Question

Define $\mathbf{UCYCLE}\mathbf{=}\mathbf{\left\{}\mathbf{⟨}\mathbf{G}\mathbf{⟩}\mathbf{\right|}\mathbf{G}$is an undirected graph that contains a simple cycle}.Show that $\mathbf{UCYCLE}\mathbf{\in }\mathbf{L}$. (Note: $\mathbf{G}$ may be a graph that is not connected.)

Short answer

We can simply count the number of vertices and edges in the graph$\mathrm{G}$ in open space using the DFS technique. We may also recollect the beginning vertex and the edge we're travelling by using free space. This demonstrates that$\mathrm{UCYCLE}\in \mathrm{L}$ is correct and proved.

Step-by-step solution

Given condition

$\mathrm{UCYCLE}=\left\{⟨\mathrm{G}⟩\right|\mathrm{G}\mathrm{is}\mathrm{an}\mathrm{unidirected}\mathrm{graph}\mathrm{that}\mathrm{accepts}\mathrm{a}\mathrm{simple}\mathrm{cycle}\}$

The condition is to proved $\mathrm{UCYCLE}\in \mathrm{L}$.

Explanation

It's possible that the graph $\mathrm{G}$utilised here isn't linked. In order to look through the tree, the edges of the graph incident on the vertex can be traversed lexicographically. When the graph's degree is a $\mathrm{x}$, entering the graph through the edge will result in exiting the graph through the ${\mathrm{x}}^{\mathrm{th}}$main edge. As a result, we enter a degree graph through the ${\mathrm{x}}^{\mathrm{th}}$edge and exit through the ${(\mathrm{x}+1)}^{\mathrm{th}}$edge.

On the tree, we're using a Depth First Search (DFS) method. We begin at an edge and work our way back to the same vertex from where we began. We must return to the beginning vertex in some way. However, if the graph contains a cycle, we will pick a vertex m and an initial edge$(\mathrm{m},\mathrm{n})$ from which to begin traversing the tree, with the objective of eventually reaching the starting vertex$\mathrm{m}$ by crossing an edge that is distinct from the edge $(\mathrm{m},\mathrm{n})$.

Final answer

We will identify all of the edges that arise from each of the vertices in the graph in order to complete the aforementioned process without being trapped in a cycle. Then we'll start crossing the tree from vertex to vertex, cutting across all of its edges. When we return to the original vertex after traversing the tree and cross an edge that is distinct from the edge from which we began the traverse, it is clear that the graph is a cycle. The number of vertices and edges of the graph $\mathrm{G}$in free space may be easily counted using this method. We may also remember the beginning vertex and the edge we are crossing by employing free space. This demonstrates that$\mathrm{UCYCLE}\in \mathrm{L}$ is a reliable source of information.