>> Yuval Peres: Okay. Good afternoon everyone. We are delighted to have Louigi AddarioBerry tell us about probabilistic aspects of minimum spanning trees.
>> Louigi Addario-Berry: Thanks very much Yuval. Thanks for the invitation. I'm starting with
this picture because it's a nice picture that's somewhat related to the subject of my talk. And so
since there was some interest, maybe I'll tell you quickly just what this is. So here we have a
hundred thousand uniformly random points in the square. We formed there a minimum
spanning tree and…
>>: [indiscernible]
>> Louigi Addario-Berry: Say again. The minimum spanning tree according to the Euclidean
distance. And we've picked some random root vertex which is somewhere around here. If I
have a laser I can do that from further away, and then colored all the vertices according to their
graft distance in the minimum spanning tree from that root vertex, so the closest vertices are
red. And then orange, yellow, green, blue, purple, violet codes, graph distance from the tree
from that root and you see all these interesting curves start to appear. So this is a…
>>: [indiscernible]
>> Louigi Addario-Berry: Uniformly at random.
>>: Uniformly at random.
>> Louigi Addario-Berry: Yeah. So this is a picture that no one can say anything about to date.
If we had used some version of this on the triangular lattice, then the people to ask would be
[indiscernible] who have a forthcoming paper on the subject.
>>: The black lines…
>> Louigi Addario-Berry: And the black lines, so the problem is there’s so many points here that
you can't see the, most of the edges of this tree, only the colors. So what I've done is drawn the
important edges and what important means is that, so an edge is drawn in bold here if on
either side of that edge lies at least 3% of the tree. So if the edge cuts at least 3000 vertices off
from the rest of the graph. So it gives you some idea of the, broadly of the structure of the
tree. Okay. But I won't be talking about two dimensions today. This is just sort of a
motivational picture. I'll be talking about minimum spanning trees in high dimensions. So just a
very brief reminder about what they are and how we build them, so if you trace back a bit you
find the earliest reference that I've been able to find is Otakar Boruvka and he asked the
question in the following sort of general form. So you are given the complete graph, so
endpoints with distinct edge weights, which we think of as distances between the points and
you're interested in forming a network so buying a network which is the unique connective
graph that minimizes the total length. Okay? So you want to connect up all of the cities and
you don't want, you want to pave a minimum number of miles of road. Okay. So we've chosen
all of the edge weights to be succinct so that there's a unique minimizer. That minimizer is
necessarily a tree. You're never going to buy a cycle if all you care about is a minimal
connectivity and this is the minimum spanning tree. So one of the standards, there's many easy
algorithms for building these things. One of the things I like about this problem is that it's
algorithmically very simple but probabilistically quite challenging. So here's one of the
methods, sort of simple greedy method often called Kruskal’s algorithm. It's really essentially
from this earlier paper. It just says order the edges of your network by increasing order of
weight. Then consider the edges one at a time in that order and add each edge unless doing so
would create a cycle. So we're always going to add the smallest weight edge in this example.
Okay, second smallest is 3. We add it. We don't add 6 because it would create a cycle. Add 7
and now we are already connected. We know we're not going to add any more. Don't add 8 or
9. Okay? So this is quite standard. I'm going over it quickly but I wanted to emphasize one
thing about this process, okay, which is the following. Anytime we decide not to add an edge,
that's because it would create a cycle, which means that its endpoints are already in the same
component. Okay? So that means that at every step if we're only interested in the connectivity
that has been formed by the edges that we've added, okay, so which, at a given stage, which
pairs of vertices are now connected up by some path, it doesn't matter whether we add all of
the edges or only the edges that this procedure adds, okay? So let's just convince ourselves of
that quite quickly. Here we add an edge, add an edge. The third edge we don't add, right, but
if we did and it we wouldn't gain any new connectivity. There was already a path between
those vertices. Okay. And so on. Here we add an edge and now any edge that we go, the last
edge we don't add, but that doesn't change the connectivity. Okay? So this gives you a
coupling between two processes. One where you only add edges that don't create cycles and
one where you add all the edges and that coupling is going to be basic for this talk. So I hope
it's clear to all. Okay. So what's…
>>: [indiscernible]
>> Louigi Addario-Berry: Say that again.
>>: That is not Boruvka’s That’s Kruskal’s.
>> Louigi Addario-Berry: This really is Kruskal’s, but I mean, in essence, in essence it's Boruvka’s
algorithm, okay? Kruskal, Kruskal really says one at a time, but Boruvka more or less says do
some greedy procedure with possibly larger clumps than single vertices, okay. So it's… I would
give the credit for this to Boruvka.
