A lower bound for the number of triangles that contain a particular edge












0












$begingroup$


If $u leftrightarrow v$ in a graph $G$, prove that $uv$ belongs to at least $d(u) + d(v) − n(G)$ triangles in $G$, where $n(G)=$ number of vertices in $G$.



In this question, I tried a lot. I know we are supposed to start with $2$ vertices and check the vertices they are adjacent to, but I'm unable to write the entire solution. Please help.






graph-theory







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  • $begingroup$
    What is $n(G)$?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 12:54










  • $begingroup$
    Number of vertices in G
    $endgroup$
    – user626673
    Dec 17 '18 at 12:59
















0












$begingroup$


If $u leftrightarrow v$ in a graph $G$, prove that $uv$ belongs to at least $d(u) + d(v) − n(G)$ triangles in $G$, where $n(G)=$ number of vertices in $G$.



In this question, I tried a lot. I know we are supposed to start with $2$ vertices and check the vertices they are adjacent to, but I'm unable to write the entire solution. Please help.






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $n(G)$?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 12:54










  • $begingroup$
    Number of vertices in G
    $endgroup$
    – user626673
    Dec 17 '18 at 12:59














0












0








0





$begingroup$


If $u leftrightarrow v$ in a graph $G$, prove that $uv$ belongs to at least $d(u) + d(v) − n(G)$ triangles in $G$, where $n(G)=$ number of vertices in $G$.



In this question, I tried a lot. I know we are supposed to start with $2$ vertices and check the vertices they are adjacent to, but I'm unable to write the entire solution. Please help.






graph-theory







share|cite|improve this question











$endgroup$




If $u leftrightarrow v$ in a graph $G$, prove that $uv$ belongs to at least $d(u) + d(v) − n(G)$ triangles in $G$, where $n(G)=$ number of vertices in $G$.



In this question, I tried a lot. I know we are supposed to start with $2$ vertices and check the vertices they are adjacent to, but I'm unable to write the entire solution. Please help.






graph-theory




graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 13:09









amWhy

1




1










asked Dec 17 '18 at 12:51







user626673



















  • $begingroup$
    What is $n(G)$?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 12:54










  • $begingroup$
    Number of vertices in G
    $endgroup$
    – user626673
    Dec 17 '18 at 12:59


















  • $begingroup$
    What is $n(G)$?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 12:54










  • $begingroup$
    Number of vertices in G
    $endgroup$
    – user626673
    Dec 17 '18 at 12:59
















$begingroup$
What is $n(G)$?
$endgroup$
– Robert Z
Dec 17 '18 at 12:54




$begingroup$
What is $n(G)$?
$endgroup$
– Robert Z
Dec 17 '18 at 12:54












$begingroup$
Number of vertices in G
$endgroup$
– user626673
Dec 17 '18 at 12:59




$begingroup$
Number of vertices in G
$endgroup$
– user626673
Dec 17 '18 at 12:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

Working in the direction you mentioned, let the set of vertices adj. to $u$ be $A$ and to $v$ be $B$. The set of vertices adjacent to both $u$ & $v$ $=Acap B$. Note that
$$|Acap B|=|A|+|B|-|Acup B|$$
$$|A|=d(u), |B|=d(v) quad and quad|Acup B|leq n(G)$$
Also, $|Acap B|=$the no. of triangles asked since they'll formed with the vertices common to $u$ and $v$.
So, $$|Acap B|=|A|+|B|-|Acup B|geq d(u)+d(v)-n(G)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
    $endgroup$
    – user626673
    Dec 17 '18 at 13:10






  • 1




    $begingroup$
    For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
    $endgroup$
    – Ankit Kumar
    Dec 17 '18 at 13:42











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1 Answer
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1 Answer
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active

