Maximum number of vertices in a simple graph
$begingroup$
In a connected simple graph $G$ with $30$ edges, what is the maximum number of vertices possible?
According to me, the answer is 31, obtained in a star structured graph. But the solution manual claims that the answer should be $9$ .
Am I correct or what am I doing wrong? Please let me know.
Thanks in advance!
combinatorics discrete-mathematics graph-theory
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
$endgroup$
add a comment |
$begingroup$
In a connected simple graph $G$ with $30$ edges, what is the maximum number of vertices possible?
According to me, the answer is 31, obtained in a star structured graph. But the solution manual claims that the answer should be $9$ .
Am I correct or what am I doing wrong? Please let me know.
Thanks in advance!
combinatorics discrete-mathematics graph-theory
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
$endgroup$
add a comment |
$begingroup$
In a connected simple graph $G$ with $30$ edges, what is the maximum number of vertices possible?
According to me, the answer is 31, obtained in a star structured graph. But the solution manual claims that the answer should be $9$ .
Am I correct or what am I doing wrong? Please let me know.
Thanks in advance!
combinatorics discrete-mathematics graph-theory
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
$endgroup$
In a connected simple graph $G$ with $30$ edges, what is the maximum number of vertices possible?
According to me, the answer is 31, obtained in a star structured graph. But the solution manual claims that the answer should be $9$ .
Am I correct or what am I doing wrong? Please let me know.
Thanks in advance!
combinatorics discrete-mathematics graph-theory
combinatorics discrete-mathematics graph-theory
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
edited Dec 26 '18 at 13:15
greedoid
42.2k1152105
42.2k1152105
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
asked Dec 26 '18 at 11:54
Abhilash MishraAbhilash Mishra
121
121
add a comment |
add a comment |
2 Answers
2
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votes
$begingroup$
Is the question asking instead for the minimum number of vertices for a simple graph with 30 edges? A simple graph of $n$ vertices has at most $frac{n(n-1)}{2}$ undirected edges...that gives 36 edges for $n=9$, but only 28 edges for $n=8$.
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
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add a comment |
$begingroup$
Idea:
You can ask what is a minimum number of vertices that graph is not connected and with 30 edges.
Then exsist partition $(A,B)$ with no edges between them. So we have $${|A|choose 2}+{|B|choose 2} geq 30$$
where $n=|A|+|B|$ the number you are looking for.
Let $x=|A|$ then $|B|=n-x$ and $0<x<n$. So $$30leq {x^2 +(n-x)^2-nover 2}leq {n-1choose 2}+{1choose 2}$$
and thus $$30leq {n-1choose 2}implies n_{min}= 9$$
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
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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
$begingroup$
Is the question asking instead for the minimum number of vertices for a simple graph with 30 edges? A simple graph of $n$ vertices has at most $frac{n(n-1)}{2}$ undirected edges...that gives 36 edges for $n=9$, but only 28 edges for $n=8$.
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
$endgroup$
add a comment |
$begingroup$
Is the question asking instead for the minimum number of vertices for a simple graph with 30 edges? A simple graph of $n$ vertices has at most $frac{n(n-1)}{2}$ undirected edges...that gives 36 edges for $n=9$, but only 28 edges for $n=8$.
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
$endgroup$
add a comment |
$begingroup$
Is the question asking instead for the minimum number of vertices for a simple graph with 30 edges? A simple graph of $n$ vertices has at most $frac{n(n-1)}{2}$ undirected edges...that gives 36 edges for $n=9$, but only 28 edges for $n=8$.
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
$endgroup$
Is the question asking instead for the minimum number of vertices for a simple graph with 30 edges? A simple graph of $n$ vertices has at most $frac{n(n-1)}{2}$ undirected edges...that gives 36 edges for $n=9$, but only 28 edges for $n=8$.
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
answered Dec 26 '18 at 12:29
SamadinSamadin
42117
42117
add a comment |
add a comment |
$begingroup$
Idea:
You can ask what is a minimum number of vertices that graph is not connected and with 30 edges.
Then exsist partition $(A,B)$ with no edges between them. So we have $${|A|choose 2}+{|B|choose 2} geq 30$$
where $n=|A|+|B|$ the number you are looking for.
Let $x=|A|$ then $|B|=n-x$ and $0<x<n$. So $$30leq {x^2 +(n-x)^2-nover 2}leq {n-1choose 2}+{1choose 2}$$
and thus $$30leq {n-1choose 2}implies n_{min}= 9$$
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
$endgroup$
add a comment |
$begingroup$
Idea:
You can ask what is a minimum number of vertices that graph is not connected and with 30 edges.
Then exsist partition $(A,B)$ with no edges between them. So we have $${|A|choose 2}+{|B|choose 2} geq 30$$
where $n=|A|+|B|$ the number you are looking for.
Let $x=|A|$ then $|B|=n-x$ and $0<x<n$. So $$30leq {x^2 +(n-x)^2-nover 2}leq {n-1choose 2}+{1choose 2}$$
and thus $$30leq {n-1choose 2}implies n_{min}= 9$$
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
$endgroup$
add a comment |
$begingroup$
Idea:
You can ask what is a minimum number of vertices that graph is not connected and with 30 edges.
Then exsist partition $(A,B)$ with no edges between them. So we have $${|A|choose 2}+{|B|choose 2} geq 30$$
where $n=|A|+|B|$ the number you are looking for.
Let $x=|A|$ then $|B|=n-x$ and $0<x<n$. So $$30leq {x^2 +(n-x)^2-nover 2}leq {n-1choose 2}+{1choose 2}$$
and thus $$30leq {n-1choose 2}implies n_{min}= 9$$
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
$endgroup$
Idea:
You can ask what is a minimum number of vertices that graph is not connected and with 30 edges.
Then exsist partition $(A,B)$ with no edges between them. So we have $${|A|choose 2}+{|B|choose 2} geq 30$$
where $n=|A|+|B|$ the number you are looking for.
Let $x=|A|$ then $|B|=n-x$ and $0<x<n$. So $$30leq {x^2 +(n-x)^2-nover 2}leq {n-1choose 2}+{1choose 2}$$
and thus $$30leq {n-1choose 2}implies n_{min}= 9$$
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
edited Dec 26 '18 at 13:13
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
answered Dec 26 '18 at 12:22
greedoidgreedoid
42.2k1152105
42.2k1152105
add a comment |
add a comment |
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