Unweighted Shortest Paths

Unweighted Shortest Paths

In some shortest path problems, all edges have the same length. For example, we may be trying to find the shortest path out of a maze. Each cell in the maze is a node, and an edge connects two nodes if we can move between them in a single step. In this problem, we simply want to minimize the number of edges in a path to an exit. We therefore say that the edges are unweighted — they contain no explicit length information, and the length of each edge is considered to be $1$.

We could of course apply Dijkstra’s algorithm to this problem, using $1$ as the length of each edge. However, if analyze what this algorithm does in this case, we find that we can optimize it to achieve significantly better performance.

The optimization revolves around the use of the min-priority queue. Note that Dijkstra’s algorithm first adds all outgoing edges from the start node u to the min-priority queue, using their lengths as their priorities. For unweighted edges, each of these priorities will be $1$. As the algorithm progresses it retrieves the minimum priority and removes an edge having this priority. If it adds any new edges before removing the next edge, they will all have a priority $1$ greater than the priority of the edge just removed.

We claim that this behavior causes the priorities in the min-priority queue to differ by no more than $1$. To see this, we will show that we can never reach a point where we change the maximum difference in priorities from at most $1$ to more than $1$. First observe that when the outgoing edges from u are added, the priorities all differ by $0\le 1$. Removing an edge can’t increase the difference in the priorities stored. Suppose the edge we remove has priority $p$. Assuming we have not yet achieved a priority difference greater than $1$, any priorities remaining in the min-priority queue must be either $p$ or $p+1$. Any edges we add before removing the next edge have priority $p+1$. Hence, the priority difference remains no more than $1$. Because we have covered all changes to the priority queue, we can never cause the priority difference to exceed $1$.

Based on the above claim, we can now claim that whenever an edge is added, its priority is the largest of any in the min-priority queue. This is certainly true when we add the outgoing edges from u, as all these edges have the same priority. Furthermore, whenever we remove an edge with priority $p$, any edges we subsequently add have priority $p+1$, which must be the maximum priority in the min-priority queue.

As a result of this behavior, we can replace the min-priority queue with an ordinary FIFO queue, ignoring any priorities. For a graph with unweighted edges, the behavior of the algorithm will be the same. Because accessing a FIFO queue is more efficient than accessing a min-priority queue, the resulting algorithm, known as breadth-first search, is also more efficient.