>>: [indiscernible] Boruvka does is every vertex is a minimal cost edge. And then you certainly
select all of those in your [indiscernible], it would be the same exponent as [indiscernible].
>> Louigi Addario-Berry: Right. So then, well, and it's not individual vertices, right? It's
components. It's every component…
>>: [indiscernible]
>> Louigi Addario-Berry: Yeah, yeah. But I mean it's essentially, it's essentially this sort of
global greedy procedure. It's slightly different from Kruskal; you're right but it's… Okay. So you
can think of various probabilistic models that you might impose if you wanted to understand
this very typical structure of these. I showed you one at the start. Given the sort of, the
description of the problem I just gave you, a natural probabilistic model it would just be to say
instead pick iid weight, iid weights or lengths, okay, and so that gives you sort of two different
probabilistic structures, one broadly Euclidean, one broadly iid or mean field, okay? And then
there's course, so that's sort of one axis of division that you can look at for research into
minimum spanning trees from a probabilistic perspective. Another is whether you study things
that actually have to do with the weights themselves or you study the sort of, metric or graph
structure of the object that you build. Okay. And there's a lot of research in both of those
camps. So on the weight functionals, so maybe the most famous classic results in this subject is
that the total weight of the minimum spanning tree if you use independent uniform weights on
the complete graph converges as 8 out of 3, so this is a famous result due to Enfreeze
[phonetic], okay but there's a lot more research in a similar direction. On the other hand, if you
want to study the global structure then there's also a huge body of work and there's all sorts of
questions you can ask. One natural probabilistic framework for asking those questions is to try
to study some sort of distributional convergence of your finite objects to some limit object
which might be an infinite volume limit ala global weak convergence or it might be some
compact limit, something to do with curves in the plane if you're in a Euclidean setting or…
There's a variety of possibilities and a lot of work on the subject including by people in this
room. Okay. So I've mostly been focused on trying to understand the iid picture and the metric
structure, okay? So there's a variety of ways you can do that. You can ask for local structure,
sort of global structure which you might get at by rescaling your tree to sort of maintain a
compact object that then gets large or you might try some other things. So now the question
you can ask is how, you know, if we manage to understand one object then, to what degree is it
a template for other objects? So if we can handle one sort of, some sorts of weight can we
learn to handle other weights or can we handle minimum spanning trees of other graphs and so
on. Okay. So that's, I'll call that the question of universality. Today I'm really going to focus on
talking about global structure. Okay. Global structure for iid edge weights, okay? And let me
just quickly make one observation that comes from the algorithm I just described. Okay. If I
take the complete graph on end vertices and I want to find, to understand the metric structure
of the minimum spanning tree, then as long as I use iid weights with some continuous
distribution, it doesn't matter what the weights are. Why is that? Well the algorithm I just
described starts by ordering the edges in increasing order of weight and as long as you have iid
continuous edge weights, then that's just starting from a uniformly random permutation of the
edges. Okay? And once you have that permutation, everything else about the procedure is
determined. So the distributional properties of the tree, if you don't care about the weights
themselves, only rely on this fact, so you're really free to choose the weights you want. We're
going to end up choosing uniform 0 1 weights to get a nice coupling, okay? So here's a,
description of the global structure of this tree. So if you take the minimum spanning tree which
I'll call Tn, so this is a random tree. And rescale it so that each edge has length n to the -1/3 in
mass 1 over n, so the total mass of the object is 1, and you call this rescaled metric space,
measured metric space Mn, then for some suitable notion of convergence and distribution Mn
converges in distribution to some random limiting metric space. Okay? So there's, to fully
explain this theorem I need to tell you sort of a few things, what this notion of convergence and
distribution is. There's sort of some work to make sense of what exactly this is telling you. For
today, I'd like to just focus on one thing about this theorem, which is this rescaling here, the
fact that the edges should be rescaled by n to the -1/3 so that the limiting object is compact,
okay? And so that statement is really equivalent to this statement. That if you look at the
diameter of the minimum spanning tree of the complete graph with these iid continuous edge
weights, then that diameter is of order n to the 1/3. So if you want to get a compact nondegenerate limit, you should rescale by n to the 1/3. So this is, this is some older work. I
should mention who I'm working with on this slide, Nicoli Broutin, Christina Goldschmidt and
Gregori Miermont and this theorem that I'm going to focus on is joint work with Nicoli Broutin
and Bruce Reed. Okay. So any questions about the result? Okay.
>>: 2006 and 2009.
