[Discrete Mathematics] Hamilton Cycles

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  • เผยแพร่เมื่อ 28 ก.ย. 2024
  • We introduce the concept of Hamilton Cycles in Graph Theory.
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    We introduce Hamilton cycles, sort of similar to Euler Circuits.
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ความคิดเห็น • 25

  • @tyroneslothdrop9155
    @tyroneslothdrop9155 9 ปีที่แล้ว +35

    You're doing great work on this channel. I hope you continue to focus on advanced mathematics in the future because that is what's most lacking in the TH-cam-verse.

    • @Trevtutor
      @Trevtutor  9 ปีที่แล้ว +20

      Tyrone Slothdrop I'm planning on moving into Linear Algebra, then hopefully some Group Theory, Rings, and Fields afterwards. We'll see how things go. Working on a side project at the moment off the channel. Big plans!

  • @imazsurve3290
    @imazsurve3290 2 ปีที่แล้ว +2

    this playlist was very helpful. Thanks a ton!!!

  • @buibui__
    @buibui__ 4 ปีที่แล้ว

    For the 8:10 is it wrong for the vertex at the last go back to the a just because it is bipartite graph which cannot gone through it's own vertex again?

  • @krishnaswaroop9293
    @krishnaswaroop9293 5 ปีที่แล้ว

    Thanx sir, the bipartite graph result was very helpful..

  • @mageboyd619
    @mageboyd619 6 ปีที่แล้ว +4

    Did u go to SFU?

    • @Trevtutor
      @Trevtutor  6 ปีที่แล้ว +5

      Just a total coincidence that the courses match MACM 101 and MACM 201. ;P

    • @mageboyd619
      @mageboyd619 6 ปีที่แล้ว +2

      LOL macm 201 final in 2 hrs thanks for all your hard work!!

  • @holly3802
    @holly3802 6 ปีที่แล้ว +1

    Hi! You say around 11:15 that the graph has a Hamilton cycle because Va = Vb. But that isn't sufficient to prove the existence of one I think. What if a bipartite graph with equal independent sets of vertices had a cut edge? I mean, I know the graph you gave has a Ham cycle, but I think it's over-generalizing to say "because Va = Vb" since that's not always true. I could be wrong though, I am new to this.

    • @TheWhinoceros
      @TheWhinoceros 6 ปีที่แล้ว

      Yeah, Trev got that one part wrong. A very simple graph with 4 vertices: o -- o -- o -- o, is bipartite and Va = Vb, but the graph clearly does not have a Hamilton cycle. I only know this now because I thought Va = Vb was sufficient to prove the existence of a Hamilton cycle, and lost marks on a quiz! Damn it Trev! Haha

  • @KansasFashion
    @KansasFashion 6 ปีที่แล้ว

    Love you man

  • @gor3947
    @gor3947 ปีที่แล้ว

    cavt tanem dzec

  • @kiana6672
    @kiana6672 7 ปีที่แล้ว

    Hi Trev, how come you don't go into more depth for this topic? The textbook has a larger scope and depth, and the questions in MACM 201 go beyond what this intro video teaches. All your other topics were much more exhaustive.

  • @gor3947
    @gor3947 ปีที่แล้ว

    lav poshmaneci(

  • @PrakashBesra
    @PrakashBesra 8 ปีที่แล้ว +5

    A Hamiltonian cycle, also called a Hamiltonian circuit, Hamilton cycle, or Hamilton circuit, is a graph cycle (i.e., closed loop) through a graph that visits each node exactly once.

    • @ARNAKLDO
      @ARNAKLDO 5 ปีที่แล้ว

      Prakash Besra what about the initial node? it is visited twice once the cycle is created does it not?

  • @grasshopperweb
    @grasshopperweb 4 ปีที่แล้ว

    my combinatorics class does all the proofs :( a little disappointed you didn't cover those.

  • @daisyd5168
    @daisyd5168 5 ปีที่แล้ว

    Hi Trev, but does |V_a|=|V_b| guarantee a Hamilton cycle in a bipartite graph? I don't think so.

    • @RichardZamoraTheArtsareAwesome
      @RichardZamoraTheArtsareAwesome 4 ปีที่แล้ว +2

      A bipartite with equal amount of nodes that have same degree I believe guarantee a hamilton cycle

  • @mikeegayon2942
    @mikeegayon2942 8 ปีที่แล้ว

    is hamiltonian cycle and hamiltonian circuit are the same?

  • @abhisheksharma-tq8xg
    @abhisheksharma-tq8xg 9 ปีที่แล้ว +1

    are hamilton cycle and hamilton path same? plz rply

    • @MrSidd12
      @MrSidd12 8 ปีที่แล้ว +2

      +abhishek sharma if the starting and ending point are same then it is a cycle otherwise it is a path.