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  • MARTEN VAN DIJK: So today, we're going

  • to talk about communication networks.

  • Communication networks is a great application

  • of graph theory.

  • So what we're going to study is, how

  • do you route packets through networks?

  • So you have the internet, which is a chaotic network.

  • It's not organized.

  • We are interested in highly structured networks

  • and you can find them, for example,

  • in parallel computers, where you want to route the data flow.

  • You can find them in certain telephone switches networks

  • and so on.

  • So we are going to talk about a few very special ones,

  • binary trees, and then slowly we will figure out

  • what all these performance measures really mean.

  • This one has to do with latency.

  • We have switches, their size, the number of them,

  • congestion, and then we will slowly get down

  • to Benes network, which is a really beautiful network

  • with beautiful parameters.

  • And we are going to prove those.

  • So let's start off with the first one, the complete binary

  • tree, and let me draw it for you.

  • In this network, we will have a root

  • and let me just draw it first We have vertices that

  • represent here the switches.

  • So these circles-- let me explain it over here-- actually

  • represent a switch.

  • And the idea is that these actually direct packets

  • through the network.

  • And these packets are fixed-size packets

  • of data, so like, I don't know, say 4,000 bytes or bits

  • or whatever the network wants you to comply to.

  • So these are fixed-size pieces of data.

  • So what we want is we want to be able from every terminal--

  • and the terminal I will denote by a square--

  • from every terminal, I want to be

  • able to reach any other terminal.

  • So what is a terminal?

  • A terminal is like a computer or something like that.

  • It's actually the source and the destination of data.

  • So what we are looking for is how can we

  • route-- how we can find a network of switches

  • that are connected through wires, fibers, or-- yeah?

  • What's the question?

  • AUDIENCE: Can you move down a bit

  • the top of the-- how it's getting cut off?

  • No, the--

  • That one.

  • MARTEN VAN DIJK: Oh, sorry.

  • AUDIENCE: All right.

  • Thank you.

  • MARTEN VAN DIJK: So what we want is

  • we want to route packets that's come from any terminal

  • to any other terminal.

  • That is what our goal is and we want to make sure

  • that that is efficient.

  • So the first one is this binary tree.

  • And let's see how this may work.

  • We may have switches that actually have

  • inputs coming from terminals.

  • And the switches may also output to terminals,

  • so here at the bottom.

  • At this site, we have a similar structure.

  • this is the root of the tree.

  • We have another switch over here.

  • We go down, we go up here, and once more, like this.

  • And again, we have-- oops.

  • We have input coming in or an output coming out

  • to their respective terminals.

  • So what is happening here is that I

  • would like to have an input-- say input zero wants

  • to travel all the way over to say,

  • the output that is present over here.

  • So let me label these.

  • So we have the output zero, input one, output one,

  • input two and output two, input three, and output four.

  • So well, I can definitely reach every single output

  • from any input so that's great.

  • So this looks like something that you are familiar with,

  • right?

  • It's just a tree.

  • It's a directed graph, but these edges

  • go in both directions, right?

  • So I have an edge that goes from here to here and back from here

  • to here.

  • So this is the kind of layout that you could try out

  • first to see whether this type of network

  • would lead to good performance.

  • So let's have a look at the different parameters

  • and see how well this behaves.

  • So here, we have a few parameters

  • that we will be talking about.

  • So first of all, let's talk about the latency

  • in this particular network.

  • So how are we going to measure this?

  • Well, we're going to look at this graph

  • and we're going to measure it by the number of wires

  • that you need to go through from an input to an output.

  • So let me write this down.

  • So the latency is the time that is required for a packet

  • to travel from an input to an output.

  • And how are we going to measure this?

  • Well, we're just going to measure this

  • by the number of wires that we need to go through.

  • So this you have seen before.

  • We can measure this by the diameter

  • of that particular graph.

  • So here, we will define it for a network.

  • So the diameter of a network is going

  • to be the length of the shortest path between the input

  • and output that are furthest apart.

  • So let's have a look at the graph above.

  • So for example, we can clearly see that, for example, input

  • and output-- so say, input zero and output one

  • are connected by just going up one step over here,

  • but just going up from here to here.

  • Then, this switch forwards the packet to this switch.

  • This switch reroutes it, forwards it over here,

  • and then it goes back to the output, output one.

  • So for example, this particular path only has 1, 2, 3, 4 edges.

  • And what we are interested in is sort of the worst-case time

  • that it requires to go from an input to an output.

  • So that means that we are interested in a diameter.

  • And a diameter is in this case, well, the shortest path

  • that you can find from an input to an output that are furthest

  • apart.

  • So what are those who are furthest apart?

  • Well, of course, you would like to go through here, right?

  • So if I connect the input zero to say, output four,

  • I will need to go all the way up through the route

  • down to the output.

  • And how many edges do we see here?

  • 1, 2, 3, 4, 5, 6-- so in this example,

  • we have a diameter that is equal to six.

  • And in general, if you are looking at n times n networks,

  • what does it mean?

  • n is the number of inputs and n is also the number of outputs.

  • So in this case, we have a four times-- well,

  • this is actually three over here--

  • we have four inputs and four outputs.

  • So this particular example depicted on the board

  • is a four times four network.

  • So if you generalize this for any size binary tree,

  • say, an n times n network, then what's

  • the diameter of such a general network?

  • Well, if we have n inputs and n outputs,

  • well, we have to go all the way up

  • through towards the root and all the way down.

  • So we actually count the length of a leaf to the root

  • here twice.