>> Louigi Addario-Berry: So there was a conference paper where we described the results and
the journal version appeared in 2009. We proved it in 2006 and we proved it again in 2009.
[laughter]
>>: [indiscernible] expectation of the other [indiscernible]
>> Louigi Addario-Berry: Ah. That was another possibility.
>>: [indiscernible] [laughter]
>> Louigi Addario-Berry: Independently and simultaneously. Okay. So here's our setup. We're
going to have the complete graph, I said, now I'm making a choice from edge weights, uniform,
iid uniform edge weights. Okay. And remember the point I made a minute ago which is in the
process where you add the edges one it time, if you are only concerned about the connectivity
structure, the set of components, the set of vertices of each component, it doesn't matter
whether you add all of the edges or just a sub, or just a, the tree edges, okay, and that provides
us probabilistically now with a coupling with the classical error training random graph tree n, p.
So I'll remind you that the error training graph process uses these uniform 0 1 edge weights and
then for a parameter p between 0 and 1, you only keep the edges that weigh at most p. Okay?
And then every edge in the resulting graph, every edge is independently present and is
probability p. And we can do the same thing for the tree, just throw away all of the edges of
the tree with weight bigger than p. That gives us some forest, in general, right, because we've
taken a tree and thrown away some edges. The point is that the connective components of the
forest are exactly the same as the connective components of the tree. Okay. So if we want to
understand the global structure of this tree, then the time at which that global structure -- if we
view the p as a time parameter, then the time at which that global structure forms will be the
same time at which the sort of, at which Gnp becomes hooked up into one connective graph.
So in particular, it's classical that the connectivity threshold for Gnp is at log n over n. p is log n
over n, so we certainly don't need to go past log n over n to understand the final minimum
spanning through Tn. But in fact, the bulk of the structure was formed much earlier, around p
as 1 over n. Okay. And I'll make that precise. So I'm reading everything of interest that occurs
when p is around 1 over n, so here's a comic sketch of what I mean. If you look at edges just a
bit less than 1 over n at .99 over n then all the connective components of Gnp have logarithmic
size, okay? And so that will also be true for the tree. So if we're trying to build something of
size order n to the third, which you can at least believe -- I'm going to convince you of that, but
if that is indeed the case, then nothing really interesting has formed. Nothing really substantial
has formed on that scale at this time. On the other hand, if we go just a bit past 1 over n then
there's already unique component of linear size and all the other components have logarithmic
size. Okay, so you can somehow imagine that if all the other components have at most
logarithmic size, then plausibly they are not going to affect the diameter of the final tree very
much even after they hook on to, to that largest component. We know at this time then there's
going to be a component of this forest of linear size and all these logarithmic little pieces
plausibly won't change things very much. I'll try to convince you of that, but, this comic picture
should at least motivate why we want to study what's happening around p is equal to 1 over n.
This is really the time at which the structure of the MST is going to be built. I'm about to focus
in on what's happening around p equals 1 over n and describe to you how we're going to
understand the structure of Tnp at that time. Before I do that, I just want to make sure that
this picture is clear. So this is already an important thing to understand, right? When we're just
below 1 over n we have components of logarithmic size. Just above we have one component of
linear size. All the others are logarithmic. Let's look at what happens at that at that breakpoint
at 1 over n. Okay. So it turns out that at this point the components of Gn1 over n are pretty
treelike. This is good news for us if we're trying to understand the minimum spanning tree by
some sort of comparison with this graph process. In fact, the graph process is built, is building
of things that are already pretty close to being trees. If you look at the largest number of
edges, the largest number of edges more than a tree that any of these components has, hey,
that's some random quantity, but it stays bounded in probability as n gets larger. In other
words, it forms a tight sequence of random variables. That's, in fact, I mean there's a, the
expectation stays bounded of the largest number of cycles. Okay. So that's good, and most of
them in fact are already trees. So a component of Gnp that happens to be a tree is also exactly
a component of Tnp, right? It's the same set of vertices. It must be the same set of edges if it's
just tree. Okay. So how big are these things and what's their diameter? Well it turns out that
they have size around n to the 2/3. This was already asserted in the classical 1960 paper of
[indiscernible] on the subject, maybe 1959. But the fact that at 1 over n the size should be
around n to the 2/3 is certainly classical result. And more importantly for this talk, the diameter
of these components, of these largest components is around n to the 1/3. Okay?
>>: What is [indiscernible]
>> Louigi Addario-Berry: So it means that if you give me an epsilon, I can give you k such that
the probability is at least n to the 2/3 over k and at most kn to the 2/3 is at least 1 minus
epsilon. So you can bound it in an interval, a multiplicative interval around n to the 2/3 with
probability as close to 1 as you like.
