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 3

Huffman codes; introduction to dynamic programming.

- Tim RoughgardenProfessor

Computer Science

So in this video we'll finally discuss Huffman's Algorithm which is a greedy

Â Algorithm that constructs the prefix free binary code minimizing the average

Â encoding length. So, let me just briefly remind you the

Â formal statement of the compositional problem that I left you with last time.

Â So, the input is just a frequency for each symbol I, coming from some alphabet

Â sigma. So the responsibility of the algorithm is

Â to get optimal compression, so to compute an optimal code.

Â So the code has to be binary. We have to use only zeros and ones.

Â We have to be prefix free meaning the encodings of any two characters, neither

Â one can be a prefix of the other, that's to facilitate unambiguous

Â decoding. And finally, the average number of bits

Â needed to encode a character, where the average is with respect to the input

Â frequencies, should be as small as possible.

Â Now remember these kinds of codes correspond to binary trees.

Â The prefix free condition just says that the symbols of sigma should be in one to

Â one correspondence with the leaves of the tree.

Â And finally remember the encoding lengths of the various symbols just correspond to

Â depths of the corresponding leaves. So we can formally express the averaging

Â coding length. Which given a legal tree capital T.

Â I'm using the notation capital L of T. So what do we do?

Â We just take the average over the symbols of the alphabet weighted by the provided

Â frequencies and we just look at the number of bits used to encode that

Â symbol. Equivalently depth of the corresponding

Â leaf in the tree T. So we want the tree that makes this

Â number as small as possible. So this task is a little bit different

Â than any we've seen so far in this course, right?

Â All we're given as an input is an array of numbers and yet we have to produce

Â this actual full blown tree. So how can we just take a bunch of

Â numbers and produce them in a sensible, principled way?

Â Some kind of tree whose leaves correspond to symbols of the alphabet.

Â So let's spend a slide just thinking about, you know, at a high level where

Â would this tree come from, how would you build it up given this unstructured

Â input. So there's certainly no unique answer to

Â this question and one idea which is very natural but that turns out to be sub

Â optimal is to build a tree using a top down approach, which you can also think

Â of as an. Initiation of the divide and conquer

Â algorithm design paradigm. The divide and conquer paradigm, you'll

Â recall, involves breaking the given sub-problem into usually multiple,

Â smaller sub-problems. They're solved recursively, and the

Â solutions are then combined into one for the original problem.

Â Because trees, the desired output here, have a recursive substructure, it's

Â natural to think about applying this paradigm to this problem.

Â Specifically you love to just lean on recursive call to construct for you the

Â left sub tree, another sub call constructing the right

Â sub tree. And then you can fuse the results

Â together under a common root vertex. So it's not clear how to do this

Â partitioning of the symbols into two groups, but one idea to get the most bang

Â for your buck, the most information out of the first bit of your encoding. You

Â might want to split them in, the symbols, into groups that have roughly, as close

Â to as possible, of 50% of the overall frequency.

Â So this topped out approach is sometimes called Fanno-Shannon coding.

Â Fanno was Huffman's graduate adviser. Shannon is the Claude Shannon inventor of

Â information theory. But Huffman in a term paper believe it or

Â not, realized the topped down is not the way to go.

Â The way to go is to build the tree from the bottom up.

Â Not only are we going to get optimal codes, but we're going to get the

Â blazingly fast greedy algorithm that constructs them.

Â So what do I mean by bottom up? I mean we're going to start with just a

Â bunch of nodes, each one labelled with one symbol of the alphabet.

Â So, in effect, we're starting with the leaves of our tree.

Â And then we're going to do successive mergers.

Â We're going to take at each step two sub-trees thus far and link them together

Â as sub-trees under a common internal node.

Â So, I think you'll see what I mean in a picture.

Â So imagine we want to build a tree where the leaves are supposed to be just A, B,

Â C, and D. So one merger would be oh, well let's

Â just take the leaves C and D and link them as siblings under a common ancestor.

