I am trying to prove that bidirectional BFS could have a worse run time than regular BFS. I am confused on how this would be possible, considered bidirectional BFS always divides the run time in two, since it is executing two searches. Would a specific order of nodes and vertices have to take place in order to give bidirectional a worse running time? Or is it a quantity issue?
Edit: Here is a graph I made. I'm assuming bi-directional BFS would infinitely search in this instance because if search one went right, and search two went left, they wouldn't find a path. Then, search one would go left, and search two would go right. Doesn't this mean that the search would run infinitely, because it doesn't stop until both have found the same path?