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 just like in the Hopcroft Ullman analysis we're going to think about the

Â total amount of work done by our union fine data structure as constituting 2

Â pieces, work done visiting good nodes and work done visiting bad nodes.

Â The previous quiz says that we have a bound on visits to good nodes, even on a

Â per operation basis. At most inverse acronym for N, good nodes

Â are visited every single operation. What remains is to have a global bound on

Â the number of visits to bad nodes. The argument will be to show, that over

Â an arbitrary sequence of m find and union operations, the total number of visits to

Â bad nodes is bounded above by big O of n, times, alpha of n.

Â So here is the crux of the argument. Here is why, when you do a visit to a bad

Â node, the subsequent path compression massively increases the gap between that

Â object's rank and the rank of the parent. So, let's freeze the data structure at

Â the moment where we found the operation and makes a visit to a bad object,

Â call that Bad Object X. Let's think about what the world with the

Â data structure must look like in this scenario.

Â So we've got our bad object X, it may or may not have nodes pointing to

Â it. We're not going to care.

Â By virtue of it being bad and meeting the third criterion we know the rank of X is

Â at least 2. It's delta value could be anything.

Â Whatever it is, let's call it. Okay.

Â So because x is not a root, it has some parent, call that parent p.

Â And not only does x have ancestors, but because it meets the fourth criterion, we

Â actually know it has an ancestor y who has the same delta value as x.

Â That is it has an ancestor y With delta of y equal to k.

Â It is possible that p and y are the same object ir they could be different.

Â It's not going to matter in the analysis. So we're calling that the statistic delta

Â is only definied for non roots, we can conclude that y is also not the

Â root. It must then have some parent p prime.

Â P prime might be a root or it might not, we don't care.

Â [SOUND] So, now let's understand of the effect of the past compression, which is

Â going to happen subsequent to this find operation.

Â X's parent points is going to be rewired to the root of this tree.

Â The root of this tree is either at P prime, or even higher than that.

Â Given that fact let's understand how much bigger the rank of X's new parent P

Â primer higher is compared to the rank of it's old parent P.

Â So the rank of x's new parent is at least the rank of P prime.

Â So if p' is in fact the root, then the rank of x's new parent is just the rank

Â of p'. Otherwise, this new parnet is even higher

Â than P prime because its ranks only increase going up the tree, that means it

Â would only be higher than the rank of P prime.

Â How does the rank of P prime compare to its child, that of its child y? Well and

Â here is a key point, because delta of y is equal to k.

Â Remember what the definition of delta is, it quantifies the gap between the rank of

Â an object, and that of its parent. We are going to use that here for y, and

Â its parent P prime. It means the rank of P prime is so big

Â It's at least the function, a sub k applied to y's rank.

Â So our third and final inequality is the same as the first one.

Â So it could be the case that y actually is the same as p, it actually is x's old

Â parent. In that case the rank of x's old parent

Â is precisely the rank of y. Otherwise, y is even higher up X's old

Â parent P and therefore, by the monotonicity of rank, the rank of Y is

Â even bigger, than the rank of X's old parent.

Â So now in this chain of inequalities, I want you to focus on the left-most

Â quantity, and the right-most quantity. What does this say, at the end of the

Â day? It says when you apply path compression to X, it acquires a new

Â parent. And the rank of that new parent is at

Â least the A sub K function applied to the rank of it's old parent.

Â Again, path compression at the very least applies the A sub K function to the rank

Â of X's parent. So now let's suppose you visit some bad

Â object X, not just once, but the same object X, while it's bad over and over

Â and over again. This argument says that every such visit

Â to the object x while it was bad applies the function A sub k to the rank of its

Â parent. So in particular, let's use r to denote

Â the rank of this bad object x and, again, by virtue of it being bad, r has to be at

Â least 2. Imagine we do a sequence of R visits to

Â this object X while it's bad. Each of those visits will result in it

Â acquiring a new parent. And that new parents rank is much bigger

Â than the rank of the previous parent. It's at least A sub K, applied to the

Â rank of the previous parent. So, after R visits,

Â that's applying the function A sub K, R times to the rank of X's original parent

Â which forces rank at least that of X, at least R.

Â Well we have another name for applying the function A sub K, R times to R.

Â Remember this is just by definition of the acronym function A sub K plus 1.

Â Applied to r. So what does this mean? Well this means,

Â that after r visits, to a bad object x, that has rank r, the rank of x's parent

Â has to have grown, so much, that that growth is reflected in our statistic

Â delta, that measures the gap in the rank, between objects, and the objects parent.

Â So remember how we defined delta of x. It's the largest value of k so that the

Â rank of x's parent is even bigger than a sub k applied to the rank of x.

Â So what this inequality shows is that every r Parent pointer updates to x allow

Â you to take k to be even bigger than before.

Â You can bump up k by 1 and still have this inequality be satisfied.

Â That is, every r visits to a bad object of rank r, delta has to increase by at

Â least 1. But there's only so many distinct values

Â of the statistic delta of X can take on. It's always non negative, it's always an

Â integer, it can only increase, and it's never bigger than the inverse Ackermann

Â function alpha of N. So therefore, the total number of visits

Â to an object X, while it's bad, over an arbitrary sequence, can not be more than

Â R, the number of visits needed to increment delta of X,

Â times the number of distinct values, which is alpha of N, the inverse

Â function. So now we've done all the hard work.

Â We're almost there, we just need 1 final slide putting it all together.

Â All right, so to see how all of this comes together, let's first recall that

Â all that we need to do is bound the total number of visits to bad objects over this

Â entire sequence of union and find operation.

Â So in the previous slide, we showed that for a bad object x, with rank r, the

Â total number of times it's visited while it's bad, is bounded above by r times the

Â inverse Ackermann function of n. So lets sum that over all of the objects

Â to get our initial bound on the total number of visits to bad objects.

