Dept. Computer Science Networks and Graphs VU University ...

Dept. Computer Science
Networks and Graphs
VU University Amsterdam 03.07.2013
Part I
BE SURE THAT YOUR HANDWRITING IS READABLE
1a Let G denote a simple graph with n > 1 vertices and m edges. For each of the following mathematical
statements, (1) translate the statement into common English and (2) tell whether it is true or false.
1. ω(G) = 1 ⇒ ∀u ∈ V (G) : δ(u) > 0
2. ∀u ∈ V (G) : δ(u) > 0 ⇒ ω(G) = 1
3. ∑ v∈V (G) δ(v) = 2m
4. ∀H ⊆ G : |E(H)| ≤ m
5. |{(u,v)|∃(u,v) − path}| = 0 ⇒ ω(G) = n
10pt
1. ω(G) = 1 ⇒ ∀u ∈ V (G) : δ(v) > 0: If G is connected, then each vertex in G has at least degree 1.
True.
2. ∀u ∈ V (G) : δ(v) > 0 ⇒ ω(G) = 1: If each vertex in G has at least degree 1, then G is connected.
False.
3. ∑ v∈V (G) δ(v) = 2m: The sum of all vertex degress is equal to twice the number of edges. True.
4. ∀H ⊆ G : |E(H)| ≤ m: Each subgraph of G has the same or less number of edges than G. True
5. |{(u,v)|∃(u,v) − path}| = 0 ⇒ ω(G) = n: If the number of paths between any two vertices in G is
zero, then G consists of n components. True.
2 Prove that if two graphs have the same degree sequence, they do not also need to be isomorphic. 6pt
See Figure 2.4 for examples.
3a Consider the original Köningsberg bridge graph, shown below. Modify the graph such that it becomes
Eulerian by (1) deleting two edges, (2) adding two edges, and (3) adding an edge and deleting
an edge.
6pt
(1) delete 〈n,w〉 and 〈s,e〉; (2) add 〈n,e〉 and 〈s,w〉; (3) add 〈n,s〉 and delete 〈w,e〉.
3b Apply Fleury’s algorithm to find an Euler tour in the following graph.
6pt

4 Prove by induction that, for a given graph G, ∑ w∈V (G) δ(w) = 2 · |E(G)|. 10pt
Prove this by induction on the number of edges. Clearly, when |E(G)| = 0 the theorem is true,
as each vertex will have degree 0. Assume it holds for any graph with k > 1 edges and assume
|E(G)| = k +1. Remove an arbitrary edge e = 〈u,v〉 from G. For the resulting graph G ∗ we know the
theorem holds. Moreover, δ G ∗(u) = δ G (u) − 1, and likewise, δ G ∗(v) = δ G (v) − 1. As a consequence
∑ w∈V (G) δ(w) = 2 + ∑ w∈V (G ∗ ) δ(w) = 2 + |E(G ∗ )| = 2 · (1 + |E(G ∗ )|) = 2 · |E(G)|.
5a Can a directed tree with n > 1 vertices be strongly connected? Explain your answer.
No. Strongly connected implies that for any two vertices u and v there exists a directed (u,v)-path
as well as a directed (v,u)-path. In a directed tree, any two vertices are connected through a single
path in the underlying undirected graph. Every edge in that path will have a single direction, making
it impossible for any orientation to be strongly connected.
5b Prove that if a directed graph with n > 1 vertices is strongly connected, every vertex will have a
degree larger than 1.
Prove by contradiction. If δ(v) = 0 then G is not even connected. Let δ(v) = 1, and adjacent to u. We
either have (1) the arc 〈u,v〉 ⃗ or (2) the arc 〈v,u〉. ⃗ In (1) u cannot be reached from v; in (2) v cannot
be reached from u, so that G can never be strongly connected. Therefore, our assumption that there
may be a vertex with degree less than 2 is false.
5c Consider a strongly connected directed graph G and its underlying undirected graph G ∗ . Prove that
λ(G ∗ ) ≥ 2.
Prove by contradiction. If λ(G ∗ ) ≤ 1, then there exists an edge e such that its removal will make
G ∗ become disconnected with two components G ∗ 1 and G∗ 2 . It is impossible to assign a direction to e
such that each vertex in G ∗ 1 can reach a vertex in G∗ 2 in G, or vice versa. Hence, λ(G∗ ) ≥ 2.
4pt
4pt
4pt
Part II
6a Compute for the following graph all the shortest paths to vertex 1 using Dijkstra’s algorithm. Be sure
to make clear how you came to your answer.
6pt
9
6
4
14
6
2
1 9 11
2
3
7
5
10
15
Dijkstra proceeds in a number of steps. Let S t (u) denote the set of vertices for which shortest paths
to vertex u have been found after t ≥ 0 steps. Set S 0 (u) = {u}. Let label (s,d) v attached to vertex v
denote the distance d to u, and s being v’s next neighbor on the shortest path to u. Let L t (u) be the
set of labels of vertices in S t (u), and ¯L t (u) labels of other nodes. We then have:
1. S 1 (1) = {1}; L 1 (1) = {(1,0) 1 }; ¯L 1 (u) = {(1,9) 2 ,(−,∞) 3 ,(−,∞) 4 ,(1,7) 5 ,(1,14) 6
2. S 2 (1) = {1,5}; L 1 (1) = {(1,0) 1 ,(1,7) 5 }; ¯L 1 (u) = {(1,9) 2 ,(5,22) 3 ,(−,∞) 4 ,(1,14) 6
3. S 3 (1) = {1,2,5}; L 1 (1) = {(1,0) 1 ,(1,9) 2 ,(1,7) 5 }; ¯L 1 (u) = {(2,20) 3 ,(−,∞) 4 ,(2,11) 6
4. S 4 (1) = {1,2,5,6}; L 1 (1) = {(1,0) 1 ,(1,9) 2 ,(1,7) 5 ,(2,11) 6 }; ¯L 1 (u) = {(2,20) 3 ,(6,20) 4
5. S 5 (1) = {1,2,3,5,6}; L 1 (1) = {(1,0) 1 ,(1,9) 2 ,(2,20) 3 ,(1,7) 5 ,(2,11) 6 }; ¯L 1 (u) = {(6,20) 4
6. S 6 (1) = {1,2,3,4,5,6}; L 1 (1) = {(1,0) 1 ,(1,9) 2 ,(2,20) 3 ,(2,20) 4 ,(1,7) 5 ,(2,11) 6 }; ¯L 1 (u) =
{}
2

