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The problem on the whiteboard appears to involve solving a minimum spanning tree (MST) using **Prim's algorithm** to determine the lowest cost for connecting a group of houses (A, B, C, D) by cabling. Let's break down the steps and find the solution.

### Steps in the image:

1. **Houses (nodes)**: A, B, C, D.
2. **Edges and weights (costs of cabling)**:
   - A–B: 1
   - A–C: 4
   - A–D: 3
   - B–C: 2
   - B–D: 5
   - C–D: 1

### Prim's Algorithm

Prim's algorithm builds the MST starting from an arbitrary node and adding the cheapest edge that connects a new node, expanding the tree until all nodes are connected.

### Step-by-step Solution

1. **Start with node A** (or any arbitrary node):
   - Consider edges from A: (A–B = 1), (A–C = 4), (A–D = 3).
   - The minimum edge is A–B = 1.
   - Add A–B to the MST.

2. **From B**, add the edge with the minimum weight connecting a new node:
   - Edges from B: (B–C = 2), (B–D = 5).
   - The minimum edge is B–C = 2.
   - Add B–C to the MST.

3. **From C**, add the minimum weight edge connecting a new node:
   - Edges from C: (C–D = 1), (A–C = 4 already ignored).
   - The minimum edge is C–D = 1.
   - Add C–D to the MST.

4. Now all nodes (A, B, C, D) are connected.

### Total Cost:
- Edges in the MST: A–B = 1, B–C = 2, C–D = 1.
- Total cost = 1 + 2 + 1 = 4.

This matches the total cost indicated on the whiteboard.

### Final Minimum Spanning Tree:
- **Edges**: A–B, B–C, C–D.
- **Total cabling cost**: 4.

Let me know if you'd like further clarification or have any questions! 

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Here are some additional related questions:

1. What is the difference between Prim's algorithm and Kruskal's algorithm?
2. How does Prim's algorithm ensure that the MST is found efficiently?
3. Can Prim's algorithm be applied to disconnected graphs?
4. How do we handle ties between edges of the same weight in Prim's algorithm?
5. How can you modify Prim's algorithm to work on directed graphs?

**Tip**: Always verify that all nodes are included in the spanning tree when solving MST problems.

Un repartidor de paquetes tiene que entregar varios cosas en un vecindario con el menor costo posible de cableado. Casas: A, B, C, D. Costo de cableado de las casas: A-B: 1, A-C: 4, A-D: 3, B-C: 2, B-D: 5, C-D: 1.

Solve a problem involving a packet delivery person using Prim's algorithm to find the minimum cost for connecting four houses (A, B, C, D) by cabling. The solution involves graph theory and minimum spanning trees. Total cable cost is 4.

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