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

So, in this video, we're going to prove Tarjan's inverse Ackermann bound in the

Â performance of the union find data structure with union by rank and path

Â compression. Frankly, I hope you're kind of excited.

Â This is one of the crown jewels in the entire history of algorithms and data

Â structures. Let me remind you what the bound says.

Â You've got a union-find data structure, we're doing lazy unions, we're doing

Â union by rank, we're doing path compression.

Â And for an arbitrary sequence of m find a union operations. The total amount of

Â work you do over the sequence of operations is bounded above by m times

Â alpha of n, where n is the number of objects in the data structure and alpha

Â is the inverse Ackermann function that we defined in the previous video.

Â I want to give a shout out here to Dexter Kozen's beautiful book the Design and

Â Analysis of Algorithms. It is his analysis that I will be

Â following closely here. So, it's true that merely stating,

Â Tarjan's bound is non-trivial. We have this entire video defining the Ackermann

Â function and it's inverse so we could make sense of the statement.

Â That said, we're well positioned to approve Tarjan's bound.

Â In fact, the template of the proof is going to very much mirror what we already

Â did for the log star analysis by Hopcroft and Ullman.

Â So in that spirit, let's review what were the two basic workhorses, the two

Â building blocks that drove forward the Hopcroft-Ullman analysis.

Â So the first building block is the rank Lemma, which dates back all the way to

Â the videos pre-path compression. And remember the rank Lemma gives us

Â control on how many objects can exist with a given rank.

Â And in particular, that upper bound is decreasing exponentially with the rank,

Â it's n/2^r objects at most with the given rank r.

Â But, of course, we need a second building block to exploit the facts that we're

Â using the path compression optimization. So, how do we quantify the progress made

Â by path compression? Well, we argue that every time a node has

Â its parent pointer updated in path compression, it acquires a new parent

Â with strictly larger rank. So, you can think of the Hopcroft-Ullman

Â analysis as a clever and essentially optimal exploitation of these two

Â building blocks. How did we use them? Well, we define the

Â notion of rank blocks and each rank block has size to raise to the size of the

Â previous rank block. Why were rank blocks useful? Well, they

Â allowed us to measure progress as we followed parent pointers.

Â We defined an object to be good, if following it's parent, catapulted you

Â into a subsequent ranked block. Because there's only a log star number of

Â ranked blocks, you can only visit log star good nodes on any given find

Â operation. So, that gave us a per operation bound on

Â the log star for visits to good nodes. And then, for visits to bad nodes, we use

Â the second building block to argue that every time you visit a bad node, the rank

Â of its parent increases. And that can only happen so many times

Â before the rank of this object's parent is so much bigger than this object's rank

Â itself that the node has to be good. You have to make a lot of progress by

Â following its parent pointer. And the point is, if we want to do better

Â than the Hopcroft-Ullman analysis, we're not going to do better by taking these

Â same two building blocks and exploiting them in a better way, that can't be done.

Â So, what we need to do is have a better version of one of the two building

Â blocks. So, the key idea in Tarjan's analysis is

Â to have a stronger version of the second building block.

Â We're going to argue that path compression doesn't merely increase the

Â rank of nodes parents by one, but in fact, it increases typically the rank of

Â an object's parents by a lot. And what's kind of amazing is the link of

Â the proof we're about to do is really basically the same as that in the

Â Hopcroft-Ullman analysis. And the steps match up almost perfectly.

Â So, the bound is even better, the argument is even more ingenious.

Â It's even harder to imagine how one would come up with this idea oneself. But, as

Â far as checking it, as far as understanding it, the complexity level is

Â roughly the same as what we've already done in the log star analysis, so here we

Â go. So, in this slide, I'm going to give you

Â a definition. And the point of the definition is to

Â measure how much progress we make in terms of ranks when we follow a node's

Â parent pointer. So, the definition here is going to play exactly the same role

Â that the definition of rank blocks played in the Hopcroft-Ullman analysis.

Â So, what I'm going to define is a number, a statistic measuring the difference, the

Â gap, between the rank of an object and the rank of its parent. So, this

Â definition is only going to make sense for non-root objects x.