>>: That means n probability. It doesn't mean [indiscernible]
>> Louigi Addario-Berry: No. [indiscernible] yeah. And I'm going to, I mean basically everything
I say in the talk will hold in the sense of improbability and also in expectation and I'm going to
sort of gloss that because everything that, in fact, happens with very high probability, so for the
most part I won't be too careful about that, but feel free to ask if something’s unclear. Okay.
So I'll sort of tried to justify this diameter into the 1/3 to you, but let me remark before I do so
that this already yields a lower bound on the diameter of the resulting tree. Okay? Because if I
take a component of the graph and it has diameter n to the 1/3, well, the tree has -- we know
that the forest has that same component, right, so the tree that spans, the minimum spanning
tree of that component, well what does it do? It takes all of the edges of the graph and then it
removes some of them and that can only increase the diameter, so that component of the tree
must also have diameter at least n to the 1/3, and on the other hand, that bit of tree is some
subtree of the final minimum spanning tree, so the minimum spanning tree must have diameter
at least n to the third. Okay. So this immediately yields the lower bound. Okay. I said that
already. Let me now explain where n to the 1/3 comes from. So let's suppose right here, here's
the vertices 1 up to n, right? And let's suppose you give me some particular set of vertices s
and you tell me that s is a component of Gn 1 over n, okay? Well we know that it's pretty
treelike. I said it has sort of at most they few edges more than a tree and it has a decent
chance of actually being a tree, okay? So suppose, so supposing s is a component of Gn 1 over
n and let's suppose that it actually is a tree. Then the symmetry of the model tells you that any
spanning tree of that set of vertices is equally likely to appear under that conditioning. So we
know that it's a tree. Which tree is it? Well for any particular tree to occur that gives you m-1
edges that have to occur and the rest of the edges have to not occur, plus no edges to the
complement. That means for any tree you are conditioning on the same number of, you're
asking for the same number of edges and non-edges and so any tree is equally likely. Okay. So
if I tell you that the components of the tree, it's a uniformly random spanning tree of its set of
vertices and uniform spanning trees, if I give you a uniformly random spanning tree of m
vertices that's the same as a uniformly random tree on m vertices, and it has m to root m,
right? That's classical. Okay. So then the diameter, that's probably too low to see, but it's
written on the slide as well. Diameter root m here m is equal to n to the 2/3; take the square
root you get n to the 1/3. That's where the n to the 1/3 comes from. And that's, that's true,
that's literally true if it happens to be a tree, if it has a few more edges than a tree then its
distribution is not too different from what you'd get if it really was a tree. That takes a little bit
of thought but it's pretty straightforward.
>>: [indiscernible] right there.
>>: [indiscernible] for…
>> Louigi Addario-Berry: It's a tree with probability, yeah, you can bound it away from 0 as n
goes to infinity. So that at least explains the lower bound plus some idea of where the scaling
comes, and the 1/3 comes from if you already believe this. Okay. So for an upper bound we
have to do some more work, right? We have, now we have, we said we have some
components of this size, right? And any given one of those components is going to have
diameter n to the 1/3 already. To get an upper bound we need to say that after all those
components look up and all the smaller component hookup, that the diameter doesn't get too
much bigger than this n to the 1/3. So you need to understand how these components are
going to connect together. Okay. So the sort of fundamental idea for how we do that is a
straightforward deterministic lemma that I'll tell you in a second. Before I tell it to you, since
we've already started talking about the structure of the critical random graph, I'd like to say a
little bit more about that because now we're going to start increasing the value of p. I told you
what this looks like at p exactly 1 over n, okay? And on the other hand my comic, what I called
the comic slide said that once we're at 1.01 over n, we've sort of already determined the global
structure, so we need to understand what's going on for p between 1 over n and 1.01 over n.
