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

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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 2

Kruskal's MST algorithm and applications to clustering; advanced union-find (optional).

- Tim RoughgardenProfessor

Computer Science

This video will prove the correctness of our greedy algorithm for clustering.

Â We'll show that it maximizes the spacing over all possible K clusterings.

Â You might have hoped that we could deduce the correctness of this greedy algorithm

Â for clustering immediately from our correctness proofs for various greedy

Â minimum spanning tree algorithms. Unfortunately that doesn't seem to be the

Â case. In the minimum cost spanning tree

Â problem, we're focusing on minimizing the sum of the edge cost.

Â Here we're looking at different objective, maximizing the spacings.

Â We do need to do a proof from scratch. That's said, you know, the arguments

Â we'll use should look familiar to you not just from the sort of exchange type

Â arguments when we prove the cut property, but also it might remind you even more,

Â going back further, to our greedy algorithms for scheduling.

Â So let's now set up the notation for the proof.

Â As usual, we're going to look at the output of r algorithm.

Â It achieves some objective function value, some spacing.

Â We're going to look at an arbitrary competitor.

Â Some other proposed scheduling. We're going to show that we're at least

Â as good, our spacing is, at least, as large.

Â So specifically, we'll denote the clusters in the output of r algorithm by

Â C1 up to CK. Our clustering has some spacing, some

Â distance between the near, closest pair of separated points.

Â Call it capital S. We're going to denote our competitor,

Â some alternative K clustering by C-Hat one of the C-Hat K, what is it that we're

Â tryin to show? We want to show that this arbitrary other

Â clustering has spacing no larger than R's, if we can show that, then because

Â this clustering was arbitrary it means the greedy clustering has spacing as

Â large as any other, so it's maximizing the spacing, that's what we want to

Â proof. But differently we want to exhibit a pair

Â of points separated by this cluster and C one-half to C1K, such that the distance

Â between those separated points is S or smaller.

Â So, let me just quickly depose of a trivial case.

Â If the C hats are the same as the C's, possibly up to a renaming, then of course

Â exactly the same pairs of points are separated into each of the clustering, so

Â that the spacing is exactly the same. So that's not a case we have to worry

Â about. The interesting case, then, is when the c

Â hats differ fundamentally from the cs, when they're not merely a permutation of

Â the clusters in the greedy clustering. And the maneuver we're going to do here

Â is similar in spirit to what we did in our scheduling correctness proof.

Â Way back in our scheduling correctness proof, we argued that any schedule that

Â differs from the greedy one, suffers from, in some sense, a local flaw.

Â We identified an adjacent pair of jobs that was, in some sense, out of order

Â with respect to the greedy ordering. The analog here is, we're going to argue

Â that, for any clustering which is not merely a permutation of the greedy

Â clustering. There has to be a pair of points which is

Â classified differently in the c hats relative to the c's.

Â By differently, I mean they're clustered together in the greedy clustering.

Â These points, p and q, belong to the same cluster, c sub i.

Â Yet, in this alternative clustering, which is not just the permutation of the

Â greedy clustering. They're placed in different clusters.

Â One, maybe p and c hat i, and q and some other c hat j.

Â So I want to now split the proof into an easy case and a tricky case.

Â To explain why the easy case is easy lets, lets observe a property that this

Â greedy clustering algorithm has. Now the algorithm's philosophy is that

Â the squeaky wheel should get the grease. That is, the separated pair of points

Â that are closest to each other are the ones that should get merged.

Â So for this reason, because it's always the closest separated pair that get

Â merged, if you look at the sequence of point pairs that get merged together,

Â that determine the spacing in each subsequent iteration, the distances

Â between these sort of worst separated points is only going up over time.

Â At the beginning of the algorithm, the closest pair of points in the entire

Â point set are the ones that get directly merged.

Â Then those are out of the picture, and now that some further away pair of

Â points are separated, it determines the spacing,

Â then they get merged. Once they've been coalesced, then there

Â is still some further away pair of points, which is now the smallest

Â separated. They get merged, and so on.

Â So if you look at the sequence of distances between the pairs of points

Â that are directly merged by the greedy algorithm, that is only going up over

Â time. And this sequence culminates with the

Â final spacing S of the greedy algorithm. At some sense, the spacing of the output

Â of the greedy algorithm is the distance between the point period that would get

Â merged if we ran the greedy algorithm one more in moderation but unfortunately

Â we're not allowed to do that. Okay?

Â So the point is, for every pair of points directly merged by the greedy algorithm,

Â they're always a distance at most S away from each other.

Â So the easy case, then, is when this pair of points, pq,

Â which, on the one hand, lie in a common greedy structure,

Â but on the other hand, in different clusters with c hats.

