Rook Chess Graph Representation
0
I'm doing an exercise where I have an adjacency matrix for the rook's graph. I need to figure out the right graph using its 3x3 adjacency matrix composed of the following values:
9 7 3
5 2 8
1 4 6
My main doubt is: Is the matrix representing a Multigraph where I have N edges between each node ? or Is the matrix representing a weighted digraph?
Thanks so much!
graph-theory adjacency-matrix
asked Nov 24 at 15:42
Mariano Mirabelli
1
add a comment |
0
I'm doing an exercise where I have an adjacency matrix for the rook's graph. I need to figure out the right graph using its 3x3 adjacency matrix composed of the following values:
9 7 3
5 2 8
1 4 6
My main doubt is: Is the matrix representing a Multigraph where I have N edges between each node ? or Is the matrix representing a weighted digraph?
Thanks so much!
graph-theory adjacency-matrix
asked Nov 24 at 15:42
Mariano Mirabelli
1
What's the source of the problem? Normally, a rook graph is a simple graph (not a multigraph) and an adjacency matrix is an $n times n$ 0-1 matrix, where $n$ is the number of vertices in the graph.
– Fabio Somenzi
Nov 24 at 16:09
Hi Fabio, Yes, but in this exercise, they give me the matrix above and ask me the following: draw the 3 × 3 rook’s graph associated to the following chessboard. For that reason, I'm not sure if I can draw the graph as a weighted digraph or as a multigraph. Thanks for the reply!
– Mariano Mirabelli
Nov 24 at 16:22
1
So, the exercise's text does not explicitly say the matrix is an adjacency... Perhaps it's just a way to give names to the squares of the chessboard, and hence to the vertices of the graph.
– Fabio Somenzi
Nov 24 at 16:28
So... each square is a node and the number is just the node's name ? Well, that is an interesting approach. So, maybe are 9 nodes and the edges are the connections based on rook movements. Yes, maybe that is the right way. Thanks a lot!
– Mariano Mirabelli
Nov 24 at 16:39
add a comment |
0
0
0
I'm doing an exercise where I have an adjacency matrix for the rook's graph. I need to figure out the right graph using its 3x3 adjacency matrix composed of the following values:
9 7 3
5 2 8
1 4 6
My main doubt is: Is the matrix representing a Multigraph where I have N edges between each node ? or Is the matrix representing a weighted digraph?
Thanks so much!
graph-theory adjacency-matrix
asked Nov 24 at 15:42
Mariano Mirabelli
1
I'm doing an exercise where I have an adjacency matrix for the rook's graph. I need to figure out the right graph using its 3x3 adjacency matrix composed of the following values:
9 7 3
5 2 8
1 4 6
My main doubt is: Is the matrix representing a Multigraph where I have N edges between each node ? or Is the matrix representing a weighted digraph?
Thanks so much!
graph-theory adjacency-matrix
graph-theory adjacency-matrix
asked Nov 24 at 15:42
Mariano Mirabelli
1
asked Nov 24 at 15:42
Mariano Mirabelli
1
asked Nov 24 at 15:42
Mariano Mirabelli
1
asked Nov 24 at 15:42
Mariano Mirabelli
1
asked Nov 24 at 15:42
Mariano Mirabelli
1
1
What's the source of the problem? Normally, a rook graph is a simple graph (not a multigraph) and an adjacency matrix is an $n times n$ 0-1 matrix, where $n$ is the number of vertices in the graph.
– Fabio Somenzi
Nov 24 at 16:09
Hi Fabio, Yes, but in this exercise, they give me the matrix above and ask me the following: draw the 3 × 3 rook’s graph associated to the following chessboard. For that reason, I'm not sure if I can draw the graph as a weighted digraph or as a multigraph. Thanks for the reply!
– Mariano Mirabelli
Nov 24 at 16:22
1
So, the exercise's text does not explicitly say the matrix is an adjacency... Perhaps it's just a way to give names to the squares of the chessboard, and hence to the vertices of the graph.
– Fabio Somenzi
Nov 24 at 16:28
So... each square is a node and the number is just the node's name ? Well, that is an interesting approach. So, maybe are 9 nodes and the edges are the connections based on rook movements. Yes, maybe that is the right way. Thanks a lot!
