1. Exam Timetabling Problem using Graph Coloring

2. OUTLINE Problem statement Graph Construction Special Properties
Graph Solutions
Conclusion

3. Problem Statement Scheduling the Exam timetable for the courses
of the CS department, so that no student shouldn’t have the same time slot for the courses he is having.

4. Problem Statement
Suppose want to schedule some final exams for CS courses
with following course numbers:
1007, 3137, 3157, 3203, 3261, 4115, 4118, 4156
Assuming that there are no students in common taking the following pairs of courses:
1007‐3137
1007‐3157, 3137‐3157
1007‐3203
1007‐3261, 3137‐3261, 3203‐3261
1007‐4115, 3137‐4115, 3203‐4115, 3261‐4115
1007‐4118, 3137‐4118
1007‐4156, 3137‐4156, 3157‐4156

5. Problem Statement
How many exam slots are necessary to schedule exams by considering the courses that are given so that students should not have the 2 exams in same slot?

6. Problem statements The exam timetabling problem is to assign exams to
specific time slot which must satisfied the hard constraints with
the objective of minimizing the soft constraints violation [3].
For this research, hard constraints that must be satisfied are:
• Exam constraint - there is only one exam for each
subject.
• Student conflict - a student cannot take two exams at
the same time or slot.
• Seating restriction - the number of students seated for
an exam cannot exceed the room capacity
Soft constraints for this problem are:-
• A student should not have more than one exam per
day
• Exams should not be split across rooms

7. Graph Construction:
One way to do this is to put edges down where students mutually excluded.

8. Graph construction
Then computing the complementary graph so that no courses have common students.

9. Graph Construction
For the convenience we are re-constructing the graph as like this.

10. Graph Solutions
The graph is obviously not 1‐colorable because there exist edges.

11. Graph Solutions
The graph is not 2‐colorable because there exists triangles.

12. Graph solutions Is it 3‐colorable? Try to color by Red, Green, Blue.
Pick a triangle and color the vertices 3203‐Red,3157‐Blue and 4118‐Green.
So that 4156 must be blue to color because it has not having a common student with the course 3157

13. Graph solutions So 3261 and 4115 must be Red.
And it is easy to color 3137 and I am coloring these courses with blue.

14. Graph solutions
Therefore we need 3 exam slots which are required for the courses to be conducted.

15. Graph solutions Days Courses Monday(blue) 3137,3157,1007,4156
Tuesday(red)
3203,3261,4115
Wednesday(green)
4118

16. Graph Properties 3-colorable graph:
G is 3-Colorable graph. A 3-coloring of a graph is an assignment of one out of three colors to each vertex such that adjacent vertices have different colors. Courses are represented by vertices.
Two vertices are connected with an edge if the corresponding courses have a student in common
The minimum number of colors used to color the graph is called chromatic number.
The algorithms that are using for this problem is RLF Algorithm.

17. Graph Properties RLF Algorithm:
The RLF algorithm combines the strategy of the Large First algorithm with the structure of the AMIS algorithm.
Like the LF algorithm , at each step in the RLF procedure a node is selected for coloring which will in some sense , leave the resulting un colored nodes colorable in as few colors as possible. As with the AMIS algorithm, the RLF procedure completes the assignment of color i before commencing assignment of color i + l.

18. Graph properties K-Colorable:
A graph  is said to be k-colorable if each of its vertices can be assigned one of k colors in such a way that no two adjacent vertices are assigned the same color. The assignment is called a coloring.
Note: K-coloring may contain fewer than k colors for k > 2.
The problem in coloring schedule is NP-Hard problem.

19. References:
Exam Timetabling Using Graph Colouring Approach Burairah Hussin, Abd Samad Hasan Basari,Abdul Samad Shibghatullah, Siti Azirah Asmai.

20. Thank You
Any queries?