Mixing
graph-theory problem about outdegree and indegree
$begingroup$
Let $G$ be a directed graph with $n$ vertices.
Assume that if we forget the directions of the edges of $G$, then we obtain the complete undirected graph $K_n$.
Let $v_1, v_2, ldots, v_n$ be the $n$ vertices of $G$. Prove that:
$(od(v_1))^2+(od(v_2))^2+ cdots + (od(v_n))^2 = (id(v_1))^2+(id(v_2))^2+ cdots + (id(v_n))^2$
Here, "od" means "outdegree" and "id" means "indegree".
graph-theory directed-graphs
edited Feb 3 at 5:53
darij grinberg
11.5k33168
asked Oct 20 '12 at 15:14
genigeni
5429
$endgroup$
add a comment |
$begingroup$
Let $G$ be a directed graph with $n$ vertices.
Assume that if we forget the directions of the edges of $G$, then we obtain the complete undirected graph $K_n$.
Let $v_1, v_2, ldots, v_n$ be the $n$ vertices of $G$. Prove that:
$(od(v_1))^2+(od(v_2))^2+ cdots + (od(v_n))^2 = (id(v_1))^2+(id(v_2))^2+ cdots + (id(v_n))^2$
Here, "od" means "outdegree" and "id" means "indegree".
graph-theory directed-graphs
edited Feb 3 at 5:53
darij grinberg
11.5k33168
asked Oct 20 '12 at 15:14
genigeni
5429
$endgroup$
1
$begingroup$
This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.
$endgroup$
– Graphth
Oct 20 '12 at 15:24
add a comment |
$begingroup$
Let $G$ be a directed graph with $n$ vertices.
Assume that if we forget the directions of the edges of $G$, then we obtain the complete undirected graph $K_n$.
Let $v_1, v_2, ldots, v_n$ be the $n$ vertices of $G$. Prove that:
$(od(v_1))^2+(od(v_2))^2+ cdots + (od(v_n))^2 = (id(v_1))^2+(id(v_2))^2+ cdots + (id(v_n))^2$
Here, "od" means "outdegree" and "id" means "indegree".
graph-theory directed-graphs
edited Feb 3 at 5:53
darij grinberg
11.5k33168
asked Oct 20 '12 at 15:14
genigeni
5429
$endgroup$
Let $G$ be a directed graph with $n$ vertices.
Assume that if we forget the directions of the edges of $G$, then we obtain the complete undirected graph $K_n$.
Let $v_1, v_2, ldots, v_n$ be the $n$ vertices of $G$. Prove that:
$(od(v_1))^2+(od(v_2))^2+ cdots + (od(v_n))^2 = (id(v_1))^2+(id(v_2))^2+ cdots + (id(v_n))^2$
Here, "od" means "outdegree" and "id" means "indegree".
graph-theory directed-graphs
graph-theory directed-graphs
edited Feb 3 at 5:53
darij grinberg
11.5k33168
asked Oct 20 '12 at 15:14
genigeni
5429
edited Feb 3 at 5:53
darij grinberg
11.5k33168
asked Oct 20 '12 at 15:14
genigeni
5429
edited Feb 3 at 5:53
darij grinberg
11.5k33168
edited Feb 3 at 5:53
darij grinberg
11.5k33168
edited Feb 3 at 5:53
darij grinberg
11.5k33168
11.5k33168
asked Oct 20 '12 at 15:14
genigeni
5429
asked Oct 20 '12 at 15:14
genigeni
5429
asked Oct 20 '12 at 15:14
genigeni
5429
5429
1
$begingroup$
This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.
$endgroup$
– Graphth
Oct 20 '12 at 15:24
add a comment |
1
$begingroup$
This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.
$endgroup$
– Graphth
Oct 20 '12 at 15:24
1
1
$begingroup$
This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.
$endgroup$
– Graphth
Oct 20 '12 at 15:24
$begingroup$
This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.
$endgroup$
– Graphth
Oct 20 '12 at 15:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(I prefer to use $deg_o$ and $deg_i$ for the out-degree and in-degree.)
HINT: Let the vertices be $v_1,dots,v_n$; you want to prove that $$sum_{k=1}^nbig(deg_o(v_k)big)^2=sum_{k=1}^nbig(deg_i(v_k)big)^2;,$$ or, equivalently, that $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=0;.$$
Now $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)Big(deg_o(v_k)+deg_i(v_k)Big);,$$ and since the undirected graph is $K_n$, you know exactly what $deg_o(v_k)+deg_i(v_k)$ is. You should now be able to reduce the problem to evaluating the sum
$$sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)=sum_{k=1}^ndeg_o(v_k)-sum_{k=1}^ndeg_i(v_k);,$$ which is very easy if you just think about what that last difference of sums really represents.
edited Feb 3 at 5:54
darij grinberg
11.5k33168
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
add a comment |
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$begingroup$
(I prefer to use $deg_o$ and $deg_i$ for the out-degree and in-degree.)
