discrete-math

[lucid marlin]

sleep well

[proud birch]

hi is anyone avail to help?

(∀x in Z)[(A(x) → R(x))∨T(x)] (∃x in Z)[T(x) → P(x)] (∀x in Z)[A(x)∧ ¬P(x)] conclusion: (∃x in Z)[R(x)]

I have to write down a derivation for this, but im having trouble starting

[pliant horizon]

@proud birch nice pfp
how familiar are you with the rules of inference?

i started with the 3rd statement to manipulate the second

[quasi finch]

just a yes or no plz 🙂

[proud birch]

im kind of familiar but also kinda confused understanding them. for the 3rd statement is it possible to use univeral instantiation?

to get,
A(x)∧ ¬P(x) (3) universal instantiation

[pliant horizon]

Ah scratch that. I used existential instantiation on (2) first

[lucid marlin]

How can you show if pj | Q then Q-(p1p2...pn) = 1

[pliant horizon]

@lucid marlin room is in use #❓how-to-get-help

[proud birch]

yeah i got
T(x) → P(x) (2) existential instan.

but i dont know how to approach it from here with lines 1 and 3 :/

[pliant horizon]

Statement 3 seems to be the most useful to start with since its an and statement rather than an or

What do (2) and (3) have in common?

[proud birch]

im guessing a(x)?

ops sorry read that wrong

p(x) are in both statements

[pliant horizon]

Right

So if we were to simplify out that 'not P(x)' statement, how could we use that with statement 2?

It's going to be one of the rules of inference

[proud birch]

i can get 'not t(x)' using modul tollens

[pliant horizon]

Yeah exactly

[proud birch]

oh i see, but is there a rule in order for me to bring out any information within universal quantifiers?

or am i able to just bring out the A(x)∧ ¬P(x) from (∀x in Z)[A(x)∧ ¬P(x)] just like that?

[pliant horizon]

It's been a while since I did universal/existential instantiation/generalization but yes im pretty sure you can just do that

[proud birch]

oh okay

hm let me try and see if i can solve this

[pliant horizon]

Alright gl

[proud birch]

`1. (∀x in Z)[(A(x) → R(x))∨T(x)]
2. (∃x in Z)[T(x) → P(x)]
3. (∀x in Z)[A(x)∧ ¬P(x)]

4.(A(x) → R(x))∨T(x) (1) Universal Instantiation
5. T(x) → P(x) (2) Existential Instantiation
6. A(x)∧ ¬P(x) (3) Universal Instantiation
7. A(x) (6) Simplification
8. ¬P(x) (6) Simplification
9. ¬T(x) (5)(8) Modus Tollens
10. (A(x) → R(x)) (4)(9) Modus tollendo
11. (∃x in Z)[R(x)] (10)(7) Modus Ponnens`

I got this, and thank you for assisting me

not sure if its entirely correct though

[pliant horizon]

Yeah np.
For one thing I think usually when you do existential instantiation you usually use some new element like a. So it would be T(a)->P(a) and you'd use a for the rest of the proof (this would also make (4) unnecessary)
And then id also have getting R(a) by modus ponens and existential generalization as separate steps.
Otherwise that looks good to me

[proud birch]

okay i will take that into consideration

thank you, helped alot 🙂

[pliant horizon]

[weary tiger]

i know this proof says that gamma proves A

but does that assume A is a tautology?

[weary tiger]

how does this make sense?

for (1)

how does A conclusion satisfy A ^ B

[lucid marlin]

@nimble cove I understand it now, thanks.

[weary tiger]

⊆

is tht subset

[stray reef]

yes that's the subset symbol

[weary tiger]

tyvm

does anyone know how to look up symbols in word quickly

this is time consuming

[stray reef]

why use word at all

[weary tiger]

my professor wants it typed

[stray reef]

latex no go?

[weary tiger]

im not familiar

im CS student

[stray reef]

anyway word has an equation editor

[weary tiger]

okay ty, ill just have to accept it for what it is

and learn latex as u said

[burnt herald]

For b), when they R-T, aren't we removing all the reflexive elements? how are we still getting the number of reflexive elements?

[nimble holly]

I am struggling with induction, any help would be appreciated

I got to like fk+2 < (13/8)^k+1

and with a bit of maths I got it to

fk+1 + fk < (13/8) * (13/8)^k

I dont know where to go from here

<@&286206848099549185> any ideas ?

[analog sonnet]

I got to like fk+2 < (13/8)^k+1
Isn't that basically what you have to prove

[nimble holly]

yeah but how do i prove it

I kept looking at it

and it looks proved

is that it ?

cause it seems confusing as hell

[weary tiger]

kept looking at it
It looks proved

[nimble holly]

yeah lmao

fk+1 + fk < (13/8) * (13/8)^k

this is what I am up to

I dont know how to go from here to proving it

and I am pretty sure I need to sub in the assumption I got at some point

[burnt herald]

For b), when they R-T, aren't we removing all the reflexive elements? how are we still getting the number of reflexive elements?
@burnt herald anyone?

[pale epoch]

how did you arrive at this without the assumption even @nimble holly

[nimble holly]

I changed fk+2 to fk+1 + fk because fibonacci

and I changed (13/8)^k+1 to (13/8) * (13/8)^k

I did not need the assumption

[pale epoch]

well

(13/8)^k+1 does not come into play yet

let's say you want to prove the claim for $f_{k+2}$

[vital dew]

Lochverstärker:

[pale epoch]

you use definition of fibonacci as you did to get $f_{k+2} = f_{k+1} + f_{k}$

[vital dew]

Lochverstärker:

[nimble holly]

okay got that

[pale epoch]

then you use the assumption

[nimble holly]

well I can apply the assumption to fk+1

[pale epoch]

i.e. you assume you have already proved the theorem for $f_l$ for $l < k+2$

[vital dew]

Lochverstärker:

[pale epoch]

indeed, but also to $f_k$

[vital dew]

Lochverstärker:

[pale epoch]

this is sometimes called strong induction

[naive mural]

💪🏿

[nimble holly]

so strong induction is needed to answer not regular induction ?