Okay, which is called the barely super critical phase. And the following is a picture of that. So
let's look at Gn1 plus epsilon over n, yeah. Epsilon’s small, and order the components by size,
and here's the picture. If epsilon is small enough then you can think of epsilon as being 0; we
are really at the critical phase. And small enough turns out to mean n to the minus 1/3. So if
epsilon is this small, then we are back on the picture of the previous page. All the components
have size around n to the 2/3 and diameter around n to the 1/3, okay; the largest components
have those properties. You can, you can easily convince yourselves that this value of epsilon is
at least plausible because if you take two components size n to the 2/3, then there is n to the
4/3 non-edges between them and once one of those non-edges comes along and becomes an
edge, they'll hook up and form a bigger component. So the characteristic time on which 2 of
the sort of largest components will hookup is around 1 over n to the 4/3 and that's precisely
what you get by plugging n to the minus 1/3 over n in there. So this is reasonable. Okay. Once
you get much bigger than n to the minus 1/3, but still smaller than n the picture is sort of in a
way much more uniform. It always looks, here we have sort of two lines to say that this is, you
know, more or less up to constant factors. Once epsilon is much bigger than n to the minus 1/3
a giant component, a unique largest component is already forming that has size around 2
epsilon n. It turns out this is sort of more recent, but also sort of well studied and well known
facts, the second largest component has size asymptotically 2 over epsilon squared log epsilon
cubed n. And so we don't, I'm not saying anything about the diameter of this one, but it is
classically known that the second largest component has this size and the second largest
diameter, which might not be the diameter of the second largest component, yeah, the largest
diameter outside of this giant guy over here is of order of 1 over epsilon log epsilon cubed n.
Okay. So you don't have to worry too much about these logs. They disappear when epsilon is n
to the minus 1/3 and for the purposes of this talk you're not going to be in bad shape if you
pretend they're not there and just treat this as one over epsilon squared and diameter one over
epsilon. So you can think of that kind of picture. I'll come back to some of this again in a
minute when I need it, but it's good at least to have seen it, right? Okay. So here's the
deterministic lemma that I promised that's going to tell us how to understand how these
components hookup. Okay? The picture that's worth drawing for you is that rather than just
looking at all of the connections as they form, we're going to do the following thing. We're
going to treat the largest component of the growing tree as special, okay? And we're gonna
look at how it increases in size and diameter in stages. So first we'll increase p a little bit and
see what new connections form that are not to this component, so that are between other
components, okay. And then once we've raised p a little bit and seen some new connections
coming over here, we will look at how those new components connect to this graph. So that's
the rough picture and that's the picture you should have in mind when I tell you this, this
lemma. Okay. So I'm defining some quantity which is just, it's LP for the longest path, the
largest number of vertices on any path in a connected graph. Okay. That's certainly an upper
bound on the diameter and it's essentially equal to the diameter if we are talking about trees.
The diameter is usually in terms of edges which is the reason for that minus 1. Okay. So if I
take, if I take some fixed graph G, yeah and 2 subgraphs of G but one bigger than the other,
okay? So I have 2 subgraphs and then spanning trees of each of those graphs. This is the
situation I have with my coupling between Gnp and Tnp, right? I got increase in graphs; that's
Gnp and spanning trees of those graphs and I say what's the diameter of T prime in terms of T?
Well an upper bound is whatever diameter I started with plus twice the length of any, the
longest length of any path outside of that tree. Okay? So let me, this is quite obvious when you
draw the picture; it's just saying that you had some spanning tree of H, right, now you have a
spanning tree of H prime. How much could the diameter have grown? Well at most you take
some path here, some path here, right, any path in T prime can be decomposed in this way and
there's some portion that's in H prime and then a path in H and then another path in H prime.
The fact that it's a tree guarantees you that you can't have, you can't have more bits than this.
Okay? So that means that that gives you this upper bound immediately. This twice the longest
path is just the length, an upper bound length of these 2 paths and this is an upper bound on
the length of the middle bit. Okay? Is that clear? Okay. So how are we going to apply that? So
let's go back to our coupling and order the components by size just like I said before. Then this
tells us that if when we increase from p to p prime for some values p and p prime, if which
component is the largest hasn't changed, right, then we can apply the lemma immediately to
get an upper bound of this form. We've got the new diameter is at most the old diameter plus
twice the longest path from out here. And now you can understand where, why I was caring
about the diameters of those, those small components. It's because they're going to come in as
some sort of a bound on this quantity. Okay? So I'll keep that. So now I want to explain in a
little more detail that picture of what's happening in the barely supercritical range that I just
gave to you. And this is, there's sort of a variety of references you can cite. One appropriate
reference is much of the work in this area is due to Thomas Luczak in the early to mid ‘90s. The
particular result that I'm saying here is essentially from a paper in 1994. So in this range that
we are interested in, okay, the largest component has size 2 epsilon n and if you remove that
component, okay, then what's left essentially looks like a sub critical random graph on the
remaining vertices. Okay? So here the parameter was one plus epsilon over n. After removing
the largest component what's left looks like a random graph with parameter one minus epsilon
over however many vertices are left. Okay? So if this, if we just remove, I'd like to explain this
a tiny bit. If rather than removing the largest component, I just removed some fixed set of k
vertices, well the independence, the model tells me exactly what the distribution of the rest is.
