CSC373 Week 2: Greedy Algorithms Karan Singh 373F19 - Karan Singh 1
Recap • Divide & Conquer Master theorem Counting inversions in 𝑃(𝑜 log 𝑜) Finding closest pair of points in ℝ 2 in 𝑃 𝑜 log 𝑜 Fast integer multiplication in 𝑃 𝑜 log 2 3 Fast matrix multiplication in 𝑃 𝑜 log 2 7 Finding 𝑙 𝑢ℎ smallest element (in particular, median) in 𝑃(𝑜) 373F19 - Karan Singh 2
Greedy Algorithms • Greedy (also known as myopic) algorithm outline We want to find a solution 𝑦 that maximizes some objective function 𝑔 But the space of possible solutions 𝑦 is too large The solution 𝑦 is typically composed of several parts (e.g. 𝑦 may be a set, composed of its elements) Instead of directly computing 𝑦 … o Compute it one part at a time o Select the next part “greedily” to get maximum immediate benefit (this needs to be defined carefully for each problem) o May not be optimal because there is no foresight o But sometimes this can be optimal too! 373F19 - Karan Singh 3
Interval Scheduling • Problem Job 𝑘 starts at time 𝑡 𝑘 and finishes at time 𝑔 𝑘 Two jobs are compatible if they don’t overlap Goal: find maximum-size subset of mutually compatible jobs 373F19 - Karan Singh 4
Interval Scheduling • Greedy template Consider jobs in some “natural” order Take each job if it’s compatible with the ones already chosen • What order? Earliest start time: ascending order of 𝑡 𝑘 Earliest finish time: ascending order of 𝑔 𝑘 Shortest interval: ascending order of 𝑔 𝑘 − 𝑡 𝑘 Fewest conflicts: ascending order of 𝑑 𝑘 , where 𝑑 𝑘 is the number of remaining jobs that conflict with 𝑘 373F19 - Karan Singh 5
Example • Earliest start time: ascending order of 𝑡 𝑘 • Earliest finish time: ascending order of 𝑔 𝑘 • Shortest interval: ascending order of 𝑔 𝑘 − 𝑡 𝑘 • Fewest conflicts: ascending order of 𝑑 𝑘 , where 𝑑 𝑘 is the number of remaining jobs that conflict with 𝑘 373F19 - Karan Singh 6
Interval Scheduling • Does it work? Counterexamples for earliest start time shortest interval fewest conflicts 373F19 - Karan Singh 7
Interval Scheduling • Implementing greedy with earliest finish time (EFT) Sort jobs by finish time. Say 𝑔 1 ≤ 𝑔 2 ≤ ⋯ ≤ 𝑔 𝑜 When deciding whether job 𝑘 should be included, we need to check whether it’s compatible with all previously added jobs 𝑗 ∗ , where 𝑗 ∗ is the last added job o We only need to check if 𝑡 𝑘 ≥ 𝑔 o This is because for any jobs 𝑗 added before 𝑗 ∗ , 𝑔 𝑗 ≤ 𝑔 𝑗 ∗ o So we can simply store and maintain the finish time of the last added job Running time: 𝑃 𝑜 log 𝑜 373F19 - Karan Singh 8
Interval Scheduling • Optimality of greedy with EFT Suppose for contradiction that greedy is not optimal Say greedy selects jobs 𝑗 1 , 𝑗 2 , … , 𝑗 𝑙 sorted by finish time Consider the optimal solution 𝑘 1 , 𝑘 2 , … , 𝑘 𝑛 (also sorted by finish time) which matches greedy for as long as possible o That is, we want 𝑘 1 = 𝑗 1 , … , 𝑘 𝑠 = 𝑗 𝑠 for greatest possible 𝑠 373F19 - Karan Singh 9
Interval Scheduling Another standard method is induction • Optimality of greedy with EFT Both 𝑗 𝑠+1 and 𝑘 𝑠+1 were compatible with the previous selection ( 𝑗 1 = 𝑘 1 , … , 𝑗 𝑠 = 𝑘 𝑠 ) Consider the solution 𝑗 1 , 𝑗 2 , … , 𝑗 𝑠 , 𝑗 𝑠+1 , 𝑘 𝑠+2 , … , 𝑘 𝑛 o It should still be feasible (since 𝑔 𝑗 𝑠+1 ≤ 𝑔 𝑘 𝑠+1 ) o It is still optimal o And it matches with greedy for one more step (contradiction!) 373F19 - Karan Singh 10
Interval Partitioning • Problem Job 𝑘 starts at time 𝑡 𝑘 and finishes at time 𝑔 𝑘 Two jobs are compatible if they don’t overlap Goal: group jobs into fewest partitions such that jobs in the same partition are compatible • One idea Find the maximum compatible set using the previous greedy EFT algorithm, call it one partition, recurse on the remaining jobs. Doesn’t work (check by yourselves) 373F19 - Karan Singh 11
Interval Partitioning • Think of scheduling lectures for various courses into as few classrooms as possible • This schedule uses 4 classrooms for scheduling 10 lectures 373F19 - Karan Singh 12
Interval Partitioning • Think of scheduling lectures for various courses into as few classrooms as possible • This schedule uses 3 classrooms for scheduling 10 lectures 373F19 - Karan Singh 13