  • So in general, we have a diameter that looks like this.

  • It's 2 times 1 plus the logarithm of n.

  • So in this lecture, we will have n is going to be a power of 2,

  • just to make calculations simple.

  • And the logarithm is always to the base two.

  • So this is a diameter of a general binary tree.

  • And well, what are the other parameters?

  • So that does not look too bad.

  • It's logarithmic in answer.

  • That sounds pretty good.

  • What about the switch sizes?

  • Well, how do I measure those?

  • It's like the number of inputs that get into it

  • and the number of outputs that get out.

  • So in this case, I will have 1, 2 inputs that

  • go into this switch and there are two outputs coming out.

  • So this is what we call a two times two switch.

  • So this will be a two times two switch.

  • But if you look at this one, for example,

  • we see one, two, three outgoing edges and three ingoing edges.

  • So this is actually a three times three switch.

  • And in a general binary tree, we will

  • see that all these intermediate nodes over here, they

  • are all three times three switches.

  • So approximately half of the switches

  • are actually three times three switches.

  • So that's the switch size.

  • Now, you may say, well, why don't I

  • use a larger-sized switch?

  • That would help me a lot, right?

  • If I could use, say, a four times four switch, then

  • I would be able to have more inputs coming in,

  • more outputs coming out, and I can actually

  • maybe use a ternary tree rather than a binary tree.

  • In a binary tree, every note at the level

  • has two children, right?

  • But we could design a tree that has

  • at every level three children.

  • So then, they can use four times four switches.

  • But if you do that, then the path from the leaf

  • up to the root is getting shorter

  • and the diameter gets smaller.

  • So if I increase the switch size-- so rather

  • than three times three, we look at four times

  • four or five times five, six times six and so on-- then

  • the diameter will actually reduce.

  • So what about having a monster switch, like I have just one

  • switch and I have my input zero all the way up

  • to input n minus 1 and then I have my outputs

  • on the other side?

  • Well, of course, the switch size is n times n

  • but the diameter is nothing, right?

  • The diameter is reduced to one.

  • You can immediately go from an input to an output

  • through the switch.

  • But this, of course, conceals the problem.

  • So what we are interested in is, well, we're

  • actually really interested in how

  • to solve the problem of routing all these inputs

  • to these outputs using smaller switches of size three

  • times three or two times two.

  • What we're really interested in is,

  • what is the internal structure in this monster switch?

  • I sort of have concealed the problem by just saying, oh,

  • I've got a big switch.

  • But what we want to solve today is

  • how do we do the routing in this case within the monster switch?

  • So we want to use just small switch sizes

  • and build up a network using these smaller ones,

  • like three times three switches or two times two switches.

  • Now, so that brings us to yet another parameter,

  • because here, we'd like to count the number or smaller

  • switches that we use and that relates

  • to the cost of the network, the amount of hardware

  • that you need to put into it.

  • So in this example, we have the switch count.

  • Well, it's pretty simple, right?

  • It's 1, 2, 3, 4, 5, 6, 7-- we have seven switches.

  • And in general, if we have n inputs-- so 1,

  • 2, 3, 4 inputs-- then the number of switches

  • that we use in the binary tree is 2 times the number

  • of inputs minus 1.

  • So let's write that down.

  • So over here, we would have 2 times n minus 1,

  • which is the number of switches that you actually use.

  • So how can you see that actually?

  • So in general, we have 1 plus 2 plus 4 plus 8 and so on plus n.

  • And it's a power of 2, according to our assumptions.

  • And if you add them all up, I think you'll-- well,

  • you can check for yourself that this is actually equal to 2

  • times n minus 1.

  • So now, we have the switches.

  • So so far, this looks pretty good, actually.

  • We use small switch sizes.

  • The number of switches is linear in n.

  • The diameter is logarithmic in n so that sounds good.

  • So what about congestion?

  • Do you any idea-- what's the problem with this graph?

  • What is the big, big problem here?

  • What can happen in a very sort of worst-case scenario

  • where the packets get routed from inputs to the outputs?

  • If they need to go to certain locations,

  • then they all may have to travel through the root.

  • So you get congestion over here.

  • We don't like that.

  • So this root is actually then overloaded.

  • Actually, you can already see that say,

  • this particular switch-- if this switch fails, then actually,

  • we will have two disjoint trees that cannot even communicate

  • to one another.

  • So this brings us to the idea of congestion.

  • And in order to define it better,

  • you will need a few definitions.

  • So to start, we will define a permutation

  • and we will use this to stipulate the requirement

  • that we want on how inputs and outputs are related

  • to another, which input needs to communicate to which output.

  • So permutation is a function pi from the set 0 to n minus 1

  • to the same set.

  • And it is such that no two numbers

  • are mapped to more than once.

  • So no two numbers are mapped to the same value.

  • So what we really want-- to put it in mathematics,

  • we want that pi of i is only equal to pi of j

  • if and only if i is equal to j.

  • So let's have an example to plug into that picture over there.

  • So a first example could be pi of i equals, say, n minus 1

  • minus i.

  • This is a proper permutation.

  • No two numbers map to the same value.

  • Another one could be the identity permutation,

  • like you map i to the same i.

  • So that's another example.

  • Now, how do we use permutations to go

  • towards the idea of congestion?

  • So permutation can be used to formulate the permutation

  • routing problem.

  • And the permutation routing problem is defined like this.

  • It's defined as follows.

  • What we want is that for each i, we

  • want to direct the packet at input i to output pi of i.

