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Lecture06_07 GREEDY ALGORITHM-DATA ANALYTICS ALGORITHMS.ppt
- 1. CSE 421 Algorithms Richard Anderson Lecture6 Greedy Algorithms
- 2. Greedy Algorithms • Solveproblems with the simplest possible algorithm • The hard part: showing that something simple actually works • Pseudo-definition – An algorithm is Greedy if it builds its solution by adding elements one at a time using a simple rule
- 3. Scheduling Theory • Tasks –Processing requirements, release times, deadlines • Processors • Precedence constraints • Objective function – Jobs scheduled, lateness, total execution time
- 4. • Tasks occurat fixed times • Single processor • Maximize number of tasks completed • Tasks {1, 2, . . . N} • Start and finish times, s(i), f(i) Interval Scheduling
- 5. What is thelargest solution?
- 6. Greedy Algorithm forScheduling Let T be the set of tasks, construct a set of independent tasks I, A is the rule determining the greedy algorithm I = { } While (T is not empty) Select a task t from T by a rule A Add t to I Remove t and all tasks incompatible with t from T
- 7. Simulate the greedyalgorithm for each of these heuristics Schedule earliest starting task Schedule shortest available task Schedule task with fewest conflicting tasks
- 8. Greedy solution basedon earliest finishing time Example 1 Example 2 Example 3
- 9. Theorem: Earliest FinishAlgorithm is Optimal • Key idea: Earliest Finish Algorithm stays ahead • Let A = {i1, . . ., ik} be the set of tasks found by EFA in increasing order of finish times • Let B = {j1, . . ., jm} be the set of tasks found by a different algorithm in increasing order of finish times • Show that for r<= min(k, m), f(ir) <= f(jr)
- 10. Stay ahead lemma •A always stays ahead of B, f(ir) <= f(jr) • Induction argument – f(i1) <= f(j1) – If f(ir-1) <= f(jr-1) then f(ir) <= f(jr)
- 11. Completing the proof •Let A = {i1, . . ., ik} be the set of tasks found by EFA in increasing order of finish times • Let O = {j1, . . ., jm} be the set of tasks found by an optimal algorithm in increasing order of finish times • If k < m, then the Earliest Finish Algorithm stopped before it ran out of tasks
- 12. Scheduling all intervals •Minimize number of processors to schedule all intervals
- 13. How many processorsare needed for this example?
- 14. Prove that youcannot schedule this set of intervals with two processors
- 15. Depth: maximum numberof intervals active
- 16. Algorithm • Sort bystart times • Suppose maximum depth is d, create d slots • Schedule items in increasing order, assign each item to an open slot • Correctness proof: When we reach an item, we always have an open slot
- 17. Homework Scheduling • Tasksto perform • Deadlines on the tasks • Freedom to schedule tasks in any order
- 18. Scheduling tasks • Eachtask has a length ti and a deadline di • All tasks are available at the start • One task may be worked on at a time • All tasks must be completed • Goal: minimize maximum lateness – Lateness = fi – di if fi >= di
- 19.
- 20. Determine the minimumlateness 2 3 4 5 6 4 5 12 Deadline Time
- 21. Greedy Algorithm • Earliestdeadline first • Order jobs by deadline • This algorithm is optimal
- 22. Analysis • Suppose thejobs are ordered by deadlines, d1 <= d2 <= . . . <= dn • A schedule has an inversion if job j is scheduled before i where j > i • The schedule A computed by the greedy algorithm has no inversions. • Let O be the optimal schedule, we want to show that A has the same maximum lateness as O
- 23.
- 24. Lemma: There isan optimal schedule with no idle time • It doesn’t hurt to start your homework early! • Note on proof techniques – This type of can be important for keeping proofs clean – It allows us to make a simplifying assumption for the remainder of the proof a4 a2 a3 a1
- 25. Lemma • If thereis an inversion i, j, there is a pair of adjacent jobs i’, j’ which form an inversion
- 26. Interchange argument • Supposethere is a pair of jobs i and j, with di <= dj, and j scheduled immediately before i. Interchanging i and j does not increase the maximum lateness. di dj di dj j i j i
- 27. Proof by BubbleSort a4 a2 a3 a1 a4 a2 a3 a4 a2 a3 a1 a4 a2 a3 a1 a1 a4 a2 a3 a1 Determine maximum lateness d1 d2 d3 d4
- 28. Real Proof • Thereis an optimal schedule with no inversions and no idle time. • Let O be an optimal schedule k inversions, we construct a new optimal schedule with k-1 inversions • Repeat until we have an optimal schedule with 0 inversions • This is the solution found by the earliest deadline first algorithm
- 29. Result • Earliest DeadlineFirst algorithm constructs a schedule that minimizes the maximum lateness
- 30. Extensions • What ifthe objective is to minimize the sum of the lateness? – EDF does not seem to work • If the tasks have release times and deadlines, and are non-preemptable, the problem is NP-complete • What about the case with release times and deadlines where tasks are preemptable?
- 31. Optimal Caching • Cachingproblem: – Maintain collection of items in local memory – Minimize number of items fetched
- 32. Caching example A, B,C, D, A, E, B, A, D, A, C, B, D, A
- 33. Optimal Caching • Ifyou know the sequence of requests, what is the optimal replacement pattern? • Note – it is rare to know what the requests are in advance – but we still might want to do this: – Some specific applications, the sequence is known – Competitive analysis, compare performance on an online algorithm with an optimal offline algorithm
- 34. Farthest in thefuture algorithm • Discard element used farthest in the future A, B, C, A, C, D, C, B, C, A, D
- 35. Correctness Proof • Sketch •Start with Optimal Solution O • Convert to Farthest in the Future Solution F-F • Look at the first place where they differ • Convert O to evict F-F element – There are some technicalities here to ensure the caches have the same configuration . . .