Can an undirected graph be disconnected?
up vote
0
down vote
favorite
This may be a rather trivial question but I am still trying to get the hang of all the graph theory terms. Nonetheless, I haven't found a source that explicitly says that an undirected graph can only be connected so is it possible to have an undirected graph that is disconnected? And if so, may I have an example one?
Much thanks!
graph-theory connectedness directed-graphs
add a comment |
up vote
0
down vote
favorite
This may be a rather trivial question but I am still trying to get the hang of all the graph theory terms. Nonetheless, I haven't found a source that explicitly says that an undirected graph can only be connected so is it possible to have an undirected graph that is disconnected? And if so, may I have an example one?
Much thanks!
graph-theory connectedness directed-graphs
if they are made of separate pieces.
– hbm
5 hours ago
4
Here's an example of (the diagram of) a disconnected undirected graph: $$huge ○,,,, ○$$
– Git Gud
5 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This may be a rather trivial question but I am still trying to get the hang of all the graph theory terms. Nonetheless, I haven't found a source that explicitly says that an undirected graph can only be connected so is it possible to have an undirected graph that is disconnected? And if so, may I have an example one?
Much thanks!
graph-theory connectedness directed-graphs
This may be a rather trivial question but I am still trying to get the hang of all the graph theory terms. Nonetheless, I haven't found a source that explicitly says that an undirected graph can only be connected so is it possible to have an undirected graph that is disconnected? And if so, may I have an example one?
Much thanks!
graph-theory connectedness directed-graphs
graph-theory connectedness directed-graphs
edited 4 hours ago
Scientifica
6,22141333
6,22141333
asked 5 hours ago
DevAllanPer
896
896
if they are made of separate pieces.
– hbm
5 hours ago
4
Here's an example of (the diagram of) a disconnected undirected graph: $$huge ○,,,, ○$$
– Git Gud
5 hours ago
add a comment |
if they are made of separate pieces.
– hbm
5 hours ago
4
Here's an example of (the diagram of) a disconnected undirected graph: $$huge ○,,,, ○$$
– Git Gud
5 hours ago
if they are made of separate pieces.
– hbm
5 hours ago
if they are made of separate pieces.
– hbm
5 hours ago
4
4
Here's an example of (the diagram of) a disconnected undirected graph: $$huge ○,,,, ○$$
– Git Gud
5 hours ago
Here's an example of (the diagram of) a disconnected undirected graph: $$huge ○,,,, ○$$
– Git Gud
5 hours ago
add a comment |
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
Undirected just mean The edges does not have direction. connected means that there is a path from any vertex of the graph to any other vertex in the graph. so take any disconnected graph whose edges are not directed to give an example. following is one:
add a comment |
up vote
4
down vote
Yes. The simplest such graph is just two vertices (no edges).
add a comment |
up vote
2
down vote
The definition of graph that I know is the following: A graph consists of two sets $(V,E)$ where $V$ is the set of vertices and $E$ is the set of edges.
The elements of $E$ are subsets (or multisets in the case of loops) of cardinality $2$ of $V$.
A graph is undirected if ${x,y}={y,x}$ where ${x,y},{y,x}in E$ and it is directed if ${x,y}neq {y,x}$.
Therefore, by taking $V={a,b,c}$ and $E={{a,b}}$, you obtain a disconnected undirected graph.
add a comment |
up vote
1
down vote
I believe, since you can define a graph $G = (E,V)$ by its edge and vertex sets, it is perfectly ok to have a disconnected graph (i.e. a graph with no path between some vertices). In fact, taking $E$ to be empty still results in a graph.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Undirected just mean The edges does not have direction. connected means that there is a path from any vertex of the graph to any other vertex in the graph. so take any disconnected graph whose edges are not directed to give an example. following is one:
add a comment |
up vote
5
down vote
accepted
Undirected just mean The edges does not have direction. connected means that there is a path from any vertex of the graph to any other vertex in the graph. so take any disconnected graph whose edges are not directed to give an example. following is one:
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Undirected just mean The edges does not have direction. connected means that there is a path from any vertex of the graph to any other vertex in the graph. so take any disconnected graph whose edges are not directed to give an example. following is one:
Undirected just mean The edges does not have direction. connected means that there is a path from any vertex of the graph to any other vertex in the graph. so take any disconnected graph whose edges are not directed to give an example. following is one:
answered 5 hours ago
mathnoob
1,247116
1,247116
add a comment |
add a comment |
up vote
4
down vote
Yes. The simplest such graph is just two vertices (no edges).
add a comment |
up vote
4
down vote
Yes. The simplest such graph is just two vertices (no edges).
add a comment |
up vote
4
down vote
up vote
4
down vote
Yes. The simplest such graph is just two vertices (no edges).
