Phase 3 IP Discrete Mathematics - Vertexes & Hierarchical Trees
Download as PPTX, PDF1 like100 views
The document describes the construction of a graph based on a given adjacency matrix, detailing the placement of nodes and edges. It outlines various simple paths between nodes and identifies the shortest path from 'a' to 'e' as 'a → b → e' with a weight of 3. Additionally, it explains a hierarchical tree structure related to college administration and concludes that introducing a direct line of communication between the college president and the head of alumni relations would prevent the structure from being a proper tree.
Phase 3 IP Discrete Mathematics - Vertexes & Hierarchical Trees
- 1. Adjacency Matrix and Shortest Path Construct a graph based on the adjacency matrix that appears below. Label all nodes with indices consistent with the placement of numbers within the matrix. a b c d e a 0 6 0 5 0 b 6 0 1 0 3 c 0 1 0 4 8 d 5 0 4 0 0 e 0 3 8 0 0 Describe the graph and why it is consistent with the matrix. As we see here, there will be 5 vertices (nodes) and 8 edges (directional lines) defining the paths from “a” to “e”. Since the matrices stipulates where the 5 nodes will be, we are able to place 5 markers with the letter A → E into the doughnuts. Now it is just a matter of assembling lines to the doughnuts using the matrix provided. Going from left to right we can assess the line weight and where each line meets up to each node. A D E B C 6 5 1 34 8 0 0
- 2. How many simple paths are there from vertex 1 to vertex 5? Explain. Which is the shortest of those paths? a b c d e a 0 6 0 5 0 b 6 0 1 0 3 c 0 1 0 4 8 d 5 0 4 0 0 e 0 3 8 0 0 Simple Paths A → B → E = 3 A → D → E = 5 A → B → D → E = 6 A → B → C → E = 15 A →D → C → E = 17 A → D → B → C → E = 14 A → B → D → C → E = 18 A → B → C → D → E = 11 As we see in the simple paths, A → B → E = 3 is the shortest route from (A → E).
- 3. TreesINSTRUCTIONS Construct and describe a tree that indicates the following: A college president has 2 employees who answer directly to him or her, namely a vice president and provost. The vice president and provost each have an administrative assistant. Three deans answer to the provost, The heads of finance and alumni relations answer to the vice president. Each dean oversees three department chairpersons, Each department chair oversees several faculty in each of their respective departments. Suppose that the professional correspondences are the same as above, with the addition that there is also a direct working relationship between the college president and the head of alumni relations (it is not necessary to draw this). Would the graph still be a tree? Why or why not? This will fall under the hierarchical tree graph. I used these many a year when making re-call rosters for our personnel. This is also known as an inverted tree. The next slide will assemble this ensemble or orchestrated positions.
- 4. Hierarchy Tree College President Provost Administrative Assistant Dean 1 Chair Person 1 Department Faculty Chair Person 2 Department Faculty Chair Person 3 Department Faculty Dean 2 Chair Person 4 Department Faculty Chair Person 5 Department Faculty Chair Person 6 Department Faculty Dean 3 Chair Person 7 Department Faculty Chair Person 8 Department Faculty Chair Person 9 Department Faculty Vice President Administrative Assistant Head of Finance Head of Alumni Relations As we see, The tree is assembled and it is definitely an open tree of dissemination of information through proper channels. If the College President and the Head of Alumni circumvent the Vice President and Administrative Assistant on the information highway, they are basically by passing two of the vertices within the dissemination process. It’s like cutting out a branch and cloning the limb to the main branch. Therefore, “NO” this would not continue to be a tree in this portion of the graph.