Simple connected graph question
$begingroup$
What is the minimum number of edges that a simple
connected graph with n vertices can have?
A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. So wouldn't the minimum number of edges be n-1? This would form a line linking all vertices.
Am I missing something? This seems too easy.
graph-theory
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
$endgroup$
add a comment |
$begingroup$
What is the minimum number of edges that a simple
connected graph with n vertices can have?
A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. So wouldn't the minimum number of edges be n-1? This would form a line linking all vertices.
Am I missing something? This seems too easy.
graph-theory
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
$endgroup$
1
$begingroup$
That is correct. A formalization of your reasoning can be done using induction.
$endgroup$
– Fimpellizieri
Apr 27 '16 at 4:27
1
$begingroup$
Yes it's easy but you do not really have a proof. Show that adding an edge reduces the number of components by at most $1$. So if you begin with $n$ isolated vertices, you have to add at least $n-1$ edges to get exactly $1$ component.
$endgroup$
– Forever Mozart
Apr 27 '16 at 4:29
$begingroup$
"Simple" is irrelevant. If the graph has the minimum number of edges for connectivity, it won't have two vertices joined by two edges.
$endgroup$
– bof
Apr 27 '16 at 4:48
add a comment |
$begingroup$
What is the minimum number of edges that a simple
connected graph with n vertices can have?
A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. So wouldn't the minimum number of edges be n-1? This would form a line linking all vertices.
Am I missing something? This seems too easy.
graph-theory
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
$endgroup$
What is the minimum number of edges that a simple
connected graph with n vertices can have?
A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. So wouldn't the minimum number of edges be n-1? This would form a line linking all vertices.
Am I missing something? This seems too easy.
graph-theory
graph-theory
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
asked Apr 27 '16 at 4:25
dibdubdibdub
93112
93112
1
$begingroup$
That is correct. A formalization of your reasoning can be done using induction.
$endgroup$
– Fimpellizieri
Apr 27 '16 at 4:27
1
$begingroup$
Yes it's easy but you do not really have a proof. Show that adding an edge reduces the number of components by at most $1$. So if you begin with $n$ isolated vertices, you have to add at least $n-1$ edges to get exactly $1$ component.
$endgroup$
– Forever Mozart
Apr 27 '16 at 4:29
$begingroup$
"Simple" is irrelevant. If the graph has the minimum number of edges for connectivity, it won't have two vertices joined by two edges.
$endgroup$
– bof
Apr 27 '16 at 4:48
add a comment |
1
$begingroup$
That is correct. A formalization of your reasoning can be done using induction.
$endgroup$
– Fimpellizieri
Apr 27 '16 at 4:27
1
$begingroup$
Yes it's easy but you do not really have a proof. Show that adding an edge reduces the number of components by at most $1$. So if you begin with $n$ isolated vertices, you have to add at least $n-1$ edges to get exactly $1$ component.
$endgroup$
– Forever Mozart
Apr 27 '16 at 4:29
$begingroup$
"Simple" is irrelevant. If the graph has the minimum number of edges for connectivity, it won't have two vertices joined by two edges.
$endgroup$
– bof
Apr 27 '16 at 4:48
1
1
$begingroup$
That is correct. A formalization of your reasoning can be done using induction.
$endgroup$
– Fimpellizieri
Apr 27 '16 at 4:27
$begingroup$
That is correct. A formalization of your reasoning can be done using induction.
$endgroup$
– Fimpellizieri
Apr 27 '16 at 4:27
1
1
$begingroup$
Yes it's easy but you do not really have a proof. Show that adding an edge reduces the number of components by at most $1$. So if you begin with $n$ isolated vertices, you have to add at least $n-1$ edges to get exactly $1$ component.
$endgroup$
– Forever Mozart
Apr 27 '16 at 4:29
$begingroup$
Yes it's easy but you do not really have a proof. Show that adding an edge reduces the number of components by at most $1$. So if you begin with $n$ isolated vertices, you have to add at least $n-1$ edges to get exactly $1$ component.
$endgroup$
– Forever Mozart
Apr 27 '16 at 4:29
$begingroup$
"Simple" is irrelevant. If the graph has the minimum number of edges for connectivity, it won't have two vertices joined by two edges.
$endgroup$
– bof
Apr 27 '16 at 4:48
$begingroup$
"Simple" is irrelevant. If the graph has the minimum number of edges for connectivity, it won't have two vertices joined by two edges.
