[0:00] Imagine you and a friend both want to draw a 
random network. I said and a friend. Come on  
[0:07] in. Hey. Hey. How you doing? You know I do know 
because I'm you. Okay. Whenever you're ready.  
[0:14] Look. Here's what you do. Okay. Both people draw 
a collection of vertices and then for every pair  
[0:21] of vertices they flip a coin to decide whether 
or not to link those two vertices with an edge.  
[0:28] Initially, the two networks look very different, 
but were these two close friends to carry on for a  
[0:35] countably infinite number of vertices, they would 
end up drawing exactly the same network. And not  
[0:43] just a network that seems similar, the identical 
network. This is the radioraph. Okay, plan two,  
[0:53] folks. I want you both to draw a new random 
graph. that's different to the previous one. Now,  
[1:00] my plan to make it different is instead of 
flipping a coin, I'm going to throw a dice  
[1:05] like on two different axes to make it random. 
And if the top number is a six, which is not,  
[1:10] I would then join the next two edges together, 
which means now there'll be far few edges. So,  
[1:16] I just did that one. That's not going to be drawn. 
So, in theory, I should end up with not a six,  
[1:22] a very different network. So my plan is I've 
numbered all the vertices and whenever I  
[1:30] pair up two of them I'm going to get the smallest 
number and find that position in the other number  
[1:35] in its binary representation and if it's a one I'm 
going to link them. If it's a zero I'm not. So the  
[1:41] first position in two is a zero. So I don't link 
them. The first position in three is a one. So I'm  
[1:47] going to link those. First position in four is a 
zero. Not linked. First position there is a one.  
[1:53] So I link those. Now I do all the second 
positions. So the second position in Well,  
[2:00] that's all well and good folks, but it turns 
out you are both once again drawing exactly the  
[2:06] same network. Unbelievable. Yes, it turns out the 
radioraph is inevitable. Hey, caught it. Thanks.  
[2:17] Even though random graphs may start off looking 
different, the longer you carry on making them,  
[2:24] the more similar they become. And should you have 
infinitely many vertices in your random graph,  
[2:32] it will always converge to be exactly the 
same. Yes, there is only one random graph.  
[2:40] It's inevitable. But no matter what you do, 
you end up with the Rado graph named after  
[2:45] mathematician Richard Rado or Rado. Anyway, people 
before Richard discovered this graph and people  
[2:51] after him have and will continue to because it 
is inevitable. So I think of it as the rando  
[2:57] graph and you won't hear graph theorists really 
talk about imagine a random graph. They will say  
[3:04] imagine the random graph. There's only one and 
you can't avoid it. It's always there lurking  
[3:11] behind you and it is the creator of all other 
graphs. Any graph you think of. So here we have  
[3:18] the Peterson graph. Big fan. You can think of your 
own favorite graph. Somewhere in the RTO graph,  
[3:24] this set of vertices exists exactly like this as 
a subgraph. And not just that one. Boom. Hersel  
[3:31] graph that's in there too. Any finite or indeed 
countably infinite graph you can imagine is in  
[3:39] there somewhere. It's inevitable and it contains 
all. If networks had a god, their deity would be  
[3:47] the radioraph. And that could be it. That could 
be the end of the video. Crazy mass fact. Isn't  
[3:54] that phenomenal? So if that's all you're here for, 
you can stop watching now. Although before you do,  
[3:58] James 3 is sponsoring the video and I want you to 
know uh internship applications for the Hong Kong  
[4:03] office are now open. I'll talk about more about 
it at the end of the video. I've been there.  
[4:06] Wonderful office. You don't have to be in Hong 
Kong to apply. Details later on. But the point is,  
[4:11] while we could end the video here, wouldn't it 
be more satisfying to prove all those things I  
[4:18] just said? Because with one additional fact 
and a little bit of thinking it through,  
[4:23] you can both convince yourself that the radioraph 
contains every other graph and you can convince  
[4:30] yourself that it's inevitable. So, we're 
going to do it. That's inevitable. Okay. Now,  
[4:37] it's just us people taking it seriously and we 
have to define some terms. I've been real fast and  
[4:42] loose on calling things either networks or graphs. 
And that's because your average human would call  
[4:49] a collection of nodes joined together by edges a 
network. Whereas mathematicians have decided we're  
[4:57] going to call this a graph. So that's why I flip 
between them. Sometimes network makes more sense,  
[5:03] but I want to use graph. So if you look things 
up afterwards, you've already got the correct  
[5:07] um terminology. And we're talking a simple graph. 
So for any pair of vertices which is what we'll  
[5:16] often call nodes because then if we call them 
vertices and edges the language is ready for  
[5:21] making this equivalent to shapes like polyhedrrons 
anyway. So for this pair of vertices you either  
[5:27] have them linked together or you don't. That's 
it. You can't have something like a double edge.  
[5:34] I mean in other situations this is fine but in 
a simple graph no none of that and none of this  
[5:42] going out from one vertex and coming back in. 
