To find the minimum number of edges in an undirected graph $G$ with 42 vertices, where each vertex has a degree of at least 4, we can proceed as follows:

Step 1: Understand the Relationship Between Degrees and Edges

In an undirected graph, the sum of the degrees of all vertices is equal to twice the number of edges. This is because each edge contributes to the degree of two vertices.

Let $E$ be the number of edges in the graph, and $n = 42$ be the number of vertices. Let the degree of vertex $i$ be $d_i$ for $i = 1, 2, \dots, 42$.

The sum of the degrees of all vertices is given by: $\sum_{i=1}^{42} d_i = 2E$

Step 2: Use the Minimum Degree Condition

We know that the degree of each vertex is at least 4. Therefore: $\sum_{i=1}^{42} d_i \geq 42 \times 4 = 168$

Step 3: Calculate the Minimum Number of Edges

Given that the sum of the degrees of all vertices is at least 168, we have: $2E \geq 168$ $E \geq \frac{168}{2} = 84$

Thus, the minimum number of edges $E$ that the graph can have is 84.

Conclusion

The minimum number of edges in the graph $G$ is $\boxed{84}$.

Would you like further details or have any questions? Here are some related questions to consider:

1. What would the maximum number of edges be in a graph with 42 vertices?
2. How does the minimum number of edges change if the minimum degree is increased to 5?
3. What if the graph were directed with similar constraints? How would the calculation change?
4. How does the concept of a complete graph relate to this problem?
5. What is the role of Euler's formula in analyzing graphs with additional properties?

Tip: Remember that in any undirected graph, the sum of all vertex degrees is always even because each edge contributes exactly two to the total sum of degrees.