dan815
  • dan815
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Compassionate
  • Compassionate
Not a single bit. Now, toss some y = mx + b at me, and we'll talk.
dan815
  • 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
  • Compassionate
yes

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dan815
  • 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
  • dan815
@zzr0ck3r @zepdrix
dan815
  • dan815
@ganeshie8
dan815
  • 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
  • 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
  • anonymous
Yeah \(C_n\) usually denotes complete graphs on \(n\) vertices.
anonymous
  • anonymous
For the first equation, simplest case is \(C_3=K_3\)...
dan815
  • dan815
ok that is true
anonymous
  • anonymous
*second equation
dan815
  • dan815
oh so do they want a solution for each relation
dan815
  • dan815
all these equations are not like suppsoed to be solved together?
dan815
  • dan815
that would make my life so much easier T_T
anonymous
  • 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
  • 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
  • dan815
thank you for the help!
anonymous
  • anonymous
Happy to help!
dan815
  • dan815
hmm C_x=K_y,z
dan815
  • dan815
im wondering if all C_2n = K_n,n would be a solution to this
anonymous
  • 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
  • dan815
oh right so C_4=k2,2 only one
anonymous
  • 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
  • dan815
P2=K1,1?
anonymous
  • anonymous
Yup
anonymous
  • anonymous
Actually, there's more than just that one...
zzr0ck3r
  • zzr0ck3r
\(K_{2,1}\) is path
anonymous
  • anonymous
Yeah I'd missed that one. As well with \(K_{1,2}\).
dan815
  • dan815
okay yep
zzr0ck3r
  • 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
  • dan815
i got K_4=W3 only one and K_2=K1,1 for the last one
dan815
  • dan815
yeah that makes sense to me

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