Â One thing I want you to remember from previous videos is that once an object is

Â not a root, it will never again be a root, and it's

Â rank will never again change. So, non-root objects have rank fixed

Â forevermore. So, we're going to use the notation delta

Â of x for the statistic. And the definition of delta of x is the largest

Â value of k such that a sub k, this is Ackermann function remember, such that a

Â sub k applied to the rank of this object x, is bounded above by the rank of x's

Â parent. So, the bigger the gap between the rank

Â of x's parent and x's rank, the bigger the value of delta of x.

Â So, let me mention, talk through some simple examples to make sure this makes

Â sense, and also review the Ackermann function a little bit.

Â So, first of all, let me just point out the delta of x is always non-negative for

Â any non-root object x. So, why is this true? Well remember, and

Â this goes all the way back to our union by rank discussion, the rank of an

Â object's parent is always at least one more than the rank of that object.

Â And the function a sub 0, recall we defined as the successor function.

Â So, for every object that's not a root, it's always true that it's parent has

Â rank at least one more than it. And that means we can always at least

Â take k to be zero and have the inequality be satisfied.

Â Now, when is it going to be true that an object x has a delta value at least equal

Â to one? Well, that is equivalent to stating we can take k to be 1 in this

Â inequality, and the inequality holds. So, let's remember what is the function a

Â sub one. In the previous video, we discovered that

Â was just the doubling function. So, we can take k=1 and have this

Â inequality satisfied if and only if the rank of the node's parent is at least

Â double the rank of that object. Similarly, an object x has a delta value

Â equal to at least 2 if and only if we can take k=2 on this right-hand side of the

Â definition and have this inequality be satisfied.

Â So, let's recall what the function a sub 2 is.

Â a sub 2 of r is just r*2^r. So, an object x has delta value equal to

Â 2 if and only if its parent's rank is substantially bigger than its own rank,

Â in the sense that if it has rank r, its parent's ranks has to be at least r*2^r.

Â And in general, the bigger the value of delta, the bigger the gap between the

Â rank of the object x and the rank of its parent.

Â And, of course, because the Ackermann function grows so ridiculously quickly,

Â the gap between this object's rank and its parent's rank is growing massively as

Â you take delta to be larger and larger. So, now that we understand this better,

Â one thing that should be clear is that the delta value of an object x can only

Â go up over time, right? So remember, we're talking only

Â about non-root objects x, so the x's rank itself, is frozen forevermore.

Â The only thing that's changing is it's acquiring potentially new parents,

Â through path compression, and each new parents rank is bigger than the previous

Â one. So, the gap in the ranks is only going

Â up, and therefore the delta value of this object x can only go up, over time.

Â So one final comment, you recall on the Hopcroft-Ullman analysis, the way we

Â defined rank blocks ensured that there weren't too many of them, there were only

Â log star. Similarly here, the way we're defining

Â the statistic delta, guarantees that there are not too many distinct delta

Â values that objects can take on. So, we've already argued that delta is a,

Â integer that's at least zero, and we can also bound it from above.

Â At least, for any object x that has a rank of at least 2, which will be good

Â enough for our purposes. If an object x has rank at least 2, its

Â value of delta can not be any larger than the inverse acromyn of n, than alpha of

Â n. And this is really just by the way we

Â define the inverse Ackermann function alpha, we defined it as the smallest

Â value of k such that if you apply a sub k to 2, that catapult to beyond end.

Â And since ranks of objects are bounded above by n, actually they're even bounded

Â above by log n but in particular by n that means you will never see a delta

Â value bigger that alpha of n for any object x has the rank at least two.

Â So, that completes the definition of the statistic data of x.

Â And, once again, the point of this statistic is to quantify progress through

Â rank space as we traverse a parent pointer, from an object.

Â So, in the Hopcroft-Ullman analysis, we used rank blocks for that purpose.

Â Here, we're using the statistic delta of x.

Â And now, we have to redefine good and bad objects to reflect this different

Â measurement for progress, for gaps between ranks of objects, and ranks of

Â their parents. So, that's what I'm going to provide for

Â you on this slide, the definition of good and bad nodes.

