Given an arbitrary yet low complexity directed graph ($\lt 50 $ nodes), is the computational complexity of a (breadth-first search)-based algorithm the same as that of a (depth-first search)-based algorithm? Intuitively I think yes.
Am I right? Not looking for a detailed proof. I'm just needing to retrieve a list of all paths in a graph of "locations". The graph is not necessarily a tree.
Btw, I am coding the solution in python, so if you have any opinion on whether BFS > DFS or vise versa here, please weigh in on that too.
Oh yeah, specifically, given two nodes, I want to grab a list of all paths between them, regardless of apparent shortness. I am aware that there could be infinitely, many, so obviously I mean all cycleless paths, whatever the term is...