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I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.

Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail. So I really hope someone could give me a hint on how to approach this problem.

7 Answers 7

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Suppose $G$ is disconnected. We want to show that $\bar{G}$ is connected. So suppose $v$ and $w$ are vertices. If $vw$ is not an edge in $G$, then it is an edge in $\bar{G}$, and so we have a path from $v$ to $w$ in $\bar{G}$. On the other hand, if $vw$ is an edge in $G$, then this means $v$ and $w$ are in the same component of $G$. Since $G$ is disconnected, we can find a vertex $u$ in a different component, so that neither $uv$ nor $uw$ are edges of $G$. Then $vuw$ is a parth from $v$ to $w$ in $\bar{G}$.

This shows that any two vertices in $\bar{G}$ have a path (in fact a path of length one or two) between them in $\bar{G}$, so $\bar{G}$ is connected.

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    I think the generally useful tactic here is to actually look at the graph and draw some pictures...I started with writing out complex formulas, but it was a wrong approach to this particular problem.2012-03-19
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If $\bar G$, the complement of $G$, is not connected, then there exists a partitioning of vertices into two disjoint sets $V_1$ and $V_2$ such that no edge of the complement is between them, i.e. for all $v_1 \in V_1$ and $v_2 \in V_2$ we have $\langle v_1,v_2\rangle \notin \bar E$. However, this means that for all $v_1$ and $v_2$ from $V_1$ and $V_2$ respectively, $\langle v_1,v_2 \rangle \in E$, hence $G$ is connected.

I hope this helps ;-)

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    Would the downvoter care to explain?2013-03-21
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I'm not very clever, and thinking gives me a headache. Let's try to solve this thing with a minimum of cleverness and hard thinking. A little brute force is OK.

Let's try for a proof by contradiction, that will always work if anything does. So we assume a disconnected graph $G$ with disconnected complement $\overline G.$

Then there are two vertices $a,b$ which can't be connected by a path in $G,$ and there are two vertices $c,d$ which can't be connected by a path in $\overline G.$

Let $H$ be the subgraph of $G$ induced by $\{a,b,c,d\},$ a graph with $2,3,$ or $4$ vertices depending on how $\{a,b\}$ and $\{c,d\}$ overlap. (Maybe I could eliminate one of those cases if I thought about it. Nuts to that.) If two vertices can't be connected by a path in the big graph, then they can't be connected by a path in the subgraph, right? So $H$ and $\overline H$ are still disconnected.

Great, that means I only have to check disconnected graphs with $2,3,$ or $4$ vertices. I can do that with brute force!

Allowing myself a tiny bit of cleverness to save work: If my disconnected graph $H$ has a disconnected complement, and if I can add edges to $H$ without making it connected, then the complement (which is losing edges) will still be disconnected! So I only have to test maximal disconnected graphs!! With just a little bit of thought, I see that a maximal disconnected graph is the sum of two complete graphs. Up to $4$ vertices, the only graphs I have to check are: $K_1+K_1,\ K_1+K_2,\ K_1+K_3,\ K_2+K_2.$ The complements of those graphs are $K_{1,1},\ K_{1,2},\ K_{1,3},\ K_{2,2}.$ The complements are all connected! That does it!!

Wait a minute, that's looking suspiciously similar to the proofs in the other answers.

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Additionally, it will be easy if simply look at the adjacency matrix of the graph.

More explanation: The adjacency matrix of a disconnected graph will be block diagonal. Then think about its complement, if two vertices were in different connected component in the original graph, then they are adjacent in the complement; if two vertices were in the same connected component in the orginal graph, then a $2$-path connects them.

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    You may be right. I didn't actually read Chris' answer :-)2013-04-19
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Suppose one of G = (V,E) and G' = (V,E) is disconnected; say G with components G., . . . .,Gk, k > 1, w.l.o.g since G = G''. Any two vertices v∉Gi and w∉Gj will be connected in G' since (a) if i 6≠j then vw ∉ E, so vw 2 E'. (b) if i = j then there must be a third vertex u in another component such that vu ∉ E and wu ∉ E. In this case, v and w would be connected in G' through edges vu,wu ∉ E'. Since any pair of vertices in G' are connected, G' is connected.

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This is another answer, showing G¯ is connected in 3 cases by showing there exists an {a,b} path in each case. Let x,y be vertices in G. Let H be the component that holds the vertex x but not y.

C1: Suppose a and b are arbitrary vertices not in H. Then x~a and x~b in G¯ so there is a unique {a,b} path in G¯, namely {a,x,b}. (~ means adjacent)

C2: Suppose H holds one of the two vertices, a and b. Without loss of generality suppose H holds a. Then x~b and a~b in G¯. So there is a unique {a,b} path in G¯, namely {a,b}.

C3: Suppose H holds both the a and b vertices. Then y~a, y~b, y~x in G¯. So there is a unique {a,b} path in G¯, namely {a,y,b}.

Therefore there is a unique {a,b} path in G¯ for arbitrary a and b vertices in all 3 cases. Therefore G¯ is connected.

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    Please use proper formatting to enhance readability; for some basic information about writing math at this site see e.g. [here](/help/notation), [here](//meta.math.stackexchange.com/q/5020), [here](//meta.stackexchange.com/a/70559) and [here](//meta.math.stackexchange.com/q/1773).2017-04-03