The primary topics in this part of the specialization are: greedy algorithms (scheduling, minimum spanning trees, clustering, Huffman codes) and dynamic programming (knapsack, sequence alignment, optimal search trees).

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From the course by Stanford University

Greedy Algorithms, Minimum Spanning Trees, and Dynamic Programming

291 ratings

Stanford University

291 ratings

Course 3 of 4 in the Specialization Algorithms

The primary topics in this part of the specialization are: greedy algorithms (scheduling, minimum spanning trees, clustering, Huffman codes) and dynamic programming (knapsack, sequence alignment, optimal search trees).

From the lesson

Week 1

Two motivating applications; selected review; introduction to greedy algorithms; a scheduling application; Prim's MST algorithm.

- Tim RoughgardenProfessor

Computer Science

For our next case study of how to use greedy algorithms, we're going to turn to

Â the application domain of scheduling. That is how do you schedule jobs on

Â shared resources in order to accomplish some objective.

Â So, the domain of scheduling there's lots of different applications of greedy

Â algorithms. We'll see two in this course.

Â we'll start for today just with the following simple scenario.

Â So, we'll assume, for today, that there's just one shared resource.

Â This resource could represent any number of things.

Â For concreteness, you can think of it as a computer processor.

Â And then, there's a lot of different things that got to get done.

Â So, for example, there's a lot of processes that have to be handled by this

Â processor. In the algprithmic question, we are going

Â to study, is, in what order should we sequence these jobs?

Â Which one should we do first, which one should we do second, and so on, all the

Â way up to which one should we do last. So, obviously to answer this question, we

Â need to pin down the mathematical model a little bit more precisely.

Â And lets start with just, you know, what is the characteristics of jobs, what

Â information do we have that might lead us to prefer one job over another.

Â But for this problem, we're going to assume that each job comes with two known

Â parameters. So, first of all, job j has what we're

Â going to call a weight w sub j. That's a non-negative real number.

Â And you should think of the weight of a job as quantifying its importance.

Â That is, jobs with a higher weight, in some sense, deserve to be processed

Â earlier than those with a lower weight. And secondly, each job j is going to come

Â with a non negative length l sub j. Depending on the application, you may or

Â may not have a good estimate of how long jobs are going to take, but for today to

Â keep things simple, let's assume we know what the length of every job is, and

Â that's l sub j, it's part of the input to are problem.

Â So. we have now defined the input to this

Â computational problem. We get n jobs each specified by a weight

Â and a length. And we know that the output is going to

Â be a sequence of these n jobs in some order.

Â So, what we have to understand now is what criterion do we want to optimize?

Â What are we trying to accomplish with this sequence?

Â To explain that, I need to tell you about completion times of jobs.

Â So, the completion time of a job is defined hopefully in exactly the way

Â you'd think. So, for the job which is scheduled first,

Â it's just the length of the job because that's how long it takes to process that

Â job. For whatever job gets scheduled second,

Â its completion time is the length of the first job and then, the length of that

Â job itself. So, in other words, it's just the total

Â time which elapses before that job gets completed,

Â okay? So, in general, the completion time of a

Â job is just the sum of the lengths of the jobs scheduled to before that job plus

Â the length of that job itself. To make sure this is clear, let's go

Â through a quick example. So, suppose there are three jobs with

Â lengths one, two, and three. I'm not going to tell you the job weights

Â because they're irrelevant for the purposes of computing the completion

Â time. And let's suppose we do the schedule

Â where we just schedule job one first, the job two, then job three.

Â So, pictorially, I'm going to represent that schedule just by stacking the jobs

Â on top of each other with the interpretation that time starts at the

Â bottom. So, time zero is where we schedule job

Â one. And then, time increases as we go from

Â the bottom to the top of the diagram. And the question then is, what are the

Â completion times of these three jobs? Okay.

Â So, the correct answer is answer C. So, for the first job, it gets scheduled

Â first so it's very happy and it just takes one unit of time to complete, so

Â its completion time is one. The second job, well, it has to wait for

Â the first job to complete so one unit of time elapses and then, it itself has to

Â complete so that's two more units so it gets to the completion time of three.

Â For the third job, it has to wait for the first two to complete, so that adds three

Â to the clock, and then plus it takes three units of time for a total of six.

Â So, that's the definition of job completion times. In some sense, we

Â obviously want completion times to be as small as possible.

Â But it's not so simple. In any given schedule, the jobs that are

Â give early on are going to have small completion times and the jobs towards the

Â end are going to have big completion times. So inevitably, we're going be have

Â to trading off the completion times between different jobs.

Â So, what is the optimal way to so that? Well, that depends on our objective

Â function, and in scheduling, there's many different objective functions you might

Â want to use. today, I'm just going to tell you about

Â one. It's not the only natural objective

Â function, but it's one of several most natural objective functions.

Â It's called minimizing the weighted sum of completion times.

Â You translate this English phrase into mathematics in the obvious way.

Â What you want to do is you want to minimize the sum over all n jobs of their

Â completion time, but then multiplied by their weight of [UNKNOWN] j.

Â Okay. So, the sum over j of w j times c j.

Â The w j is the weight and c j is the completion time as defined on the

Â previous slot. If you think about it for a second,

Â you'll realize this is equivalent to minimizing the weighted average of the

Â completion times with the weights given as in the input.

Â So, just to make sure this makes sense, let's go back to the example that we saw.

Â In that example, we had jobs with lengths one, two, and three, and we thought about

Â to schedule or we scheduled them in that order.

Â To evaluate the subjective function, I'd have to tell you their weights, so let's

Â suppose their weights are three, two, and one, respectively.

Â In this case, the weighted sum of completion times in the schedule,

Â in the previous slide, well first, we begin with a, the first

Â job, which has weight three. Its completion time, remember, was one.

Â Then, we have the second job with weight two, its completion time is three.

Â Then, we have the third job with weight one, its completion time was six.

Â So, we sum up the weighted completion times and we get a total of fifteen.

Â And I'll let you verify that, in fact, all of the three factorial or six

Â schedules in that example, this is, in fact, the schedule that minimizes the

Â weighted sum of completion times. And the algorithmic question we're going

Â to study next, is how do we do this in general?

Â Given arbitrary input in jobs, weights, and lengths, what is the sequence that

Â minimizes this sum over all n factorial sequences you might consider?

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