Why DFS and not BFS for finding cycle in graphs

Question

Predominantly DFS is used to find a cycle in graphs and not BFS. Any reasons? Both can find if a node has already been visited while traversing the tree/graph.

Answer 1

If you place a cycle at a random spot in a tree, DFS will tend to hit the cycle when it's covered about half the tree, and half the time it will have already traversed where the cycle goes, and half the time it will not (and will find it on average in half the rest of the tree), so it will evaluate on average about 0.5*0.5 + 0.5*0.75 = 0.625 of the tree.

If you place a cycle at a random spot in a tree, BFS will tend to hit the cycle only when it's evaluated the layer of the tree at that depth. Thus, you usually end up having to evaluate the leaves of a balance binary tree, which generally results in evaluating more of the tree. In particular, 3/4 of the time at least one of the two links appear in the leaves of the tree, and on those cases you have to evaluate on average 3/4 of the tree (if there is one link) or 7/8 of the tree (if there are two), so you're already up to an expectation of searching 1/2*3/4 + 1/4*7/8 = (7+12)/32 = 21/32 = 0.656... of the tree without even adding the cost of searching a tree with a cycle added away from the leaf nodes.

In addition, DFS is easier to implement than BFS. So it's the one to use unless you know something about your cycles (e.g. cycles are likely to be near the root from which you search, at which point BFS gives you an advantage).