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Design and analysis of
algorithms
Presented by: Aoun-Haider
FA21-BSE-133@cuilahore.edu.pk
Articulation point in graph:
• Articulation point is such a point in a graph which can cause dividing
of graph into different graphs by deleting it.
Cont.
It is important to detect articulation point because a network will be disconnected if articulation
point fails. If articulation point is detected, we can apply security features on that point to prevent
such failures.
0
1 2
3
4 5
6
7 8
Find articulation point in the given
graph??
0
1
2
3
4
5
6
7
8
Step#01: Apply DFS (Depth first search)
0
1
2
3
4
5
6
7
8
Step#01: Draw dotted line where graph
connection is available.
0
1
2
3
4
5
6
7
8
Assign order to each node in which
Order DFS was applied
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L
0
1
2
3
4
5
6
7
8
Compute lower index from child to
parent. For example, 1 to 0 path
is 1 -> 3 -> 2 -> 0 and minimum
value of ‘d’ is 1
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[3] >= d[1]
1 >= 2
False!!
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[4] >= d[3]
3 >= 3
True!!
So, 3 is the articulation point.
Continue..
There can be more articulation points
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[2] >= d[3]
1 >= 3
False!!
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[5] >= d[4]
3 >= 4
False!!
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[6] >= d[5]
6 >= 5
True!!
So, 5 is also articulation point.
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[7] >= d[6]
6 >= 6
True!!
So, 6 is also articulation point.
0
1
2
3
4
5
6
7
8
Step#02: determine point
u = parent
v = child
If L[v] >= d[u]
articulation_point = u
d = 1
d = 2
d = 3
d = 4
d = 5
d = 6
d = 7
d = 8
d = 9
0 1 2 3 4 5 6 7 8
d 1 2 9 3 4 5 6 7 8
L 1 1 1 1 3 3 6 6 6
L[8] >= d[7]
6 >= 7
False!!
Conclusion:
• Articulation points, also known as cut vertices, are critical elements in
graph theory. An articulation point is a vertex in a graph whose removal
would increase the number of connected components in the graph. In
other words, if an articulation point is removed, it will disconnect the graph
or split it into multiple disconnected parts.
• Detection of this point important for:
- Connectivity analysis
- Network reliability
- Graph portioning
- Route planning
- Graph visualization

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10c -Articulation_point.pptx

  • 1. Design and analysis of algorithms Presented by: Aoun-Haider FA21-BSE-133@cuilahore.edu.pk
  • 2. Articulation point in graph: • Articulation point is such a point in a graph which can cause dividing of graph into different graphs by deleting it.
  • 3. Cont. It is important to detect articulation point because a network will be disconnected if articulation point fails. If articulation point is detected, we can apply security features on that point to prevent such failures.
  • 4. 0 1 2 3 4 5 6 7 8 Find articulation point in the given graph??
  • 6. 0 1 2 3 4 5 6 7 8 Step#01: Draw dotted line where graph connection is available.
  • 7. 0 1 2 3 4 5 6 7 8 Assign order to each node in which Order DFS was applied d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L
  • 8. 0 1 2 3 4 5 6 7 8 Compute lower index from child to parent. For example, 1 to 0 path is 1 -> 3 -> 2 -> 0 and minimum value of ‘d’ is 1 d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6
  • 9. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6
  • 10. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[3] >= d[1] 1 >= 2 False!!
  • 11. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[4] >= d[3] 3 >= 3 True!! So, 3 is the articulation point. Continue.. There can be more articulation points
  • 12. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[2] >= d[3] 1 >= 3 False!!
  • 13. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[5] >= d[4] 3 >= 4 False!!
  • 14. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[6] >= d[5] 6 >= 5 True!! So, 5 is also articulation point.
  • 15. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[7] >= d[6] 6 >= 6 True!! So, 6 is also articulation point.
  • 16. 0 1 2 3 4 5 6 7 8 Step#02: determine point u = parent v = child If L[v] >= d[u] articulation_point = u d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 d = 8 d = 9 0 1 2 3 4 5 6 7 8 d 1 2 9 3 4 5 6 7 8 L 1 1 1 1 3 3 6 6 6 L[8] >= d[7] 6 >= 7 False!!
  • 17. Conclusion: • Articulation points, also known as cut vertices, are critical elements in graph theory. An articulation point is a vertex in a graph whose removal would increase the number of connected components in the graph. In other words, if an articulation point is removed, it will disconnect the graph or split it into multiple disconnected parts. • Detection of this point important for: - Connectivity analysis - Network reliability - Graph portioning - Route planning - Graph visualization