It's distributed like Gn-kp, right, and if you want to write p in this form then now you have to
use something over n-k instead of something over n. Okay? So if you just write, rewrite p in
terms of n prime you get essentially one minus epsilon instead of one plus epsilon when this is
the number of vertices that you remove. Okay? So that's the content of this is really that even,
so this would be true even if we removed some deterministic set of vertices and it's still true
even if you remove this particular random set of vertices. Okay? Good. The nice thing about
this is that that tells us that if we look at, if we remove the largest component what we have is
essentially a sub critical random graph, so we can even increase p a little bit on what's left and
we'll still have a sub critical random graph. So in particular, if I take p prime to be one minus
alpha p, okay, then what’s outside of the giant component now is going to look like a random
graph on the remaining vertices with this parameter, 1 minus alpha times what we had before.
Okay. In particular, if I take alpha small enough that I'm still subcritical, right, so alpha small
enough that one plus alpha times 1 minus epsilon is still less than 1 epsilon over 2, okay, then I
can apply the proceeding, the earlier bounds on the length of the longest path outside the giant
that came from that picture. Okay? So the point, again, what this is saying is you were already
at the point where, so this was my comic sketch. We're already to the point where there's one
big guy. Out here we know that the largest diameter was around one over epsilon with some
logarithmic correction that I told you not to worry about, and that would be true even after we
do the thing like I described which is scatter a few extra edges out here, okay? And these edges
come precisely from increasing the diameter from p to one plus alpha times p, okay? So even
after you scatter those the diameter is still small; still you get a bound of the same order.
Good. Okay. Well that more or less is all we need, in fact. Okay? And here's why. So let me
just, and this is really the key point, so let me keep it for a second. Okay? So here's what we're
going to do. We'll start, we know what's going on at one over n, right, I described to you that's
how we got a lower bound. We know what's going on at one over n. We even know what's
going on at one over n if we add in a little bit, okay, of this form. Okay? And to be careful you
need to care about how this lambda changes things, but you'll have to believe me that even if I
pick some large fixed lambda the diameter is still only n to the 1/3. And that you can prove
using classical facts. Okay. So we'll start from here and then we'll start increasing the value of
p in this way that I described. We'll add some edges out here and see how they connect to the
largest component. Add some more edges, see how they connect to the largest component.
And we'll do that in the geometric way. Okay? So this is my starting point epsilon 0, and then
I'll just let epsilon i be one plus alpha to the i times epsilon 0. Okay? So each time I multiply
epsilon by one plus alpha to get the next one. Okay? And I'll keep doing that almost all the way
up through the critical window. Okay. So I'm thinking when I say lambda large I want you to
think of that as saying large enough that we can start pretending that this picture is really the
correct picture, that there's one large component and all the rest are sort of moderately well
behaved. And that holds until we get almost to epsilon equals one. It's only really epsilon
equals little o of one, but let's pretend sort of that one over lambda, this large lambda is a proxy
for small, so we can continue until we get almost up to 1 .01. Our new 1 .01 is 1 plus 1 over
lambda, okay? So what happens, this fact in blue up here, plus the bound from the corollary on
the diameter outside tells you that when you go from stage i to stage i plus 1 the diameter
increases at most this quantity with the value epsilon i that we just wrote. Okay? Which as I
said if you want you can pretend it's just 1 over epsilon i. Okay, good, because we know what
we start with and we know what the difference is at each step, right? And I've just rewritten
that using the value of epsilon. It's n to the 1/3 and if you keep track of the log you get i over
one plus alpha to the i. Okay? So just sum over the whole range. This sum is just, this sum is
dominated by some geometric sum, right, and so even if we sum to infinity these terms we get
something bounded, and so that gets us almost all the way through the critical window and we
still haven't lost more than an extra factor n to the 1/3 in the diameter, so the diameter may
have increased by a constant, maybe even a large constant factor, but not by anything more.
Okay. Great. So I mean, this, if you believe this, it should already convince you that n to the
1/3 is the right answer because we said that at time 1.01, right, one plus one over lambda, all
the other components are already logarithmic, so they shouldn't, it'd be surprising if they had
any very large effect on the diameter, but let me really prove that to you as well.
>>: [indiscernible] order epsilon…
>> Louigi Addario-Berry: No, no. I mean alpha, I think you can take alpha as equal to 5/4, for
example. It's really some, all you need is that each time…
>>: Oh you mean [indiscernible]
>> Louigi Addario-Berry: I mean alpha is a quarter, yeah, so remember…
>>: Didn't you have to [indiscernible] alpha times one?