[naive mural]

They’re equivalent

Just a different name

[nimble holly]

well it is what it is

[pale epoch]

then you get $f{k+2} = f{k+1} + f_{k} < (\frac{13}{8})^k + (\frac{13}{8})^{k-1} $

eh

[nimble holly]

yeah i understand that part

I wrote it down

I just thought I cant use it

[vital dew]

Lochverstärker:

[pale epoch]

bad tex, but this

[nimble holly]

because I never proved fk< (13/8)^k-1

[pale epoch]

you get what i mean

this is part of the induction hypothesis

you can assume the statement is already true for all $l < k+2$

in this case

[vital dew]

Lochverstärker:

[pale epoch]

and if it then follows that it is true for k+2

the statement is proven

because

you showed it for f_0 or f_1 "by hand"

then to prove f_2, you assume f_0 and f_1 (which is proven) and it works

then for f_3, you already have proven f_2, f_1 and f_0

so it works

and so on

if that explanation doesn't work, look up 'strong induction'

[nimble holly]

Oh god

[pale epoch]

it's equivalent to 'normal' induction and i bet there are some videos explaining why

[nimble holly]

dont I need to prove that $(\frac{13}{8})^k + (\frac{13}{8})^{k-1} $

[vital dew]

Chh:

[pale epoch]

well, you need to show that this is < (13/8)^(k+1)

[nimble holly]

is less than (13/8)^k+1 ?

[pale epoch]

yes

[nimble holly]

How do I tackle that then ?

wait dont I need to prove its more ?

if a>c and b>a therefore b>c

isn't that the entire point ?

[pale epoch]

🤔

you want $f_{k+2} < \dots \leq \left( \frac{13}{8}\right)^{k+1}$

[vital dew]

Lochverstärker:

[nimble holly]

tbh I confused my self

isn't it possible to prove it by the fact

that (13/8)^n increases alot faster

than any fibonacci would ?

[pale epoch]

that's what you are proving

[nimble holly]

well without induction

I mean

[pale epoch]

how would you prove that

[nimble holly]

since fibonacci would be limited to the being less than 2x the previous fibonacci number

where 13/8= 1.625 which is more than the golden ratio

I probably am struggling to get my point accross

[pale epoch]

well, that would take a lot more machinery

to prove it that way rigorously

[nimble holly]

tbh I just dont want to do induction...

but it is what it is

[lucid marlin]

For the <- direction of the proof how does that prove A is a subset B? I get letting x be in A and by the assumption you know x in A^c Union B, and therefore in B, so you have x in A and x in B but how is that enough to show x in A subset B

[vital dew]

N/𝔄:

[lucid marlin]

couldn't you also argue this also says for all x in B we have x in A and thus B is a subset of A?

because we have x in A and x in B how do I know either is a subset of the other

[weary tiger]

Would that matter?

[lucid marlin]

yes the proof wants A subset B

[weary tiger]

If A=B then B ⊆ A and A ⊆ B

[lucid marlin]

All I know is x in A and x in B

how do I know if its A subset B or subset A

[weary tiger]

They didn't say x ∈ B

Did we read the same proof?

They assumed x ∈ A and concluded x ∈ B

[lucid marlin]

it says x in A^c Union B, so therefore x in B since x is not in A^c since x in A

[weary tiger]

Yeah

[lucid marlin]

so x in A and x in B

but x in B and x in A

so not sure how to show A subset B

or if this means also B subset A

[weary tiger]

x ∈ A which is a subset of U. Therefore x ∈ U. Since U = A^c ∪ B we have x ∈ A^c ∪ B

But since x ∈ A, therefore x ∉ A^c, so x ∈ B

[lucid marlin]

yeah, how do i go from x in A and x in B to x in A subset B

does x in A and x in B mean both A subset B and subset A?

i dont think so

so not sure how to get A subset B

because I cant tell the subset direction just off x in A and x in B

[weary tiger]

yeah, how do i go from x in A and x in B to x in A subset B
This is not what we are doing

We are not "going from x in A and x in B"

All we assume is that x ∈ A

And that U=A^c ∪ B (the condition from the iff)

All we assume is that x ∈ A
We show that from this assumption it follows that x ∈ B

[lucid marlin]

x ∈ A and x ∈ A^c ∪ B -> x ∈ B, do you agree?

therefore x ∈ A and x ∈ B

[weary tiger]

No

p->q does not mean that p ∧ q is true

[lucid marlin]

x ∈ A and x ∈ A^c is a contradiction

[weary tiger]

Yes that is a contradiction but not what is stated

What is stated is that x ∈ A^c ∪ B

[lucid marlin]

x ∈ A and x ∈ A^c ∪ B -> x ∈ B so you get x ∈ A -> x ∈ B

[weary tiger]

Yes assuming x ∈ A^c ∪ B.

x ∈ A ⇒ x ∈ B does not imply x ∈ A ∧ x ∈ B

[lucid marlin]

are you allowed to just let x ∈ A? because the assumption is x ∈ A^c ∪ B

[weary tiger]

Consider the possibility that A is the empty set

[lucid marlin]

x ∈ A and x ∈ A^c ∪ B -> x ∈ B so you get x ∈ A -> x ∈ B therefore A subset B <--- is that the proof

[weary tiger]

That is not the proof but very roughly the way this was concluded yes.

[lucid marlin]

2 questions 1) are you allowed to say x ∈ A if the assumption is only x ∈ A^c ∪ B ? if so, why? this is more a meta-questions about what is allowed in proofs 2) what would your version of what i wrote above proof be (i.e: x ∈ A and x ∈ A^c ∪ B -> x ∈ B so you get x ∈ A -> x ∈ B therefore A subset B )

[weary tiger]

1. doesn't make sense

It follows that x ∈ A^c ∪ B when x ∈ A

You don't have to assume it

Or "be allowed to say it"

[last timber]

It follows that x ∈ A^c ∪ B when x ∈ A

how

[weary tiger]

Clearly since A ⊆ U and U=A^c ∪ B, it follows from x ∈ A that x ∈ U=A^c ∪ B

For question 2, you are missing details I gave a complete proof above

[last timber]

sorry, i am being dumb

how A can be subset of A^c U B (or it is provided that A also is in B)?

[weary tiger]

x ∈ A which is a subset of U. Therefore x ∈ U. Since U = A^c ∪ B we have x ∈ A^c ∪ B
But since x ∈ A, therefore x ∉ A^c, so x ∈ B

It is provided that U is the "universal set" here and the iff assumption is U=A^c ∪ B

So since A is a subset of this "universal set" (otherwise complement doesn't make sense and this is stated in the problem), A ⊆ U

[lucid marlin]

okay, so you have A subset U, x in A^c Union B

you can just say 'let x in A"

or are you deducing x in A

im a bit confused

[weary tiger]

You are trying to prove this proposition: ∀x (x ∈ A ⇒ x ∈ B)

[lucid marlin]

yes

[weary tiger]

To prove an implication P ⇒ Q you have to assume P and prove Q

[lucid marlin]

so x in A is another assumption

and thats fine because A subset U

[weary tiger]

No that is not why it is fine

To prove an implication P ⇒ Q you have to assume P and prove Q

[lucid marlin]

P is x in A^c Union B

[weary tiger]

P is x ∈ A and Q is x ∈ B

[lucid marlin]

im kind of confused about the x in A justification, it is just an assumption?

not a deduction

our assumption is x in A^c Union B

not x in A

but we are allowed to say x in A as an additional assumption?

