Eigenvalues of the Laplacian of the directed De Bruijn graph Planned maintenance scheduled...
Eigenvalues of the Laplacian of the directed De Bruijn graph
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$begingroup$
We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.
We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.
We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:
https://core.ac.uk/download/pdf/82810454.pdf
Thanks!
co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics
edited 4 hours ago
Fedor Petrov
52.5k6123241
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
$endgroup$
add a comment |
$begingroup$
We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.
We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.
We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:
https://core.ac.uk/download/pdf/82810454.pdf
Thanks!
co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics
edited 4 hours ago
Fedor Petrov
52.5k6123241
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
$endgroup$
add a comment |
$begingroup$
We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.
We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.
We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:
https://core.ac.uk/download/pdf/82810454.pdf
Thanks!
co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics
edited 4 hours ago
Fedor Petrov
52.5k6123241
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
$endgroup$
We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.
We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.
We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:
https://core.ac.uk/download/pdf/82810454.pdf
Thanks!
co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics
co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics
edited 4 hours ago
Fedor Petrov
52.5k6123241
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
edited 4 hours ago
Fedor Petrov
52.5k6123241
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
edited 4 hours ago
Fedor Petrov
52.5k6123241
edited 4 hours ago
Fedor Petrov
52.5k6123241
edited 4 hours ago
Fedor Petrov
52.5k6123241
52.5k6123241
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
asked 5 hours ago
Serge the ToasterSerge the Toaster
592
592
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
add a comment |
$begingroup$
I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
add a comment |
$begingroup$
The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
add a comment |
$begingroup$
The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
answered 2 hours ago
Richard StanleyRichard Stanley
29.4k9118191
29.4k9118191
add a comment |
add a comment |
$begingroup$
I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
$endgroup$
add a comment |
$begingroup$
I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
$endgroup$
add a comment |
$begingroup$
I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
$endgroup$
I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
edited 4 hours ago
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
answered 5 hours ago
Dima PasechnikDima Pasechnik
9,44311952
9,44311952
add a comment |
add a comment |
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