Â Now in the second step let's merge the leaf B with the sub-tree we got from the

Â previous merge, the sub-tree comprising the nodes C, D, and their common

Â ancestor. So now of course we have no choice but to

Â merge the only two sub-trees we have left,

Â and then that gives us a single tree. Which is in fact exactly the same

Â lopsided tree we were using in the previous video as a running example.

Â So let me explain what I hope is clear and what is maybe unclear at this

Â juncture. I hope what's intuitively clear is that

Â the bottom of approaches, a systematic way to build trees that have a prescribed

Â set of leaves. So what do we want?

Â We want trees whose leaves are labeled with the symbols of the alphabet sigma.

Â So if we have an alphabet with n symbols, we're going to start with just the N

Â leaves. What does a merger do?

Â Well, there's two things. First of all, it introduces a new

Â internal node, an unlabelled node. And secondly, it takes two of our old

Â subtrees and fuses them into one, merges them into one.

Â We take two subtrees, we make one the left child of this new internal node, the

Â other, the right child of this new internal node.

Â So that drops the number of subtrees we're working with by one.

Â So if we start with the N leaves and we do N minus one successive mergers, what

Â happens? Well on the one hand, we introduce N

Â minus one new unlabeled internal nodes. And on the other hand, we construct a

Â single tree. And the leaves of this tree that we get

Â are in one to one correspondence with the alphabet letters as desired.

Â Now what I don't expect you to have an intuition for is what should we be

Â merging with what and why? Forget about, you know, how do we get an

Â optimal tree at the end of the day. I mean, even just if we wanted to design

Â a greedy algorithm. If we just wanted to make a myopic

Â decision, that looks good right now, how would we even do that?

Â What's our greedy criteria that's going to guide us to merge a particular pair of

Â trees together? So we can re-frame this quandary in the

Â same kind of question we asked for minimum cost spanning trees and really

Â more generally with greedy algorithms. When you're making irrevocable decisions

Â which strikes fear in your heart, is that this decision will come back and haunt

Â you later on. You'll only realize at the end of the

Â algorithm that you made some horrible mistake early on in the algorithm.

Â So just as for MST's, we ask, you know, when can we be sure that including an

Â edge irrevocably is not a mistake, it's safe in the tree that we're building?

Â Here we want to ask, you know, we have to do a merger, we want to do successive

Â mergers and how do we know that a merger is safe?

Â That it doesn't prevent us from eventually computing an optimum solution.

Â Well here's the way to look at things that will at least give us an intuitive

Â conjecture for this question. We'll save the proof for the next video.

Â So what are the ramifications when we merge two subtrees, each containing a

Â collection of symbols? Well, when we merge two subtrees, we

Â introduce a new internal node which unites these two subtrees under them, and

Â if you think about it, at the end of the day on the final tree, this is yet

Â another internal node that's going to be on the root to leaf path,

Â for all of the leaves in these two sub trees.

Â That is, if you're a symbol and you're watching your subtree get merged with

Â somebody else. You're bummed out, your like, man that's

Â another bit in my encoding. That's yet one more node I have to pass

Â through to get back to the root. I think this will become even more clear

Â if we look at an example. So naturally, we'll use our four symbol

Â alphabet A, B, C, D. And initially, before we've merged

Â anything, each of these is just its own leaf A, B.

Â C, D. So there's no internal nodes above 'em.

Â In the sense, everybody's encoding length at the beginning is zero bits.

Â So now, imagine we've merged C and D, we introduce a new internal node, which is

Â the common an-ancestor of these two leaves.

Â And as a result, C and D are bummed out. They said, well, there's one bit that

Â we're going to, to have to incur our encoding length, there's one new internal

Â node we're always going to have to pass through enroute back to the root of the

Â eventual tree. Now suppose next we merge B with the

Â subtree containing both C and D. Well everybody but A is bummed out about,

Â about this merger because we introduce another internal node, and it contributes

Â one bit to the encoding of each of B, C, and D.

