Introduction to Line Graphs and 2-sections (Hypergraph Episode 5)

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  • เผยแพร่เมื่อ 26 ธ.ค. 2024

ความคิดเห็น • 13

  • @indusuresh7097
    @indusuresh7097 10 หลายเดือนก่อน +1

    Hi, thank you for making this video which is very useful in my studies. But i have one doubt in line graph that is what if the intersection of hyperedges contain more than one vertices? for example let e_1 and e_2 are hyperedges and there intersection is more than 1 vertices, so then what about the line graph?

    • @VitalSine
      @VitalSine  10 หลายเดือนก่อน

      Hi, great question. If the intersection is more than 1 vertex, in the line graph e_1 and e_2 are still made adjacent. The definition of adjacency in a line graph only requires that the edges share at least one incident vertex.

  • @judahdsouza9196
    @judahdsouza9196 2 ปีที่แล้ว +1

    i NEED the video on cliques of hypergraphs and 2-section graphs!

    • @VitalSine
      @VitalSine  2 ปีที่แล้ว

      A conformal hypergraph video is coming soon, we'll look at a lot of concepts surrounding cliques of 2-sections and the hypergraph edges 👍

  • @pswjt1266
    @pswjt1266 ปีที่แล้ว +1

    Hi, this series is very useful thank you for making this! I have a very specific technical question if you don't mind me asking. I work on an algorithm which analyzes the 3-line graph of our hypergraph structured data, which is the line graph except you only include edges for hyperedge intersections which have at least 3 vertices. I have found that if you associate every edge in the 3-line graph with the vertices that form the corresponding connection between hyperedges, you can define each set of these vertices as a hyperedge and construct another hypergraph, which seems to have very interesting properties. Does that make sense? Is there some hypergraph theory that describes such an operation? I have not been able to find it thus far. Thank you again!

    • @VitalSine
      @VitalSine  ปีที่แล้ว

      You're very welcome, and interesting question! If I understand correctly, the sets of vertices that make up an (intersection with 3+ vertices) of hyperedges in your original hypergraph become your new hyperedges, and the vertices of the new hypergraph are still the same as the vertices of the original hypergraph?

    • @pswjt1266
      @pswjt1266 ปีที่แล้ว

      @@VitalSine Yes exactly!

    • @VitalSine
      @VitalSine  ปีที่แล้ว +1

      ​@@pswjt1266 Hello again! I looked around but could not find anything matching the operation you described, but this is similar and may be of interest: arxiv.org/abs/1901.06292
      I think you could perform their "edge intersection hypergraph" operation and then weakly delete any edges with exactly 2 vertices to get the same result as your new operation, except you would miss all of the submaximal intersecting sets of vertices of size 3+ present in your original hypergraph.

    • @pswjt1266
      @pswjt1266 ปีที่แล้ว +1

      ​@@VitalSine This does indeed seem close to what I have in mind! Thank you so much!

    • @VitalSine
      @VitalSine  ปีที่แล้ว

      @@pswjt1266 You're very welcome, and thank you for sharing this new operation with me.

  • @frankng7554
    @frankng7554 2 ปีที่แล้ว +1

    Hi i really enjoy watching your videos on hypergraph, which are so informative and intuitive. Recently, I got confused about the concept "similarity" and the similarity function in hypergraph. Will you cover this topics in your future videos?

    • @VitalSine
      @VitalSine  2 ปีที่แล้ว +1

      Hello, thanks for reaching out, I'm glad that you enjoy my videos! I'm not familiar with the similarity concept regarding hypergraphs, could you please direct me to the paper/papers discussing this topic? I can try to look it over and cover the topic in a future video.

    • @frankng7554
      @frankng7554 2 ปีที่แล้ว +1

      ​@@VitalSine Thanks for your reply! The relevant content is in Chapter 1.3.2 of this book: "Bretto, A. (2013). Hypergraph theory. An introduction. Mathematical Engineering. Cham: Springer". Looking forward to seeing your new videos in this series!