$endgroup$
– bof
Apr 27 '16 at 4:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Fact: if the degree of all points is 2 or more, there is a cycle $v_1 rightarrow v_2 rightarrow ldots v_1$ in the graph.
(Proof sketch: start anywhere and keep walking, till you meet a vertex you've already seen (which will happen as there only finitely many): there are no dead ends as long as we meet new points because of the degree condition)).
This sets you up for induction: assume it's true for $n$ ($n=1$ is trivial), then take a minimally connected graph of size $n+1$. If all degrees are $2$ or more, we have a cycle, and we can remove an edge from the cycle and keep connectedness of the total graph, contradicting minimality. So we have a point of degree $1$ (degree $0$ cannot happen in a connected graph), and we remove this 1 edge and 1 vertex and apply the induction assumption...
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f1760584%2fsimple-connected-graph-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Fact: if the degree of all points is 2 or more, there is a cycle $v_1 rightarrow v_2 rightarrow ldots v_1$ in the graph.
(Proof sketch: start anywhere and keep walking, till you meet a vertex you've already seen (which will happen as there only finitely many): there are no dead ends as long as we meet new points because of the degree condition)).
This sets you up for induction: assume it's true for $n$ ($n=1$ is trivial), then take a minimally connected graph of size $n+1$. If all degrees are $2$ or more, we have a cycle, and we can remove an edge from the cycle and keep connectedness of the total graph, contradicting minimality. So we have a point of degree $1$ (degree $0$ cannot happen in a connected graph), and we remove this 1 edge and 1 vertex and apply the induction assumption...
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
Fact: if the degree of all points is 2 or more, there is a cycle $v_1 rightarrow v_2 rightarrow ldots v_1$ in the graph.
(Proof sketch: start anywhere and keep walking, till you meet a vertex you've already seen (which will happen as there only finitely many): there are no dead ends as long as we meet new points because of the degree condition)).
This sets you up for induction: assume it's true for $n$ ($n=1$ is trivial), then take a minimally connected graph of size $n+1$. If all degrees are $2$ or more, we have a cycle, and we can remove an edge from the cycle and keep connectedness of the total graph, contradicting minimality. So we have a point of degree $1$ (degree $0$ cannot happen in a connected graph), and we remove this 1 edge and 1 vertex and apply the induction assumption...
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
Fact: if the degree of all points is 2 or more, there is a cycle $v_1 rightarrow v_2 rightarrow ldots v_1$ in the graph.
(Proof sketch: start anywhere and keep walking, till you meet a vertex you've already seen (which will happen as there only finitely many): there are no dead ends as long as we meet new points because of the degree condition)).
This sets you up for induction: assume it's true for $n$ ($n=1$ is trivial), then take a minimally connected graph of size $n+1$. If all degrees are $2$ or more, we have a cycle, and we can remove an edge from the cycle and keep connectedness of the total graph, contradicting minimality. So we have a point of degree $1$ (degree $0$ cannot happen in a connected graph), and we remove this 1 edge and 1 vertex and apply the induction assumption...
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
Fact: if the degree of all points is 2 or more, there is a cycle $v_1 rightarrow v_2 rightarrow ldots v_1$ in the graph.
(Proof sketch: start anywhere and keep walking, till you meet a vertex you've already seen (which will happen as there only finitely many): there are no dead ends as long as we meet new points because of the degree condition)).
This sets you up for induction: assume it's true for $n$ ($n=1$ is trivial), then take a minimally connected graph of size $n+1$. If all degrees are $2$ or more, we have a cycle, and we can remove an edge from the cycle and keep connectedness of the total graph, contradicting minimality. So we have a point of degree $1$ (degree $0$ cannot happen in a connected graph), and we remove this 1 edge and 1 vertex and apply the induction assumption...
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
answered Apr 27 '16 at 5:48
Henno BrandsmaHenno Brandsma
112k348120
112k348120
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.
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%2f1760584%2fsimple-connected-graph-question%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
1
$begingroup$
That is correct. A formalization of your reasoning can be done using induction.
$endgroup$
– Fimpellizieri
Apr 27 '16 at 4:27
1
$begingroup$
Yes it's easy but you do not really have a proof. Show that adding an edge reduces the number of components by at most $1$. So if you begin with $n$ isolated vertices, you have to add at least $n-1$ edges to get exactly $1$ component.
$endgroup$
– Forever Mozart
Apr 27 '16 at 4:29
$begingroup$
"Simple" is irrelevant. If the graph has the minimum number of edges for connectivity, it won't have two vertices joined by two edges.
$endgroup$
– bof
Apr 27 '16 at 4:48