Sorted Edges Circuit 566E43

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 circuit that visits all vertices exactly once and returns to the start, minimizing total cost. 2. **List the edges with weights:** Edges and weights are: $AB=8$, $BC=15$, $CD=2$, $DE=11$, $EA=1$, $BE=9$, $DA=3$, $CE=4$, $BD=6$, $CA=13$. 3. **Sort edges by increasing weight:** $$EA=1, CD=2, DA=3, CE=4, BD=6, AB=8, BE=9, DE=11, CA=13, BC=15$$ 4. **Apply the sorted edges algorithm:** - Start adding edges from the smallest weight, avoiding creating a smaller circuit before all vertices are included and avoiding vertices with degree more than 2. - Add $EA=1$ (connects E and A). - Add $CD=2$ (connects C and D). - Add $DA=3$ (connects D and A). Now A has degree 2, D has degree 2. - Add $CE=4$ (connects C and E). Adding this would create a cycle C-D-A-E-C with 4 vertices, but we have 5 vertices total, so allowed. - Add $BD=6$ (connects B and D). D already has degree 2, so cannot add this edge. - Add $AB=8$ (connects A and B). A has degree 2, cannot add. - Add $BE=9$ (connects B and E). E has degree 2, cannot add. - Add $DE=11$ (connects D and E). Both have degree 2, cannot add. - Add $CA=13$ (connects C and A). Both have degree 2, cannot add. - Add $BC=15$ (connects B and C). C has degree 2, cannot add. We see that vertex B is not yet connected in the circuit. We need to reconsider edges involving B. 5. **Reconsider edges involving B:** - Add $BD=6$ instead of $DA=3$ or $CE=4$ to include B. Let's try again: - Add $EA=1$ - Add $CD=2$ - Add $BD=6$ - Add $CE=4$ - Add $AB=8$ Check degrees: - A connected to E and B (degree 2) - B connected to D and A (degree 2) - C connected to D and E (degree 2) - D connected to C and B (degree 2) - E connected to A and C (degree 2) This forms a circuit visiting all vertices. 6. **Calculate total cost:** $$1 + 2 + 6 + 4 + 8 = 21$$ **Final answer:** The total cost of the resulting circuit is $21$.