  • So you want to do that for all i.

  • So let's have a look at this particular example, where we

  • look at identity permutation.

  • So if you do that, we can easily route this, right?

  • So I want to send a packet from input zero to output zero.

  • So I can simply go into this direction.

  • I just go towards this switch and it

  • gets routed back to this one.

  • I can go like this and this one can go like this

  • and this one goes like that.

  • Now, if you look at the other permutation,

  • the picture looks very different.

  • Now, we want to route input zero to output three.

  • In order to do this, I will actually

  • need to go all the way through here and then all the way down

  • to this particular output.

  • And now, the picture gets into a big mess because for input one,

  • we have to go to output two.

  • So for input one, well, we go all the way like this,

  • we again go through the root, and then

  • we go down to this particular output.

  • And as you can see, for input two, well,

  • we need to connect to output one.

  • So again, we go all the way up and we go all the way down.

  • And for this one, we will again go all the way up and all

  • the way down to input zero.

  • So now, you can see that this particular switch over here

  • has to serve packets from all the inputs.

  • All the four packets have to travel

  • through this particular node here.

  • So this leads us to the following definition

  • of congestion.

  • So the congestion-- oh, before we continue,

  • let me first define a path.

  • So for i, we direct a packet at input i to output pi of i.

  • And the path that corresponds to this route

  • is actually denoted as follows.

  • So the path taken is denoted by P i pi i.

  • So now, we can define the congestion

  • of a set of such paths.

  • So the congestion of the path corresponding to P zero

  • to P pi zero and so on and we go all the way up

  • to the n minus 1 input that needs to be

  • mapped to pi of n minus 1.

  • So the congestion is now defined as the largest

  • number of paths that pass through a single switch.

  • So in our example, we saw that in the case of the blue arrows

  • here for the identity permutation,

  • well, this switch only needs to transmit one packet

  • and all those actually zero packets.

  • So actually, the congestion here is equal to 1.

  • And for this particular permutation,

  • well, we had to direct all the packets through the root

  • and it's the most accessed switch.

  • And that switch has congestion four, right?

  • So the congestion over here is equal to 4.

  • Now, this does not look so good because for a binary tree,

  • we always have this vulnerable root that

  • is right here in the center connecting the left side

  • to the right side.

  • So we can always find a permutation-- actually,

  • this permutation over here-- that leads

  • to this worst-case congestion.

  • So what we're interested in is the maximum congestion,

  • which is sort of the worst-case scenario.

  • And we'll define it as follows.

  • The maximum congestion is actually

  • equal to the maximum over all permutations pi.

  • So this is kind of the worst-case routing problem

  • that I can imagine and it may occur in practice.

  • So in the worst case, how can I solve it the best?

  • So I want to find the minimum of the congestion

  • of a path over here and the minimum

  • is over these types of paths.

  • So actually, this is our solution

  • to this routing problem.

  • We want to find the best kind of solution

  • for this worst-case scenario-- so

  • the minimum over all solutions for these paths

  • So well, for this particular tree structure,

  • this permutation is really the worst-case scenario

  • that you can have because every packet needs to be routed

  • through the center over here.

  • And it means that our maximum congestion for an arbitrary

  • tree is actually equal to n.

  • So that looks really bad, actually.

  • So we don't like this at all.

  • So let's find out where we can do a little bit better

  • and we come to look at the two-dimensional array

  • and see what that would lead up to.

  • And its structure is as follows.

  • We essentially have inputs on the left

  • and the outputs are on the bottom

  • and they are in a grid structure.

  • So we have input zero, input one, input two, input three.

  • They all connect to their terminals.

  • We have switches, four of those, and they are all

  • connected in this grid.

  • And at the very bottom, we will have the outputs, the output

  • terminals.

  • So this is output zero and here, we

  • will have output one, output two, and output three.

  • So notice that my circle start to resemble my squares,

  • but these are all the switches right here in the center.

  • So how does this work?

  • Well, do we have a better parameter?

  • So let's look at it together.

  • So we need to first of all figure

  • out what the diameter is.

  • So what's the diameter of this particular network?

  • So what's the shortest path between the furthest input

  • and output?

  • So if you look at that, we can see

  • that if I go all the way from here

  • and I go all the way down to this corner,

  • that looks like the largest path and I

  • need to cross all these wires.

  • And in general, for any n, we will have

  • that the diameter is 2 times n.

  • Now, what about the switch size?

  • It looks a little bit smaller, right?

  • Because over here, we had three inputs coming in

  • and three outputs coming out but over here,

  • we see that every single switch is only

  • two inputs and two outputs.

  • So that makes the size two times two.

  • Now, the number of switches is pretty bad, right,

  • because we have n squared switches.

  • So that's really horrible.

  • That's a lot.

  • We would like to do much better.

  • And what about the congestion?

  • Do you have any idea what the congestion could

  • be in this particular case?

  • We will prove a theorem on that.

  • For any permutation, is there a way to route

  • the inputs to the outputs in such a way that the switches

  • get almost not congested?

  • So in the binary tree, we had a congestion

  • of n, which is linear in the switches.

  • But over here, we can do much better.

  • We will show that the congestion of an n-input array

  • is actually equal to 2.

  • So that's great.

  • So I'll prove it in a moment, but that

  • looks really fantastic.

  • And so it's way better than the binary tree.

  • Now, this is really not so good and this is also much larger,

  • but still-- we will start to think next

  • after we show this particular property how

  • to combine these two and see how we can come up

  • with another network that's able to combine in some ways

  • these two properties.