[weary tiger]

What do you suppose x ∈ A ⇒ x ∈ B means?

[lucid marlin]

A subset B

[weary tiger]

Yes but that is not what I mean

What do you suppose P ⇒ Q means?

[lucid marlin]

if P then you prove Q

[weary tiger]

No

It means that whenever P is true, Q is true

So to prove this statement we assume P is true, to show Q is true

[lucid marlin]

so we assume x in A is true?

[weary tiger]

As a hypothetical

[lucid marlin]

Okay, this makes much more sense

[weary tiger]

If x ∈ A was true, then x ∈ B follows

[lucid marlin]

Thank you for helping me understand this proof

x in A is a hypothetical along with our given assumptions, we show if x in A then x in B

so in these subset proofs we can look at our assumptions and then use hypotheticals, and show if these hypotheticals are true along with our given assumptions, then we can prove a conclusion we wish to show?

we arent strictly required to only use the assumptions stated

but can introduce such hypotheticals to get the proof going

?

[weary tiger]

Yes

[lucid marlin]

that is a key insight that will help me get unstuck with susbet proofs. i had incorrectly assumed you were only allowed to use the stated assumptions

this makes a lot more sense, you cleared up a lot of bugs in my thinking. thank you

[weary tiger]

Np

[valid wave]

with boolean implies

how am I supposed to know if its true or false

like boolean and for example

if I have T and T the result would be T

how would I do that with boolean implies

😦

[weary tiger]

bool thing = true; by default

[valid wave]

im not following

how does it apply to boolean implies though

[pale epoch]

implies is just defined as false when "true -> false" and true in all other cases

those are definitions

[valid wave]

thanks

but if implies mean if then

if something is false

then true?

how is that true

[pale epoch]

because it's defined that way

[valid wave]

so if something is false then its true

[pale epoch]

what

no

[valid wave]

idk im struggling to understand that

if implies means

if then

if something is false then something else is true

how is that statement true

[pale epoch]

yes, but there are 2 different statements

because out of falsehood anything follows

its the principle of explosion

the formal explanation is, take a true statement P

now assume (not)P

then P or Q is true for any Q

and because (not)P is true, Q must be true

but that doesn't really matter, you can just treat the symbol "=>" as defined that way

[valid wave]

ah ok

thanks!

[narrow epoch]

I don't get why they use P(n-1) => P(n) here,

[nimble cove]

@narrow epoch that's probably an error

its meant to be used after it has been shown

that P(n-1) => P(n)

[narrow epoch]

I think it holds, asked some other folks

you can also show P(n-1) implies P(n)

I just didnt know

[weary tiger]

that's the same thing

[stray reef]

does there exist a graph which has 20 edges, is 3-regular and has girth 5, but is not isomorphic to the dodecahedron graph?

[scenic needle]

since groups have a single closed operation, can it be addition or multiplication or only addition?

sorry if this isnt the right channel for that

[weary tiger]

only one operation

you can have a multiplcative group or additive group, for example respectively reals with normal multiplication and integers with normal addition

[pale epoch]

it can be neither

a group is just a set with some operation

[scenic needle]

okay thats what i thought i just found a diagram online and it described a group as only being closed under addition

also what is the general term for these types of things like fields, rings, and groups?

for example fields, rings, and groups are all examples of ________

[pale epoch]

algebraic structures

[weary tiger]

is there a way i can distrubute ~ to A->B

im trying to get ~A->~B

[nimble cove]

is ~ the negation

?

[weary tiger]

yeah

\not

[honest barn]

That just... doesn’t follow

So you can’t

[weary tiger]

im trying to prove A->B == ~B->~A

[nimble cove]

~ (A-> B) <=> A->~B

[weary tiger]

with hilbert proof

[honest barn]

Yeah that one’s true, but you can’t derive ~A -> ~B from A -> B.

[weary tiger]

oh

[honest barn]

Is all I pointed out

[weary tiger]

my mistake

[honest barn]

Oh lol, you typed the original thing backwards?

[weary tiger]

trying to prove A->B == ~B->~A
1) A->B <hyp>
2) ~A v B <~v Axiom>
3) B v ~A <v associtiavity axiom>
4) ~B v A <idk what to put herE>

im stuck on 4

the <...> are annotations for hilbert proof steps

[honest barn]

Can’t you just do the reverse of what got you from 1 to 2?

Err

[weary tiger]

i have to start from a hypothesis

the hypothesis here is A->B

[honest barn]

Oh

That step isn’t even true

[weary tiger]

o

[honest barn]

B v ~A doesn’t imply ~B v A

So I’m not 100% certain but

[weary tiger]

ur right

[honest barn]

But umm

You know what ~(A -> B) is tight?

[weary tiger]

~(~A v B)?

[honest barn]

Yeah it should be

A ^ ~ B

[weary tiger]

demorgans law

?

i think

[honest barn]

I’m not sure, I only know that in the context of sets

[weary tiger]

oh yeah its called demorgans law 1

[honest barn]

Oh wait

I’m dumb

You can finish after 3

[weary tiger]

rly

[honest barn]

Just do the reverse of what got you from 1 to 2

That gets you ~B -> ~A

[nimble cove]

Can’t you just do the reverse of what got you from 1 to 2?
@honest barn

[honest barn]

What?

Am I wrong?

