Stuck need help!The class I'm taking is computer science discrete structures. Problem is attached. please view attachment before answering. Really struggling with this concept. Thank you so much. **Question:**For which \( n \) does the graph \( K_n \) contain an Euler circuit? Explain.---**Explanation:**An Euler circuit in a graph is a circuit that visits every edge exactly once and returns to the starting vertex. For a connected graph to have an Euler circuit, every vertex must have an even degree.In a complete graph \( K_n \), every vertex is connected to every other vertex, meaning each vertex has a degree of \( n-1 \).For \( K_n \) to have an Euler circuit, \( n-1 \) must be even for all vertices, meaning \( n \) must be odd. Therefore, a complete graph \( K_n \) contains an Euler circuit if \( n \) is odd.For example, \( K_3 \), \( K_5 \), \( K_7 \), etc., all contain Euler circuits.

An undirected graph has Eulerian cycle if following two conditions are true. a) All vertices with non-zero degree are connected. We don’t care about vertices with zero degree because they don’t belong to Eulerian Cycle or Path (we only consider all edges). ….b) All vertices have even degree.An undirected multigraph has an Euler cycle if and only if it is connected and has all vertices of even valence. Thus, we want to find Kn where each vertex will be of even valence. Given n vertices, Kn will have n(n−1)/2 edges. If the total number of valences in Kn divided by n vertices, the result will be even, thus the Kn having an Euler cycle. We also know the total number of valences of any graph is twice the number of edges. Thus, ∑:valences Kn = 2[n(n − 1)/2]⇒ n(n − 1) is the total number of valences ∀ Kn graph.Now we take the total number of valences, n(n − 1) and divide it by n vertices ∀ Kn graph and the result is n − 1. n − 1 is the valence each vertex will have in any Kn graph.Thus, for a Kn graph to have an Euler cycle, we want n − 1 to be an even value. But we already know in terms of complete graphs, if the number of n vertices is odd, we will have an even n − 1 value and if n is even, we will have an odd n − 1 value. In other words, For n =odd Kn has an Euler cycleeven Kn does not have an Euler Cycle