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)

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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)

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

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http://prntscr.com/8jsnet here is the question beside the other questions so you have an idea of what this question is asking
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
it simply doesnt make sense if its a cycle though would it, like there can be no graph that satisfies all the conditions then
Yeah \(C_n\) usually denotes complete graphs on \(n\) vertices.
For the first equation, simplest case is \(C_3=K_3\)...
ok that is true
*second equation
oh so do they want a solution for each relation
all these equations are not like suppsoed to be solved together?
that would make my life so much easier T_T
I think so. It's not apparent whether you're asked to solved them simultaneously or not. If yes, this seems pretty hard.
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
thank you for the help!
Happy to help!
hmm C_x=K_y,z
im wondering if all C_2n = K_n,n would be a solution to this
Not true for \(K_{3,3}\), each vertex would have 3 adjacent vertices. All cycles' vertices must have degree no greater than 2.
oh right so C_4=k2,2 only one
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.
P2=K1,1?
Yup
Actually, there's more than just that one...
\(K_{2,1}\) is path
Yeah I'd missed that one. As well with \(K_{1,2}\).
okay yep
But anything higher that \(2,2\) is a graph with all verts with deg greater than 1, so that there is a cycle
i got K_4=W3 only one and K_2=K1,1 for the last one
yeah that makes sense to me

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