6b Provide a minimal spanning tree T for the graph from (a). Compared to the tree found in (a), T ’s
weight will be the same or less. Why is that so?
T consists of the edges 〈1,2〉,〈1,5〉,〈2,6〉,〈4,6〉,〈3,4〉. By construction, a shortest-path tree may
contain paths that are shortest in distance to a specific node, but that contain edges that are simply
not good for all nodes. In our example, the edge 〈2,3〉 is good for getting quickly from 3 to 1, but is
not useful for all nodes.
4pt
7a Consider a real-world network G with 1000 vertices and 7500 edges. To which random network
should G be compared?
The basic idea is that G should be compared to an ER(n, p) network with n = 1000 vertices and the
same number of edges. The average degree of an ER(n, p) network is p(n − 1), so that ∑δ(v) =
n · p · (n − 1) = 2m. With n = 1000 and m = 7500, we have that p = 15,000/(1000 · 999) ≈ 0.015.
7b Assume that the clustering coefficient of G from (a) is equal to 0.05. Is this high? Be sure to explain
your answer.
Yes, it’s high. We should compare G to an ER(1000,0.015) network, for which we know that the
clustering coefficient is equal to 0.015. That of G is more that three times this value.
6pt
4pt
8a Consider a Watts-Strogatz graph WS(n,k, p). Explain what the parameters n, k, and p stand for.
n is the total number of nodes; k is the initial degree of each node, and p is the probability that an
edge will be rewired after connecting a vertex to its k/2 left-hand and k/2 right-hand neighbors.
8b What will happen to the average path length in a WS(n,k, p) graph when p increases? And what
about the clustering coefficient?
The average path length will drop rapidly (virtually exponentially), while the clustering coefficient
will also drop, but much more gradually.
8c Prove that for any two vertices u and v in a WS(n,k, p) graph, the length of a shortest path between
u and v is less or equal than n/k.
The worst case is when u and v belong to a WS(n,k,0) graph, and are placed opposite each other on
the ring that is used to construct the graph. Vertex u is joined with its k/2 left-hand neighbors, and
those neighbors are joined with their left-hand neighbors, and so on. As a consequence, it will take
less than (n/2)/(k/2) = n/k “hops” to reach v from u.
3pt
4pt
8pt
9a Give the definition of betweenness centrality and an informal description of what it tries to measure.
If S(x,u,y) is the collection of shortest paths between x and y that passes through u (u ≠ x ≠ y),
and S(x,y) is the collection of all shortest paths between x and y, the betweenness centraility of u is
|S(x,u,y)|
defined as ∑ u≠x≠y |S(x,y)|
. It tries to capture the importance of vertex u by considering the effect on
information flow through the network when u is removed.
9b Compute the betweenness centrality for vertex 2 in the graph from (6a). Explain your answer.
The trick is that we simply need to count the number of shortest paths on which vertex 2 lies for any
pair of vertices, out of the total number of shortest paths for such a pair.
1 2 3 4 5 6
1 - - 1/1 1/1 0/1 1/1
2 - - - - - -
3 - - - 0/1 0/1 1/1
4 - - - - 1/2 0/1
5 - - - - - 1/1
6 - - - - - -
which leads a betweenness centrality of 6.5.
9c Consider a tree with n vertices {1,2,...,n} in which vertex 1 is joined to all other vertices. What is
the betweenness centrality of vertex 1?
Simply the number of triangles at 1, which is equal to ( n−1) 2 .
6pt
6pt
3pt
3

Final grade: (1) Add, per part, the total points. (2) Let T denote the total points for the midterm exam
(0 ≤ T ≤ 50); D1 the total points for part I; D2 the total points for part II. The final number of points E is
equal to max{T,D1} + D2.
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