  • And maybe we can find a good solution that way.

  • It turns out we will not immediately be able to do that.

  • We will need to make another step

  • and come to the last network.

  • It really has good parameters.

  • So what about the theorem?

  • So if you prove this, well, how do we start?

  • You just start with any permutation.

  • If I want to prove something about the congestion,

  • it's defined as the maximum of all permutations.

  • So let's take one of them and see what we can prove.

  • So let us define the paths for this permutation.

  • So what we really want to do is we take any permutation

  • and we want to find a really good solution for the routing.

  • If that gives us a very low congestion, we are very happy.

  • So the way to do this is well, maybe you have an idea already.

  • So how would I route this?

  • So I want to connect an input i, say, 1, 2,

  • output two, for example.

  • How can I do this?

  • Any suggestions?

  • So of course, I could go any path,

  • but somehow, I want to have some uniform structure that

  • hopefully helps me to prove that the congestion in every switch

  • is very small.

  • So how could I think about this?

  • Well, if I make sure that, say, a packet that

  • goes from one to output two is only

  • going to be participating in the wires

  • off the i-th throw and the P-i-th column,

  • then I know that every wire will only get traveled over twice

  • by a packet.

  • This could either be a packet that goes into this direction

  • or-- so a switch will be accessed at most twice.

  • A switch can either receive a packet from this direction

  • or receive a packet from the upper part.

  • So that will be a really good idea.

  • So let's define that.

  • So we say that in our solution, we will design it such

  • that the path from input i is actually

  • going to be rightward to column pi i

  • and then downward to the output-- so downward to output

  • by i.

  • So this is a really good solution to the routing problem

  • because now, we can continue our proof as follows.

  • We just say, well, if you look at the switch in row i

  • and column pi i, well, this one actually

  • transmits at most two packets because a packet can only

  • come from the left or it's going to go from the top.

  • So either one of the two-- at most, those two packets

  • will go through the switch.

  • So this shows that we have a congestion of at most two

  • for any permutation.

  • And in order to prove equality, because that's really

  • what the theorem says, we also have

  • to show that there exists a permutation that

  • achieves a congestion of two.

  • And that is pretty straightforward.

  • We can, for example use a specific permutation

  • that maps zero to zero and maps n minus 1 to n minus 1.

  • Well, for this particular permutation,

  • when we look at the picture over here,

  • we see that input zero needs to go to output zero.

  • We also see that this lowest input, input three,

  • needs to travel all the way up to here.

  • But it's clear that the packet that needs to go over here

  • needs to travel through that switch in the lower left bottom

  • corner.

  • And the input three also needs to travel through that.

  • So here, we clearly see that we you have a congestion of two.

  • So now, the proof is complete because we

  • have shown this upper bound.

  • So for any permutation, the congestion is at most two

  • and we see that this specific permutation achieves

  • this congestion.

  • So this is the end of this proof.

  • So that's great.

  • So now, what we'd like to do is we'd

  • like to combine these two networks

  • and see what we can learn from both.

  • So now, we'll be taking out a lot of chalk over here.

  • So the idea is to construct a butterfly network

  • and I will draw it in such a way that you can

  • see the recursive structure.

  • The idea is to do the following thing.

  • So let me see how I can do this the best.

  • So I will just do the top line first and I have the spacing.

  • So we have input zero, a terminal, we have a switch,

  • we have a switch, we have a switch, and another one,

  • and here, we have the output zero.

  • So the whole idea is that I'm going

  • to combine every two outputs by using a small butterfly

  • structure.

  • So we have two, output three, output four-- actually,

  • I need a little bit more space.

  • Do it once more, output one, two, three, four, five,

  • six, and a last one, seven.

  • This is going to be pretty tight on the board.

  • So what's happening is this.

  • So these are all connected, of course, to switches.

  • The switches output those.

  • And the idea is that we create the following structure.

  • This switch can either forward it over here

  • or it can cross it over to this particular line.

  • And this switch can either forward it or cross it over

  • to this line.

  • So this is a very small butterfly structure.

  • Here, we have two inputs and two outputs.

  • And we will repeat this process and we'll do the same

  • on each of these other levels.

  • So we forward those or we cross them, like this.

  • And now that we have constructed all these smaller butterfly

  • structures, we can start to combine two butterfly

  • structures together in the bigger one.

  • So here, we had two outputs that we combined

  • in a butterfly structure.

  • Now, we use two butterfly structures

  • that we put into a bigger version.

  • So how do we do this?

  • Well, we have that the upper half over here

  • can either forward those packets or cross them over

  • to the bottom part butterfly structure.

  • So for these, we can either forward them straight on

  • or we can go to the top butterfly.

  • So you see that these two inputs, these two switches,

  • either can forward packets to this sub-butterfly network

  • or to the top butterfly network.

  • Now, we'll continue this process and for these, you'll

  • do the same.

  • So we can either go straight or we go down.

  • And over here, we can go straight

  • or we can go to the top butterfly network.

  • Well, now we have the final part where we combine essentially

  • these two butterfly networks.

  • We have two butterfly networks created here now composed again

  • of smaller ones and these two are

  • being composed to this bigger butterfly network.

  • Again, we take these four switches.

  • They can route their packets forward

  • to the top butterfly sub-network or to the bottom one.

  • So they can either go straight ahead

  • or this one can connect to the first over here,

  • this one to the second, to the third, and this to the fourth.

  • And in the same style, these can forward them straight like this

  • and then go up like this.

