The document describes the Dijkstra's algorithm for finding the shortest paths between vertices in a graph. It provides an example graph with 8 vertices (A-H) and runs the algorithm on it. The algorithm works by first assigning a distance of 0 to the source vertex A and infinity to others. It then iteratively finds the closest unvisited vertex and updates distances until all vertices are visited. The example runs the algorithm on the graph in 4 steps, updating distances and parents at each step until it converges to the final shortest paths.

1. Zain Zahid (bscs)
Air university, islamabad
zainzahid99@gmail.com
*

2. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
Vertices Distance Parent
A 0 0
B ∞ 0
C ∞ 0
D ∞ 0
E ∞ 0
F ∞ 0
G ∞ 0
H ∞ 0 ∞
0
∞
∞
∞
∞
∞∞
2
For demonstration purpose, we would consider the following graph:
• Considering A as the source, assign it the distance zero.
• For all vertices except A assign a distance infinity.

3. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
∞
∞
∞
∞
∞∞
Vertices Distance Parent
A 0 0
B 8 A
C
D
E
F
G
H
2

4. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
∞
6
∞
∞
∞∞
Vertices Distance Parent
A 0 0
B 8 A
C
D
E 6 A
F
G
H
2

5. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
∞
6
∞
∞
∞14
Vertices Distance Parent
A 0 0
B 8 A
C 14 B
D
E 6 A
F
G
H
2

6. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
∞
6
9
∞
∞14
Vertices Distance Parent
A 0 0
B 8 A
C 14 B
D
E 6 A
F 9 E
G
H
2

7. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
∞
6
9
8
∞14
Vertices Distance Parent
A 0 0
B 8 A
C 14 B
D
E 6 A
F 9 E
G 8 E
H
2

8. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
∞14
Vertices Distance Parent
A 0 0
B 8 A
C 14 B
D 9 G
E 6 A
F 9 E
G 8 E
H
2

9. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
1814
Vertices Distance Parent
A 0 0
B 8 A
C 14 B
D 9 G
E 6 A
F 9 E
G 8 E
H 18 C
2

10. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
1814
Vertices Distance Parent
A 0 0
B 8 A
C 14 B
D 9 G
E 6 A
F 9 E
G 8 E
H 18 C
2
STEP-2 START CHECKING EACH VERTEX AGAIN
B D E F G H remains same

11. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
187
Vertices Distance Parent
A 0 0
B 8 A
C 7 G
D 9 G
E 6 A
F 9 E
G 8 E
H 18 C
2
Change will be only in C

12. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
117
Vertices Distance Parent
A 0 0
B 8 A
C 7 G
D 9 G
E 6 A
F 9 E
G 8 E
H 11 C
2
STEP-3 START CHECKING EACH VERTEX AGAIN
Change will be only in H

13. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
117
Vertices Distance Parent
A 0 0
B 8 A
C 7 G
D 9 G
E 6 A
F 9 E
G 8 E
H 11 C
2
And B C D E F G remains same

14. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
117
Vertices Distance Parent
A 0 0
B 8 A
C 7 G
D 9 G
E 6 A
F 9 E
G 8 E
H 11 C
2
STEP-4 START CHECKING EACH VERTEX AGAIN
This time there is no change and all vertices remain same
So we can stop here

15. A
E
F
G
C
D
B
H
6
-1
4
6
2
3
1
-2
6
8
8
0
9
6
9
8
117
Vertices Distance Parent
A 0 0
B 8 A
C 7 G
D 9 G
E 6 A
F 9 E
G 8 E
H 11 C
2
Shortest Paths are:
B=AB=8
C=AEGC=7
D=AEGD=9
E=AE=6
F=AEF=9
G=AEG=8
H=AEGCH=11

16. *There are two cases in iterations
*We might not notice any change in distance for any vertex in
later iterations. It would be good to stop at that point for
making the algorithm efficient.
*We might endlessly notice changes in distances of one or the
other vertex in later iterations. This will be the case of negative
weight cycles. Hence, we need to stop at some point and that
would be the after |V| – 1