[nimble cove]

u said that earlier

[honest barn]

No but like

[nimble cove]

no youre probably right

[honest barn]

That was from step 4

Which would have got you B -> A

Since actually 3 to 4 was bad

[weary tiger]

man im sorry

idk what u mean by reverse

[honest barn]

Like

From step 1 to step 2

You used this thing you called ~v axiom

Presumably this says that A -> B is equivalent to ~A v B

[weary tiger]

yeah

[honest barn]

Use the other direction of this on B v ~A

Which says this is equivalent to ~B -> ~A

[weary tiger]

ohhhhh right on

[honest barn]

That’s why I said reverse haha

[weary tiger]

my bad

[honest barn]

It’s just the other part of thag axiom

No worries

I wasn’t precise

[weary tiger]

i really appreciate ur time and patience

thank you

[honest barn]

No problem, I’m glad I could help

[weary tiger]

😄

[errant bear]

no problem 🙂

[lucid marlin]

How is 3k+2 equivalent to 3k-1, k int.. this is a line in a proof im reading and dont see it

its a proof proving there are infinite number of primes of the form 3k+2

[last timber]

@lucid marlin 3k-1=3(k-1)+2

[lucid marlin]

ah i see thanks @last timber

[last timber]

yw

[coral lava]

does someone here have an infinite patience to explain to me how the fuck i'm supposed to solve 5x = 3 mod 23

[unreal stump]

Substitute

[coral lava]

you w0t

what do i substitute and where

and how do i find out what to substitute

[unreal stump]

Well,you have 23 possibilities

x=0 mod (23),x=1 mod (23)... x=22(mod 23)

[coral lava]

wait hwhat

so i'm basically saying "let x be a number such that the remainder is r when divided by 23" and i take it from there?

i hate online learning, this makes no sense to me

what am i supposed to do if i'm dealing with bigger numbers, say mod 2384

do i just say "welp time to waste 10 years of my life" and just start from 0?

[unreal stump]

You multiply both sides with the multiplicative inverse

[coral lava]

multiplicative inverse of what?

[unreal stump]

Did you understand why there are 23 possibilities?

[coral lava]

yeah i get why

because mod 23

it's cyclical and all that makes sense

[unreal stump]

Multiplicative inverse(wrt mod 23) of a is b such that (ab)=1 mod 23

[coral lava]

ok so i'm trying to find the multiplicative inverse of what?

[unreal stump]

Of 5

Let that be a

So, a(5x) =a(3) mod 23

Which is to say x=3a mod 23

And you are done

[coral lava]

wait wha

one sec

how did you get from a(5x) =a(3) mod 23 to x=3a mod 23\

[unreal stump]

Because a5=1 mod 23

By definition of inverse

[coral lava]

oh ok

didn't see the regrouping on the left for some reason

so how do I find a?

[unreal stump]

Do you know how to find integers a and b such that ax+by=1 when x and y are coprime?

[coral lava]

nope

[unreal stump]

Do you know Euclid division algorithm?

[coral lava]

heard the prof mention it in class

i know bits and pieces

but it makes no sense to me

[unreal stump]

Ok,If you don't know that, you have to brute force this

[coral lava]

well how do I do it?

[unreal stump]

Euclid's algorithm?

If you are talking about this question,just cycle through x=1 mod 23,x=2 mod 23..?

[coral lava]

yeah the algorithm

[unreal stump]

Given a number a,and a number b. There are unique integers q and r satisfying 0<=r<q and a=bq+r

[coral lava]

bro what

[unreal stump]

Do you know divison

This is literally that

[coral lava]

the thing i know is the one where you go like
23 = 5*4 + 3
5 = 3*1 + 2
3 = 2*1 + 1

[unreal stump]

Yea,That

[coral lava]

how are those the same thing

[unreal stump]

In 23=5 * 4 +3 ,4 and 3 are unique

That is there are no other numbers (a,b) satisfying 23=5*a+b and 0<=b<5

[coral lava]

oh ok

so i go through it and get to 3 = 2*1 + 1 then what do I do?

or do I go one more and do 2 = 1*1 + 1?

[unreal stump]

Now you write 2=5-3*1

And 3=23- 5*4

And 1=3-2*1

So,it becomes 1= (23-5* 4) -(5- 3* 1)

Or 1=(23-5* 4)-(5 - (23-5* 4)*1)

[coral lava]

what am i looking for here

[unreal stump]

Or 1= 2* 23 - 9* 5

You are looking for 2 and -5 here

So,now you can write (2)23+(-9)5=1

Now take mod 23 on both sides

[coral lava]

wait one sec

[unreal stump]

Gives us (-9)5 (mod 23) =1 (mod 23)

Or (14)(5)=1(mod 23)

[coral lava]

im trying to figure out the rearranging so that i can get the 2 and -9

either way my brain's not having it today

oh wait

signs

yeah those are important

[unreal stump]

Or just write 1=2* 23 -5* 9

46-45

[coral lava]

oh ok i got there

so then I have -9 and 2

which one do i use

[unreal stump]

So,now take mod 23 on both sides

Of the equation 1=2(23)+(-9)5

[coral lava]

so i get 1(mod 23) = (2*23 + -9*5)) (mod 23)?

[unreal stump]

Yes

[coral lava]

i just don't get what I do with this info

everything I know is telling me to just reduce it but then I just end up with 1 (mod 23) = 1 (mod 23)

[unreal stump]

(2* 23 + -9* 5) mod 23 reduces to (2 * 23) mod 23+ (-9 * 5 ) mod23

Which is 0 mod 23 + (-9*5) mod 23

Which tells us (-9*5) mod 23=1 mod 23

[coral lava]

i didn't know you could do that

[unreal stump]

(a+b) mod 23 is a mod 23 + b mod 23

[coral lava]

so i have my 1 (mod 23) = -9*5 (mod 23), is that supposed to tell me that a = -9?

[unreal stump]

Yes

Actually , it's -9+23k

[coral lava]

so imma use 14

[unreal stump]

Yes

That's it

[coral lava]

so i get x = 52 mod 23 which is basically just 6 mod 23?

[unreal stump]

14*3 is 42

[coral lava]

math.exe has left the chat

attempt no. 2 with correct math

19 mod 23?

[unreal stump]

Yes

[coral lava]

time to try it on another problem

[abstract ravine]

So i have this Graph, which is supposed to represent distance in meters I need to find the shortest path using Djkstra algorithm

``Use the steps of Dijkstras algorithm to find the minimal distance spanning tree from A to all other locations`

i get confused when i reach E
So i starts off like this:

Step 1: I visit all the vertices from A, the smallest distance is A -> F so we assign F = 30, G = 38, B = 31.

Step 2: I start from F (since its the smallest distance from A) if we look at F -> G the distance is 30+42 = 72 which is larger than G's current value '38', G stays 38. Now we look at F -> E (currently E is infinity) so since F = 30 it becomes 30+20 = 50, which is smaller than infinity.

Now we have E = 50 and G = 38

[abstract ravine]

<@&286206848099549185>

[mystic moss]

what's your question

[abstract ravine]

where do i go from there?

[mystic moss]

you go to the closest unvisited vertex

[abstract ravine]

Ok so we go to E from here, then E -> D = 80, E -> G = 76

[mystic moss]

E? why?

