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draw the edges needed in order to make the following graph complete. (i…

Question

draw the edges needed in order to make the following graph complete.

(image of a graph with vertices c, b, d, a, e)

clear all draw: line segment

find any hamiltonian circuit on your complete graph. give your answer as a list of vertices, starting and ending at the same vertex. example: abca

(text box for answer)

question help: video 1 video 2

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Explanation:

Step1: Understand Complete Graph

A complete graph with \( n \) vertices has an edge between every pair of distinct vertices. Here, vertices are \( A, B, C, D, E \). First, identify missing edges. From the graph, check connections: \( A \) to \( E \) is missing, \( B \) to \( D \)? Wait, no—wait, original graph: let's list all pairs. Vertices: \( A, B, C, D, E \). All pairs: \( AB, AC, AD, AE, BC, BD, BE, CD, CE, DE \). Check existing: \( AB \) (yes), \( AC \) (yes? Wait the graph: \( C \) to \( B \), \( C \) to \( D \), \( C \) to \( E \)? Wait maybe better to first complete the graph by adding edges like \( AE \), \( BE \)? Wait no, let's see the given graph: \( A \) is connected to \( B, D \)? Wait the initial graph: \( C - B \), \( C - D \), \( C - E \), \( D - B \)? Wait maybe I missee. But for Hamiltonian circuit, in a complete graph, any permutation of vertices that starts and ends at the same vertex, visiting each once. So a possible circuit: \( A - B - C - D - E - A \), but wait, need to check edges. Wait in complete graph, all edges exist, so any cycle that includes all vertices once and returns. So a simple one: \( ABCDEA \) (visits \( A, B, C, D, E \) and back to \( A \)). Or \( ADEBCA \), etc. Let's pick a simple one: start at \( A \), go to \( B \), \( C \), \( D \), \( E \), back to \( A \). So \( ABCDEA \).

Step2: Verify Hamiltonian Circuit

A Hamiltonian circuit visits each vertex exactly once (except start/end) and returns to start. So \( A \to B \to C \to D \to E \to A \) (or any order, as long as each vertex is visited once and edges exist in complete graph). So the circuit can be \( ABCDEA \) (or other permutations, like \( AEDCBA \), etc.).

Answer:

\( ABCDEA \) (Other valid answers like \( ADEBCA \), \( ABDCEA \), etc. are also correct as long as each vertex is visited once and returns to \( A \).)