4
$\begingroup$

I feel like the answer is no but I'm not sure. I think it's commonly accepted that DFS runs in $O(|V| + |E|)$ time. I've read a few explanations and they all make sense if the neighbour traversal for any given vertex can be done in arbitrary order.

But I've noticed a commonly suggested DFS behavior is to traverse the neighbors in alphabetical order (i.e. CLRS exercise 22.3-2), and I don't see how this can be done in $O(|V|+|E|)$ time. This became evident to me when actually trying to implement this in runnable code.

I see two ways to do it:

  1. I can keep the list of vertices and each vertex's adjacency list sorted as I'm constructing the graph. However this means $O(V)$ insertion time for each new vertex in the graph, which means a total of $O(|V|^2)$ insertion time over $|V|$ nodes. And $O(|E'|)$ insertion time for a new edge where $|E'|$ is the number of neighbours in a particular vertex's adjacency list, meaning $O(|E'|^2)$ time to construct the adjacency list for any given vertex.

OR

  1. Construct the graph and insert the vertices and adjacency elements in arbitrary order, but then sort them before running DFS. But comparison-based sorts are $\Omega(n\log n)$ so I'd have $O(|V|\log|V|)$ sort time for the list of vertices, and $O(|E'|\log|E'|)$ for each adjacency list.

In either case it doesn't seem like the runtime is $O(|V| + |E|)$ any more. Can someone confirm or refute? Thanks.

$\endgroup$
2
$\begingroup$

No, there is no linear time algorithm (for worst case).

Note for a list of $n$ positive integers, you can construct a tree with a root and $n$ children encoded by these integers respectively and traverse this tree in alphabetical order to sort these integers. So your problem is at least as hard as sorting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.