Â And, this definition will play exactly the same role as it did in the

Â Hopcroft-Ullman analysis. For the good nodes, we'll be able to just

Â get a per operation bound for how many visits we make to them. And then, we'll

Â be stuck with, and this will be the main part of the proof,

Â a global analysis arguing that the total number of visits to bad nodes over a

Â sequence of operations cannot be too big. So, the bad objects will be those that

Â meet the following set of four criteria. If an object fails to meet any of these

Â four criteria, it will be called good. So, the first two 2 criteria will be

Â familiar from the Hopcroft-Ullman analysis.

Â So, first of all, roots get a pass, and direct descendants of roots also get a

Â pass. So, to be bad, it must be the case you're

Â neither a root nor are you the child of a root.

Â So, what's the motivation for these two criteria? Well, it's exactly the same as

Â it was in the Hopcroft-Ullman analysis. It's just to ensure that bad nodes, after

Â they're visited on a find, get their parent pointer updated.

Â So remember, when you do path compression, after you've done a find

Â from an object x all the way up to its corresponding root, everybody's parent

Â pointer gets updated except for the root and except for the child of the root.

Â Those two nodes keep the same parent pointer.

Â So, to make sure we have progress, we want to exclude the root and the directed

Â sentence of the root from being bad, we'll acount for them separately.

Â The third criterion is not conceptually important, its just for technical

Â convenience sort of an out of factor the way I've defined the inverse arithmetic

Â function. We're going to give nodes that have rank

Â zero or one also with free paths. So, in order to bad, we're going to worry

Â about the case where the rank is at least 2.

Â So, the final criterion is really the ingenious one and it's the one that

Â ensures that when you do path compression typically objects parents ranks will

Â increase extremely rapidly, not just by 1.

Â And this criterion says, for an object x to be bad, it has to be the case that is

Â has some ancestor y. So, an object you get to by traversing

Â parent pointers from x, that has exactly the same value of delta.

Â So, if an object x has delta of x equal to 2, for it to be bad, there has to be

Â some ancestor object y that also has a delta value equal to 2.

Â And again, if x fails to meet any of these four conditions, then x gets a

Â pass, x is called a good object.

Â Now, I said that the role of this definition is going to be to split the

Â work done into two groups and that we're going to be able to easily bound, even on

Â a per operation basis, the number of visits to a good node.

Â So, the next quiz asks you to think about that carefully.

Â Alright. So, the correct answer is B. It's theta of the inverse Ackermann

Â function of n. And this is really one of the points, of

Â how we define good and bad nodes. We guarantee that there can not be too

Â many good nodes whenever we do a fine operation.

Â So, to see why this is true, let's just count up the number of nodes along a path

Â that can fail to meet each of the four criteria.

Â Well, so the first criterion was we give roots a free pass.

Â So there's at most one node, that's good because it's a root.

Â Similarly, on a given path there's at most one object on that path, it's a

Â direct descendant of the root. Because ranks always strictly go up when

Â you follow parent pointers that can be at most one object of rank zero and at most

Â one object of rank one. So, we, and most two objects get a free

Â pascals of the third criterion. Well, such thing about the interesting

Â case nodes that good be because they failed to meet the fourth criterion.

Â How many can that be? Well, let's focus on a particular value of delta.

Â So, let's say, think about all of the objects along a path that have delta of x

Â equal to 3. There may well be lots of objects on this

Â path with delta of x equal to 3, may be there's like 17 of them.

Â Call the highest to the closest to the root, such object z. Of these 17 objects

Â with delta of x equal to 3, z is going be the only one that is good.

Â The other 16 objects with delta of x equal to 3 are going to be bad.

Â Why? Well, any object other than z has an ancestor, namely z, with exactly the same

Â delta value. So that, those 16 objects do meet the

Â fourth criterion they will be classified as bad.

Â Summarizing for each of the alpha of n possible values, of delta, only the

Â upper-most object, the object closest to the root with that particular delta

Â value, can possibly be classified as good.

Â So, that bounds at alpha of n, the number of objects that fail to meet this fourth

Â criterion. That gives us the total bound of the

Â number of good nodes on any path.

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