[abstract ravine]

because we were at vertex F

if we go to F - > G

it ends up as = 72

[mystic moss]

right, why are you starting at F

[abstract ravine]

F -> E is 50

i just wrote why in the above text

in the Steps

Step 1 and Step 2

[mystic moss]

you don't need to go to adjacent vertices in consecutive steps

[abstract ravine]

ohh shit wait

[mystic moss]

you go to the

closest unvisited vertex

[abstract ravine]

you don't need to go to adjacent vertices in consecutive steps
@mystic moss What does this mean

from what i understand

you are saying that since F doesnt have good options i should leave F

[mystic moss]

i'm saying your options aren't E and G, they're E, G, and B

(technically your options are the vertices other than A and F, but these three have distances less than infinity so you're really only choosing among them)

[abstract ravine]

(technically your options are the vertices other than A and F, but these three have distances less than infinity so you're really only choosing among them)
@mystic moss from F my options are only E and G

[mystic moss]

you don't "start" from F

[vital dew]

Lostboy:

[paper oracle]

how do i make a finite state machine that only accepts the binary strings 10, 1010, 101010... and 01, 0101, 010101...

[long nimbus]

The chance that a seed will hatch immediately after it has been planted is about 97%. Calculate this probability to one decimal accurate.

[stray reef]

@paper oracle strings of even length which alternate ones and zeros strictly?

do you want a possible solution or do you want general tips

[weary tiger]

Make a machine with 4 states one state represents 1 and one 0, one of the two is final the other two are for dealing with possible errors in the sequence.

[wintry schooner]

hey guys how do i go about doing this question? im quite confused here

[quaint mirage]

B ?

[lethal lodge]

can anyone explain this to me?

[drifting sail]

@quaint mirage can't be B since 2 is not related to 1, i.e. neither (1,2) or (2,1) are in the set R. (also, please post in related channels even if you don't get an answer immediately; it's clearly not a #probability-statistics question)

[quaint mirage]

my bad , so only post such questions in discrete math ?

[drifting sail]

questions such as that should, yeah

[quaint mirage]

@drifting sail so do you think D is correct answer or A ?

[drifting sail]

it's A since (0,1) and (1,0) are in R. (check the definition of equivalence relation in whichever book or course notes you're using)

[stray reef]

@weary tiger i came up with a 6-state machine for moshe's question

[hazy aspen]

your zeroes are very perfect

[quaint mirage]

@drifting sail Thanks, it's correct .

@drifting sail should i delete my question now ?

[drifting sail]

no need to

[quaint mirage]

ok

[wanton sable]

can anyone help me understand how they did b)? i'm pretty lost on this one
i thought the answer would just be something like k!/(n!m!) so i'm pretty confused as to why there's a sum involved

[honest barn]

so like

you do cases over all possible k-letter combinations

you can have 0 as and k bs

1 a and k - 1 bs

2 as and k - 2 bs...

so the total number is equal to the number of combinations for each of these, just summed up

err

okay sorry that would be for the total number

when you say that the number of bs is <= number of as we say "okay let n be the number of bs"

and m the number of as

then we know m + n = k and we need n <= m

so we can do cases by just looking at how many there are for each possible combination of those

in that case if you look at a) n + m = k so that's what's on the top and n = k - m and that's what's on the bottom

[wanton sable]

ooohh okay that makes a whole lot more sense, thank you :D

[honest barn]

Also fwiw I think MAYBE

what you said counts the total number of combinations

without the requirement the number of b-s is <= a-s

oh hmm

no that's when you say there's like n b-s and m a-s

I think

Oh yeah that's true, that's literally exaclty k choose k - m

since k - m = n

[vital dew]

Kees:

[weary tiger]

How you solve this problem? x is real number

[stray reef]

break into cases based on the parity of floor(x)

[mystic moss]

@wanton sable @honest barn if you've understood how they've gotten the answer, could you explain to me why the answer for evens is 2^(k-1) - (k choose k/2)/2? shouldn't it be + (k choose k/2)/2?

[weary tiger]

@stray reef its mean odd or even? x is not only integer

[stray reef]

i didn't say x had to be an integer

floor(x) is

[opal cedar]

I understand that this is true, but I'm not sure how I show it, anyone have any ideas?

[nimble cove]

@golden talon k is an integer so floor(k) = k

k < k+1/2 <k+1

so floor(k+1/2) = k

[golden talon]

Oh thanks

[pale epoch]

use the definition of even

[nimble cove]

@golden talon n = 2k for an integer k

[pale epoch]

how did you define "even"

truth tables

or use rules of inference that you know

[sour brook]

Is it a coincidence that logical identities are almost parallel to set identities? It’s just a matter of different symbols and notations, isn’t it

[pale epoch]

no, it's not

union and intersection is defined via logical and and logical or

so you can think of it as translating logical operations into the language of sets

[wise dew]

Can someone help me with some discrete math hw

1. How many strings, as a function of n, are in the language (a + aa + aaa) n ? Prove your answer by induction on n. Hint: This regular expression contains all strings of just a with lengths between n and 3n inclusive.

2. , by induction for all naturals n, that the number of strings of length n in the language (a + ab) ∗ is exactly Fn+1, where Fn is the Fibonacci sequence (F0 = 0, F1 = 1, Fn+1 = Fn + Fn−1 for n > 1). Hint: This regular expression was used as an example in the lecture 28 slides.

[unkempt cosmos]

Is this survey of math?

[vale folio]

guys one quick question

[errant bear]

no

[vale folio]

what is n^2 + n^2/2+n^2/4

• n

[timber nexus]

so i need to prove that AxorB = (AuB)-(AnB). can i consider this to be the same as (A⋃B)⋀¬(A⋂B)?

[frail agate]

can i get a second opinion on this question, i feel like i got the proof right, but the teacher wanted it in a specific way

although maybe i did mess up, but i feel like it came out correct

this is post exam btw

wait, no, i think i understand what's wrong

[lethal lodge]

can someone explain me this question

(b)

[delicate ridge]

@lethal lodge look up "proof by induction"

there's nothing different here to the standard technique

[potent oracle]

how do I solve this?

[weary tiger]

idk where to even get started

do I use these

[charred forge]

can someone explain what the runtime rules of thumb are to me?

I have no clue what they are and my professor is going to quiz us on it and he didnt even post any notes or examples

[lyric pumice]

@weary tiger Yes.

It would be educational to show that the identities are true.

[lyric pumice]

@quaint mirage An equivalence class is a set of elements from R that are equivalent by the property of the relation.

[rapid lily]

when doing binomial theorem,

does x(9x^2 - 2x^-1)^14

turn into

(9x^3 - 2)^14

or

(9x^16 - 2x^13)^14

trying to find coefficient of x^-4 in expansion of x(9x^2 - 2x^-1)^14

[quaint mirage]

@quaint mirage An equivalence class is a set of elements from R that are equivalent by the property of the relation.
@lyric pumice {0,1,2,3} ?

