Compute Shortest Path

If $v\in T$

d[v] = shortest path length from s to v on the tree T

If $v\notin T$

$d[v]=min_{\{u\in T~{}\&~{}(u,v)\in E\}} (d[u]+w(u,v))$

$d[v]=\infty$ if no $(u,v)\in E$ for any $u \leftarrow T$

1. Let T contains only $s$
2. $d[s]\leftarrow D$ and $d[v]\leftarrow w(s,v)$
3. While T does not cover all vertices
   1. FInd $v\notin T$ that minimize $x = y$. Assume $(u, v)$ is the edge that minimize $d[v]$
   2. $T\rightarrow T\cup \{(u,v)\}$
   3. For each neighbour x of v, $d[x]\leftarrow \min{\{d[x],d[v]+w(v,x)\}}$

Improved version (using priority queue)

1. Delete-min: $O(\log n)$
2. Decrease-key: $O(\log n)$

Therefore, the overall complexity is
$O(n\cdot \log n + m\cdot \log n) = O((n+m)\cdot\log n)$

The priority queue implementation is also called Dijkstra's Algorithm

Bellman-Ford Algorithm

$D[v,k]:=$ shortest path weight that uses at most $k$ edges

$basecase \begin{cases} D[v,0]=\infty\\ D[s,0]=0 \end{cases}$ $D[v,k]=\min_{(u,v)\in E}(D[u,k-1]+w(u,v))$

$D[s,0] = 0,D[v,0]=\infty$ for $s\neq v$

D[s, 0] = 0
D[v, 0] = infinity for s != v
for k = 1 to n - 1
   for v in V
      D[v, k] = min{(u, v) in E} (D[u, k-1] + w(u, v))

D[*, n-1] contains the answers

Complexity

Time: $O(n\cdot m)$ Space: $O(n^2)$

Alternative Implementation (Better space efficiency)

d[v], p[v]
d[s] <- 0, d[v] <- infinity
p[v] <- null
for k = 1 to n - 1
   for each edge (u, v) in E
      if d[u]+w(u, v) < d[v]
         d[v]=d[u]+w(u,v)
         p[v]=u

   shortestdist <= d[v] <=