Â If they were, at some point, not merely in the same cluster, but actually

Â directly merged by the greedy algorithm. If, at some iteration, they determined

Â the spacing, and were picked by the greedy algorithm to have their, clusters

Â merged. Then we just argued that the distance

Â between p and q is no more than the space in capital s of the greedy clustering.

Â And since p and q lie in different clusters of the c hats.

Â It's separated by the C hats and therefore they upper bound the spacing of

Â the C hats. Maybe there's some even closer separated

Â pair by the C hats. But the very least P and Q are separated

Â so they upper bound the spacing of the C hat clustering.

Â So that's what we wanted to prove. We wanted to show that this alternative

Â spacing didn't have better spacing than our greedy spacing.

Â It had to be at most as big. It had to be at most capital S.

Â So in this easy case, when P and Q are directly merged by the greedy algorithm,

Â we're done. So the tricky case is when P and Q are

Â only indirectly merged, and you may be wondering at the moment, what does that

Â mean? How did two people wind up in the same

Â cluster if they weren't, at some point, directly merged?

Â So let's draw a picture and see how that can happen.

Â So the issue is that two points P and Q might wind up in a common greeting

Â cluster, not because the greedy algorithm ever

Â explicitly considered that point pair, but rather because of a path or cascade

Â of direct mergers of other point pairs. Imagine, for example, that at some

Â iteration of the greedy algorithm the point P was considered explicitly along

Â with the point A1, where here A1 is meant to be different than Q.

Â So that's a direct merger, and P and A1 wind up in the same cluster. Their

Â clusters are merged. Maybe the same thing happened to the

Â point Q at some point A sub L which is different than P.

Â Sooner or later maybe, you know, at some other time, some totally unrelated pair

Â of points A2 and A3 are directly merged and then at some point A1 and A2 are

Â considered by the greedy algorithm. Algorithm, because the other closest pair

Â of separated points, and, they get merged.

Â And so on. So the edges in this picture are meant to

Â indicate direct mergers, pairs of points that are explicitly fused because they

Â determine the spacing of some point of the greedy iteration.

Â But at the end of the day the greedy clustering is going to have the results

Â of all of these mergings. So in case you're feeling confused, let

Â me just point out that we really saw this exact same exact thing going on when we

Â were talking about minimum spanning trees in Kruskal's rhythm.

Â So, at an intermediate point in Kruskal's rhythm, after it's added some edges, but

Â before it's constructed a spanning tree. As we discussed, the intermediate state

Â is a bunch of different connected components.

Â And there are vertices that Have an edge chosen between them.

Â They, of course, are going to be in the same kinetic component.

Â But then again, a kenetic component could have long paths in it.

Â So you could have vertices that are in the same kinetic component in an

Â intermediate state of Kruskal's Algorithm, despite the fact that we've

Â haven't chosen an edge directly between them.

Â There's rather, a path of chosen edges between them.

Â It's exactly the same thing going on here.

Â Now, what we have going for us is that, if a pair of points, as discussed, was

Â directly merged, we know they're close. The distance between them is, at most,

Â this spacing, capital S. We really don't know anything, frankly,

Â about the distance between pairs of vertices that were not directly merged.

Â They just, sort of, accidentally wound up in a common cluster.

Â But this turns out to be good enough. This is actually sufficient to argue that

Â this competitor clustering with the c-hat has spacing no more than s?

Â No better than ours. Let's see why.

Â So given that P and Q are in a common greedy cluster it must mean there was a

Â path of direct mergers that forced them to be in the same cluster.

Â So let's let the intermediate points involved in that path denoted A1 of two

Â AL. So here's the part of the proof where we

Â basically reduce the tricky case to the easy case.

Â So we've got this pair of points, PQ. Now, remember, not, not only are they in

Â a common greedy cluster. But they're in different clusters in our

Â competitor in the c hats. So the point p is in some cluster.

Â Call it c hat i. And Q is in something else.

Â In particular, it's not in c hat i. Now, imagine you go on a hike.

Â You start at the point p, and you hike along this path.

Â You traverse these direct mergers toward q.

Â Now, you're starting inside c hat I, and you end up outside.

Â So at some point on your hike, you will traverse the boundary.

Â You will, for the first time, escape from c hat I, and wind up in some other

Â cluster. So that has to happen.

Â And let's call ai and ai+1 the consecutive pair of points at which you

Â go from inside this cluster to outside this cluster.

Â And now we're back in the easy case. Now we're dealing with a separated pair

Â that would directly merge by the greedy algorithm.

Â Remember that we set up this path to be a path of direct mergers so in particular,

Â AJ and AJ + one were direct mergers, therefore their distance is at most S.

Â And again, by virtue of being direct mergers, their distance is at most the

Â spacing of the greedy clustering and yet as a separated point by the C hats.

Â It's also an upper bound on the spacing of the C hats.

Â This means the spacing S of our greedy clustering is as good as the competitor.

Â Is the competitor was arbitrary or optimal.

Â That completes the proof.

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