Interval Partitioning • Let’s go back to the greedy template! Go through lectures in some “natural” order Assign each lecture to a compatible classroom (which?), and create a new classroom if the lecture conflicts with every existing classroom • Order of lectures? Earliest start time: ascending order of 𝑡 𝑘 Earliest finish time: ascending order of 𝑔 𝑘 Shortest interval: ascending order of 𝑔 𝑘 − 𝑡 𝑘 Fewest conflicts: ascending order of 𝑑 𝑘 , where 𝑑 𝑘 is the number of remaining jobs that conflict with 𝑘 373F19 - Karan Singh 14
Interval Partitioning • At least when you assign each lecture to an arbitrary feasible classroom, three of these heuristics do not work. • The fourth one works! (next slide) 373F19 - Karan Singh 15
Interval Partitioning 373F19 - Karan Singh 16
Interval Partitioning • Running time Key step: check if the next lecture can be scheduled at some classroom Store classrooms in a priority queue o key = finish time of its last lecture Is lecture 𝑘 compatible with some classroom? o Same as “Is 𝑡 𝑘 at least as large as the minimum key?” o If yes: add lecture 𝑘 to classroom 𝑙 with minimum key, and increase its key to 𝑔 𝑘 o Otherwise: create a new classroom, add lecture 𝑘 , set key to 𝑔 𝑘 𝑃(𝑜) priority queue operations, 𝑃(𝑜 log 𝑜) time 373F19 - Karan Singh 17
Interval Partitioning • Proof of optimality (lower bound) # classrooms needed ≥ maximum “depth” at any point o depth = number of lectures running at that time We now show that our greedy algorithm uses only these many classrooms! 373F19 - Karan Singh 18
Interval Partitioning • Proof of optimality (upper bound) Let 𝑒 = # classrooms used by greedy Classroom 𝑒 was opened because there was a schedule 𝑘 which was incompatible with some lectures already scheduled in each of 𝑒 − 1 other classrooms All these 𝑒 lectures end after 𝑡 𝑘 Since we sorted by start time , they all start at/before 𝑡 𝑘 So at time 𝑡 𝑘 , we have 𝑒 overlapping lectures Hence, depth ≥ 𝑒 So all schedules use ≥ 𝑒 classrooms. QED! 373F19 - Karan Singh 19
Interval Graphs • Interval scheduling and interval partitioning can be seen as graph problems • Input Graph 𝐻 = (𝑊, 𝐹) Vertices 𝑊 = jobs/lectures Edge 𝑗, 𝑘 ∈ 𝐹 if jobs 𝑗 and 𝑘 are incompatible • Interval scheduling = maximum independent set (MIS) • Interval partitioning = graph colouring 373F19 - Karan Singh 20
Interval Graphs • MIS and graph colouring are NP-hard for general graphs • But they’re efficiently solvable for interval graphs Interval graphs = graphs which can be obtained from incompatibility of intervals In fact, this holds even when we are not given an interval representation of the graph • Can we extend this result further? Yes! Chordal graphs o Every cycle with 4 or more vertices has a chord 373F19 - Karan Singh 21
Minimizing Lateness • Problem We have a single machine Each job 𝑘 requires 𝑢 𝑘 units of time and is due by time 𝑒 𝑘 If it’s scheduled to start at 𝑡 𝑘 , it will finish at 𝑔 𝑘 = 𝑡 𝑘 + 𝑢 𝑘 Lateness: ℓ 𝑘 = max 0, 𝑔 𝑘 − 𝑒 𝑘 Goal: minimize the maximum lateness, 𝑀 = max ℓ 𝑘 𝑘 Total lateness minimization is NP-complete • Contrast with interval scheduling We can decide the start time All jobs must be scheduled on a single machine 373F19 - Karan Singh 22
Minimizing Lateness • Example Input An example schedule 373F19 - Karan Singh 23
Minimizing Lateness • Let’s go back to greedy template Consider jobs one-by- one in some “natural” order Schedule jobs in this order (nothing special to do here, since we have to schedule all jobs and there is only one machine available) • Natural orders? Shortest processing time first: ascending order of processing time 𝑢 𝑘 Earliest deadline first: ascending order of due time 𝑒 𝑘 Smallest slack first: ascending order of 𝑒 𝑘 − 𝑢 𝑘 373F19 - Karan Singh 24
Minimizing Lateness • Counterexamples Shortest processing time first o Ascending order of processing time 𝑢 𝑘 Smallest slack first o Ascending order of 𝑒 𝑘 − 𝑢 𝑘 373F19 - Karan Singh 25
Minimizing Lateness • By now, you should know what’s coming… • We’ll prove that earliest deadline first works! 373F19 - Karan Singh 26
Minimizing Lateness • Observation 1 There is an optimal schedule with no idle time 373F19 - Karan Singh 27
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