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Connectivity 1 Ma/CS 6b Class 5: Graph Connectivity By Adam Sheffer A Connectivity Problem Prove. The vertices of a connected graph 퐺 can always be ordered as 푣1, 푣2, … , 푣푛 such that for every 1 < 푖 ≤ 푛, if we remove 푣푖, 푣푖+1, … , 푣푛 and the edges adjacent to these vertices, 퐺 remains connected. 푣3 푣4 푣1 푣2 푣5 Proof Pick any vertex as 푣1. Pick a vertex that is connected to 푣1 in 퐺 and set it as 푣2. Pick a vertex that is connected either to 푣1 or to 푣2 in 퐺 and set it as 푣3. … Communications Network We are given a set of routers and wish to connect pairs of them to obtain a connected communications network. The network should be reliable – a few malfunctioning routers should not disable the entire network. ◦ What condition should we require from the network? ◦ That after removing any 푘 routers, the network remains connected. 푘-connected Graphs An graph 퐺 = (푉, 퐸) is said to be 푘- connected if 푉 > 푘 and we cannot obtain a non-connected graph by removing 푘 − 1 vertices from 푉. Is the graph in the figure ◦ 1-connected? Yes. ◦ 2-connected? Yes. ◦ 3-connected? No! Connectivity Which graphs are 1-connected? ◦ These are exactly the connected graphs. The connectivity of a graph 퐺 is the maximum integer 푘 such that 퐺 is 푘- connected. What is the connectivity of the complete graph 퐾푛? 푛 − 1. The graph in the figure has a connectivity of 2. Hypercube A hypercube is a generalization of the cube into any dimension. ◦ In 푑-dimensions, we consider the vertices of the hypercube as the points whose coordinates are 0’s and 1’s. ◦ Two vertices are adjacent if they differ in a single coordinate. (0,1) (1,1) 1 0 (0,0) (1,0) Hypercube Properties The points of a 푑-dimensional hypercube correspond to the 푑-dimensional vectors with 0-1 entries. ◦ Thus, it has 2푑 vertices. Two vertices are adjacent if they have 푑 − 1 common coordaintes. ◦ Thus, there are 2푑−1푑 edges. (0,1) (1,1) 1 0 (0,0) (1,0) Hypercube graphs The hypercube graph 푄푑 corresponds to the 푑-dimensional hypercube. ◦ A vertex for each vertex of the hypercube. ◦ An edge for each edge of the hypercube. 푑 푑−1 ◦ The graph 푄푑 has 2 vertices and 2 푑 edges. It is 푑-regular. 4d-hypercube Hypercube graph Properties Problem. Is 푄푑 a bipartite graph? Answer. Yes! ◦ On one side we place every vertex with an even number of 1-coordinates. On the other, the vertices with an odd number of 1- coordintes. 4d-hypercube More Properties Problem. What is the connectivity of 푄푑? Answer. ◦ We can disconnect 푄푑 by removing the 푑 neighbors of any vertex of 푄푑. So the connectivity is at most 푑. ◦ We prove by induction that it is exactly 푑. 푄4 Proof by Induction We prove by induction that the connectivity of 푄푑 is 푑. ◦ Induction basis. Easy to check for 푑 = 1,2. ◦ Induction step. We can consider 푄푑 as two ′ copies 푄, 푄 of 푄푑−1 with a perfect matching between their vertices. ′ 푄 푄 푄푑 Completing the Proof 푆 – a minimum disconnecting set of 푄푑. To complete the proof, we claim that 푆 = 푑. ◦ If after removing 푆 both 푄 and 푄′ connected, it must disconnect 푄 from 푄′, and must be of size at least 2푑−1 ≥ 푑. ◦ By the induction hypothesis, disconnecting 푄 (or 푄′) requires removing 푑 − 1 vertices of 푄. To disconnect 푄 in 푄푑, we must remove at least one additional vertex from 푄′. 푘-edge Connected Graphs An graph 퐺 = (푉, 퐸) is said to be 푘-edge- connected if 푉 > 1 and we cannot obtain a non-connected graph by removing at most 푘 − 1 edges from 퐸. Is the graph in the figure ◦ 1-edge-connected? Yes. ◦ 2-edge-connected? Yes. ◦ 3-edge-connected? Yes. 푘-edge-connectivity Which graphs are 1-edge-connected? ◦ These are exactly the connected graphs. The edge-connectivity of a graph 퐺 is the maximum integer 푘 such that 퐺 is 푘- connected. What is the edge-connectivity of the complete graph 퐾푛? 