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

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