NetSci 04-3 Katz Centrality

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  • เผยแพร่เมื่อ 26 ก.ย. 2024
  • Katz centrality sits between eigenvector centrality and PageRank on the spectrum of network centrality measures. Each vertex has an inherent importance, as well as gaining importance from the quality and quantity of its neighbors. This measure has some advantages over eigenvector centrality for directed networks.

ความคิดเห็น • 6

  • @souravdey1227
    @souravdey1227 3 ปีที่แล้ว +1

    Something is wrong with the TH-cam recommendation algorithms. And I am not joking when I say that my research will be on this.
    Great stuff

  • @debasismondal3619
    @debasismondal3619 3 ปีที่แล้ว +1

    This video is great but underrated if one to understand the concepts of Eighenvector and Katz centrality I hardly came across any videos which explains it in so much detail.

  • @az362951847
    @az362951847 2 ปีที่แล้ว

    Is this Katz centrality the same as the statement in Wiki? The close form of Katz centrality in Wiki has an additional term "-1" compared to your result (the case of beta=0).

    • @andrewbeveridge7476
      @andrewbeveridge7476  8 หลายเดือนก่อน

      They are equivalent. My working definition comes from Mark Newman's "Networks" book. On the Wikipedia page, the Alpha centrality section explains how the two definitions give the same ordering of the vertices.

  • @qimingcui4952
    @qimingcui4952 2 ปีที่แล้ว

    cannot understand 7:10, it seems that v1 is not an eigenvector of A transpose but A itself, corresponding to lambda 1...

    • @andrewbeveridge7476
      @andrewbeveridge7476  8 หลายเดือนก่อน

      This is a subtle part of the explanation. At 7:10, the goal is to show that when alpha=lambda1, the matrix (I - 1/lambda1 * A^T) is not invertible. Therefore, we take the eigenvector v1 for A^T corresponding to lambda1. So (I - 1/lambda1 * A^T)v1 = v1 - v1 = 0.
      The definition of Katz centrality requires the inverse of (I - alpha *A^T). But when alpha = 1/lambda1, this inverse is not defined.
      Therefore, we typically to take alpha to be a little smaller than 1/lambda1. Perhaps a good choice is 85% of this value.