1. EMIS 8374 Dijkstra’s Algorithm Updated 18 February 2008

2. Dijkstra’s Algorithm
Network may have cycles, but all arc lengths must be nonnegative.
Maintains a partition of N into two subsets:
Set P: permanently labeled nodes
Set T: temporarily labeled nodes
Moves nodes from T into P one at a time in non-decreasing order by shortest-path distance from the source node

3. Dijkstra’s Algorithm (pg 109)
begin
P := {}; T := N;
d(i) :=  for each node i in N;
d(s) := 0 and pred(s) := 0;
while |P| < n do
pick i in T with minimum d(i) value;
move i from T to P;
for each (i, j) in A(i) do
if d(j) > d(i) + cij then
d(j) := d(i) + cij; pred(j) := i;
end;

4. Dijkstra’s Algorithm Example   2 4 5 4 3 1 1 6  2 1 6 7 2 3 2  
P = {}, T = {1, 2, 3, 4, 5, 6}

5. Dijkstra’s Algorithm Example 4    2 4 5 4 3 1 1 6  2 1 6 7 2 3 2   7 
P = {1}, T = {2, 3, 4, 5, 6}

6. Dijkstra’s Algorithm Example4 4  6   2 4 5 4 3 1 1 6  2 1 6 7 2 3 2   7 5  
P = {1,4}, T = {2, 3, 5, 6}

7. Dijkstra’s Algorithm Example4 6 2 4 4 5 3 1 1 6 11   2 1 6 7 2 3 2  7  5 5
P = {1,4,3}, T = {2, 5, 6}

8. Dijkstra’s Algorithm Example4 6 6 2 4 4 5 3 1 1 6 11 9 2 1 6 7 2 3 2  7  5
P = {1,4,3,5}, T = {2, 6}

9. Dijkstra’s Algorithm Example4  6 6 2 4 5 4 3 1 6  1 11 9 2 1 6 7 2 3 2  7  7 5
P = {1,4,3,5,2}, T = {6}

10. Dijkstra’s Algorithm Example4  6 6 2 4 5 4 3 1 1 6 9 9 2 1 6 7 2 3 2   7 5
P = {1,4,3,5,2,6}, T = {}

11. Shortest Path Tree 4  6 6 2 4 5 4 3 1 1 6 9 2 1 6 7 2 3 2   7 5

12. Shortest Path Tree 4  6 6 2 4 5 4 3 1  1 6  11 9 9 2 1 6 7 2 3 2  11 9 9 2 1 6 7 2 3 2   7 5

13. Correctness of Dijkstra’s Algorithm
At the end of any iteration the following statements hold:
The distance label d(i) is optimal for any node i in the set P. That is, d(i) = d*(i) for all i in P.
The distance label d(i) for any node i in T is the length of a shortest path from the source to i such that all internal nodes are in P.

14. Statement 2 Example: d(6) after scanning node 34 6 2 4 4 5 3 1 1 6 11 2 1 6 7 2 3 2 7 5
Shortest path in Statement 2 uses internal nodes 1, 4, and 3

15. Statement 2 Example: d(6) after scanning node 54 6 2 4 4 5 3 1 1 6 9 2 1 6 7 2 3 2 7 5
Shortest path in Statement 2 uses internal nodes 1, 4, and 5

16. Proof Statements 1 and 2 are clearly true after the first iteration.
Assume they are true after iteration k and prove that they hold after iteration k+1

17. Base Cases for Proof by Induction
After the first iteration the only node in P is the source with d(s) = 0.
After the first iteration there are two cases for a node j in T:
If j is in A(s), then d(j) = csj
If j is not in A(s), then d(j) = .

18. Illustration of Base Cases14  2 4 5 14 3 1 1 6  2 1 6 7 2 3 2  7
P = {1}, T = {2, 3, 4, 5, 6}

19. Proof that Statement 1 Holds after Iteration k+1
Let i be the node moved from T to P in iteration k+1
Need to show d(i) = d*(i)
Idea
Let B be a path from s to i
Let L(B) be the length of B
Show that d(i) ≤ L(B)

20. Proof that Statement 1 Holds after Iteration k+1
There are two cases to consider for B
All internal nodes of B are in P
At least one internal node of B is in T
Case 1: Statement 2 implies that d(i) ≤ L(B)

21. Proof that Statement 1 Holds after Iteration k+1
Case 2:
Let j be first internal node of B that is in T
By Statement 2, the length of the subpath of B from s to j is at least d(j)
Since all arc lengths are non-negative, the length of the subpath of B from j to i is at least zero
Thus, L(B) ≥ d(j)
Since node i was picked to move to P, d(i) ≤ d(j)
Therefore, d(i) ≤ L(B)

22. Diagram for Case 2 P i s j
Since Statement 2 holds for the first k iterations, d(j) ≤ length of the subpath of B from s to j.

23. Diagram for Case 2 P i s j
Since the arc lengths are non-negative, the length of the subpath of B from j to i is at least zero.

24. Diagram for Case 2 P i s j
The length of path B from s to i is at least d(j).

25. Diagram for Case 2 P i s j
Since d(i) ≤ d(j), the length of path B from s to i is at least d(i).

26. Diagram for Case 2 P i s j
Since d(i) ≤ L(B) for any path B from s to i , d(i) = d*(i).

27. Proof of Statement 2
When node i moves from T to P, the algorithm changes pred(j) to i and sets d(j) = d(i) + cij for all j in T such that d(j) > d(i) + cij .
Thus, the path defined by the predecessor indices from j to s satisfies Property 4.2.
Therefore, d(j) is the length of a shortest path from s to j subject to the restriction that each internal node in the path is in P.

28. Diagram for Statement 2 j s i T P pred(j) Move node i from T to P
If d(j)  d(i) + cij then Statement 2 still holds for node j.

29. Diagram for Statement 2 j s i T P old pred(j) new pred(j)
Move node i from T to P
If d(j) > d(i) + cij then d(j) gets reset to d(i) + cij

30. Complexity of Dijkstra’s Algorithm
Node selections
Must compare the distance labels of all nodes in T: O(n + (n-1) + (n-2) + … + 1) = O(n2)
Total work for scanning nodes is O(m)
Overall complexity is O(n2 + m) = O(n2)
Can be improved to O(m + n log n) with Fibonacci heaps