

yeah, but my prof would've subtracted marks

[pale epoch]

@glossy relic your transition function takes in one state and one symbol from the alphabet and outputs a single state

so you can't have 2 arrows with the same symbol leading from 1 state to other states

but you can have 2 arrows with same symbol leading into a single state

[glossy relic]

okay that's good news thanks

[pale epoch]

left is ok, right is not

[stray reef]

the right is like

[glossy relic]

[stray reef]

"ok i'm in this state and i see a 1 but where do i go "

[pale epoch]

it's possible for NFAs

[stray reef]

yeah but this is a DFA

DFAs can't handle this shit

[pale epoch]

yes, because the transition functions codomain is the set of states (as opposed to the powerset of all states)

[glossy relic]

This was my plan at first but then I realized that q2 would ruin everything for me after I reach q6,q9 or q8 since I would get an even number (for example 0011)

So i‘m just focusing on how to solve this problem right now

my teacher has not explained us what a dfa properly has to follow so if you find any actual understanding errors from my dfa, please tell me

[stray reef]

wait

so what are you doing rn

the "even 0s odd 1s" thing?

[glossy relic]

yeah

[stray reef]

,rccw

[stray reef]

wow uh

those are some very uninformatively labeled states

[glossy relic]

yeah but i was more focused on actually solving it than properly naming it

the naming would come afterwards

[stray reef]

if you want my thought process

[pale epoch]

so, technically this is a DFA

[glossy relic]

nice , at least we got a dfa

[pale epoch]

but this will not solve the problem

i see multiple trap states

and this makes no sense

[stray reef]

there are two things to keep track of here: the parity of the count of 1s and the parity of the count of 0s
and that's all we care about
that gives us 4 states, one for each combination of odd/even for the count of 0s and the count of 1s

[pale epoch]

because you can always create an accepted string from an unaccepted one by concatenating either a 0, a 1 or both

[glossy relic]

i made those trap states only because each state needs an output for each element right?

[pale epoch]

they do, yes

but what is reaching a trap state supposed to signify

in the exercise, its impossible to prematurely know whether a string will be accepted or not

[glossy relic]

well for me it signifies that reaching it will result in an incorrect concatenation of elements so he has to go back

[pale epoch]

you always have to read the whole string

there is no going back from a trap state

[glossy relic]

@stray reef eh , my thought process is completely different xd

[stray reef]

if you reach a trap state, it means that every string that starts with what you've read so far will automatically be rejected

[pale epoch]

Ann's approach is the better one

[stray reef]

i can repost my solution

[pale epoch]

there are 4 states, you start in (even, even)

[stray reef]

which actually falls out quite naturally

from what i said

[glossy relic]

yeah i don't need those concatenation of elements that reach a trap state either way

[glossy relic]

so isn't it fine?

[stray reef]

here

[pale epoch]

because the empty string has even amount of both 0's and 1's

[stray reef]

one stick means odd, two sticks mean even

the start state is (even, even) bc zero is even

and the accept state is,,, whatever parity combination the problem listed. idr what it was

does this dfa make sense to you yatori

well. it's incomplete bc i haven't labeled the start and end states but still

[glossy relic]

is B the only end state?

in your dfa

[stray reef]

what's B

[glossy relic]

oh

it's because i remodelled it to understand it with a start state and end state

[stray reef]

what's an end state

[glossy relic]

i mean 0 even 1 odd

[stray reef]

do you mean accept state?

[glossy relic]

yes

[pale epoch]

i have a different question: is this written on an ipad, ann

[stray reef]

no

this is written in Sai using a graphics tablet

i got myself one recently for art purposes

[pale epoch]

ok, nice

[glossy relic]

is this one the only accepting state in your dfa?

[stray reef]

was "even number of 0s, odd number of 1s" the thing the problem was looking for

if so then yes

[glossy relic]

yes it is

i can't manage to think smartly and quickly , i always do big stuff , even with my code , that's a big yikes for me

thanks for the help guys , appreciate it

[glossy relic]

regarding self loops in dfa , are they considered outgoing or ingoing edges?

[stray reef]

both

also, incoming*?

[glossy relic]

yeah i meant incoming

okay thanks

so instead of doing state a has an outgoing element 1 and an outgoing element 0 , i can do a self loop 1 and an outgoing element 0 right?

[pale epoch]

sure

[glossy relic]

Does this seem correct for „all strings with at least one 1 and exactly two zeroes“?

[pale epoch]

,rcw

[pale epoch]

is A the start and D only accepting state?

[glossy relic]

yes

ah nvm

i get 000 too

instead of element 0 from F to D

it should be element 1

i just mistyped when i corrected it

does that seem correct?

