Proof: Connected Graph of Order n Has at least n-1 Edges | Graph Theory

แชร์
ฝัง
  • เผยแพร่เมื่อ 30 ก.ค. 2020
  • A connected graph of order n has at least n-1 edges, in other words - tree graphs are the minimally connected graphs. We'll be proving this result in today's graph theory lesson!
    We'll be using a special type of contradiction proof called a proof by minimum counterexample. If we assume that there is a graph contradicting our claim, we can consider one such graph of minimum order, then we will show that this "minimum order counterexample" is actually not a minimum order counterexample, producing a contradiction and proving our claim.
    Connected graphs with one less edge than vertices NEVER have cycles, and connected graphs with no cycles are called tree graphs; check out these lesson on them...
    Intro to Tree Graphs: • Intro to Tree Graphs |...
    Proof that a tree graph of order n has size n-1: • Proof: Tree Graph of O...
    Proof that a connected graph with n vertices and n-1 edges is a tree graph: • Proof: Graph with n Ve...
    ◆ Donate on PayPal: www.paypal.me/wrathofmath
    ◆ Support Wrath of Math on Patreon: / wrathofmathlessons
    I hope you find this video helpful, and be sure to ask any questions down in the comments!
    +WRATH OF MATH+
    Follow Wrath of Math on...
    ● Instagram: / wrathofmathedu
    ● Facebook: / wrathofmath
    ● Twitter: / wrathofmathedu
    My Music Channel: / seanemusic

ความคิดเห็น • 25

  • @PunmasterSTP

    Minimal counterexample? More like "Magnificent lectures, with quality that's more than ample!"

  • @LearningCS-jp4cb
    @LearningCS-jp4cb 16 ชั่วโมงที่ผ่านมา

    Didn't get the minimum order part of proof, but luckily i can understand alternate contradiction using same argument.

  • @tchappyha4034
    @tchappyha4034 ปีที่แล้ว +1

    Thank you very much for your elegant proof.

  • @-Manasa-ij4bb
    @-Manasa-ij4bb 3 ปีที่แล้ว +1

    Consider an undirected graph G that has n distinct vertices where each vertex has degree 2. Assume n≥3. What is the maximum number of circuits that G can contain? If all the vertices of G are contained in a single circuit, what is the maximum number of vertices that can be contained in an independent set?

  • @imwatchingyou9975
    @imwatchingyou9975 3 ปีที่แล้ว +6

    I don't know what to say

  • @tifawine9113
    @tifawine9113 3 ปีที่แล้ว +3

    BUT the triangel has 3 sides. Why should be just 2 edges?

  • @sufiyanbudye4275
    @sufiyanbudye4275 4 ปีที่แล้ว +1

    Neatly explained 👌

  • @parasmittal3697
    @parasmittal3697 3 ปีที่แล้ว +2

    Great proof , thanks a lot for this !!

  • @rahulkhatav9089
    @rahulkhatav9089 3 ปีที่แล้ว +3

    Good explanation. Hope it saves me for the exams 😅

  • @abhistyadav9421
    @abhistyadav9421 ปีที่แล้ว

    How do you prove that after removing an end vertex the graph still remains connected

  • @eddieadnan1
    @eddieadnan1 ปีที่แล้ว

    Hello can you make a vdieo on Prufer code

  • @Salman15474
    @Salman15474 3 ปีที่แล้ว

    what a clever neat proof this was

  • @tybargky461
    @tybargky461 ปีที่แล้ว +2

    I am still confused lol

  • @kanikasihag9881
    @kanikasihag9881 3 ปีที่แล้ว

    Hey!!could you pls make a video on the que:

  • @mariyamsanayam1867
    @mariyamsanayam1867 4 ปีที่แล้ว

    Nice class

  • @dogkansarac4889
    @dogkansarac4889 3 ปีที่แล้ว

    dont you think using induction is way easier than this proof?

  • @monke9865
    @monke9865 3 ปีที่แล้ว

    Really appreciate your videos.