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| greedy_algorithm (g, m) // g[1:m] contain m input { soln := 0; // compute the unique answer for j=1 to m do { y= select(g) if feasible(soln, y) then soln = union(solnn, y); } return soln; } |
Firstly, sorting activities by the finish time. Then, starting with the first, option the next feasible activity that is compatible with previous one.
For example, assume the activity numbers and start and finish times are as follows:
| Activity | Start | Finish |
| 1 | 1 | 4 |
| 2 | 3 | 5 |
| 3 | 0 | 6 |
| 4 | 5 | 7 |
| 5 | 3 | 8 |
| 6 | 5 | 9 |
| 7 | 6 | 10 |
| 8 | 8 | 11 |
| 9 | 8 | 12 |
| 10 | 2 | 13 |
| 11 | 12 | 14 |
Greedy algorithm 1st choose ‘1’, then ‘2’, ‘3’ are out, so ‘4’ is chosen, the ‘5’, ‘6’, ‘7’ not compatible with (1, 4) so ‘8’ is chosen, then ‘9’, ‘10’ not good, ‘11’ is chosen. finally: (1, 4, 8, 11).
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