[pale epoch]

seems ok, if you correct the outgoing arrows of F

[glossy relic]

yeah

removing the F self loop and changing the outgoing edge arrow to 1 should fix it

thanks man

[stray reef]

my solution, if anyone's interested

[pale epoch]

that would've been also my solution

[glossy relic]

ann's having fun with dfas too

[stray reef]

double circle denotes an accepting state rigt

[pale epoch]

ye, this is standard notation

[stray reef]

can any of y'all throw a criterion at me to try and build a dfa for

[pale epoch]

arrow from nothing into start state and double circle accepting states

[glossy relic]

well i can just give you the dfas that i have to do

[stray reef]

sure

[stray reef]

i can spoiler my sols if you need me to

[glossy relic]

yeah i'd appreciate that!

i'm going to continue with those now and i'll probably come back to verify my answers or ask a question

[pale epoch]

i have slightly harder questions

[stray reef]

2.4a

2.4b

wait 2.4a is wrong, ||"0" should also be an accepting state||

[glossy relic]

What do you think of this for 2.4a?

[stray reef]

shit i forgot to spoiler the one for 2.4b

[glossy relic]

My 2.4b

Dw i didn’t look at yours ^^

S is my starting state

anything you guys think is important and should be added here?

[proud mason]

I have this mixed graph. What would be the degree of A?

I know the outdegree is at least one because it has a directed connection with C.

[robust mango]

hm, I think 3. In/out degree with B is 2 and out degree with C is 1

[proud mason]

So that undirected connection between A and B is equivalent to a bidirected connection between them?

[robust mango]

im not sure tbh, quite some time since i did these graphs. sorry

[proud mason]

No problem.

This is really confusing me lol.

[robust mango]

It could be in any direction so I think we take both

urgh!

[proud mason]

I can represent that as G = ({A, B, C}, {{A, B}}, {(A, C)}), right?

Not sure, but I think this might help http://combinatoricswiki.org/wiki/The_Degree/Diameter_Problem#Mixed_graphs.

No.

[glossy relic]

so i did this:
Let K be the set of all elements in the alphabet.
Let K followed by “-“ and a number mean that we have a set K without that number.

i just have to continue doing this for every number until 9 included, what do you think of that?

[glossy relic]

and a self loop can loop the element 0 to infinity amount of times right?

[pale epoch]

what does that even mean

also your picture will just accept all words, because it always ends in state L

unless input is empty string

unless

not sure how you introduced NFAs and what missing arrows mean

[azure lichen]

missing arrow just means that branch of the computation halts rejectingly, no?

and if there’s no path to an accepting state then it rejects

so empty string would reject

or are there other formalisms?

[pale epoch]

i'm not sure actually, been a while

but yeah, then what i said is wrong

but then this pic will reject 12

[glossy relic]

which pic do you mean? my latest pic?

[pale epoch]

yes

[glossy relic]

i made state L in case it's only one digit

because the first digit is a digit that never appeared before

i forgot to loop P1 and P2

but i would add a loop like K-numberthatwon'tappearattheend

for each P

[pale epoch]

yeah, then it should be fine i guess

[glossy relic]

Regarding the loop issue, I meant that this nfa would accept 0,01,011 and so on right ?

just an example

[pale epoch]

yes

[glossy relic]

perfect thanks a lot !

can't i just use a loop on my starting state with all the elements asked here ?

would be cheating i guess

[pale epoch]

the transition function is still a symbol, not a string

also not sure what "recognize" means

but like you can't mark an arrow with "abc"

only 1 symbol allowed

[glossy relic]

i meant a loop with a,b,c

wouldn't that work?

like it's three different symbols

[pale epoch]

then what

[glossy relic]

that's it

[pale epoch]

with starting state accepting?

[glossy relic]

yeah

[pale epoch]

then it accept all words that include an a, b or c

[glossy relic]

yes

my teacher doesn't anwer my emails

[pale epoch]

as a first letter, i guess

[glossy relic]

so i can't even ask what recognize means for him

wait, wouldn't a starting state which is also an accepting state with a loop of a,b,c accept all words

like aabbc

or cbaa

[pale epoch]

reading an a lands you in starting state

you read another a, that lands you in starting state

you read a b, land in starting state

and so on

so you accept aabbc

[glossy relic]

yes perfect

so basically i just need a state

for all 3 subexercices

[pale epoch]

recognize means one of 2 things

either: only accept this one string, rejects all others

or accept all strings that include given string as a substring

[glossy relic]

yeah and we don't know which one he means

so i'm assuming right now the second one which is the easiest version of the two but i'll do both versions and just tell him to pick the one he wants based on his understanding of the word

[pale epoch]

the NFAs are rather similar

except for the latter solution you would need to know the alphabet

[glossy relic]

i have to create the alphabet myself since he didn't give me one, i guess

[pale epoch]

i mean, in general you need to know that anyways, but 🤷

your professor is probably enjoying his time off to get some actual work done

[glossy relic]

but like , teachers usually should answer emails (we are at a fachhochschule so we are only 30 students) and we don't have any hollidays right now so there are no excuses for not answering

but like it's only been a day since i wrote my mail and i have time till saturday, i just want to get rid of it since i have so many other things to do

[pale epoch]

dunno about you, but here everyone is still busy in trying to suddenly plan a whole semester to be online only

so profs dont care about anything

[glossy relic]

oh , my college prepared some stuff in advance and since we're a informatics bachelor , online doesn't change that much for us

all my other teachers respond pretty quickly

but well , with the current times , i just hope he's safe and sound , i don't mind waiting more

i just want to know so badly which word he actually meant instead of writing two abc xd

