A triangle in an undirected graph is a $\mathbf{3}-\mathbf{clique}$. Show that$\mathbf{TRIANGLE}\in \mathbf{P}$ , where $\mathbf{TRIANGLE}=\left\{<\mathbf{G}>|\mathbf{G}\mathbf{contains}\mathbf{a}\mathbf{triangle}\right\}.$

$ClassP:P$s a class of languages that are decidable in polynomial time on a deterministic single – tape Turing – machine.

Specified language, a triangle is an undirected graph is a$3-clique$.Now we have to show that $TRIANGLE\in P$

Let $AbetheTuringmachinethatdecidesTRIANGLE$ is polynomial time$A$can be described as follows:

:$A=\text{'}\text{'}oninput$$G<V,E>$

$V$ denotes set of vertices of the graph $G$.

$E$denotes set of edges of the graph$G$.

For $u,v,w\in V$and $u<v<w$we enumerate all triples$<u,v,w>$.Check whether all three edges $\left(u,v\right)(v,w)$exist in $E$or not. If exist then accept.Otherwise reject.”

Enumeration of all triple require$0\left(\right|v{|}^3)$time. Checking whether all three edges belong to $E$ take $0\left(\right|E\left|\right)$time.Overall time is $0\left(\right|V{|}^3\left|E\right|)$which is polynomial in the length of the inlets. Therefore $TRIANGLE\in P$. Three edges connection over base on the theory of polynomial show$G$the length of inplet..