Graph theory,
http://prntscr.com/8jsfnv
you guys know what this question is asking?
\[k_{m,n}\]
is a bipartite graph
\[k_n\]
is a complete graph of n vertices
\[W_y\] is a wheel with y+1 vertices where y vertices are arranged in a cycle and the y+1th vertex in adj to all the other y vertices
therefore a wheel W_n has an average degree of (4*n)/(n+1)

- dan815

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- Compassionate

Not a single bit. Now, toss some y = mx + b at me, and we'll talk.

- dan815

they have some equations where bipartite graphs are equaling a complete graph that should only be possible when its a 1-factor like a complete graph of only 2 vertices

- Compassionate

yes

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## More answers

- dan815

http://prntscr.com/8jsnet
here is the question beside the other questions so you have an idea of what this question is asking

- dan815

@zzr0ck3r @zepdrix

- dan815

@ganeshie8

- dan815

im a little confused if C_n supposed to be Cycle of n vertices or its just a random graph, I know K_n is the compelte graph and Kn,m is the bipartite complete graph and W_n is the wheel n+1 graph that is confirmed for sure

- dan815

it simply doesnt make sense if its a cycle though would it, like there can be no graph that satisfies all the conditions then

- anonymous

Yeah \(C_n\) usually denotes complete graphs on \(n\) vertices.

- anonymous

For the first equation, simplest case is \(C_3=K_3\)...

- dan815

ok that is true

- anonymous

*second equation

- dan815

oh so do they want a solution for each relation

- dan815

all these equations are not like suppsoed to be solved together?

- dan815

that would make my life so much easier T_T

- anonymous

I think so. It's not apparent whether you're asked to solved them simultaneously or not. If yes, this seems pretty hard.

- dan815

at this point im just gonna do that ive wasted needless time on this question all the other questions were very quick and simple, so im going to assume thats what they want for this too

- dan815

thank you for the help!

- anonymous

Happy to help!

- dan815

hmm C_x=K_y,z

- dan815

im wondering if all
C_2n = K_n,n would be a solution to this

- anonymous

Not true for \(K_{3,3}\), each vertex would have 3 adjacent vertices. All cycles' vertices must have degree no greater than 2.

- dan815

oh right so C_4=k2,2 only one

- anonymous

For \(P_x=K_{y,z}\) it looks like you can only have one easy solution, assuming \(P_x\) is a path on \(x\) vertices.

- dan815

P2=K1,1?

- anonymous

Yup

- anonymous

Actually, there's more than just that one...

- zzr0ck3r

\(K_{2,1}\) is path

- anonymous

Yeah I'd missed that one. As well with \(K_{1,2}\).

- dan815

okay yep

- zzr0ck3r

But anything higher that \(2,2\) is a graph with all verts with deg greater than 1, so that there is a cycle

- dan815

i got K_4=W3 only one
and K_2=K1,1 for the last one

- dan815

yeah that makes sense to me

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