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Data Structures and Algorithms
Key to Homework Assignment 7
- (a) Draw the adjacency matrix for the graph of Figure 7.
1 2 3 4 5 6 1 0 10 0 20 0 2 2 10 0 3 5 0 0 3 0 3 0 0 15 0 4 20 5 0 0 11 10 5 0 0 15 11 0 3 6 2 0 0 10 3 0 (b) Draw the adjacency list representation for the same graph.
1 2 3 4 5 6 (^2 10 4 20 )
(^1 10 3 3 )
2 3 5 15
(^1 20 2 5 5 11 6 )
(^3 15 4 11 6 )
1 2 4 10 5 3
(c) If a pointer requires 4 bytes, a vertex label requires 2 bytes, and an edge weight requires 2 bytes, which representation requires more space for this graph? Both graphs need to store a vertex label for each of the 6 vertices:
6 × 2 bytes = 12 bytes.
In addition the adjacency matrix will store 36 edge labels for a total of
36 × 2 bytes + 12 bytes = 84 bytes.
Data Structures and Algorithms, Key to Homework 7 2
There are 9 edges, and the adjacency list will store each edge weight twice. So this will increase the total for the list to
18 × 2 bytes + 12 bytes = 48 bytes.
In addition, for each of the 18 edge nodes, the list will store a pointer:
18 × 4 bytes + 48 bytes = 120 bytes.
Finally, the list will require 6 pointers for the heads of the lists, giving a grand total of
6 × 4 bytes + 120 bytes = 144 bytes.
So the adjacency list requires more space for this graph.
- Find the DFS tree for the graph of Figure 7.25, starting at Vertex 1. This is just a linear list. If the vertices during a call to DFS are visited in increasing order, then the list will be 1 → 2 → 3 → 5 → 4 → 6.
- Find the BFS tree for the graph of figure 7.25, starting at Vertex 1. If the vertices are visited in increasing order, then the tree will be
1
2 4 6
3 5
- Use mathematical induction on the number of vertices n to prove that a free tree has n − 1 edges. Recollect that a free tree is a connected, undirected graph with no simple cycles. Base Case. If a free tree has no vertices (n = 0), the result is false. So suppose T is a free tree with a single vertex. Since edges in undirected graphs must join distinct vertices, T can’t have any edges. So in this case we have |E| = n − 1. Induction Hypothesis. Suppose that n = n 0 ≥ 1 is a nonnegative integer such that every free tree with n nodes has n − 1 edges. Induction Step. Suppose now that T is a free tree with n = n 0 + 1 ≥ 2 vertices. Using a familiar argument, we can show that T must have at least one vertex v 0 with at most one incident edge. (v 0 is the analog of a leaf in an ordinary tree.) Furthermore, since T is connected and T contains at least 2 vertices, v 0 must have exactly one incident edge e 0. Consider the subgraph S obtained from T by removing v 0 and e 0.