>> Louigi Addario-Berry: I had 1 plus alpha times 1 minus epsilon is less than 1 minus epsilon
over 2.
>>: [indiscernible] epsilon
>> Louigi Addario-Berry: Oh, sorry, yeah. So that was, this is what I meant.
>>: [indiscernible]
>> Louigi Addario-Berry: So here we have, so we have a random graph with p as around 1
minus epsilon over n, right, and so we can afford 1 minus epsilon over 2 over n, yeah. So I can
take alpha, yeah, I can take alpha like epsilon 1 over 2, yeah. And let me write this
>>: [indiscernible]
>> Louigi Addario-Berry: Say that again.
>>: [indiscernible]
>> Louigi Addario-Berry: Let me write this as 1 plus -- p is 1 over n plus lambda over n to the
4/3. Okay. Then, then to get from, yeah. So I can -- that's fine. So I can, so epsilon is -- so I can
take epsilon like lambda over 2 over n to the 4/3 at the -- so I can, I can increase, right, from 1
minus lambda over n to the 4/3 to 1 minus n minus lambda over 2 n to the 4/3 in the first step.
Yeah? And when I do that, the parameter p increases, right, so p prime then is like 1 plus 5
lambda over 4, 1 over n plus 5 lambda over 4 over n to 4/3. Right? Okay.
>>: [indiscernible]
>> Louigi Addario-Berry: So…
>>: [indiscernible]
>> Louigi Addario-Berry: So, I mean the point is that epsilon needs to geometrically increase,
right? Epsilon needs to geometrically increase so that this sum can geometrically decrease. So
I may have mis-parameterized alpha in this statement, but the point, but even if alpha is epsilon
over 2, right, then we still get a geometric increase in the parameter epsilon i if we add these, if
we add these epsilon over 2s at each step, right? So then epsilon i plus 1 is really epsilon i times
1 plus…
>>: [indiscernible] I don't see how it can work the way you wrote it at 1 plus alpha because I
know than the alpha [indiscernible] becomes too small [indiscernible]. Maybe I'm missing
something. But if you basically are choosing each side of your alpha to be the current epsilon
over 2…
>> Louigi Addario-Berry: Yes.
>>: Then only the first few alphas are small and then…
>> Louigi Addario-Berry: Yeah, and then they grow geometrically.
>>: [indiscernible] formula [indiscernible] because then you have a fixed alpha.
>> Louigi Addario-Berry: No. You're correct, so that came from hastily preparing the slides and
not looking at my old paper, but, no. You're right. So this looks, alpha needs to be small when
epsilon is small, but if you take alpha, I mean in the first step epsilon gets a constant factor
bigger than you can take a alpha constant factor bigger, right? And the point is that then these,
I mean the key point is that these epsilon i -- if you do that, if you start with if alpha i is of the
form constant greater than 2 to the i divided by n to the 4/3, right, then these, then the
corresponding epsilon i are still decreasing geometrically quickly. And so…
>>: [indiscernible]
>> Louigi Addario-Berry: Yeah, the epsilon [indiscernible] are increasing geometrically.
>>: [indiscernible] so now you instead of alpha, you want alpha i?
>> Louigi Addario-Berry: Yes.
>>: [indiscernible] so how do you [indiscernible] to finish?
>> Louigi Addario-Berry: So [indiscernible]
>>: [indiscernible]
>> Louigi Addario-Berry: Let's take that alpha i equals 5/4, 5/4 to the i, alpha and i. Let's…
>>: [indiscernible] increase geometrically [indiscernible]
>> Louigi Addario-Berry: So I think this is fine.
>>: [indiscernible] epsilon done to alpha?
>>: 2 alpha.
>> Louigi Addario-Berry: I mean, you know, and each…
>>: It's still [indiscernible] epsilon [indiscernible] is alpha i times [indiscernible]
>> Louigi Addario-Berry: Epsilon i plus 1 is 1 plus alpha i and that's [indiscernible]
>>: So that means that epsilon i is occurring exponentially and i squared [indiscernible]
>>: No, because of the 1 plus…
>> Louigi Addario-Berry: So sorry for the confusion, is the picture now clear? So let's, let's go
on from here though. I mean, so we can essentially repeat the same idea, right, starting now
from the barely super critical phase to go all the way up to the threshold for connectivity, okay?
