I am given the following graph G:
and asked to determine if its planar or not, and if it is not, give a $K_{3, 3}$ or $K_{5}$ that is a subgraph of G
I have determined G is not planar. I have done this by using the circle chord method.
According to the theorem of Kuratowski:
A graph is planar if and only if it does not contain a subgraph that is a K5 or K3,3 configuration.
I have spent hours trying to find the subgraph $k_{3, 3}$ or $k_{5}$ and cant find neither.
Do these subgraphs really not exist? And if thats the case, then G must be planar, but my circle chord method has repeatedly shown that it is not planar. I have therefore seemingly hit a paradox. Can anyone help me resolve this?
As it mentioned in comment, it doesn't need to have a $K_5$ or $K_{3,3}$ as subgraph. Make $d \to e$, you'll have graph $G' \sim G$ with $K_5$ as subgraph.