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active

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active

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1












$begingroup$

Working in the direction you mentioned, let the set of vertices adj. to $u$ be $A$ and to $v$ be $B$. The set of vertices adjacent to both $u$ & $v$ $=Acap B$. Note that
$$|Acap B|=|A|+|B|-|Acup B|$$
$$|A|=d(u), |B|=d(v) quad and quad|Acup B|leq n(G)$$
Also, $|Acap B|=$the no. of triangles asked since they'll formed with the vertices common to $u$ and $v$.
So, $$|Acap B|=|A|+|B|-|Acup B|geq d(u)+d(v)-n(G)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
    $endgroup$
    – user626673
    Dec 17 '18 at 13:10






  • 1




    $begingroup$
    For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
    $endgroup$
    – Ankit Kumar
    Dec 17 '18 at 13:42
















1












$begingroup$

Working in the direction you mentioned, let the set of vertices adj. to $u$ be $A$ and to $v$ be $B$. The set of vertices adjacent to both $u$ & $v$ $=Acap B$. Note that
$$|Acap B|=|A|+|B|-|Acup B|$$
$$|A|=d(u), |B|=d(v) quad and quad|Acup B|leq n(G)$$
Also, $|Acap B|=$the no. of triangles asked since they'll formed with the vertices common to $u$ and $v$.
So, $$|Acap B|=|A|+|B|-|Acup B|geq d(u)+d(v)-n(G)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
    $endgroup$
    – user626673
    Dec 17 '18 at 13:10






  • 1




    $begingroup$
    For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
    $endgroup$
    – Ankit Kumar
    Dec 17 '18 at 13:42














1












1








1





$begingroup$

Working in the direction you mentioned, let the set of vertices adj. to $u$ be $A$ and to $v$ be $B$. The set of vertices adjacent to both $u$ & $v$ $=Acap B$. Note that
$$|Acap B|=|A|+|B|-|Acup B|$$
$$|A|=d(u), |B|=d(v) quad and quad|Acup B|leq n(G)$$
Also, $|Acap B|=$the no. of triangles asked since they'll formed with the vertices common to $u$ and $v$.
So, $$|Acap B|=|A|+|B|-|Acup B|geq d(u)+d(v)-n(G)$$






share|cite|improve this answer









$endgroup$



Working in the direction you mentioned, let the set of vertices adj. to $u$ be $A$ and to $v$ be $B$. The set of vertices adjacent to both $u$ & $v$ $=Acap B$. Note that
$$|Acap B|=|A|+|B|-|Acup B|$$
$$|A|=d(u), |B|=d(v) quad and quad|Acup B|leq n(G)$$
Also, $|Acap B|=$the no. of triangles asked since they'll formed with the vertices common to $u$ and $v$.
So, $$|Acap B|=|A|+|B|-|Acup B|geq d(u)+d(v)-n(G)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 13:03









Ankit KumarAnkit Kumar

1,352219




1,352219












  • $begingroup$
    Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
    $endgroup$
    – user626673
    Dec 17 '18 at 13:10






  • 1




    $begingroup$
    For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
    $endgroup$
    – Ankit Kumar
    Dec 17 '18 at 13:42


















  • $begingroup$
    Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
    $endgroup$
    – user626673
    Dec 17 '18 at 13:10






  • 1




    $begingroup$
    For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
    $endgroup$
    – Ankit Kumar
    Dec 17 '18 at 13:42
















$begingroup$
Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
$endgroup$
– user626673
Dec 17 '18 at 13:10




$begingroup$
Thank you, it was a very clear solution. However, my question was recently edited to be "A lower bound for the number of triangles that contain a particular edge ". I'm just wondering, is there any upper bound to it to?
$endgroup$
– user626673
Dec 17 '18 at 13:10




1




1




$begingroup$
For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
$endgroup$
– Ankit Kumar
Dec 17 '18 at 13:42




$begingroup$
For that, we'll have to minimise $|A cup B|$, which is the no. of vertices common to u or v, which is $d(u)+d(v)$, if I'm not wrong. You can get the corresponding upper bound then.
$endgroup$
– Ankit Kumar
Dec 17 '18 at 13:42


















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