HINT: Let the vertices be $v_1,dots,v_n$; you want to prove that $$sum_{k=1}^nbig(deg_o(v_k)big)^2=sum_{k=1}^nbig(deg_i(v_k)big)^2;,$$ or, equivalently, that $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=0;.$$
Now $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)Big(deg_o(v_k)+deg_i(v_k)Big);,$$ and since the undirected graph is $K_n$, you know exactly what $deg_o(v_k)+deg_i(v_k)$ is. You should now be able to reduce the problem to evaluating the sum
$$sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)=sum_{k=1}^ndeg_o(v_k)-sum_{k=1}^ndeg_i(v_k);,$$ which is very easy if you just think about what that last difference of sums really represents.
edited Feb 3 at 5:54
darij grinberg
11.5k33168
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
add a comment |
$begingroup$
(I prefer to use $deg_o$ and $deg_i$ for the out-degree and in-degree.)
HINT: Let the vertices be $v_1,dots,v_n$; you want to prove that $$sum_{k=1}^nbig(deg_o(v_k)big)^2=sum_{k=1}^nbig(deg_i(v_k)big)^2;,$$ or, equivalently, that $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=0;.$$
Now $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)Big(deg_o(v_k)+deg_i(v_k)Big);,$$ and since the undirected graph is $K_n$, you know exactly what $deg_o(v_k)+deg_i(v_k)$ is. You should now be able to reduce the problem to evaluating the sum
$$sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)=sum_{k=1}^ndeg_o(v_k)-sum_{k=1}^ndeg_i(v_k);,$$ which is very easy if you just think about what that last difference of sums really represents.
edited Feb 3 at 5:54
darij grinberg
11.5k33168
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
add a comment |
$begingroup$
(I prefer to use $deg_o$ and $deg_i$ for the out-degree and in-degree.)
HINT: Let the vertices be $v_1,dots,v_n$; you want to prove that $$sum_{k=1}^nbig(deg_o(v_k)big)^2=sum_{k=1}^nbig(deg_i(v_k)big)^2;,$$ or, equivalently, that $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=0;.$$
Now $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)Big(deg_o(v_k)+deg_i(v_k)Big);,$$ and since the undirected graph is $K_n$, you know exactly what $deg_o(v_k)+deg_i(v_k)$ is. You should now be able to reduce the problem to evaluating the sum
$$sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)=sum_{k=1}^ndeg_o(v_k)-sum_{k=1}^ndeg_i(v_k);,$$ which is very easy if you just think about what that last difference of sums really represents.
edited Feb 3 at 5:54
darij grinberg
11.5k33168
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
(I prefer to use $deg_o$ and $deg_i$ for the out-degree and in-degree.)
HINT: Let the vertices be $v_1,dots,v_n$; you want to prove that $$sum_{k=1}^nbig(deg_o(v_k)big)^2=sum_{k=1}^nbig(deg_i(v_k)big)^2;,$$ or, equivalently, that $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=0;.$$
Now $$sum_{k=1}^nleft(big(deg_o(v_k)big)^2-big(deg_i(v_k)big)^2right)=sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)Big(deg_o(v_k)+deg_i(v_k)Big);,$$ and since the undirected graph is $K_n$, you know exactly what $deg_o(v_k)+deg_i(v_k)$ is. You should now be able to reduce the problem to evaluating the sum
$$sum_{k=1}^nBig(deg_o(v_k)-deg_i(v_k)Big)=sum_{k=1}^ndeg_o(v_k)-sum_{k=1}^ndeg_i(v_k);,$$ which is very easy if you just think about what that last difference of sums really represents.
edited Feb 3 at 5:54
darij grinberg
11.5k33168
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
edited Feb 3 at 5:54
darij grinberg
11.5k33168
edited Feb 3 at 5:54
darij grinberg
11.5k33168
edited Feb 3 at 5:54
darij grinberg
11.5k33168
11.5k33168
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
answered Oct 20 '12 at 15:30
Brian M. ScottBrian M. Scott
461k40518920
461k40518920
add a comment |
add a comment |
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$begingroup$
This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.
$endgroup$
– Graphth
Oct 20 '12 at 15:24