Â It contributes an extra one to the encoding of C and D, and it contributes a

Â zero to the encoding of B. So all of their encodings in some sense

Â inc, get incremented, go up by one, compared to how things were before.

Â Now in the final iteration, we have to merge everything together.

Â We have no choice, there's only two sub-trees left.

Â So here, everybody's encoding length gets bumped up by one.

Â So, A finally picks up a bit at zero to encode it and B, C, and D each pick up an

Â additional bit of one, which is prepenned into their encodings thus far.

Â And, in general what you'll notice is that the final encoding length of a

Â symbol, is precisely the number of mergers that its subtree has to endure.

Â Every time your subtree gets merged with somebody else you pick up another bit in

Â your encoding, and that's because there's one more internal node that you're going

Â to have to traverse enroute to the root of the final tree.

Â In this example, the symbols C and D, well they got merged with somebody else

Â in each of the three iterations. So, they're the two symbols that wind up

Â with the encoding length of three. The symbol B, it didn't get merged with

Â anybody in the first iteration, only the second two.

Â That's why it has an encoding length of two.

Â So this is really helpful. So this, lets us relate, the operations

Â that our algorithm actually does, namely mergers back to the objective function

Â that we care about, namely the average encoding length.

Â Mergers increase the encoding lengths of the participating symbols by one.

Â So this allows us to segway into a design of a greedy heuristic for how to do

Â mergers. Let's just think about the very first

Â iteration. So we have our N original symbols, and we

Â have to pick two to merge. And remember the consequences of a merge

Â is going to be an increase in the encoding length by one bit, whichever two

Â symbols we're going to pick. Now we want to do is minimize the average

Â encoding length with respect to the frequencies that were given.

Â So which pair of symbols are we the least unhappy to suffer an increment to their

Â encoding length, was going to be the symbols that are the least frequent.

Â That's going to increase your averaging poding length by the least.

Â So that's the green merging hiuristic. Somethings gotta increase.

Â Pick the ones that are the least frequent to be the ones that get incremented.

Â So that seems like a really good idea of how to proceed in the first iteration.

Â The next question is, how are we going to recurse?

Â So, I'll let you think about that in the following quiz.

Â So let's agree that the first iteration of our greedy heuristic is going to merge

Â together the two symbols that possess the lowest frequencies.

Â Let's call those two symbols little A and little B.

Â The question is then how do we make further progress?

Â What do we do next? Well, one thing that would be really nice

Â is if we could somehow recurse on a smaller subproblem.

Â Well, which smaller subproblem? Well, what does it mean that we've merged

Â together the symbols A and B? Well, If you think about it, in the tree

Â that we finally construct by virtue of us merging A and B, we're forcing the

Â algorithm to output a tree in which A and B are siblings, in which they have

Â exactly the same parent. So what does it mean for the encoding

Â that we compute that A and B are going to be siblings with the same parent?

Â It means that their encodings are going to be identical,

Â save to the lowest order bits. So A will get encoded with a bunch of

Â bits followed by a zero, B will be encoded by exactly the same prefix of

Â bits followed by A1. So they're going to have almost of the

Â same encoding, so for our recursion let's just treat them as the same symbol.

Â So let's introduce a new meta symbol, let's call it AB, which represents the

Â conjunction of A and B. So it's meant to represent all of the

Â frequencies of either one, all of the occurrences of either one of them.

Â But remember the input to the computational problem that we're studying

Â is not just an alphabet, but also frequencies of each of these symbols of

Â that alphabet. So my question for you is when we

Â introduce this new meta symbol A B. What should be the frequency that we

Â define for this meta symbol? All right, so hopefully your intuition

Â suggested answer C, that for this recursion to make sense, for it to

Â conform to our semantics of what this merging does.

