1 cs lunchblerner/cs312/slides/slides08.pdf · cs lunch learn about the cs department and cs major...

CS Lunch

Learn about the CS Department and CS major

Wednesday, 12:15 to 1:00

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Midterm

No homework this week

Midterm on Feb. 25

Covers chapters 1-3

Will post some sample questions

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Slides08 - IntervalScheduling.key - February 18, 2019

New topic!4-1

New topic!

Greedy Algorithms

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New topic!

Greedy AlgorithmsGeneral approach:

1. Sort the data2. Consider each data item in order

and make a local decision.

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Interval SchedulingJob j starts at sj and finishes at fj.Two jobs compatible if they don't overlap.Goal: find maximum subset of mutually compatible jobs.

Time0 1 2 3 4 5 6 7 8 9 10 11

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Interval Scheduling: Greedy SolutionGreedy template. Consider jobs in some order. Take each job provided it's compatible with the ones already taken.

Idea 1: Earliest start time. Consider jobs in ascending order of start time sj.

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Idea 1: Earliest start time. Consider jobs in ascending order of start time sj.

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Earliest Start Time: Bad Example

Earliest start time picks single yellow jobBetter solution is the 4 short blue jobsOther ideas?

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Idea 2: Shortest interval. Consider jobs in ascending order of interval length fj - sj.

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Idea 2: Shortest interval. Consider jobs in ascending order of interval length fj - sj.

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Shortest Interval: Bad Example

Shortest interval picks single yellow jobBetter solution is the 2 blue jobsOther ideas?

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Idea 3: Fewest conflicts. For each job, count the number of conflicting jobs cj. Schedule in ascending order of conflicts cj.

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Fewest Conflicts: Bad Example

Fewest conflicts picks the 3 yellow jobsBetter solution is the 4 jobs on the top rowOther ideas?

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Idea 4: Earliest finish time. Consider jobs in ascending order of finish time fj.

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Idea 4: Earliest finish time. Consider jobs in ascending order of finish time fj.

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Earliest Finish Time - Optimal Solution

Sort jobs by finish times so that f1 ≤ f2 ≤ ... ≤ fn.

A ← {}for j = 1 to n { if (job j compatible with A) A = A ∪ {j}}return A

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Hallmark of a Greedy Algorithm

Sort data according to some criteria

Consider each piece of data in sorted order and make a local decision

Result is globally optimal (if this problem is amenable to a greedy solution!)

Complexity is generally no better than O(n log n) due to the sort

Important to prove that the solution is optimal

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Interval PartitioningLecture j starts at sj and finishes at fj.Goal: find minimum number of classrooms to schedule all lectures so that no two occur at the same time in the same room.

Time9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30

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Interval Partitioning Lower Bound

The depth of a set of intervals is the maximum number that contain any given time.Key observation. Number of classrooms needed ≥ depth.

Time9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30

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Interval Partitioning Lower BoundExample: Depth of schedule below = 3Question: Does there always exist a schedule equal to depth of intervals?

Time9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30

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Time9 9:30 10 10:30 11 11:30 1212:30 1 1:30 2 2:30

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Interval Partitioning: Greedy Solution

Sort intervals by starting time so that s1 ≤ s2 ≤ ... ≤ sn.d ← 0 // Number of classrooms

for j = 1 to n { if (lecture j is compatible with some classroom k) schedule lecture j in classroom k else allocate a new classroom d + 1 schedule lecture j in classroom d + 1 d ← d + 1 }

Complexity?

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1 cs lunchblerner/cs312/slides/slides08.pdf · cs lunch learn about the cs department and cs major...

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