Say someone has found a way to create a graph given a CNF boolean expression in O(n^3) time, and that any spanning tree of this special graph will be a solution to the CNF equation.
The scenario seems to be hinting that the someone has found a solution to the SAT problem and solved P=NP by reducing the SAT problem to the Spanning Tree problem with a gadget (reduction) that runs in only O(n^3) time.
But what if the graph that their algorithm creates has n! or 2^n nodes and edges?
In that scenario, while a spanning tree algorithm such as DFS or BFS may run in linear time on the number of nodes/edges, it would NOT be running in poly time on the number of inputs to the boolean expression. And so the person would not have found an efficient algorithm to the SAT problem since running the full solution would take n! time to evaluate.
Is this reasoning correct?
