# Connected graphs ¶

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A graph is connected (connexe) if we only have one connected component (composante connexe). Otherwise, we are calling the graph disconnected (non connexe).

A connected component $C$, is a subgraph of $G$ in which every vertex inside is connected to at least one other vertex inside $C$. A strong component (composante connexe maximale) is a connected component in which we can't add more vertex inside.

Algorithm

• pick a vertex
• start with the first connected component $C_i$, $i=0$
• while there are vertex remaining
• for each vertex
• if $C_i$ is empty or this vertex is adjacent to a vertex inside $C_i$
• else: we go to the next vertex
• i++

When iterating the vertices, you should do it by looking at the vertex in the edges that incident to a vertex inside your connected component.

## Super-connectivity ¶

A graph is super-connected Forte connexité/f-connexe, if, from any vertex, we can go to any other vertex.

Algorithm

• pick a vertex
• mark it "+" and "-"
• mark all vertex we can reach with "+"
• mark all vertex we can be reached from "-"
• you got a first super-connected component (All nodes with "+" and "-")
• if there are remaining edges, restart from one of them

Note: a complete graph is super-connected.

## Terminology ¶

• reduced graph (Graphe réduit)

If $C_1, C_2, C_3$ are super-connected components, then a graph having the nodes $C_1, C_2, C_3$ is called Graphe réduit. If $A \in C_1$ was adjacent to $B \in C_2$ then we have $C_1$ is adjacent to $C_2$.

• Bridge (isthme)

An edge that, once removed, will disconnect the graph.

• Articulation point (Point d'articulation)

A vertex that, once removed, will disconnect the graph.

## Example 1 - Connected graph ¶

Let $G$, the following graph 1. What are the connected components?
2. Is the graph connected?
3. Create a subgraph $G'$ with $\text{{a,b,c,d}}$.
4. Is $G'$ connected? And super-connected?
1. $C_1={e,f,g}$ and $C_2={a,b,c,d,h,i}$.

Let's apply our connected algorithm 1. No, we got more than one connected component
2. simply extracting the vertex and their edges 1. We got only one component, so the graph is connected. The graph does seem to be super-connected.

Let's apply the super-connected algorithm The graph is super-connected.

## Example 2 - Transitive closure and Connectivity ¶

Is the following graph $G$ super-connected? Tip: use the transitive closure. Using Roy-Warshall's algorithm, we got As you may notice, this is a complete graph $K_{6}$. Since the transitive closure is a complete graph, then $G$ is super-connected.

This wasn't the goal of this example, but here is Roy-Warshall's algorithm The complete algorithm (text)

• picking A
• $s=C$
• $p=B$, creating (B,C)? yes
• $p=E$, creating (E,C)? yes
• $p=F$, creating (F,C)? yes
• picking B
• $s=A$
• $p=C$, creating (C,A)? yes
• $p=F$, creating (F,A)? no
• $p=D$, creating (D,A)? yes
• $s=C$
• $p=F$, creating (F,C)? no
• $p=D$, creating (D,A)? no
• picking C
• $s=B$
• $p=A$, creating (A,B)? yes
• $p=D$, creating (D,B)? no
• $p=E$, creating (E,B)? yes
• $p=F$, creating (F,B)? no
• $s=E$
• $p=A$, creating (A,E)? yes
• $p=B$, creating (B,E)? yes
• $p=D$, creating (D,E)? yes
• $p=F$, creating (F,E)? yes
• picking D
• $s=A$
• $p=E$, creating (E,A)? no
• $s=B$
• $p=E$, creating (E,B)? no
• $s=C$
• $p=E$, creating (E,C)? no
• $s=E$
• $s=F$
• $p=E$, creating (E,F)? no
• picking E
• $s=A$
• $s=A$
• $p=B$, creating (B,A)? no
• $p=C$, creating (C,A) ? yes
• $p=D$, creating (D,A)? no
• $p=F$, creating (F,A)? no
• $s=B$
• $p=A$, creating (A,B)? no
• $s=B$
• $p=C$, creating (C,B)? no
• $p=D$, creating (D,B)? no
• $p=F$, creating (F,B)? no
• $s=C$
• $p=A$, creating (A,C)? no
• $p=B$, creating (B,C)? no
• $s=C$
• $p=D$, creating (D,C)? no
• $p=F$, creating (F,C)? no
• $s=D$
• $p=A$, creating (A,D)? yes
• $p=B$, creating (B,D)? yes
• $p=C$, creating (C,D)? yes
• $s=D$
• $p=F$, creating (F,D)? yes
• $s=F$
• $p=A$, creating (A,F)? yes
• $p=B$, creating (B,F)? yes
• $p=C$, creating (C,F)? yes
• $p=D$, creating (D,F)? no
• $s=F$
• picking F
• done