What is the “common neighborhood” of a single vertex in a graph?












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In the paper "On finding bicliques in bipartite graphs: a novel algorithm and its application to the integration of diverse biological data types" the authors propose an improvement to an algorithm, by sorting candidate vertices by "common neighborhood size" (page 8 at left).



What is the "common" neighborhood for a single vertex?










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  • A "neighborhood" of a vertex is the set of vertices it is adjacent to, so "common neighborhood size" would most likely mean "vertices of the same degree."
    – Math1000
    Dec 24 '18 at 22:11
















0














In the paper "On finding bicliques in bipartite graphs: a novel algorithm and its application to the integration of diverse biological data types" the authors propose an improvement to an algorithm, by sorting candidate vertices by "common neighborhood size" (page 8 at left).



What is the "common" neighborhood for a single vertex?










share|cite|improve this question
























  • A "neighborhood" of a vertex is the set of vertices it is adjacent to, so "common neighborhood size" would most likely mean "vertices of the same degree."
    – Math1000
    Dec 24 '18 at 22:11














0












0








0







In the paper "On finding bicliques in bipartite graphs: a novel algorithm and its application to the integration of diverse biological data types" the authors propose an improvement to an algorithm, by sorting candidate vertices by "common neighborhood size" (page 8 at left).



What is the "common" neighborhood for a single vertex?










share|cite|improve this question















In the paper "On finding bicliques in bipartite graphs: a novel algorithm and its application to the integration of diverse biological data types" the authors propose an improvement to an algorithm, by sorting candidate vertices by "common neighborhood size" (page 8 at left).



What is the "common" neighborhood for a single vertex?







graph-theory terminology






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edited 2 days ago









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asked Dec 24 '18 at 18:09









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  • A "neighborhood" of a vertex is the set of vertices it is adjacent to, so "common neighborhood size" would most likely mean "vertices of the same degree."
    – Math1000
    Dec 24 '18 at 22:11


















  • A "neighborhood" of a vertex is the set of vertices it is adjacent to, so "common neighborhood size" would most likely mean "vertices of the same degree."
    – Math1000
    Dec 24 '18 at 22:11
















A "neighborhood" of a vertex is the set of vertices it is adjacent to, so "common neighborhood size" would most likely mean "vertices of the same degree."
– Math1000
Dec 24 '18 at 22:11




A "neighborhood" of a vertex is the set of vertices it is adjacent to, so "common neighborhood size" would most likely mean "vertices of the same degree."
– Math1000
Dec 24 '18 at 22:11










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Given two vertices $x$ and $y$, $N(x, y) = N(x) cap N(y)$ is the common neighbourhood of those two vertices where the size would be denoted as $|N(x) cap N(y)|$.




What is "common" neighborhood for a single vertex?




It seems a bit superfluous to use the term "common neighbourhood" when referring to a single vertex since the neighbours that a vertex has in common with itself is all of its neighbours.



$$
N(x, x) = N(x) cap N(x) = N(x) tag{Idempotent law}
$$



I think the authors of the paper are primarily concerned with comparing distinct vertices in partition $V$. This is covered in section "Candidate selection" which describes why selecting candidates in non-decreasing order of common neighbourhood size might reduce the number of non-maximal subsets that the algorithm has to generate. So in Figure 5 for graph $G_4$, they are sorting based on $|N(v_i, v_{j})|$, which in this example results in the algorithm not picking candidate vertex $v_1$ first.



enter image description here






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    1 Answer
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    1 Answer
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    active

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    active

    oldest

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    1














    Given two vertices $x$ and $y$, $N(x, y) = N(x) cap N(y)$ is the common neighbourhood of those two vertices where the size would be denoted as $|N(x) cap N(y)|$.




    What is "common" neighborhood for a single vertex?




    It seems a bit superfluous to use the term "common neighbourhood" when referring to a single vertex since the neighbours that a vertex has in common with itself is all of its neighbours.



    $$
    N(x, x) = N(x) cap N(x) = N(x) tag{Idempotent law}
    $$



    I think the authors of the paper are primarily concerned with comparing distinct vertices in partition $V$. This is covered in section "Candidate selection" which describes why selecting candidates in non-decreasing order of common neighbourhood size might reduce the number of non-maximal subsets that the algorithm has to generate. So in Figure 5 for graph $G_4$, they are sorting based on $|N(v_i, v_{j})|$, which in this example results in the algorithm not picking candidate vertex $v_1$ first.



    enter image description here






    share|cite|improve this answer


























      1














      Given two vertices $x$ and $y$, $N(x, y) = N(x) cap N(y)$ is the common neighbourhood of those two vertices where the size would be denoted as $|N(x) cap N(y)|$.




      What is "common" neighborhood for a single vertex?




      It seems a bit superfluous to use the term "common neighbourhood" when referring to a single vertex since the neighbours that a vertex has in common with itself is all of its neighbours.



      $$
      N(x, x) = N(x) cap N(x) = N(x) tag{Idempotent law}
      $$



      I think the authors of the paper are primarily concerned with comparing distinct vertices in partition $V$. This is covered in section "Candidate selection" which describes why selecting candidates in non-decreasing order of common neighbourhood size might reduce the number of non-maximal subsets that the algorithm has to generate. So in Figure 5 for graph $G_4$, they are sorting based on $|N(v_i, v_{j})|$, which in this example results in the algorithm not picking candidate vertex $v_1$ first.



      enter image description here






      share|cite|improve this answer
























        1












        1








        1






        Given two vertices $x$ and $y$, $N(x, y) = N(x) cap N(y)$ is the common neighbourhood of those two vertices where the size would be denoted as $|N(x) cap N(y)|$.




        What is "common" neighborhood for a single vertex?




        It seems a bit superfluous to use the term "common neighbourhood" when referring to a single vertex since the neighbours that a vertex has in common with itself is all of its neighbours.



        $$
        N(x, x) = N(x) cap N(x) = N(x) tag{Idempotent law}
        $$



        I think the authors of the paper are primarily concerned with comparing distinct vertices in partition $V$. This is covered in section "Candidate selection" which describes why selecting candidates in non-decreasing order of common neighbourhood size might reduce the number of non-maximal subsets that the algorithm has to generate. So in Figure 5 for graph $G_4$, they are sorting based on $|N(v_i, v_{j})|$, which in this example results in the algorithm not picking candidate vertex $v_1$ first.



        enter image description here






        share|cite|improve this answer












        Given two vertices $x$ and $y$, $N(x, y) = N(x) cap N(y)$ is the common neighbourhood of those two vertices where the size would be denoted as $|N(x) cap N(y)|$.




        What is "common" neighborhood for a single vertex?




        It seems a bit superfluous to use the term "common neighbourhood" when referring to a single vertex since the neighbours that a vertex has in common with itself is all of its neighbours.



        $$
        N(x, x) = N(x) cap N(x) = N(x) tag{Idempotent law}
        $$



        I think the authors of the paper are primarily concerned with comparing distinct vertices in partition $V$. This is covered in section "Candidate selection" which describes why selecting candidates in non-decreasing order of common neighbourhood size might reduce the number of non-maximal subsets that the algorithm has to generate. So in Figure 5 for graph $G_4$, they are sorting based on $|N(v_i, v_{j})|$, which in this example results in the algorithm not picking candidate vertex $v_1$ first.



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        EdOverflowEdOverflow

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