1. Single Source Shortest Paths
Definition
Algorithms
Bellman Ford
DAG shortest path algorithm
Dijkstra

2. Definition Input Task Weighted, connected directed graph G=(V,E)
Weight (length) function w on each edge e in E
Source node s in V
Task
Compute a shortest path from s to all nodes in V

3. Comments If edges are not weighted, then BFS works
Optimal substructure
A shortest path between s and t contains other shortest paths within it
No known algorithm is better at finding a shortest path from s to a specific destination node t in G than finding the shortest path from s to all nodes in V

4. Negative weight edges
Negative weight edges can be allowed as long as there are no negative weight cycles
If there are negative weight cycles, then there cannot be a shortest path from s to any node t (why?)
If we disallow negative weight cycles, then there always is a shortest path that contains no cycles

5. Relaxation technique
For each vertex v, we maintain an upper bound d[v] on the weight of shortest path from s to v
d[v] initialized to infinity
Relaxing an edge (u,v)
Can we improve the shortest path to v by going through u?
If d[v] > d[u] + w(u,v), d[v] = d[u] + w(u,v)
This can be done in O(1) time

6. Bellman-Ford Algorithm
Bellman-Ford (G, w, s)
Initialize-Single-Source(G,s)
for (i=1 to V-1)
for each edge (u,v) in E
relax(u,v);
if d[v] > d[u] + w(u,v)
return NEGATIVE WEIGHT CYCLE

7. Running Time for (i=1 to V-1) The above takes (V-1)O(E) = O(VE) time
for each edge (u,v) in E
relax(u,v);
The above takes (V-1)O(E) = O(VE) time
if d[v] > d[u] + w(u,v)
return NEGATIVE WEIGHT CYCLE
The above takes O(E) time

8. Proof of Correctness
If there is a shortest path from s to any node v, then d[v] will have this weight at end
Let p = (e1, e2, …, ek) = (v1, v2, v3, …, v,+1) be a shortest path from s to v
s = v1, v = vk+1, ei = (vi, vi+1)
At beginning of ith iteration, during which we relax edge ei, d[vi] must be correct, so at end of ith iteration, d[vi+1] will e correct
Iteration is the full sweep of relaxing all edges
Proof by induction

9. Negative weight cycle for each edge (u,v) in E
if d[v] > d[u] + w(u,v)
return NEGATIVE WEIGHT CYCLE
If no negative cycle, d[v] <= d[u] + w(u,v) for all edges (u,v)
If there is a negative weight cycle, for some edge (u,v) on the cycle, it must be the case that d[v] > d[u] + (u,v).

10. DAG shortest path algorithm
DAG-SP (G, w, s)
Initialize-Single-Source(G,s)
Topologically sort vertices in G
for each vertex u, taken in sorted order
for each edge (u,v) in E
relax(u,v);

11. Running time Improvement
for each vertex u, taken in sorted order
for each edge (u,v) in E
relax(u,v);
We only do 1 relaxation for each edge: O(E) time
In addition, O(V+E) for the sorting and we get O(V+E) running time overall

12. Proof of Correctness
If there is a shortest path from s to any node v, then d[v] will have this weight at end
Let p = (e1, e2, …, ek) = (v1, v2, v3, …, v,+1) be a shortest path from s to v
s = v1, v = vk+1, ei = (vi, vi+1)
Since we sort edges in topological order, we will process node vi (and edge ei) before processing later edges in the path.

13. Dijkstra’s Algorithm Dijkstra (G, w, s)
/* Assumption: all edge weights non-negative */
Initialize-Single-Source(G,s)
Completed = {};
ToBeCompleted = V;
While ToBeCompleted is not empty
u =EXTRACT-MIN(ToBeCompleted);
Completed += {u};
for each edge (u,v) relax(u,v);

14. Running Time Analysis While ToBeCompleted is not empty
u =EXTRACT-MIN(ToBeCompleted);
Completed += {u};
for each edge (u,v) relax(u,v);
Each edge relaxed once: O(E)
Need to decrease-key potentially once per edge
Need to extract-min once per node

15. Running Time Analysis cont’d
O(E) edge relaxations
Priority Queue operations
O(E) decrease key operations
O(V) extract-min operations
Three implementations of priority queues
Array: O(V2) time
decrease-key is O(1) and extract-min is O(V)
Binary heap: O(E log V) time assuming E >= V
decrease-key and extract-min are O(log V)
Fibonacci heap: O(V log V + E) time
decrease-key is O(1) amortized time and extract-min is O(log V)

16. Proof of Correctness s u v Only nodes in S
Assume that Dijkstra’s algorithm fails to compute length of all shortest paths from s
Let v be the first node whose shortest path length is computed incorrectly
Let S be the set of nodes whose shortest paths were computed correctly by Dijkstra prior to adding v to the processed set of nodes.
Dijkstra’s algorithm has used the shortest path from s to v using only nodes in S when it added v to S.
The shortest path to v must include one node not in S
Let u be the first such node. s u v
Only nodes in S

17. Proof of Correctness
The length of the shortest path to u must be at least that of the length of the path computed to v.
Why?
The length of the path from u to v must be < 0.
No path can have negative length since all edge weights are non-negative, and thus we have a contradiction. s u v
Only nodes in S

18. Computing paths (not just distance)
Maintain for each node v a predecessor node p(v)
p(v) is initialized to be null
Whenever an edge (u,v) is relaxed such that d(v) improves, then p(v) can be set to be u
Paths can be generated from this data structure