[slow socket]

im trying to Expand wy to sum-of-minterms form wyz+wyz'

i have this so far, idk what to do

[mystic moss]

(also @wanton sable update: there's a small consensus that it's supposed to be 2^(k-1) + (k choose k/2)/2 so if this is for a course you might want to ask your professor about that)

[weary tiger]

any ideas on how to do this?

I tried to do P(8,4)*2*5, but that was too big

that gives you 16800, but the max number of subsets is 1024

[burnt herald]

can someone explain why is it 2 raised to nC2?

isn't there just n reflexive relations?

[weary tiger]

@worthy palm do I do C(8,4)

the 5 locations 3 or 4 can be

and 2 because there is 3 or 4

wait

order doesn't matter

so 2*C(8,4)

140?

yeah

@worthy palm did you get 140?

ok thanks

Given that a and b are odd numbers where a != b. Show that there is a
unique integer c such that | a - c | = | b - c |. how you prove this >

[charred forge]

how would i go about this A program takes time proportional to nlogn where n is the input size. If the program takes 20 milliseconds for input size 100, how many seconds does it take for input size 10,000?

[weary tiger]

how you solve this ?

[lyric pumice]

@quaint mirage {0,1,2,3} is not an equivalence class for the relation.

[quaint mirage]

then what is ?

[lyric pumice]

{(1+4a, 1+4b) | a and b are integers} is an equivalence class for the relation.

[weary tiger]

@worthy palm wait why is it not 2*C(9,4)-C(8,3)?

i got C(9,4) from assuming 3 is in the set and the rest is the combo of the remaining 9

and multiply by 2 bc there is 3 and 4

then subtract C(8,3) bc that is when 3 and 4 are included together

[last sigil]

@weary tiger Hint: draw a number line and try some test cases for a & b. what is c?

[weary tiger]

@last sigil i found it but stuck in profing the equation

[last sigil]

wdym

please write out what you found

[slow socket]

How can i figure out how many 2 input AND gates and 2 input OR gates are needed to build a circuit given a function?

[unreal stump]

Depends on the function,simplify it if you want to know that

[slow socket]

Z = AB + CDB + BC. It's this. do i just try and draw it?

is it like that?

[stray reef]

the CDB is redundant i think

so you can save yourself an AND

and then you have AB + BC which you can simplify to B(A+C) saving yourself another AND

[slow socket]

so i would have 2 two input AND gates?

im having trouble picturing how i would draw what you wrote

[stray reef]

you would have a single AND and a single OR

the OR would take A and C as inputs, and the AND would take B and the OR's output as inputs

[unreal stump]

Isn't every and gate 2 input?

[slow socket]

this?

isnt this B(A+C)

[stray reef]

yeah

[slow socket]

how is that Z = AB + CDB + BC then

[unreal stump]

AB+CDB+BC=B (A+C(D+1))

Which reduces to B(A+C)

[slow socket]

but its asking how many 2 input AND gates and 2 input OR gates are needed

not to do it in the minimum number of gates

wait nvm lol

[unreal stump]

Well, You can add an arbitary number of extra AND and OR gates

So atleast one AND and one OR gate

[slow socket]

ok what about this, to see if i understand it. How many 2 -input AND gates are needed to build a circuit for Z = AB + CD + AD + C

i first factor it out?

[stray reef]

staying true to the expression as written you would have 4 ANDs and 3 ORs

it might be possible to do better tho

[slow socket]

i managed to get to 3 ANDs

Z = AB + CD + AD + C
i can do this?
A(B+D) + C(D+1)

[stray reef]

D+1 is just 1

[slow socket]

A(B+D) + C

1 AND?

[weary tiger]

can someone explain this to me?

[slow socket]

(A + B)(A B)
will this be reduced to AB ?

[stray reef]

yes

[slow socket]

what do those dots mean? NOR?

[stray reef]

don't think so

they're just decorative dots

[slow socket]

the dot has the be open and in front of the gate correct?

what if there is an open dot before the gate

like this

idk if that open dot on B does anything. this just a NOR gate?

[stray reef]

uh

these open dots i think are nor gates but im really not sure

[slow socket]

i looked these up and the dot on the right means is NOR gate, but i've never seen the open dot on the left

[stray reef]

sorry

i meant not gates

stupid autocorrect

[slow socket]

not gates are the triangles

[stray reef]

triangles with a dot in front.

[brave lava]

can anyone help with the middle one

nvm i got it

it was complex

[potent oracle]

how do I solve this?
@potent oracle

[burnt herald]

is this correct?

How do i link to (A-b) U (B-A)?

[lyric pumice]

@potent oracle That can be solved with arithmetic (add, subtract, multiply, divide), but it is tricky.

[rugged pebble]

does anyone know how to get the inverse of thi s

f: S -> S, f(a1a2a3a4) = a4a3a2a1

[weary tiger]

Assuming S consists of words consisting of four letters, this function is its own inverse.

[rugged pebble]

what do you mean by that?

the function takes a string and just reverses it

how would I show that

[unreal stump]

Just apply the function twice

[rugged pebble]

sorry, i dont get it

[unreal stump]

what happens if you reverse a reversed string?

[rugged pebble]

its just the original

[unreal stump]

Which means f o f=id

Which means f is it's own inverse

[rugged pebble]

so I can show it like

that

[unreal stump]

Yes

[rugged pebble]

so just f^-1 = a1a2a3a4

[unreal stump]

No,It's f

[weary tiger]

lol.

[rugged pebble]

just f

not f^-1 = f

or

like, it says to write what f^-1 =

[weary tiger]

Why is it so foreign to you that if you reverse a string and reverse it again, the final result will be the original string?

[rugged pebble]

its not

i fully understand that

its just formally writing it out

if the inverse is the original string, which it is. Then to write it out it should be f^-1 = a1a2a3a4

[karmic falcon]

Will anyone be so kind to go over it with me? I already have a solution, just want to understand/make sure it's correct.

[weary tiger]

if the inverse is the original string, which it is. Then to write it out it should be f^-1 = a1a2a3a4
@rugged pebble you assert that f^-1 = f and prove ut by showing that the identity f(f^-1(x))=x=f^-1(f(x)) holds.

[hasty ibex]

How to prove this uncountable?

[silk coyote]

I am totally stuck with one exercise, I am trying to find the general term for the sequence: 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, ....

The recursive sucession is the following one

But I need a general explicit term that does not use recursion at all.

