Question

draw a graph with four odd vertices and at least one loop.
choose the correct graph below.
a.
b.
c.
d.
e. the described graph is impossible.

Explanation:

Step1: Recall Euler's handshaking lemma

In any graph, the sum of the degrees of all vertices is equal to twice the number of edges (so it is even). A loop adds 2 to the degree of its vertex, which keeps the parity (even/odd) of the vertex's degree the same as if the loop did not exist.

Step2: Analyze odd vertex count

We are asked for a graph with 4 odd vertices. The sum of 4 odd numbers is $odd + odd + odd + odd = even + even = even$, which satisfies the handshaking lemma. However, check the given options:

• Option A: Vertices A(1, odd), B(2, even), C(2, even), D(1, odd) → 2 odd vertices, does not match.
• Option B: Loop on A (degree 3, odd), B(2, even), C(2, even), loop on D (degree 3, odd) → 2 odd vertices, does not match.
• Option C: Vertices A(2, even), B(2, even), C(2, even), D(2, even) → 0 odd vertices, does not match.
• Option D: Loop on C (degree 3, odd), A(2, even), B(2, even), D(1, odd) → 2 odd vertices, does not match.

Step3: Verify feasibility of the graph

While a graph with 4 odd vertices is theoretically possible (e.g., a graph with two disjoint edges, but we need at least one loop: adding a loop to any vertex in that graph would change its degree parity, resulting in 3 odd vertices (sum is odd, invalid) or if we add a loop to a vertex in a graph with 4 odd vertices, we would get 2 odd vertices. To get 4 odd vertices with at least one loop, we need to add loops in pairs (since each loop preserves parity), but none of the options meet this, and the question asks to choose from the given graphs.

Answer:

E. The described graph is impossible.