Sorted Edges 3155F3

1. **State the problem:** We need to apply the sorted edges algorithm (also known as the greedy algorithm for the Traveling Salesman Problem) to find a Hamiltonian circuit starting and ending at vertex $A$. 2. **List all edges with weights and sort them in ascending order:** - $BA = 1$ - $CA = 2$ - $CB = 3$ - $AD = 5$ - $AE = 7$ - $DE = 8$ - $CE = 10$ - $BE = 11$ - $BD = 12$ - $CD = 15$ 3. **Apply the sorted edges algorithm:** - Start with the smallest edge $BA = 1$, add edge $A-B$. - Next edge $CA = 2$, add edge $A-C$. - Next edge $CB = 3$, adding $B-C$ would create a cycle $A-B-C-A$ too early, so skip it. - Next edge $AD = 5$, add edge $A-D$. - Next edge $AE = 7$, adding $A-E$ would create a vertex $A$ with degree 3 (not allowed), so skip it. - Next edge $DE = 8$, add edge $D-E$. - Next edge $CE = 10$, adding $C-E$ would create a cycle $C-A-D-E-C$ too early, so skip it. - Next edges $BE = 11$, $BD = 12$, and $CD = 15$ are skipped because they would create cycles or vertices with degree > 2. 4. **Check the path formed:** Edges included: $A-B$, $A-C$, $A-D$, $D-E$. Vertices $B$ and $C$ have degree 1, $A$ has degree 3 (which is invalid for a Hamiltonian circuit), so we must adjust. 5. **Adjust to form a valid Hamiltonian circuit:** Since $A$ has degree 3, we must remove one edge connected to $A$ and add another edge to connect the graph properly. Try removing $A-C$ and adding $C-B$: - Remove $A-C$ (weight 2) - Add $C-B$ (weight 3) Edges now: $A-B$, $B-C$, $A-D$, $D-E$ Check degrees: - $A$: degree 2 ($A-B$, $A-D$) - $B$: degree 2 ($A-B$, $B-C$) - $C$: degree 1 ($B-C$) - $D$: degree 2 ($A-D$, $D-E$) - $E$: degree 1 ($D-E$) Still $C$ and $E$ have degree 1, so add edge $E-C$ (weight 10) to connect them: Edges: $A-B$, $B-C$, $C-E$, $E-D$, $D-A$ 6. **Form the circuit:** Starting at $A$: $A-B-C-E-D-A$ 7. **Final answer:** The Hamiltonian circuit using the sorted edges algorithm is $ABCDEA$ with the path $A-B-C-E-D-A$. This path visits every vertex exactly once and returns to the start vertex $A$.