hennnice_help | Mathematics | Page 1

lean glacierBOT Mar 20, 2024, 5:45 PM

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robust drift Mar 20, 2024, 7:22 PM

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this question is really vague. the constraints aren't defined well

Generally, when talking about bracelets in combinatorics, two n-tuples are equivalent bracelets if a combination of rotations and reflections yields the same n-tuple.

By this definition, I see no way that we get to 18

I only see 9:
RGBG, RBGB, GRBR //two of one color, 1 of each of the other
CNCN //two of one color, two of another alternating 3C2 = 6 ways to pick colors

I'd love to help but first step is understanding wtf the question is even asking.

river nacelle Mar 20, 2024, 9:10 PM

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i agree, 18 is not obvious at all, i can't get 18

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it's 18 if there's like a knot on it or something

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then you get 4 of each RGBG, RBGB, GRBR and 2 of each CNCN
4×3 + 2×3 = 18

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wait that doesn't help to distinguish in the second case

river nacelle Mar 20, 2024, 10:10 PM

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okay, so not a bracelet at all

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taylor approved tattoos

robust drift Mar 20, 2024, 10:19 PM

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river nacelle taylor approved tattoos

plus the assumption that one bead is oriented up

river nacelle Mar 20, 2024, 10:24 PM

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it's too hard i can't solve it

robust drift Mar 20, 2024, 10:25 PM

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unfortunately, ths dyslexic AI doesn't trust most hand writing and really has trouble with cursive.

for n=3, I'd think we gotta be RGB permutated --> 3! = 6

for n=2, any two beads should work --> 3! = 6

for n=1, I'd think it 0 because the bead is adjacent to itself

robust drift Mar 20, 2024, 10:32 PM

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river nacelle it's too hard i can't solve it

don't give up! It's just graduate level combinatorics.

river nacelle Mar 20, 2024, 10:34 PM

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if it can't be solved by imagining the whole thing it's not worth this AI's time

robust drift Mar 20, 2024, 10:57 PM

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I got it

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now i get to be all Socratic and lead you to the answer

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ok, lets insert a bead between the top bead and the one counter clockwise to the left? How many colors could it be? Is this all the ways to construct the tattoos?

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not necessarily, lets look at RGRG, chopping G would give RGR which would be an illegal tattoo. To construct something like RGRG, we're going to have to look at 2 bead tattoos

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look at 2 bead tatoos

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RG, RB, GR, GB, BR, BG

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We gotta make RGRG and RGRB

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@desert smelt when you cite me as having assisted you, make sure my name is not capitalized. Not totally kidding, for graduate level work you really should be citing everybody who helps you with anything.

robust drift Mar 20, 2024, 11:53 PM

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it's not a necklace or bracelet either

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I used R instead of W

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your list has 20 items

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at least these 2 are duplicates:

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c(1) is 18; c(0) is 6; c(-1) = 6
I said this before

sick orchidBOT Mar 21, 2024, 12:01 AM

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samenn.

robust drift Mar 21, 2024, 12:01 AM

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it really doesn't matter but ok, n = -1 is a much simpler way to think of this problem

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look again at my comments after I said "I got it"

sick orchidBOT Mar 21, 2024, 12:28 AM

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samenn.

robust drift Mar 21, 2024, 12:41 AM

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choping 2 from 4 beads is analagous to a construction from 2 beads

c(0) = 6; c(-1) = 6;
c(0) != 3 + c(-1)

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why?

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?

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RGB, RBG, GRB, GBR, BRG, BGR

what else?

sick orchidBOT Mar 21, 2024, 1:08 AM

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samenn.

robust drift Mar 21, 2024, 1:13 AM

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nope

how many ways we can construct a tattoo of length n?
how are we constructing them?

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forget c0 for now

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no, with restriction

Also I didn't BGRR, you did. I just did find and relace w-->R

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how many ways can I construct a string of length n from a string of lenght n-1?

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1 = (number of colors) - 2

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how many ways can I construct a string of length n from a string of lenght n-2?

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not sure what you're saying; can you give an example construction?

sick orchidBOT Mar 21, 2024, 1:43 AM

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samenn.

robust drift Mar 21, 2024, 1:49 AM

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not sure what L(N) is but you showed, you can make 1 string of n from every string of n-1

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so we got
c(n) = 1*c(n-1) + ???

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why is 2 the coefficient of c(n-2)?

robust drift Mar 21, 2024, 2:46 AM

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2 is the coefficient of c(n-2)
a.k.a. there are 2 ways we can construct a string of length n from a string of length n-2
what are those two ways?

sick orchidBOT Mar 21, 2024, 2:58 AM

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samenn.

robust drift Mar 21, 2024, 3:04 AM

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GB -->GBGB
GB -->GBGR

these are the 2 construction paths from 2->4 that include GBG (which is not a legal 3 bead solution)

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hence the coefficient of c(n-2) is 2

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you tell me. If that's your answer, you're going to need to explain it

robust drift Mar 21, 2024, 3:39 AM

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neither

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you tell me. If that's your answer, you're going to need to explain it

robust drift Mar 21, 2024, 4:36 AM

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seems like a really confusing explanation. I just hope you understand it.

robust drift Mar 21, 2024, 5:11 AM

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the reason we need to construct n from length n-2 as well as n-1 is because:
G....GR and G...GB are legal in n
but G....G is not legal in n-1

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❤️

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I graduated a long time ago. I joked I was an AI that couldn't read cursive.

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sure, I like combinatorics

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@ me on any problem you find interesting

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I'm semi-retired now

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Sorry if it seemed like I was giving you a hard time. I don't like handing people the answers.

desert smelt Mar 21, 2024, 5:22 AM

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ch_help

#hennnice_help