Shortest Path

Before we see how to find the shortest path from vertex $a$ to vertex $b$ we need to define a few things.

$D(a,b)=$the length of the shortest path from vertex $a$ to vertex $b$. If no such path exists, the length is $\infty$.

If the graph is unweighted the length of a path is the number of edges it takes to get from $a$ to $b$. If however the graph is weighted the length is the sum of the edge weights.

Single source shortest path (SSSP) are all the shortest paths from vertex $s$ to all other vertices.

For Unweighted Graphs

In an unweighted graph we can just use a BFS as this starts with all vertices with distance 1 then 2 etc.

void BFS(Vertex s) {
 Queue<Vertex<K>> R = new LinkedList<Vertex<K>>();
 s.dist = 0;
 R.add(s);
 while(!R.isempty()) {
  Vertex v = R.remove();
  for(Vertex w : v.adjList) {
   if (w.dist == Integer.MAX_VALUE) {
    w.dist = v.dist + 1;
    R.add(w);
   }
  }
 }
}

For weighted graphs

For a weighted graph this is slightly trickier as the shortest path isn't necessarily the path with the least edges.

Dijkstra's algorithm

1. Assign all the vertices the attributes "finished", "Distance and „Via/Predecessor“. Initialize the distance of the root vertex as 0 and all others as $\infty$.
2. While there are unvisited nodes. So finished=false and distance $\leq \infty$.
   1. Choose the vertex $v$ with the smallest distance.
   2. Set $v.finished = true$
   3. For all vertices $w$ that have and edge between $v$ and $w$
      1. Set int d = v.dist + edge weight between $v$ and $w$.
      2. if(d < w.dist) w.dist = d; w.via = v;

The time complexity of this algorithm is as followed. For 1 While loop step 1 takes $O(n)$, step 2 takes $O(1)$, step 3 takes $O(outdeg(v))$

For n While loops this the becomes $O(n^2 + m)$

Improvements

We could save the vertices that are not finished in a Set so we don't have to look through the entire table. This however doesn't have an effect on the time complexity.

We could save the vertices that are not finished in a Min-Heap. Init=$=(n)$ step 1 then becomes deleteMin() which is $O(log n)$ and step 3 becomes decreaseKey outdeg times O(log n). With these improvements our time complexity is O((n+m)log n).