  • And these are all connected because in this example,

  • let's just have an eight by eight network,

  • butterfly network.

  • We have input zero to seven.

  • So this is the butterfly network.

  • In a way, what you can see here is

  • you can see sort of the two-dimensional structure,

  • like we have rows and columns.

  • At the same time, we can also see this binary sort of tree

  • feeling we get from it, which is that a switch can forward

  • sort of its packets to either, say, the top butterfly

  • or the bottom butterfly.

  • So there's a split in two.

  • The same for this one, right?

  • This one goes either to this butterfly network

  • or it goes to this butterfly network.

  • So you have this tree structure sort of

  • embedded in this two-dimensional structure.

  • So what are the properties of this one?

  • So let me first define in more formal mathematics

  • how the switches route their packets,

  • so how the connections are.

  • So in order to do that, we are going to label each switch.

  • And the idea is that we're going to label it by its row

  • and by its column.

  • So we will have-- the columns are

  • numbered by level zero, level one, level two, level three,

  • yes?

  • And the rows are these integers, but we

  • are going to represent them by binary numbers.

  • So zero would be 000, 001, 010, 011-- oops-- 100, 101,

  • and then we got 110 and 111.

  • So for example, this particular switch would be labeled

  • by these three bits, 001, and the integer number, 1.

  • This one would be 011 and its column is indexed by integer 2.

  • So a switch is uniquely identified

  • by its row and column.

  • We will have b1 up to b logarithm

  • of n, which are the number of bits

  • to represent the row in digits, and to finally

  • have an integer l and this we will call the level.

  • So this particular switch either directs or routes a packet

  • to the switch that is indexed by b one up to--

  • and then we get b, l plus 1 and we take its complement.

  • So instead of if b, l plus n would be 1,

  • we would have a 0 here.

  • If it would be a 0, we will have a 1 over here.

  • But we repeat all the other bits and we get to b log n.

  • And it routes us back to the next level.

  • So we will have l plus 1.

  • Another possibility because there are two outgoing edges

  • is if we have just b1 and we just copy b, l plus 1,

  • essentially.

  • We route a packet straightforward.

  • We don't do anything special.

  • We get b log n over here and then to the next level.

  • So for example, let's see where we can see how this works.

  • So for example, take this particular switch.

  • We have 010.

  • So it can either go straight on to the next level.

  • It would go to 010 but then instead of level one,

  • we have level two, which is the right edge over there.

  • The other one is if this one goes up, well,

  • we will need to switch the first bit over here, a 1.

  • We swap it into 0 and then we go to the three zeros over here

  • and we go to the next level and that

  • would be this particular rule.

  • So what we can do here is to-- so when we see this,

  • we can start to figure out how we can direct inputs

  • to outputs.

  • So let's do this.

  • So suppose I want to route a packet from a certain input,

  • one of these, all the way to one of the outputs over here.

  • So the way to do this is as follows.

  • We can just start-- for example, I want to go from switch x1

  • up to x log n comma 0.

  • So I start completely at the left over here

  • and I want to go somewhere of my choice to the right.

  • So I want to somehow move all the way to some other row, y1

  • indexed by y by the bit pattern, y1 up to y log n,

  • but now at the very last level, which is log n.

  • Well, how do I do it?

  • Well, this switch, I can use that rule up there

  • and simply change x1 to y1.

  • I can either leave x1 as it is if it's the same as y1

  • or I can swap it to its complement

  • if that's the value of y1.

  • So what I can do is I can just simply route it to y1.

  • And then, I leave all the other bits the same,

  • which are x2, x3, all the way up to x log n.

  • And we will have reached the first level.

  • Now, this one can go to-- well, now I'm

  • going to swap the second bit into the bit of my choice.

  • So I leave all the other bits the same, y1 the same,

  • x3, all the others the same.

  • I just swap x2 into y2.

  • So we leave all those equal and we go to the second level.

  • And then, we go all the way to the final level

  • and we one by one swap all these bits.

  • So let's have an example.

  • Suppose I want to connect, let's say,

  • this one to for example, well, let's say

  • this particular output.

  • So what's the binary for this one?

  • This is actually 101.

  • So if the first bit is different, I need to cross.

  • And otherwise, I need to pass straight on.

  • So let's do this.

  • So over here, I'm in 011.

  • I need to go to 101 so we need to change the zero into a one.

  • So I need to go down.

  • I need to cross.

  • Now, if I look at the second bit,

  • I also need to change it to a zero so again,

  • I need to cross, which is over here.

  • Now, the third bit is equal to 1 and it's the same.

  • So now, I can go straight ahead.

  • I do not cross and I end up at this output.

  • So what did I do?

  • For every bit that is different, I cross

  • and for the bits that are the same, I go straight ahead.

  • So this is how I can route packets from one input

  • to another output.

  • So let's look at the parameters.

  • First of all, if you look at the diameter,

  • well, it turns out that that's approximately

  • equal to the number of levels, which is the logarithm of n.

  • And to be precise, it's actually equal to 2

  • plus the logarithm of n.

  • So that's great.

  • That's a good scaling.

  • Again, it's back to the logarithm of n.

  • So we have the best of these two parameters.

  • The switches that we see have two inputs and two outputs.

  • So we again have a two times two switch.

  • The number of switches is the number

  • of rows times the number of columns.

  • The number of columns is the logarithm of n and number

  • of rows is equal to n.

  • And to make it a little bit, precise,

  • it's 1 plus the logarithm of n.

  • So that's somewhere in between those two.