Â So now, we have to be a little careful here, because there are n objects x.

Â and ranks can be as large as log n. So a naive-bound would give us something

Â like n times LogN times alpha of n, and that's not what we want.

Â We really want n times alpha of n. So we need to use the fact that not all

Â big nodes can have big rank. And that's exactly what the rank Lemma

Â says. So to rewrite this, in a way that we can

Â apply the rank Lemma, bounding a number of objects with a given rank, lets bucket

Â the objects according to their rank. So rather than summing over the objects,

Â we're going to sum over possible ranks r, and then we're going to multiply times

Â the number of objects that have that rank R.

Â While I'm at it, let me pull the alpha of n factor, which doesn't depend on the

Â object out in front of the sum. For every value of R, the rank Lemma

Â tells us there are at most N/(2^R) objects with that rank.

Â So this factor of N is independent of the rank R, so let me pull that out in front

Â of the sum. And when the dust settles we get n times

Â the inverse acronym function of n times the sum of non-negative integer r of r

Â divided by 2 to the r. So, we've seen this kind of sum without

Â the r in the numerator, just a geometric sum 1/2^r.

Â We know that's bounded by a constant, and more generally throwing an r in the

Â numerator, well that's no match for this exponentially decreasing denominator.

Â So, this sum still evaluates to, a universal constant.

Â I'll let you check that in the privacy of your own home.

Â And so that's, give us a bound of O(n) * the inverse Ackermann function of n, on

Â the total number of visits to bad objects, since we also have a per

Â operation bound of alpha(n) on the number of visits to good nodes.

Â Combining the work of the two cases we get O(n) m+n times inverse Ackermann

Â function of n. And again, the interesting case here is

Â when m is omega of n. Otherwise, you can just do this analysis

Â separately for each 3 in the data structure.

Â And, there you have it. You now understand in full detail one of

Â the most amazing results in the history of algorithms and data structures.

Â So, Tarjans bound is unimagenably close to being linear without actually being

Â linear, it's off by this inverse Acromon function factor.

Â Now, from a practical perspective for any imagenable value of N, alpha of N is

Â almost 4, so this gives you a linear time bound for Imaginable values of n.

Â In fact, even the Hopcroft-Ullman log starbound, logs stars at most 5, for any

Â imaginable value of n. So that also is in, in essence linear

Â time bound for all practical purposes. But theorotically speaking, we have not

Â proven a linear time bound, on the amount of work done by the union find Data

Â structures. So the Pavlovian response, of any

Â algorithms researcher worth their salt, would be to ask, can we do better? And,

Â we could ask that question in a couple different senses.

Â The first sense would be; well maybe we could have a sharper analysis, of this

Â exact data structure. Maybe, union by rank, and path

Â compression is sufficient To gaurantee a linear amount of work.

Â After all, we didn't need to change the data structure, we only needed to change

Â the analysis, to sharpen the log* bound, to an alpha of n bound.

Â So this question, remarkably, Tarjan answered in the negative, in his original

Â paper. He showed that, if you use this exact

Â data structure, union by rank and path compression, then they are arbitrarily

Â large Sequences of unions and finds of arbitrarily large collections of objects,

Â so that this data structure actually performs asymptotically, M the number of

Â operations times the inverse Ackermann function, N the number of objects amount

Â of work. That is keeping the data structure fixed,

Â this analysis cannot be asymptotically improved.

Â This data structure fundamentally has Worst case performance, governed by this

Â insane inverse Ackermann function. So this is already a mind boggling fact

Â that indeed Tarjan in the conclusion of his paper, notes that it's remarkable to

Â have such a simple algorithm with such a complicated running time.

Â But you could also ask the question, could we do better perhaps by improving

Â or changing the data structure. After all, by adding path compression, we

Â got a qualitative improvement in the average operation time.

Â That dropped from log to alpha of n. Perhaps there's yet another optimization

Â out there waiting to be discovered that would drop the amount of work per

Â operation over a sequence down to constant time per operation, linear work

Â overall. Tarjan, in his paper made the bold

Â conjecture that there is no linear time method, no matter how clever you are.

Â And remarkably, that conjecture has since been proved.

Â It was proved for certain classes of data structures both in Tarjan in his original

Â paper and in And a follow-up paper in '79.

Â But the final version, of this conjecture, was proven by Friedman and

Â Sachs, back in 1989. No matter how clever you are, no matter

Â what union-find data structure you come up with, there will be arbitrarily long

Â sequences of operations, so that the average amount of work you do per

Â operation Is, the inverse Ackerman function of, the number of objects, in

Â the data structure. Pretty unbelivable,

Â so let me just wrap up with a historical comment.

Â So, it's a full disclosure, I wasn't quite alive when this result came out

Â but, in talking to, in reading about it, and talking to senior researchers about,

Â my sense is that it was really sort of a water-shed moment, for the field of data

Â structures and And algorithms. Specifically it confirmed the belief of,

Â people working in algorithms and data structures, that the field posessed

Â surprising depth, and beauty. There had, of course, been earlier

Â glimpses of this, we mentioned in the optional, material in part 1, Kanuth's

Â analysis of linear probing, back in the early 1960's.

Â But this was really something for the worst case analysis, of algorithms and

Â data structures. So the fact that such a practical and

Â naturally arising problem, in algorithms and data structures, requires,

Â necessarily, the understanding of the Ackermann function and it's inverse.

Â A function, mind you, which was first Proposed and defined back in 1920s in the

Â service of recursion theory, almost 10 years before Tarjan was doing his work on

Â models of computation. It was a real eye opener and it showed

Â that this field is something that will keep many generations of scientists quite

Â busy for many years to come.

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