[sour arrow]

@hasty ibex
There's a pretty easy bijection between that and R

[safe scroll]

I have a discrete question

ok

so does anyone know linear homogenous recurrence relations?

if so

lets say that I get that some solution to one is this

then this means that I can turn it into this

and then I can combine coefficients

and get this

can I combine the two coefficients into one and just make it a1?

or do I need both?

[vital dew]

Viburnum:

[safe scroll]

hmmm

I see

ok I see

so if I have this

I would need it to be

$a_1 2^n + a_2 n 2^n + a_3n^2 2^n$

oh shoot

ya mb

ok cool

thanks @worthy palm

I see

one last thing

so I would only need to append these n^x or whatever for roots that are the same right

like

if I have two double roots

then I would have two terms that have an n in them right

ok cool

thanks man

[frank zinc]

I've got this homework assignment, but I feel that I am completely missing what my professor is getting at in part d here

Those last two photos contain my stuff, but the question overall seems odd for both c and d

I feel like the work I did is useless and not really giving much insight.

[drifting sail]

I think the idea is that you can find the value of S explicitly that way

[frank zinc]

If I only leave it as the infinite summation in part c I don't see where the sum just "falls out" in part d

[drifting sail]

more generally if $|r|<1$, then $S=\sum_{k=0}^\infty r^k$ is a convergent series and
$$
S-rS=\sum_{k=0}^\infty r^k-r\sum_{k=0}^\infty r^k=\sum_{k=0}^\infty r^k-\sum_{k=1}^\infty r^k=1,
$$
therefore $S(1-r)=1$ and hence $S=\dfrac{1}{1-r}$. Your exercise is the particular case $r=\dfrac 1 3$.

[vital dew]

derivada.schwarziana:

[frank zinc]

ah I see

[drifting sail]

I think you used the actual value in c) when in principle you could've gotten it in d) (and prove the sequence is convergent by other means in b), e.g. by proving it's a Cauchy sequence)

[frank zinc]

Maybe there still is something im missing. Is there a way I could have gotten it in d) without using the 1/(1-r) relationship or evaluating the infinite sum?

[last timber]

what was r?

1/3?

[frank zinc]

yeah

[last timber]

then 1/3S is just another geometric series

[frank zinc]

yeah I understand that, it just seems that the problem is highlighting something odd

[last timber]

but anyway

you anyway will have to eval

[vital dew]

Commander Vimes:

[last timber]

oh lol sequence term is another

kek

[frank zinc]

i believe i understand

[weary tiger]

Why would $|X_i|$ be even? I don't see how that follows in the proof.

[vital dew]

zd:

[weary tiger]

Here's a scrawling I made of a specific case of this

Textbook is trying to use the handshaking lemma for the red bit on the left as well as the fact that there are an odd number of red vertices to somehow imply directly from that, that there must be an odd number of green vertices on the left

Drawing it brought 0 extra intuition for why this might be true and I can't see it algebraically either

nvm the proof is just not recalling that every graph has an even number of odd vertices

[karmic falcon]

Does anyone have a similar problem to this I can work on or know how to solve this? Any help would be appreciated

[fluid zealot]

@karmic falcon I am not sure if I can help, but I have been working on the question since you post it here

[karmic falcon]

I see, how's it looking? :/

Or how did you go about it

[fluid zealot]

I have been trying to check the reflexive equivalence by the definition

I assume R° to be the converse of R which is{(y,x)|xRy}

So R∪R° and Q∪Q° would be quite similar

[karmic falcon]

[fluid zealot]

Then I try to check the equivalence relation by its reflexive, symmetric and transitive properties on R∪R° and Q∪Q° separately

Does this help?

[karmic falcon]

It is a little bit

But yes Im understanding what you're saying

My conclusion to the question was that RuR QuQ would not be reflexive @fluid zealot

[fluid zealot]

For your reference

[weary tiger]

https://puu.sh/GIp9H/e92e985d7b.png

I'm not quite sure which other one I'm missing

[bronze cosmos]

can i ask a graph theory question here?

[naive mural]

Yes

[bronze cosmos]

okay. is every grid graph a a parity graph?

[naive mural]

No

Why would it be

[bronze cosmos]

i make visual sec

[naive mural]

Unless grid graph just refers to rectangular grids

[bronze cosmos]

uh ok so i guess i’m unclear on the definition of grid graph

p sure a grid graph is the graph cartesian product of two paths

[naive mural]

So it’s always going to be rectangular

You can colour it right, chessboard colour it

[bronze cosmos]

ig. maybe it could also be infinite in any direction? idk

yeah that’s what i was thinking

[naive mural]

That’s okay too

[bronze cosmos]

so the answer is yes, then, right?

[naive mural]

But you can show it explicitly by partitioning it

[bronze cosmos]

yeah so you would partition the set of vertices into two subsets, call them “red” and “blue”

[naive mural]

Right, so it’s secretly a bipartite graph

[bronze cosmos]

and i guess explicitly you could say that if you plotted the points on a graph then red = points with x+y even and blue = points with x+y odd

holy shit that’s the sneakiest bipartite graph i’ve ever seen

[naive mural]

Yeah that’s a nice way to do it

[bronze cosmos]

it’s like, you can move the red vertices towards you and the blue vertices away from you

[naive mural]

I’m assuming a path graph has a countable number of nodes though so you can list them

Like if you had an uncountable number of nodes then this makes no sense

[bronze cosmos]

wait

[naive mural]

I’m semi-joking

[bronze cosmos]

doesn’t this not even make sense for countable infinity

like, can’t you make an infinitely long path?

[naive mural]

Countable infinite works fine

But uncountable infinite makes no sense

[bronze cosmos]

w8. semi unrelated question. is there such a thing as an infinite path?

[naive mural]

Yeah

[bronze cosmos]

can you have an infinite path that has a starting an ending vertex?

[naive mural]

I don’t think so...

But mathematics is about being creative

You just tweak the definitions a little and use your imagination

[bronze cosmos]

true. can you have an uncountably infinite graph? like does that even exist

[naive mural]

Yeah you can but not in a line I think

[bronze cosmos]

so you can’t view the real number line as a graph?

[naive mural]

You can in terms of vertices

And you can join the numbers together almost arbitrarily

But you can’t have a nice line of connected vertices I believe

[pale epoch]

what's a nice line?