  • But if you're thinking about it, it's

  • much better than n squared.

  • It's almost linear except for a logarithmic factor.

  • For the congestion-- and we are not

  • going to talk about it here, but you have a problem set

  • assignment that will ask you to solve

  • this-- is that actually, the congestion

  • is the square root of n or it's equal to the square root of n

  • over 2, depending on whether n is an even power

  • or n is an odd power.

  • Now, we're not going to prove that here

  • because we want to step forward to this particular network.

  • It's very exciting.

  • And you will prove this in your problem set.

  • So this one is somewhere in between, somewhere

  • in between these two extremes.

  • Now, it will be really fantastic if we can somehow

  • transform this network with a trick to,

  • again, have a really great congestion

  • of just a constant, like two or three or whatever or maybe

  • even one.

  • So for this particular network, in the 1960s,

  • Benes, a Bell Labs researcher, had the great idea

  • to use a butterfly network and attach to it, again,

  • a butterfly network, back to back sort of.

  • So what was his idea?

  • His idea was to do the following.

  • So the butterfly network as we have it right now

  • is this particular part over here.

  • And the idea is now to start up mixing all those outputs

  • that we got here together again using a similar rule.

  • So what do we do?

  • We are going to essentially repeat

  • this particular structure on this side.

  • So how do we do it?

  • Well, we go either straightforward

  • or we start to mix them again.

  • So it's like this output, this particular switch,

  • can either go straight ahead or can cross

  • to the lower part over here.

  • It goes over here and this one goes over.

  • So as you can see, we have repeated this part.

  • It's exactly the same as this structure over here.

  • We'll do the same for this part.

  • So we can either cross or we can go straight ahead.

  • Oh, we also have, of course, that these switches

  • can go straight ahead or can cross to the top.

  • I forgot about that.

  • So we have this-- oops-- as well.

  • So as you can see, this particular structure

  • repeats itself again and we slowly

  • start to build up in mixing all the outputs again

  • or the possibility, at least, to route them to any other row.

  • So how do we do this?

  • Well, we continue this particular structure now

  • over here.

  • So all these can either go straight ahead.

  • That's a possibility.

  • Or they can go all down.

  • So this switch can either go straight ahead

  • or can go to the lower half.

  • And for these, we have a similar structure.

  • We can either go straight ahead or such a switch can cross over

  • to the top over here.

  • So that's this.

  • So this is Benes network and then over here, of course,

  • we have the outputs, zero, one, and all the way down to seven.

  • So as you can see over here, the structure again

  • has a recursive nature to it.

  • You can see that this big Benes network over here

  • consists of two smaller ones that

  • are right here in the middle, this one that

  • goes all the way up to here-- so maybe I should

  • put a color boundary around it.

  • Let me check I want to do this-- right.

  • So this particular part, is again

  • a Benes network and the top part in the same picture,

  • the top subnetwork is also a Benes network, this part.

  • And if you look within those, we again

  • see a top part and a bottom part.

  • And over here, we see a top part and also a bottom part.

  • So you see this recursive nature again reappearing.

  • It turns out that with this trick,

  • we can completely eliminate congestion

  • and we can get it to only one, which is really surprising.

  • And that what we're going to prove here.

  • So this is a great invention at the time.

  • It's really, really beautiful.

  • So let me put in the other parameters.

  • So they stay approximately the same

  • up to that the diameter is about twice as

  • large because we added another sort of whole butterfly

  • structure to it.

  • The switch size stays the same.

  • We, again, have about two times more switches

  • so they sort of stay about the same up to a linear factor,

  • like a constant factor.

  • And the congestion, however, completely dropped down to one.

  • So that's what we're going to prove now.

  • And in order to get some intuition,

  • well, let me first write down the theorem.

  • Actually, let me put this over here.

  • So in order to get some insight into this,

  • we are going to use this recursive nature.

  • So we're going to use induction and we're

  • going to say, oh, for any permutation,

  • I can find really good routing for say, this red subnetwork

  • and for this blue subnetwork.

  • So I know that.

  • So what I need to do is, if I have my bigger Benes network,

  • like this one, I would need to somehow map

  • these inputs-- I need to route them

  • to either the top and the bottom subnetwork, one of the two,

  • in such a way that there will be absolutely

  • no congestion, because we want to keep this one.

  • So a switch should only see one packet coming in.

  • So that means, for example-- and we'll come back to that--

  • that for example, for this switch,

  • it should not receive a packet from both this input

  • and from this input.

  • So the intuition that we are going to create

  • is we're going to list our constraints, the constraints

  • that we need to satisfy, like the zero and the fourth input

  • should not both be mapped to this top subnetwork and so on.

  • So we will get into that and then we

  • will gain a lot of intuition on how to solve this.

  • So what's the theorem?

  • So the theorem is that the congestion of the n-input Benes

  • network is actually equal to 1.

  • And we will prove this for n equal to a power of 2.

  • We have assumed that at the start

  • that we had with all the other networks, as well.

  • And in this case, we will use induction on a.

  • So that's the method that we will

  • do because that's also the recursive structure

  • of the Benes network itself.

  • So we will use induction on a and we

  • are going to define the induction hypothesis simply

  • as, "The theorem is true for a."

  • Now, let us do the base case.

  • We always start with the base case

  • and that should be pretty easy because this is

  • the most basic Benes network.

  • So n equals 2 to the power of 1.

  • We essentially have two inputs, an input zero and an input one.

