INPUT: undirected graph, s, tOUTPUT: connectivity of s and tI perform BFS on s AND t, each taking turns to make one traversal.When a vertex exists in both s and t's BFS tree, we can assume it is connected.When one tree is done traversing but the other is not, s and t are not connected.Does such an algorithm exist or am I making stuff up?
Starting BFS at s and t
algorithms;graphs;search algorithms;search trees;graph traversal
Yes, this is a known thing. It's called bidirectional search.  It will find shortest paths in unweighted graphs and, in graphs of where every vertex has degree $d$, it will visit at most $2d^{\ell/2}=2\sqrt{d^\ell}$ vertices (where $\ell$ is the length of the shortest path), whereas ordinary BFS could visit up to $d^\ell$ vertices.