[bronze cosmos]

yeah like obviously you can’t partition R in the red and blue sense

[naive mural]

Each node has 2 neighbours

And it’s all path connected

[pale epoch]

well order the positive real numbers and the negative real numbers

define edges according to the well order

connect the least elements of both sets to 0

[naive mural]

If only

[pale epoch]

it works

[naive mural]

There is no least element

[pale epoch]

there is according to the well ordering theorem

[bronze cosmos]

he said well-ordering

[pale epoch]

not like you can write down the order

but it exists

[naive mural]

So how would you use this to get a straight line

I guess I buy it

You just squint your eyes and imagine it

But it’s weird cause I don’t think it satisfies the standard definition of path connected

Like something goes wrong

[pale epoch]

it's a connected graph

and you can 2 color it

[naive mural]

A path is defined as a sequence of edges, i feel like that implies all paths are countable

That’s why I think you just have to tweak the definition

[pale epoch]

just index it by real numbers

[naive mural]

I’ve only seen definitions which at most allow a sequence to be indexed by Z

I’m for extending the definition to any ordered thing

[pale epoch]

you can also just throw topology at it probably

a graph really is just a topological space of isolated verteces

the edges are just unit intervals

and you glue

then a path is a subspace that is homeomorphic to the unit interval

[bronze cosmos]

that’s creative

[pale epoch]

or well, this would lead to what i defined not being a path

unless by unit interval i also allow the open one

[naive mural]

I think restricting yourself by formal definitions is kinda dumb anyway

[bronze cosmos]

this hurts my brain to think about

[pale epoch]

this gives you 3 kinds of paths: homeomorphic to [0, 1], [0, 1) and (0,1)

[naive mural]

If you want an infinite path, make it even if it isn’t officially one

The first one is the one which is scary

[bronze cosmos]

u guys wanna see the context why i asked this question?

[naive mural]

Sure

[bronze cosmos]

i’m tryna solve this puzzle for fun:

these are some example snakes

these are some nonexample snakes

i been tryna throw graph theory at it but to no avail

[naive mural]

This seems like a fun puzzle, is there a link to a website where I can try?

[bronze cosmos]

https://cracking-the-cryptic.web.app/sudoku/MQbhMHdtr9

[naive mural]

I presume I’m covering the whole grid in snakes too

[bronze cosmos]

yeah i mean 49 cells total 7 snakes 7 each

[naive mural]

I didn’t even see that they were of length 7, that makes it easier now

[bronze cosmos]

yeah. things is, is you partition this “grid graph” into red and blue (as i described earlier) each circle is the same color

so there’s nothing you can deduce about two particular circles not being on the same snake

EXCEPT FOR some circle that are very far apart

ie these two circles can’t be a part of the same snake bc they are too far apart

anyway that’s as far as i’ve gotten in my hours working on this puzzle, lol

[naive mural]

You know something about the bottom left too

Like a little chunk of the snake else it doesn’t work

[bronze cosmos]

hm?

[bronze cosmos]

o good point

[naive mural]

The graph theory thing is cool tho

Like there HAS to be 4 odd snakes and 3 even ones

So you know the black dots are connected

So if your snake starts with a circle it has to end with a circle

[lyric pumice]

It looks like a problem of searching for polyominos.

So, it is likely to be an open problem.

[bronze cosmos]

why 4 odd snakes and 3 even

[naive mural]

It’s a casual puzzle you’ll find in a train station bookshop

Colour your grid red-blue

We have 1 more red than blue say

Where the bottom right corner is red

Call a snake red if it’s head is red

[bronze cosmos]

O:

[naive mural]

Each red snake covers 1 more red square than blue square

So in order to cover the whole grid we need one more red snake than blue snake

Since there are 8 red circles they must all be joined

[bronze cosmos]

so i guess u meant 3 blue snakes and 4 red snakes

[naive mural]

Yeah

[bronze cosmos]

they’re all the same parity

[naive mural]

Right and since there is 8, they must all be heads and tails

We need to connect them all

And that is hugely restrictive

And the puzzle falls apart now

[bronze cosmos]

wonderful observation

no wonder you’re Euler 2, lol

[bronze cosmos]

i solved it!

but only bc of your observation lol ur a genius

[naive mural]

I wouldn’t have thought it if you didn’t point out they were all on the same colour squares

I solved it as well but I just smashed it out using intuition+logic rather than pure logic

Like quickly checking a few cases for a bit and seeing it doesn’t work

[pale epoch]

what was your profs intended solution

if not "0"?

[near prawn]

seems ur proff simply doesnt like you

[pale epoch]

seems like a semantic issue

[near prawn]

which is the same as having a 0 coeff

[pale epoch]

if you need the point

[bronze cosmos]

@naive mural yeah me too

logic to limit the possibilities, then trial and error. it’s like a math competition problem

[brazen tendon]

does this mean that 0 number of b's is acceptable?

finite automata btw

[karmic falcon]

Anyone help me solve this above?

[ionic aurora]

@karmic falcon since set inclusion is preserved under unions, from $X \subseteq Y$ we have $X \cup A \subseteq Y \cup A$

[vital dew]

catch-22:

[ionic aurora]

and using $X=B-A, Y=C-A$ with the fact that $(Z-A) \cup A = Z \cup A$ (not hard to prove), the problem is solved

[vital dew]

catch-22:

[karmic falcon]

So would this work,

1) AuB = A u (B\A)
2) AuC = A u (C\A)

Since B-A ⊂ C-A

-> A u (B-A ⊂ A u (C-A)

-> AuB ⊂ AuC

Basically

@ionic aurora

[ionic aurora]

yes!

[karmic falcon]

Thank you

[bleak pollen]

can someone help with part 3?

<@&286206848099549185>

[robust tulip]

^ Good question Tekado

<@&286206848099549185>

[stray reef]

@iron crescent yeah "no term" and "zero coefficient" are lit the same thing lol

[quartz hound]

Guys here’s a really stupid question

How do we calculate the possible configurations of a 2*2 Rubik’s cube?

[minor dagger]

(8! * 3^7)/24

[quartz hound]

(8! * 3^7)/24
@minor dagger Thx! Can you plz elaborate a bit?

all the rotations and stuff screwed me over

btw symmetry is not considered for simplcity

[minor dagger]

ok so 8! is based off of the number of permutations of corners

3^7 is based off the orientations of the corners

8th corner is fixed

24 is the ways positions can appear as dupes

you can also do 7! * 3^6

probably easier that way

because there are 7! perms when one corner is fixed

and six of the corners can be oriented in 3 ways

thus 3^6

both equal 3,674,160

[quartz hound]

24 is the ways positions can appear as dupes
@minor dagger Hi can u plz explain this bit?

sorry am stoopid

Is this basically taking the way you could rotate the cube into account?

If so, does this have anything to do w/ group theory? XP