Â We should define the frequency of this new meta symbol to be the sum of the

Â frequencies of the two symbols that it's replacing.

Â That's because, remember this meta symbol AB is meant to represent all occurrences

Â of both A and B. So it should be the sum of their

Â frequencies. So I'm now ready to formally describe

Â Huffman's greedy algorithm. Let me first describe it on an example

Â and then I think of the general code will be, self evident.

Â So let's just use our usual example, so we're going to have letters A, B, C, D,

Â with frequencies 60, 25, 10, and 5. So we're going to use the bottom up

Â approach, so we begin with each symbol just as its own node, A, B, C, D.

Â I'm annotating in red the frequencies 60, 25, 10, 5.

Â The greedy heuristic says initially we should merge together the two nodes that

Â have the smallest frequencies. So that would be the C and the D with

Â their frequencies of 10 and 5. Now based on the idea of the last slide

Â when we merged C and D, we replaced them with a meta symbol cd whose frequency is

Â the sum of the frequencies of C and D, namely fifteen.

Â Now we just run another iteration of the greedy algorithm meaning we merge

Â together the two nodes, the two symbols that have the smallest frequencies.

Â So this is now B 25, and CD 15. So now we're down to just two symbols,

Â the original symbol A, which still has frequencies 60 and the in some sense,

Â meta, meta symbol BC, D who is now cumulative frequency is 40.

Â So when we're down to two nodes, that's going to be the point in Huffman's

Â algorithm where we hit the base case and the recursion begins to unwind.

Â So if you just have two symbols to encode there's pretty much only, one sensible

Â way to do it. One is a zero, and one is a one.

Â So at this point, the final recursive call just returns now a tree with two

Â leaves corresponding to the symbols A and BCD.

Â And now is the recursion on lines we in effect undo the mergers.

Â So for each merge we do a corresponding split of the meta node and we replace

Â that with an internal node. And then two children corresponding to

Â the symbols that were merged to make that meta node.

Â So for example, when we want to undo the merge of the B and the CD, we take the

Â node labeled BCD and we split it into two.

Â The left, left child being B, the right child being CD.

Â So the original outer most call, what it gets from its rehersive call is this tree

Â with three nodes and it has to undue it's merger, it merged C and D.

Â So it takes the leaf labeled with symbol CD and splits it into two.

Â It replaces it with a new unlabeled internal node, left child C, right child

Â D. So how does the algorithm work in

Â general? Well just as you'd expect given the discussion on this concrete example.

Â I'm going to take as the base case when the alphabet has just two symbols, in

Â that case the only thing to do is encode one with a zero, the other with a one.

Â Otherwise, we take the two symbols of the alphabet that have the smallest

Â frequencies. You can break ties arbitrarily, it's not

Â going to matter, it's going to be optimal either way.

Â We then replace these two low frequency symbols A and B with meta symbol AB,

Â intended to represent both of them in some sense.

Â As we just discussed with those semantics, we should be defining the

Â frequency of the meta symbol AB as the sum of the frequencies of the symbols

Â that it comprises. We now have a well defined, smaller sub

Â problem. It has one fewer symbol than the one we

Â were given. So, we can recursively compute a solution

Â for it. So, what the recursive call returns is a

Â tree who's leaves are in one to one correspondence with sigma prime.

Â That is, it does not have a leaf labeled A, it does not have a leaf labeled B.

Â It does have a leaf labeled AB, so we want to extend to this tree T prime

Â to be one who's leaves correspond to all of sigma.

Â And the obvious way to do that is to split the leaf labeled AB, replace that

Â with a new internal unlabeled node with two children which are labeled A and B.

Â The resulting tree capital T with leaves and correspondents to the original

Â alphabet sigma is then the final output of Huffman's Algorithm.

Â As always, with a greedy algorithm we may have intuition, it may seem like a good

Â idea. But we can't be sure without a rigorous

Â argument. That is the subject of the next video.

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