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dan815

  • one year ago

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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  1. Compassionate
    • one year ago
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    Not a single bit. Now, toss some y = mx + b at me, and we'll talk.

  2. dan815
    • one year ago
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    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

  3. Compassionate
    • one year ago
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    yes

  4. dan815
    • one year ago
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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

  5. dan815
    • one year ago
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    @zzr0ck3r @zepdrix

  6. dan815
    • one year ago
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    @ganeshie8

  7. dan815
    • one year ago
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    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

  8. dan815
    • one year ago
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    it simply doesnt make sense if its a cycle though would it, like there can be no graph that satisfies all the conditions then

  9. anonymous
    • one year ago
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    Yeah \(C_n\) usually denotes complete graphs on \(n\) vertices.

  10. anonymous
    • one year ago
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    For the first equation, simplest case is \(C_3=K_3\)...

  11. dan815
    • one year ago
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    ok that is true

  12. anonymous
    • one year ago
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    *second equation

  13. dan815
    • one year ago
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    oh so do they want a solution for each relation

  14. dan815
    • one year ago
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    all these equations are not like suppsoed to be solved together?

  15. dan815
    • one year ago
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    that would make my life so much easier T_T

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

  17. dan815
    • one year ago
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    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

  18. dan815
    • one year ago
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    thank you for the help!

  19. anonymous
    • one year ago
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    Happy to help!

  20. dan815
    • one year ago
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    hmm C_x=K_y,z

  21. dan815
    • one year ago
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    im wondering if all C_2n = K_n,n would be a solution to this

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

  23. dan815
    • one year ago
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    oh right so C_4=k2,2 only one

  24. anonymous
    • one year ago
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    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.

  25. dan815
    • one year ago
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    P2=K1,1?

  26. anonymous
    • one year ago
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    Yup

  27. anonymous
    • one year ago
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    Actually, there's more than just that one...

  28. zzr0ck3r
    • one year ago
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    \(K_{2,1}\) is path

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

  30. dan815
    • one year ago
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    okay yep

  31. zzr0ck3r
    • one year ago
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    But anything higher that \(2,2\) is a graph with all verts with deg greater than 1, so that there is a cycle

  32. dan815
    • one year ago
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    i got K_4=W3 only one and K_2=K1,1 for the last one

  33. dan815
    • one year ago
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    yeah that makes sense to me

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