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?
graph-theory terminology
edited Jan 5 at 19:46
EdOverflow
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asked Dec 24 '18 at 18:09
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3921415
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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?
graph-theory terminology
edited Jan 5 at 19:46
EdOverflow
2119
asked Dec 24 '18 at 18:09
DimsDims
3921415
$endgroup$
$begingroup$
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."
$endgroup$
– Math1000
Dec 24 '18 at 22:11
add a comment |
$begingroup$
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
edited Jan 5 at 19:46
EdOverflow
2119
asked Dec 24 '18 at 18:09
DimsDims
3921415
$endgroup$
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
graph-theory terminology
edited Jan 5 at 19:46
EdOverflow
2119
asked Dec 24 '18 at 18:09
DimsDims
3921415
edited Jan 5 at 19:46
EdOverflow
2119
asked Dec 24 '18 at 18:09
DimsDims
3921415
edited Jan 5 at 19:46
EdOverflow
2119
edited Jan 5 at 19:46
EdOverflow
2119
edited Jan 5 at 19:46
EdOverflow
2119
2119
asked Dec 24 '18 at 18:09
DimsDims
3921415
asked Dec 24 '18 at 18:09
DimsDims
3921415
asked Dec 24 '18 at 18:09
DimsDims
3921415
3921415
$begingroup$
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."
$endgroup$
– Math1000
Dec 24 '18 at 22:11
add a comment |
$begingroup$
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."
$endgroup$
– Math1000
Dec 24 '18 at 22:11
$begingroup$
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."
$endgroup$
– Math1000
Dec 24 '18 at 22:11
$begingroup$
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."
$endgroup$
– Math1000
Dec 24 '18 at 22:11
add a comment |
1 Answer
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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.
answered Jan 5 at 18:46
EdOverflowEdOverflow
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1 Answer
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$begingroup$
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.
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
$endgroup$
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.
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
answered Jan 5 at 18:46
EdOverflowEdOverflow
2119
2119
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$begingroup$
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."
$endgroup$
– Math1000
Dec 24 '18 at 22:11