– Mariano Mirabelli
Nov 24 at 16:39
add a comment |
What's the source of the problem? Normally, a rook graph is a simple graph (not a multigraph) and an adjacency matrix is an $n times n$ 0-1 matrix, where $n$ is the number of vertices in the graph.
– Fabio Somenzi
Nov 24 at 16:09
Hi Fabio, Yes, but in this exercise, they give me the matrix above and ask me the following: draw the 3 × 3 rook’s graph associated to the following chessboard. For that reason, I'm not sure if I can draw the graph as a weighted digraph or as a multigraph. Thanks for the reply!
– Mariano Mirabelli
Nov 24 at 16:22
1
So, the exercise's text does not explicitly say the matrix is an adjacency... Perhaps it's just a way to give names to the squares of the chessboard, and hence to the vertices of the graph.
– Fabio Somenzi
Nov 24 at 16:28
So... each square is a node and the number is just the node's name ? Well, that is an interesting approach. So, maybe are 9 nodes and the edges are the connections based on rook movements. Yes, maybe that is the right way. Thanks a lot!
– Mariano Mirabelli
Nov 24 at 16:39
What's the source of the problem? Normally, a rook graph is a simple graph (not a multigraph) and an adjacency matrix is an $n times n$ 0-1 matrix, where $n$ is the number of vertices in the graph.
– Fabio Somenzi
Nov 24 at 16:09
What's the source of the problem? Normally, a rook graph is a simple graph (not a multigraph) and an adjacency matrix is an $n times n$ 0-1 matrix, where $n$ is the number of vertices in the graph.
– Fabio Somenzi
Nov 24 at 16:09
Hi Fabio, Yes, but in this exercise, they give me the matrix above and ask me the following: draw the 3 × 3 rook’s graph associated to the following chessboard. For that reason, I'm not sure if I can draw the graph as a weighted digraph or as a multigraph. Thanks for the reply!
– Mariano Mirabelli
Nov 24 at 16:22
Hi Fabio, Yes, but in this exercise, they give me the matrix above and ask me the following: draw the 3 × 3 rook’s graph associated to the following chessboard. For that reason, I'm not sure if I can draw the graph as a weighted digraph or as a multigraph. Thanks for the reply!
– Mariano Mirabelli
Nov 24 at 16:22
1
1
So, the exercise's text does not explicitly say the matrix is an adjacency... Perhaps it's just a way to give names to the squares of the chessboard, and hence to the vertices of the graph.
– Fabio Somenzi
Nov 24 at 16:28
So, the exercise's text does not explicitly say the matrix is an adjacency... Perhaps it's just a way to give names to the squares of the chessboard, and hence to the vertices of the graph.
– Fabio Somenzi
Nov 24 at 16:28
So... each square is a node and the number is just the node's name ? Well, that is an interesting approach. So, maybe are 9 nodes and the edges are the connections based on rook movements. Yes, maybe that is the right way. Thanks a lot!
– Mariano Mirabelli
Nov 24 at 16:39
So... each square is a node and the number is just the node's name ? Well, that is an interesting approach. So, maybe are 9 nodes and the edges are the connections based on rook movements. Yes, maybe that is the right way. Thanks a lot!
– Mariano Mirabelli
Nov 24 at 16:39
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What's the source of the problem? Normally, a rook graph is a simple graph (not a multigraph) and an adjacency matrix is an $n times n$ 0-1 matrix, where $n$ is the number of vertices in the graph.
– Fabio Somenzi
Nov 24 at 16:09
Hi Fabio, Yes, but in this exercise, they give me the matrix above and ask me the following: draw the 3 × 3 rook’s graph associated to the following chessboard. For that reason, I'm not sure if I can draw the graph as a weighted digraph or as a multigraph. Thanks for the reply!
– Mariano Mirabelli
Nov 24 at 16:22
1
So, the exercise's text does not explicitly say the matrix is an adjacency... Perhaps it's just a way to give names to the squares of the chessboard, and hence to the vertices of the graph.
– Fabio Somenzi
Nov 24 at 16:28
So... each square is a node and the number is just the node's name ? Well, that is an interesting approach. So, maybe are 9 nodes and the edges are the connections based on rook movements. Yes, maybe that is the right way. Thanks a lot!
– Mariano Mirabelli
Nov 24 at 16:39