Yes. The simplest such graph is just two vertices (no edges).
answered 5 hours ago
platty
2,707218
2,707218
add a comment |
add a comment |
up vote
2
down vote
The definition of graph that I know is the following: A graph consists of two sets $(V,E)$ where $V$ is the set of vertices and $E$ is the set of edges.
The elements of $E$ are subsets (or multisets in the case of loops) of cardinality $2$ of $V$.
A graph is undirected if ${x,y}={y,x}$ where ${x,y},{y,x}in E$ and it is directed if ${x,y}neq {y,x}$.
Therefore, by taking $V={a,b,c}$ and $E={{a,b}}$, you obtain a disconnected undirected graph.
add a comment |
up vote
2
down vote
The definition of graph that I know is the following: A graph consists of two sets $(V,E)$ where $V$ is the set of vertices and $E$ is the set of edges.
The elements of $E$ are subsets (or multisets in the case of loops) of cardinality $2$ of $V$.
A graph is undirected if ${x,y}={y,x}$ where ${x,y},{y,x}in E$ and it is directed if ${x,y}neq {y,x}$.
Therefore, by taking $V={a,b,c}$ and $E={{a,b}}$, you obtain a disconnected undirected graph.
add a comment |
up vote
2
down vote
up vote
2
down vote
The definition of graph that I know is the following: A graph consists of two sets $(V,E)$ where $V$ is the set of vertices and $E$ is the set of edges.
The elements of $E$ are subsets (or multisets in the case of loops) of cardinality $2$ of $V$.
A graph is undirected if ${x,y}={y,x}$ where ${x,y},{y,x}in E$ and it is directed if ${x,y}neq {y,x}$.
Therefore, by taking $V={a,b,c}$ and $E={{a,b}}$, you obtain a disconnected undirected graph.
The definition of graph that I know is the following: A graph consists of two sets $(V,E)$ where $V$ is the set of vertices and $E$ is the set of edges.
The elements of $E$ are subsets (or multisets in the case of loops) of cardinality $2$ of $V$.
A graph is undirected if ${x,y}={y,x}$ where ${x,y},{y,x}in E$ and it is directed if ${x,y}neq {y,x}$.
Therefore, by taking $V={a,b,c}$ and $E={{a,b}}$, you obtain a disconnected undirected graph.
answered 5 hours ago
Karen
805
805
add a comment |
add a comment |
up vote
1
down vote
I believe, since you can define a graph $G = (E,V)$ by its edge and vertex sets, it is perfectly ok to have a disconnected graph (i.e. a graph with no path between some vertices). In fact, taking $E$ to be empty still results in a graph.
add a comment |
up vote
1
down vote
I believe, since you can define a graph $G = (E,V)$ by its edge and vertex sets, it is perfectly ok to have a disconnected graph (i.e. a graph with no path between some vertices). In fact, taking $E$ to be empty still results in a graph.
add a comment |
up vote
1
down vote
up vote
1
down vote
I believe, since you can define a graph $G = (E,V)$ by its edge and vertex sets, it is perfectly ok to have a disconnected graph (i.e. a graph with no path between some vertices). In fact, taking $E$ to be empty still results in a graph.
I believe, since you can define a graph $G = (E,V)$ by its edge and vertex sets, it is perfectly ok to have a disconnected graph (i.e. a graph with no path between some vertices). In fact, taking $E$ to be empty still results in a graph.
answered 5 hours ago
Stuartg98
586
586
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026333%2fcan-an-undirected-graph-be-disconnected%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
if they are made of separate pieces.
– hbm
5 hours ago
4
Here's an example of (the diagram of) a disconnected undirected graph: $$huge ○,,,, ○$$
– Git Gud
5 hours ago