Assignment 1 - Answer Key - Indian Statistical Institute

Assignment 1 - Answer Key
Debasis Mishra ∗
February 19, 2009
Due: February 19, 2009
1. Prove that every cycle of a bipartite graph has even number of edges.
Answer: Consider a cycle of the bipartite graph of the following form: (i 1 , i 2 , . . . , i k , i 1 ).
If the partition of vertices of the bipartite graph is B and L, then we know that alternating
vertices belong to B and L. Suppose i 1 ∈ B, then i k ∈ L and every i j ∈ L if j
is even. So k is even. Hence, number of edges is even.
2. Show whether a matching exists or not in the bipartite graphs in Figure 1. In each
case, give explanation for your answer.
1
a
1
a
2
b
2
b
3
c
3
c
4
d
4
d
Figure 1: Two Bipartite Graphs
Answer: A matching in the left graph is: (1, a), (2, c), (3, b), (4, d). In the right graph,
2, 3, and 4 demand {a, b}. Hence, 2, 3, and 4 cannot be matched. So, no matching
exists.
∗ Planning Unit, Indian Statistical Institute, 7 Shahid Jit Singh Marg, New Delhi 110016, India, E-mail:
dmishra@isid.ac.in
1

3. Let (N, T ) be a tree of an undirected graph G = (N, E). Show the following:
∑ [ ]
2 − d(i) = 2,
i∈N
where d(i) is the degree of vertex i in tree (N, T ).
Answer: We know that ∑ i∈N
d(i) = 2#T . But #T in a tree equals n−1. Substituting,
∑
i∈N d(i) = 2(n − 1). Rearranging, ∑ [ ]
i∈N 2 − d(i) = 2.
4. Show that every tree which has a vertex of degree m ≥ 2 has at least m vertices of
degree 1.
Answer: Let i be the vertex of degree m. Let {i 1 , i 2 , . . . , i m } be the set of vertices
with which i shares an edge. Consider m paths of the form: P 1 = (i, i 1 , . . . , j k 1
),
P 2 = (i, i 2 , . . . , j k 2
), . . . , P m = (i, i m , . . . , j km ) such that j k 1
, . . . , j km have degree 1.
Clearly, m such paths exist. Moreover, any pair of such paths have only one vertex
in common, which is i (else, we get a cycle in the tree). Hence, we get m vertices of
degree 1.
5. Suppose G is an undirected weighted connected graph. An edge is called a default
edge if it has the minimum weight over all edges in G. Show that a default edge
belongs to some minimum cost spanning tree of G.
Answer: Let {i, j} be a default edge. Consider the cut ({i}, N \ {i}). The edge {i, j}
is a light edge of this cut. Hence, by our result, it must be a safe edge, and hence be
part of some minimum cost spanning tree.
6. Find a minimum cost spanning tree of the graph in Figure 2.
a
2 b 4 c 1
d
5 4
4
6 2
3
3
e
5
f
3
g
2
4
3
5
7
4
2
h
3
i
6 3
j
k
Figure 2: A connected graph with weights
Answer: Apply the greedy algorithm discussed in the class.
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7. Suppose G is a directed weighted complete graph. Let the weights of edges of G
satisfy the following triangle inequality: for any three edges (i, j), (j, k), and (i, k)
we have w(i, j) + w(j, k) ≥ w(i, k). Show that a potential exists in G if and only if
w(i, j) + w(j, i) ≥ 0.
Answer: In the presence of triangle inequality, shortest path from i to j is the
edge (i, j). Hence, s(i, j) = w(i, j) for all i, j ∈ N. Now, consider a cycle C =
(i 1 , i 2 , . . . , i k , i 1 ). Suppose for all i, j ∈ N, we have w(i, j) + w(j, i) ≥ 0. But
l(C) ≥ s(i 1 , i k ) + w(i k , i 1 ) = w(i 1 , i k ) + w(i k , i 1 ) ≥ 0. Hence, a potential exists. If
a potential exists, then w(i, j) + w(j, i) is the length of cycle (i, j, i), and hence it
should be non-negative.
8. Find a feasible solution or determine that there are no feasible solutions for the following
system of difference inequalities.
x 1 − x 2 ≤ 4
x 1 − x 5 ≤ 5
x 2 − x 4 ≤ −6
x 3 − x 2 ≤ 1
x 4 − x 1 ≤ 3
x 4 − x 3 ≤ 5
x 4 − x 5 ≤ 10
x 5 − x 3 ≤ −4
x 5 − x 4 ≤ −8.
Answer: No solution exists. Check that a negative cycle exists of the form: (1, 4, 2, 3, 5, 1).
9. For the directed graphs in Figure 3, verify if a potential exists. If a potential exists,
then identify at least one potential.
Answer: In both graphs, potential exists. Fix vertex a in the first graph, and assign
it potential zero. Find shortest path from a to every other vertex to get the other
potentials. For the second graph, one can fix any vertex, and apply the same shortest
path method.
3

3
b
2
c
−3
−3
−2
4
a
−2
1
d
−2
2
1
−1
1
−1
2
2
e
1
Figure 3: Two directed graphs
−2
4