[pale epoch]

oh

lol

[glossy relic]

i made one for this one with three states but it could recognize other stuff too..

i mean if he actually means only those and nothing else , i just need to take like 5 states and i'm done

[smoky sandal]

Suppose I want to send a suitcase with oranges to Victoria. My suitcase has two combination locks. Of course, I can come up with some numbers (my key) to lock the suitcase. BUT I will not reveal my key to her. So, how can I send Victoria oranges in this suitcase without telling her my key. I do want her to have oranges.

thats my question

[pale epoch]

put oranges in, lock it with one lock, send it to her, tell her to use the other lock, send it back to you, you unlock your lock, send it back to her, she can now unlock her lock and enjoy tasty oranges

[weary tiger]

the number of edges in the complete graph of K(10) is 52 right?

whoop nvm supposed to be 45

[azure lichen]

it’s 10 choose 2 = 45

[weary tiger]

thanks hey also

is a graph a sub-graph of itself?

i need to find all possible connected sub-graphs of a basic square

vertices a, b, c, d edges {ab, ac, bd, cd}

so far ive got 3 sub-graphs

ab, ac, and {ab, ac}

im assuming since it says connected you cant have the graph itself because a and d dont have a path

and its an undirected graph so to be a connected graph it has to have a path between vertices

[short juniper]

Hi, I am giving two set A and B, the cardinality of $|A| =2 \text{ and } |B|=1$, I am told to write out the realtion. My question is does the relation between set A and B, refers to the cartesian product of set A and B?

[vital dew]

joseph2531:

[short juniper]

<@&286206848099549185>

[weary tiger]

is it n(n-1)/2 possible connected sub-graphs for an n-vertices graph?

[rugged forge]

the number of connected subgraphs of a graph depends on the graph, not just on the number of vertices

[weary tiger]

so for a square

i got 6 total

would that be right?

you sort of unwrap each side or something

[weary tiger]

wth determines strongly connected components, like if there's no out-going arrow?

[weary tiger]

how do i determine if K(10) has a euler path or a euler cycle and same with K(10, 10)

and also the hamilton path/cycle for those same graphs

[stray reef]

what's an euler path again? every edge exactly once, right

yeah you need at most two odd-deg vertices for that to happen and K10 has degree 9 on all of its vertices

[weary tiger]

yea

ok so k10 has no path nor cycle

k15 has degree 14 on all vertices so it has a cycle right

@stray reef

[stray reef]

euler yes

[weary tiger]

ok and how about k(10, 10)

that has a cycle too i believe

[stray reef]

yes

[weary tiger]

because each vertex is degree 10?

[stray reef]

all its vertices are degree 10

yes

[weary tiger]

yea

and k(10, 15) would be no because all 15?

and also how does hamilton work for those

k10, k15, k(10, 10) and k(10, 15)

i think they all have hamilton cycles

because each vertex has degree at least n/2 for each one?

[stray reef]

i don't think k(10,15) has a ham cycle

or even a ham path

k10 and k15 do. you can make a ham cycle by literally going wherever

k(10,10) too

as long as it's along the edges

k(m,n) has a ham cycle iff m=n i think

bc any ham cycle has to alternate between the two parts

[proud mason]

How does vertex degree and classification work in mixed graphs?

Like, in https://cdn.discordapp.com/attachments/496785905474994186/692026154361815131/unknown.png

What would be the degree of A?

Also, is C a sink?

If it is, that would mean A is the source, right?

But it's connected to B, so...

[obtuse lance]

why

[faint narwhal]

thats not even a function so

if you want to make any function

you need to "use up all your x's "

[faint narwhal]

its just bigger parentheses

[glossy relic]

i got for these subquestions respectively , 6 states , 6 states and 5 states, does that seem correct? (recognize means that it only accepts those 3 and nothing more)

[stray reef]

what's an epsilon-transition?

nvm

[weary tiger]

There are 4 players, A,B,C,D, each draws 13 cards. What's the probability player A gets exactly 3 clubs, 4 spades, 2 hearts, 4 diamonds?

I thought its $\frac{1}{4} \frac{\binom{13}{3}\binom{13}{4}\binom{13}{2}\binom{13}{4}}{\binom{52}{13}}$

[vital dew]

Godel:

[stray reef]

why 1/4

[weary tiger]

ok yeah thats where I wasnt sure

I thought 1/4 because we pick a player

[stray reef]

no we don't

we know who A is

[weary tiger]

ure right thx

ok in the other one whats the probability player N gets exactly 2 aces

[stray reef]

who's N

[weary tiger]

A sry

I thought its 4C2 times 50C11?

[stray reef]

$\frac{\binom{4}{2} \binom{48}{11}}{\binom{52}{13}}$ surely?

[vital dew]

Ann:

[weary tiger]

48*

[stray reef]

48, not 50. there aren't 50 non-aces, there are 48.

[weary tiger]

yeyeyeye

and in the one im not quite sure is that players A and B get all aces

so 3 possibilites right? one gets 4, one gets 1 other gets3 and both get 2