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)

Question

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)

compassionate · Answer

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

dan815 · Answer

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 · Answer

yes

dan815 · Answer

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

dan815 · Answer

@zzr0ck3r @zepdrix

dan815 · Answer

@ganeshie8

dan815 · Answer

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 · Answer

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 · Answer

Yeah $C_n$ usually denotes complete graphs on $n$ vertices.

anonymous · Answer

For the first equation, simplest case is $C_3=K_3$...

dan815 · Answer

ok that is true

anonymous · Answer

*second equation

dan815 · Answer

oh so do they want a solution for each relation

dan815 · Answer

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

dan815 · Answer

that would make my life so much easier T_T

anonymous · Answer

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

dan815 · Answer

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 · Answer

thank you for the help!

anonymous · Answer

Happy to help!

dan815 · Answer

hmm C_x=K_y,z

dan815 · Answer

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

anonymous · Answer

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

dan815 · Answer

oh right so C_4=k2,2 only one

anonymous · Answer

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 · Answer

P2=K1,1?

anonymous · Answer

Yup

anonymous · Answer

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

zzr0ck3r · Answer

$K_{2,1}$ is path

anonymous · Answer

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

dan815 · Answer

okay yep

zzr0ck3r · Answer

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

dan815 · Answer

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

dan815 · Answer

yeah that makes sense to me