  • They are connected to these switches over here that

  • can either forward them or can cross them over

  • and then they go directly to the output.

  • Notice that in this case, we just

  • have the most elementary butterfly network.

  • It's the same.

  • So we have output zero and output one.

  • So this corresponds in this picture

  • to these little small things over here,

  • this one and this one and this one over here

  • and the fourth one over here.

  • So now, let's take any permutation.

  • We want to show that we can route it in such a way

  • that there's only a congestion of one.

  • So let's do this.

  • So there are essentially only two permutations.

  • Either zero is mapped to zero and one

  • is mapped to one or zero is mapped to one

  • and one is mapped to zero.

  • So in both cases, we can just route them

  • through their own switches.

  • So we have that either pi of 0 equals 0 and pi of 1

  • equals 1, in which case we just direct them straight through

  • and we go straight through and every switch only

  • sees a packet once.

  • So for this particular permutation,

  • we have a congestion of one.

  • Now, the other permutation that we can have

  • is if zero is mapped to one and if one is mapped to zero.

  • Well, in that case, we just route

  • this cross over to the bottom row

  • and here we go from this switch to the top row.

  • Again, every switch only sees a packet once.

  • So in this case, in the base case, we are done.

  • We are happy.

  • We have shown that the congestion is equal to one.

  • So now, it gets to the harder part

  • because for the inductive step, we

  • are going to assume, of course, that it

  • holds true for a smaller Benes network.

  • So we assume that P a is true and well, let's

  • try to gain some insight here.

  • So we know from our induction hypothesis,

  • within each subnetwork, we can solve any routing problem

  • with congestion one and for this subnetwork, the same.

  • That's our induction hypothesis.

  • So how do we go ahead?

  • We need to somehow map these inputs

  • according to the permutation of our choice.

  • So that could be for some input zero goes

  • to output five or input one goes to output two, et cetera.

  • So somehow, we need to choose where

  • we are going to map this particular input to.

  • So packet zero that comes from this input

  • should either go to the red network

  • or it should go to the blue network.

  • And for each of these inputs, we can make such a choice.

  • But we have to be very smart about it

  • because we need to avoid any congestion.

  • So the intuition is that we're going

  • to set up a constraint graph, a graph that represents all

  • the constraints that we need to satisfy in order

  • to achieve congestion of one.

  • So let's do an example so that we

  • can figure out what's going on.

  • Actually, let me put it over here.

  • So just take an example permutation

  • and we'll go through this example

  • and then see how the proof works.

  • So let's as an example have pi of zero maps

  • to one, pi of one maps to five, pi of two goes to four,

  • input three goes to seven, four maps to three,

  • five to six, six to zero, and seven to two.

  • So this is just an arbitrary permutation.

  • So what do we see?

  • We want to make sure that, for example, this switch is only

  • seeing one packet.

  • So it cannot see a packet both coming from input zero as well

  • as from input four.

  • I cannot see that.

  • I do not want that to happen.

  • Similarly, for this one, I do not

  • want to see a packet coming from one or one from five.

  • So let me define a constraint graph that sort of represents

  • this.

  • So the constraint graph that we are interested in

  • is defined as follows.

  • If two packets must pass through different networks,

  • subnetworks-- so in our case, the red and blue subnetwork--

  • then we'll actually have an edge between those two.

  • So then, there is an edge between them.

  • So for this example, we're going to set up this constraint

  • graph.

  • So I was just talking about this particular switch.

  • It cannot see one coming from four and a packet from zero.

  • So what he have, we have an edge between zero and four.

  • In the same way, we have an edge from one to five.

  • Why?

  • Because a packet that comes from input one and a packet that

  • comes from input five cannot both be routed through

  • the switch because then the switch would see two packets

  • and then the congestion would not be one, but two, right?

  • So one and five also have an edge in between.

  • And in the same way, we have two and six and seven and three.

  • So two and six is this constraint, like two and six

  • over here.

  • And three and seven is the other constraint.

  • So if I have those constraints in place, well then,

  • I know that the routing that goes from level zero

  • to level one will not violate my congestion of one.

  • So that's great.

  • Then, I hope to be able to use the induction hypothesis

  • and I get a proper routing within the red subnetwork

  • and one within the blue network.

  • And then, I need to map all these to these outputs.

  • So I also have constraints on these outputs

  • because, well, For example, take this particular switch.

  • It should not see a packet coming from this particular one

  • and one from this one.

  • So how do I code that up?

  • So let me first write out what we did here

  • and then we'll do the same for the last level over there.

  • So-- oh no, that's not really necessary.

  • So at the output side over here, we

  • have similar constraints as we did over here.

  • And in this particular example, just as an example,

  • suppose we look at the packet that

  • is destined for output zero.

  • Well, what is this packet?

  • Well, I know that's pi of 6 is equal to 0,

  • according to my example.

  • So packet six is destined for this particular output

  • zero over here and goes through this particular switch.

  • So this packet and also the packet

  • for output four, which is if you look at the mapping, pi of 2

  • is equal to 4.

  • So that's packet number two.

  • Well, both of these packets cannot pass through the same

  • subnetwork.

  • So why is this?

  • So let's look at this particular example.

  • So output zero, well, comes from packet six,

  • somewhere over there.

  • Now suppose packet six was routed through the red network

  • and at the same moment also, output four--

  • the packet that is destined for output four, which

  • is packet number two-- suppose packet two was also

  • going through the red network.

  • Well, then I notice that both of these packets

  • must arrive at this particular switch in order for one

  • to be routed to output zero and the other one

  • to be routed to output four.