So now, right, we want to go, for example, from 1 plus 1 over, 1 over n plus 1 over lambda n to
1 over n +5/4 over lambda n, right, and that increase in the parameter, the duality principle still
tells us that what is outside of the largest component looks like a subcritical random graph with
parameter 1 over n prime minus alpha over n prime. Okay? And the point is that once epsilon
is of the order of 1 is not much smaller than 1, okay, the sort of 1 plus epsilon 1 minus epsilon
dichotomy can't literally be true, because for epsilon greater than 1, it's sort of trivially,
obviously false. Okay? But there's still a duality between the supercritical, between the
supercritical graph and some sub critical random graph after you remove the largest
component, okay? And in that subcritical random graph all of the components will have
logarithmic size even after you increase the parameter a little bit. Okay? By some constant
factor and so now starting from 1 over lambda n we only need to increase the parameter by a
constant factor logarithmically many times to get all the way to 1, okay, so even if at each step
in what's left we, we do the worst thing which is increase the diameter by an additive factor log
n. At the and we've only increased things by at most log squared n. Okay? So I mean, you can
in fact write these 2 arguments in, these 2 steps in a unified way. I mean in both parts you're
just supplying this diameter bound for a judiciously chosen sequence of increasing sequence of
probabilities. Okay? But sort of conceptually things look a little different when you're in this
critical window and you're above the critical window so I prefer to present it like that. Okay. So
I think that's all I have prepared for today. Thank you. [applause]
>> Yuval Peres: Questions?
>>: Is the diameter of the -- the suppose I want to look at [indiscernible] spanning tree again
[indiscernible] very small?
>> Louigi Addario-Berry: Yeah. I think that you can get log or even log log n if you were willing
to spend a tiny bit more and I think there's some more on that.
>>: We should know [indiscernible] [multiple speakers]
>> Louigi Addario-Berry: There may even be a poster [indiscernible] [laughter]
>>: That wasn't [indiscernible] [laughter] [multiple speakers]
>>: [indiscernible] object. That was a flattery question. You subliminally absorbed it.
[indiscernible]
>> Louigi Addario-Berry: So this one? [laughter]
>> Louigi Addario-Berry: So this is like spring electron. I think it's called spring electrical
embedding. It's some mathematical package. The one on the first slide is just uniform points in
the in the plane. This one is really a [indiscernible] of the complete graph with some
embedding in the plane, so the colors in this picture represent edge weights. So all of the black
edges have weights at most one over n and then again, colors, this time colors code the weights
of edges starting from red and increasing to violet.
>>: And the edges have the [indiscernible]
>> Louigi Addario-Berry: The edge is not of the appropriate line. The edges have lines given by
embedding of the complete graph as [indiscernible].
>>: [indiscernible]
>> Louigi Addario-Berry: You'd have to ask [indiscernible] that.
>>: Do you also have concentration of the diameter around the 2/3?
>> Louigi Addario-Berry: So you… 1/3?
>>: [indiscernible]
>> Louigi Addario-Berry: So you won't have -- I don't expect you to have anything to good on
the lower tail, though I'd have to think about it.
>>: What did you mean by [indiscernible]
>> Louigi Addario-Berry: I mean, we can show, we can show in a different paper that in fact the
diameter divided by n to the 1/3 [indiscernible] distribution and that the limiting random
variable has a density, so it's not a, it's not a [indiscernible]
>>: What's the [indiscernible]
>> Louigi Addario-Berry: Yeah, I mean it's, we don't find an explicit formula for it or anything.
>>: [indiscernible]
>> Louigi Addario-Berry: I expect that if you were, you could get some sort of at least
exponential and possibly Gaussian upper tail bounds.
>>: [indiscernible]
>> Louigi Addario-Berry: And, he didn't explicitly state those, but I expect, I expect you could at
least extract exponential [indiscernible] so I mean the, the point is, you know, everything that
this -- one question for answering that is how, what sort of tail bounds do you have on the
diameter, the largest diameter outside of the largest component and there you can get quite
strong bounds which I believe are Gaussian, so you need to understand how these, how these
increments contribute to some final bound, but I expect with some work you could again get
Gaussian tail bound out of that.
>>: So maximum [indiscernible]
>> Louigi Addario-Berry: I'm not sure.
>>: [indiscernible]
>> Louigi Addario-Berry: No. I don't think [indiscernible] maximum…
>>: [indiscernible]
>>: [indiscernible]
>> Louigi Addario-Berry: I mean, there's some, there's some explicit formula for the degree of
the root which comes from some mixture of poisson distributions, and you should essentially
have independence between different parts of the tree. So a guess based on that is that it
should be like log n over log log n. But I haven't proved that, yeah.
>>: [indiscernible] has to be [indiscernible] has to be given [indiscernible]
>> Yuval Peres: Okay, let's thank Louigi again. [applause]