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My question is about parts c) and d) of this question:

Let $G = (V,E)$ be an oriented graph. For every pair of vertices $u,v \in V$, we write $u \leftrightarrow v $ iff there exists a path from $u$ to $v$ and a path from $v$ to $u$.

graph

In a) I was asked to show that it is an equivalence relation. So I went on and explained reflexivity, symmetry and transitivity.

In b) I was asked to show the 'strongly connected' classes that have the relation $ \leftrightarrow $.

I have that $\{a,b\}$ and $\{f,h\}$ are strongly connected.... is this ok ?

Now in c). Let $P$ be a partition $P = \{V_1,V_2,...,V_k\}$ of the set of vertices $V$ of a graph $G$. We define the quotient graph as the graph $H$ of which the set of vertices is $P$ such that there exists an arc from $V_i$ to $V_j$ if $i \ne j $ and there exists an arc in $G$ from a vertex of $V_i$ to a vertex of $V_j$.

Calculate the quotient graph from the picture, for the partition given by the equivalence relation $\leftrightarrow$

I can't even understand it. Sorry if it's not clear, I have trouble translating it.

Where do I start with c)?

Also in d) I am asked to show that every quotient graph is acyclic...

Nick Matteo
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Dave
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    You are being asked to draw a new graph with one vertex for each set of equivalent vertices. You are then asked to prove that any graph obtained this way will not have any closed loops. – John Douma Apr 20 '13 at 19:25

1 Answers1

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Your answer for b) is correct but incomplete. Note that a path can contain many edges.

$\leftrightarrow$ partitions the graph into the following sets: $\{a,b\}$, $\{c\}$ and $\{d,e,f,g,h\}$. Two vertices of the graph are 'strongly connected' iff they are in the same set of this partition.

Your quotient graph will thus have $3$ vertices, given by $\{a,b\}$, $\{c\}$ and $\{d,e,f,g,h\}$. Can you figure out the edges between them?

Abel
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  • Why is c in the partitions sets , so I have a-c-e ? – Dave Apr 22 '13 at 16:42
  • No, you have: $a\leftrightarrow b$ and $d\leftrightarrow e\leftrightarrow f\leftrightarrow g\leftrightarrow h$. If $x\leftrightarrow c$, then $x=c$ because the set of the partition containing $c$ is ${c}$. – Abel Apr 26 '13 at 07:22