  • So in order to avoid congestion in this particular switch

  • over here, we need to have a constraint.

  • The constraint says that the packet for packets two and six,

  • that those two cannot go through the same subnetwork.

  • So essentially have another edge over here--

  • we already had the constraint but it's just the same edge.

  • So let's look at the other constraints that we have.

  • Well, let's look at a different example.

  • So for example, if I look at this switch,

  • well, if a packet goes through here

  • that needs to end up at one and a packet

  • that's goes to five, if those two packets are routed

  • through the same red subnetwork, they

  • have to end up here in order to go to both here and to there.

  • So we have congestion of two.

  • So what are those packets?

  • Well, what does pi map to to one and five?

  • Let's look over here.

  • We see that pi 0 is equal to 1 and pi 1 is equal to 5.

  • So packets zero and one are actually mapped to output

  • one and five and they should not go

  • both through the same subnetwork.

  • So we have another edge over here.

  • And now, we can continue this and we have five and seven.

  • So just have a look over there.

  • See, five and seven, they map to the outputs two and six.

  • Again, we have two and six.

  • If they are both mapped to the same network, this one,

  • for example, then I will have a problem.

  • So the other edge is over here.

  • So what did we do here?

  • We started to write out the constraints on this side

  • and we wrote out the constraints on this side.

  • So I only looked at the red subnetwork.

  • That's what I realize now.

  • I could also have looked at the blue network.

  • So let's do that also just to make the picture complete.

  • So for example, let's look at this particular example.

  • The packet six and two should not both

  • be routed through the blue network

  • because then they would both have

  • to go through this switch, one going up

  • to output zero and one going to the right to output four.

  • So in order to avoid congestion at all costs,

  • we have this constraint graph.

  • So now, we come to the key insight.

  • And the key insight is to use a two-coloring of this graph.

  • So the key insight is a two-coloring

  • of the constraint graph, which will

  • lead to a best solution for the routing problem.

  • So let's do this.

  • So we will color this one blue.

  • As you can see, this is an even cycle,

  • blue, red, blue, red, and blue and red.

  • We will make this one blue and this one red.

  • Well, it turns out that we can now start our routing process.

  • So for example, actually, I will draw a new graph

  • to make that really clear.

  • So I have my blue and my red chalk over here

  • to demonstrate what I mean.

  • So what do I do?

  • I have zero, one, two, three, four, five, six, and seven.

  • I have the switches that correspond to those.

  • Well, if it's colored red-- so zero over here is colored red--

  • I will direct it to the red subnetwork.

  • So where is this red subnetwork?

  • It's really contained over here and the blue one--

  • so this is the red one and the blue one is right here.

  • And over here, we have the outputs ranging from zero, one,

  • two, all the way to seven.

  • So input zero is colored red.

  • We go straight ahead.

  • We want to go to the red network.

  • Input one is colored blue.

  • It goes, therefore, to the blue network.

  • So this is the only way how to do it.

  • Input two is colored red.

  • Go straight ahead.

  • Input three is also colored red.

  • Go straight ahead.

  • Input five-- oh, input four is colored blue--

  • goes to the blue network.

  • Input five goes up to the red network

  • and input six goes straight ahead to the blue network.

  • It's colored blue and input seven is also colored blue.

  • Let's look at the outputs.

  • So for example, well, let's have a look at output zero.

  • so output zero-- which packet is mapped to output zero?

  • It's packet number six.

  • So six was mapped into the blue network

  • and then it needs to be mapped to output zero.

  • So there's only one edge that goes

  • from the blue network to output zero, which

  • is this particular one.

  • And then somehow, this one needs to be

  • mapped to this one over here.

  • Now, we can continue like this.

  • Output one should receives a packet from-- let's look

  • at the permutation-- from five.

  • No, sorry, output one-- pi of 0 is equal to 1

  • so packet zero needs to go to this particular output.

  • Now, packet zero is in the red network

  • so there's only one edge that goes from the red network

  • to this output.

  • So we need to have a connection over here.

  • Now, we can continue this and note and demonstrate--

  • and you can test it for yourself,

  • too-- that output four needs to receive

  • a packet from the red network.

  • Actually, it should be this particular one, which happens

  • to be packet number two.

  • And then, we have this one, right?

  • So let me just finish it.

  • We have this and we have these two and we have this one.

  • We have this one and we have this one.

  • This one goes straight ahead.

  • This one goes all the way up and this one goes all the way up.

  • So what do we see?

  • We see that packets over here, that these switches only

  • see a packet once and these ones,

  • as well, these ones also and these ones also.

  • So we have directed the packets, routed the packets

  • to the red and the blue subnetworks in such a way

  • that the congestion at the last level and at the first level

  • is still equal to one.

  • Now, we use our induction hypothesis

  • and we conclude that we can map the route that's

  • going to have a routing from packets from here

  • to here such that the congestion within the subnetworks

  • is only one, so within the blue as well as in the red.

  • So this is the insight into how this works and I notice

  • I am running out of time.

  • So the formal proof we will have to postpone until recitation,

  • but that's actually really a very simple thing

  • to do that right now.

  • So just keep this key insight and then

  • you can easily prove the theorem.

  • But this is the real insight.

  • Thank you.

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Lec 9|MIT 6.042J コンピュータサイエンスのための数学 2010年秋学期 (Lec 9 | MIT 6.042J Mathematics for Computer Science, Fall 2010)

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    Dou Lin に公開 2021 年 01 月 14 日
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