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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Stanford University

292 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 2

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

- Tim RoughgardenProfessor

Computer Science

So, welcome to this optional sequence of videos on state of the art implementations

Â of the union to find data structure.

Â Now as advanced in optional material,

Â let me make a few comments before we get started.

Â So the first comment is I'm going to assume that you're in a particularly

Â motivated mood.

Â Now, no one's forcing you to take these algorithm courses, so

Â I'm always assuming that you're very motivated individual,

Â but that's going to hold even more true than usual in these advanced videos.

Â And the way that plays out is while I'm going to hold myself to the same exacting

Â standards of clarity that I usually do, I'm going to ask a little bit more of you,

Â the viewer.

Â I'm going to perhaps dot a few less Is, cross a few less Ts than I usually do.

Â So you may find yourself periodically needing to pause and

Â think through some of the details that I'm glossing over.

Â The second comment is I'm going to give is sort of short shrift to applications

Â of the union find data structure.

Â That's not because there aren't any, there's plenty of applications of this

Â data structure but for these videos we're really just going to immerse ourselves in

Â the beauty and the depth of ideas behind the design of the data structures and

Â especially the analysis of their performance.

Â The final comment is to keep in mind that this is some seriously

Â next level material.

Â It is totally normal that the first time you see this stuff, you find it confusing.

Â You find it difficult to understand.

Â So confusion should not discourage you.

Â It does not represent any intellectual failing on your part.

Â Rather, keep in mind, it represents an opportunity to get even smarter.

Â So, with those comments in mind, let's go ahead and

Â proceed to a different approach to the Union-find data structure via lazy unions.

Â So let's have one slide with a quick review of what we already know about

Â the Union-find data structure.

Â Recall we discussed this in the context of a fast implementation of

Â Kruskal's minimum spanning tree algorithm The raison d'etre of a union-find

Â data structure is to maintain a partition of a universe of objects.

Â So in the context of Kruskal's algorithm, the objects were the vertices, and

Â the groups we wanted to maintain were the connected components with respect to

Â the edges we committed ourselves to so far.

Â The data structure should support two operations.

Â No prizes for guessing the names of those two operations.

Â First is the find operation.

Â This is given an object from the universe return the name of that object's group.

Â So for example if X was in the middle of this square that I put

Â on the right representing the universe capital X we would return a C3.

Â The name of the group that contains X.

Â So in the context of Kruskal's algorithm, we use find operation to check for cycles.

Â How did we know whether adding a new edge would create a cycle with respect to

Â the edges we've already chosen?

Â Well, it would be if and

Â only if the end points of that edge were already in the same connected component.

Â That is, if and only if the two find operations returned the same answer.

Â In the Union Operation you're given out one object but you call them x and y,

Â and then the responsibility of the operation is to merge the groups

Â that contain x and y.

Â So for example in the picture y was in C4, x was as before in C3.

Â Then your job is to fuse the groups C3 and C4 into a single group.

Â In the context of Kruskal's algorithm,

Â we needed this operation when we added a new edge.

Â This fused together two of the existing connecting components into one.

Â So that was exactly the union operation.

Â So we all ready went through one implementation of union and find.

Â That was order to get a blazingly fast implementation of Kruskal's algorithm.

Â So let's just review how that works.

Â With each group, we associated a linked structure, so

Â each object had one pointer associated with it.

Â And the invariant was in a given group,

Â there was some representative leader object.

Â And everybody in that group pointed to the leader of that group.

Â So for example, on the right I'm showing you a group with three objects, x, y, and

Â z, and x would be the leader of this group.

Â All three of the objects point directly to x, and

Â that would just be the output of the find operation, the leader of the group.

Â We use that as the group name.

Â Now the part which is cool and

Â obvious about this approach is our find operations take constant time.

Â All you do is return the leader of the given object.

Â Now the tricky part was analyzing the cost of Union operations.

Â So the problem here is that to maintain the invariant that

Â every object of a group points to its leader.

Â When you fuse two groups, you have to update a bunch of the objects' leaders.

Â We'd only have one leader for the one new group.

Â The simple, but totally crucial, optimization that we discussed was

Â when two groups merge you update the leader pointers of all objects in

Â the smaller group to point to the leader of the bigger group.

Â That is the new fused group inherits the leader

Â from the bigger of its constituent parts.

Â If we do that optimization, it's still the case that a single union might take linear

Â time fade event time but a sequence of n unions takes only big o of n log in time.

