patternMinor
Find proportion of 'important' edge among all edges in a graph
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amongedgesgraphallproportionedgeimportantfind
Problem
I met this problem when dealing with a social network problem. It is a subproblem of that. The problem is like that:
We look at a graph $G(V,E)$, where $V$ is the set of vertices and $E$ set of edges. We define $x \in V$ as an 'important' vertex if at least $\frac 1 3 deg(x)$ ($deg(x)$ refers to the degree of $x$)of the vertices in $x$'s neighborhood have degree no more than degree of $x$. We then define 'important ' edge as the edge when either one of the two vertices of the edge or both are 'important' vertices. We define the set of 'important' edge as $E_1$.
We want to show $|E_1| \geq \frac 1 2 |E|$. I cannot really think out any method. Could anyone provide some suggestions?
Many thanks.
We look at a graph $G(V,E)$, where $V$ is the set of vertices and $E$ set of edges. We define $x \in V$ as an 'important' vertex if at least $\frac 1 3 deg(x)$ ($deg(x)$ refers to the degree of $x$)of the vertices in $x$'s neighborhood have degree no more than degree of $x$. We then define 'important ' edge as the edge when either one of the two vertices of the edge or both are 'important' vertices. We define the set of 'important' edge as $E_1$.
We want to show $|E_1| \geq \frac 1 2 |E|$. I cannot really think out any method. Could anyone provide some suggestions?
Many thanks.
Solution
The problem you asked can be actually generalized to an arbitrary total order $\mathcal{O}$ on the vertices: a vertex $x$ is important iff at least $\frac{1}{3}$ of its neighbors are smaller than $x$ according to $\mathcal{O}$. For the order on degrees we can break the ties arbitrarily, which will only make important vertices and thus important edges less.
We prove it by the adjacency matrix $M$. Suppose the vertices are ordered: $x_1j\},\quad U=\{M_{i,j}\mid i|E|-\frac{1}{2}\sum_{i=1}^n|U\cap R_i|=\frac{1}{2}|E|.
\end{align*}$$
We prove it by the adjacency matrix $M$. Suppose the vertices are ordered: $x_1j\},\quad U=\{M_{i,j}\mid i|E|-\frac{1}{2}\sum_{i=1}^n|U\cap R_i|=\frac{1}{2}|E|.
\end{align*}$$
Context
StackExchange Computer Science Q#66258, answer score: 4
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