푛 − 1. The graph in the figure has an edge-connectivity of 3. Edge Connectivity and Minimum Degree What is the relation between the minimum degree and the edge- connectivity of a graph? ◦ Can they be equal? Yes ◦ Can the edge connectivity be smaller? Yes ◦ Can the minimum degree be smaller? Edge Connectivity and Minimum Degree (cont.) Claim. In any graph, the edge connectivity is at most the minimum degree. Proof. ◦ Consider a vertex 푣 of minimum degree, and denote this degree as 푑. ◦ By removing the 푑 edges that are adjacent to 푣, we disconnect the graph. Connectivity and Edge-Connectivity What is the relation between the connectivity and the edge-connectivity of a graph? ◦ Can they be equal? Yes ◦ Can the connectivity be smaller? Yes ◦ Can the edge-connectivity be smaller? Connectivity and Edge-Connectivity Claim. In any graph 퐺 = 푉, 퐸 , the connectivity is at most the edge connectivity. Proof. ◦ Let 퐹 ⊂ 퐸 be a minimum set of edges whose removal disconnects 퐺. ◦ We need to find a set of at most 퐹 vertices that disconnects 퐺. ◦ We divide the analysis into two cases: A vertex 푣 ∈ 푉 not adjacent to any edge of 퐹. Every vertex of 푉 is adjacent to an edge of 퐹. Analysis of Case 1 Assume that a vertex 푣 ∈ 푉 is not adjacent to any edge of 퐹. ◦ By definition, the removal of 퐹 disconnects 퐺. ◦ Let 퐶 be the connected component in 퐺 − 퐹 that contains 푣. ◦ Removing the vertices 푉′ in 퐶 that are adjacent to an edge of 퐹 disconnects 푣 from the vertices of 퐺 − 퐶. ◦ Since 퐹 is minimal, each 퐹 edge of 퐹 has at most 푣 one endpoint in 퐶, so 푉′ ≤ 퐹 . Analysis of Case 2 Assume that every vertex of 푉 is adjacent to an edge of 퐹. ◦ If 퐺 is a complete graph, the connectivity is 푉 − 1 and edge-connectivity is 푉 − 1. Analysis of Case 2 (cont.) Assume that every vertex of 푉 is adjacent to an edge of 퐹. ◦ If 퐺 is not complete, there exists 푣 ∈ 푉 such that deg 푣 < 푉 − 1. Removing the set 푉′ of neighbors of 푣 disconnects 퐺. ◦ 퐶 – the connected component of 푣 in 퐺 − 퐹. ◦ Every vertex of 푉′ is either in 퐶 or connected to 푣 by an edge of 퐹. ◦ Since no edge of 퐹 is between two vertices of 퐶, we have 퐹 푉′ ≤ |퐹|. 푣 Connectivity in 3-regular Graphs Claim. If 퐺 = 푉, 퐸 is a connected 3- regular graph, then the connectivity and the edge-connectivity of 퐺 are equal. Proof 푘푣 – (vertex) connectivity. 푘푒 – edge connectivity. We already know that 푘푣 ≤ 푘푒. ◦ Thus, it suffices to find a set of 푘푣 edges that disconnects 퐺. ◦ 푆 – a minimum set of 푘푣 vertices whose removal disconnects 퐺 into 퐶1 and 퐶2 (which might not be connected themselves) Proof (cont.) Every vertex of 푣 ∈ 푆 is connected to both 퐶1 and 퐶2, since 푆 is a minimum disconnecting set. ◦ Due to 3-regularity, either 푣 is adjacent to a single vertex of 퐶1, or to a single vertex of 퐶2. ◦ That is, by removing a single edge, we can disconnect 푣 from either 퐶1 or from 퐶2. ◦ By removing such an edge from each 푣 ∈ 푆, we obtain a disconnecting set of 푘푣 edges. Recap We proved that in every graph: Edge Minimum Connectivity ≤ ≤ connectivity degree This implies that small minimum degree implies small connectivity. ◦ Does large minimum degree imply large connectivity. ◦ No! Highly Connected Subgraphs Recall. The average degree of a graph 퐺 = 푉, 퐸 is 1 2 퐸 deg 퐺 = deg 푣 = . |푉| 푉 푣∈푉 Theorem (Mader `72). Let 푘 be a positive integer and let 퐺 = 푉, 퐸 be a graph such that deg 퐺 ≥ 4푘. Then there exists a 푘 + 1 -connected (induced) subgraph 퐻 ⊂ 퐺 such that deg 퐻 > 푑/2. Proof Set 푑 = deg 퐺 ≥ 4푘. Let 퐻 = (푉′, 퐸′) be a subgraph of 퐺 such 푑 that 푉′ ≥ 2푘 and 퐸′ > 푉′ − 푘 . 2 ◦ Among the subgraphs that satisfy the above, we take as 퐻 one that minimizes 푉′ . Such subgraphs exist since 퐺 is one: ◦ Since 푑 ≥ 4푘, there exists a vertex of degree at least 4푘, and thus 푉 ≥ 4푘 + 1. 2 퐸 푉 푑 푑 퐸 = ⋅ = 푉 > 푉 − 푘 . 푉 2 2 2 Proof (2) Set 푑 = deg 퐺 ≥ 4푘. ′ ′ 퐻 = (푉 , 퐸 ) – subgraph of 퐺 such that ′ ′ 푑 ′ 푉 ≥ 2푘 and 퐸 > 푉 − 푘 . 2 If 푉′ = 2푘, we get the contradiction 푑 푑 2푘 퐸′ > 푉′ − 푘 = 푘 ≥ 2푘2 > 2 2 2 푉′ = ≥ 퐸′ . 2 Proof (3) 퐻 = (푉′, 퐸′) – the minimal subgraph of 퐺 such that ′ ′ 푑 ′ 푉 ≥ 2푘 and 퐸 > 푉 − 푘 . 2 ◦ We actually have 푉′ ≥ 2푘 + 1. ◦ Removing a vertex of degree ≤ 푑/2 decreases 푑 ′ 푑 ′ 푑 푉 − 푘 by and 퐸 by at most . 2 2 2 ◦ That is, both 푉′ ≥ 2푘 and ′ 푑 ′ 퐸 > 푉 − 푘 remains valid, 2 contradicting the minimality of 퐻. Proof (4) ′ ′ 퐻 = (푉 , 퐸 ) – the minimal subgraph of 퐺 푑 such that 푉′ ≥ 2푘 and 퐸′ > 푉′ − 푘 . 2 ◦ We proved that every vertex of 푉′ is of degree larger than 푑/2 in 퐻. ◦ We need to show that 퐻 is (푘 + 1)-connected. ◦ Assume, for contradiction, that the removal of a set 푆 of 푘 vertices disconnects 퐻.
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