Â And that's because each object endures at most a logarithmic number of leader

Â updates, cause every time it's leader pointer gets updated the population

Â of the group that it inhabits doubles.

Â So in this sequence of videos we are going to discuss a different approach to

Â implementing the union find data structure.

Â And let's just go all in and

Â try to get away with updating only a single pointer in each union.

Â All right, well how are we going to do that?

Â Well, I think if we look at a picture,

Â it'll be clear what the approach is going to be.

Â So let's look at a simple example.

Â There's just six objects in the world and currently they're in two groups.

Â One, two, and three are in one group with leader one.

Â Four, five, and six are in the second group with leader four.

Â So, with our previous implementation of union find and the two groups have equal

Â size, so we pick one arbitrarily to represent the new leader.

Â Let's say four is going to be the new leader.

Â And now we update objects one, two, and three so

Â that they point directly to their new leader, four.

Â And so the new idea is very simple let's just sort of as a short hand for

Â re-wiring all of one, two and three to point to four,

Â let's just re-wire one pointer to point to four and then it's understood that two and

Â three as descendants of one also now have the new leader four as well.

Â So as a result, we do again get a directed tree afterwards,

Â but we don't get one with just depth of one.

Â We don't get as shallow, pushy a tree.

Â We get a deeper one that has two levels below the root.

Â So, let me give you another way of thinking about this in terms of an array

Â representation.

Â And here I also want to make a point that why conceptually it's going

Â to be very useful to think of these union find data structure in terms of these

Â directed trees.

Â You're actually implementing this data structure this is not how you'd do it.

Â You wouldn't bother with actually pointers between the different objects.

Â You'd just have a simply array indexed by the nodes.

Â And in the entry corresponding to a node i you would just store the name

Â of i's parent.

Â For instance if we reframe this exact same example in an array representation

Â before we do the union what do we got?

Â Well objects one, two, and three all have parent one.

Â Objects four, five, and six all have parent four.

Â So in the array we'd have a one.

Â In the first three entries and a four in the second three entries.

Â So in the old solution we have to update the parent pointers of objects one two and

Â three, they're all going to point directly to four, so

Â that gives us an array entirely of fours.

Â Whereas in the new solution the only node whose parent

Â pointer gets updated is that of object one, it gets updated from itself to four.

Â And in a particular objects two and

Â three still point to object one not directly to object four.

Â So that's how union works so in simple example.

Â How does it work in general?

Â Well, in general, you're given two objects, each belongs to some group.

Â You can think of these two groups conceptually as directed trees.

Â If you follow all the parent corners, all parent corners lead to the root vertex.

Â We identify those root vertices with the leaders of these two groups.

Â Now when you have to merge the two trees together, how do you do it?

Â Well, you pick one of the groups and

Â you look at its roots currently it points to itself.

Â You can change its parent pointer to point to the other groups leader.

Â That is you install one of the two roots as a new child of the other trees root.

Â So I hope the pros and cons of this alternative approach to the union find

Â data structure are intuitively clear.

Â The win, the pro, comes in the union operation, which is now very simple.

Â Now you might be tempted to just say union takes constant time,

Â because all you do is update one pointer, It's actually not so simple, right?

Â Because remember you're given two objects, x and y, and

Â in general you're not linking x and y together.

Â They might be somewhere deep in a tree.

Â You're linking roots together.

Â So what is true is that the union operation reduces two

Â two indications of find.

Â You have to find x's root.

Â R sub 1, you have to y's root r sub 2,

Â and then you just do constant time linking either r1 to r 2 or vice versa.

Â Now the issue with this lazy union approach is that it's not at all obvious,

Â and in fact it's not going to be true, that find operation takes constant time.

Â Remember, previously when we actually did this hard work of updating all these

Â leader pointers every time we had a union it guaranteed that whenever there is

Â a find boom, we just look in the field, we return the leader pointer, we're done.

Â Here, parent pointers do not point directly to roots rather

Â you have to traverse in general a sequence of parent pointers from a given object x

Â to go all the way up to the root of the correspondence tree.

Â So because of these trade offs.

Â These pros and cons, it's really not obvious at all whether this

Â lazy union approach to the union find data structure is a good idea.

Â Whether it's going to have any pay offs.

Â So that's going to take really some quite subtle analysis which is the main

Â point of the lectures to come.

Â It's also going to take a couple of optimizations and

Â the next video will start with the first one.

Â You might be wondering, okay, so when you do a union and

Â you have to install one root under the other, how do you make that choice?

Â So the right answer is something called union by rank.

Â That's coming up next.

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