No none of that. For every pair there either is  
[5:48] an edge or there is not an edge. And I've been 
referring to it being an infinite graph. What  
[5:54] I strictly mean is a countably infinite graph of 
indistinguishable vertices. So that means first  
[6:01] of all for every whole number all the countable 
numbers from one up to infinitely many there is a  
[6:06] vertex for that. So there's more than seven there 
would be infinitely many of these with an exact  
[6:11] correspondence to the counting numbers and they're 
indistinguishable. So while I have numbered these  
[6:17] they're just temporary labels to help us keep 
track. There's nothing inherently for about  
[6:22] that vertex. At any point we could re label them. 
The labels are just a convenience. Technically,  
[6:28] you cannot distinguish between the vertices 
other than what edges there are and there  
[6:33] are not. And that's just it. Um the board can 
can go away. Look at that magic. I'm using my  
[6:39] foot. Uh so with that in mind, as long as you're 
happy with dealing with the notion of infinity,  
[6:45] which can be a little annoying, but that's okay. 
Check out my video about infinitely many $1 bills,  
[6:50] and you're okay following the logical steps of 
graph theory. There's no other prerequisites.  
[6:57] We've got everything we need. We can prove these 
things. You can convince yourself everything I've  
[7:02] said is true. We can achieve all of our hopes 
and dreams that pertain to this specific problem  
[7:09] using a single property of an infinite graph. 
It's something called the extension property.  
[7:14] And what the extension property means is you can 
take any collection of vertices you want from an  
[7:20] infinite graph. And let's call them collection 
A. You can then take a second collection of any  
[7:25] other vertices you want and call them collection 
B. And the only restriction is there cannot be a  
[7:32] vertex that is in A and B. So you pick your own A 
and B arbitrary vertices as many or as few as you  
[7:38] want. The extension property means that out 
there somewhere is at least one more vertex  
[7:43] which is not in A or B and it has the interesting 
characteristic that it shares an edge in common  
[7:50] with every single verticy in A but absolutely none 
of them in B. So you can pick any collection of  
[7:58] vertices in A that you do want to all be connected 
to this new vertex and any vertices in B that you  
[8:04] don't want to be connected and you're guaranteed 
to have at least one vertex which fulfills your  
[8:09] wishes. And that's it. If we can show that a 
graph has the extension property, not only does  
[8:15] that mean that it definitely contains every other 
graph as a subgraph somewhere, but it also means  
[8:22] that any two graphs that both have the extension 
property are actually the same graph. We'll deal  
[8:28] with the coin and the dice graphs in a moment, 
but first of all, we'll do the one where we use  
[8:33] the binary digits to decide if we're going to link 
two edges or not. And here I've got my categories  
[8:38] A and B and they're numbered. And again, the 
numbers are arbitrary. We've just numbered all  
[8:44] the vertices in some order. Doesn't matter what. 
A quick recap on the binary method for edge rules.  
[8:52] You have two vertices. You pick whichever one has 
the lowest albeit arbitrary number. That number  
[8:58] will be our place position value. We then look at 
that place position in the binary version of the  
[9:05] other vertex's bigger number. And if that position 
in the bigger number is one, then we join those  
[9:12] two vertices together with an edge. And we don't 
if it isn't. And what I'm about to do next is show  
[9:18] for these two sets, we can always find by a method 
of construction a vertex outside of them that  
[9:26] fulfills the extension property requirements. And 
we look at all the numbers in here and we think,  
[9:32] okay, there's going to be some other vertex out 
here, number n that links to these, but not those.  
[9:40] So, first of all, we just guarantee n's going to 
be bigger than any of these. So, it's going to be  
[9:44] bigger than six. Six has three digits in binary. 
So, long as this has more than three digits in  
[9:50] binary, it's going to be bigger. And then we go 
through and we construct our number. You know  
[9:55] what? I'm actually going to give it six digits in 
binary. And then we go through and we fill in the  
[10:01] numbers to make it do what we want it to do. So we 
want it to link to one, so we put a one there. We  
[10:07] want it to link to three and four, so one's there 
and there. We don't want it to link to two, so we  
[10:12] put a zero there. We don't want it link to six, 
we put a zero there. And then the rest, whatever  
[10:18] we want. So we happen to not care about the fifth 
one. I don't know. Uh, let's make that zero. Let's  
[10:24] be a little bit lazy. which means now uh we've 
got the number 13. So it turns out n equals 13.  
[10:33] Vertex number 13 if we were joining vertices up 
with that binary matching rule would fulfill our  
[10:41] extension property requirements. And that would 
work for any set of numbered vertices in A and B.  
[10:49] You could definitely find another numbered verticy 
which links to everything in A but nothing in B.  
[10:54] and you're doing that by construction. The 
probability ones a little more hazy. Okay,  
[11:01] for the coin flip graph, we're going to start with 
the same two categories of A and B. And then we're  
[11:07] just going to pick some vertex out here somewhere. 
Let's uh say that's vertex, no, vertex I. There's  
[11:14] a probability vertex I is the one we're after. 
There's a probability it's not. It's probably  
[11:20] not you. We could say actually the probability 
of the vertex we're after not being I is equal  
[11:31] to some value X and X is between zero and one. It 
can't equal one because there is a chance it's the  
[11:41] one we're after. It's just probably not. So we'll 
look at another one. Here's a second vertex. We'll  
[11:47] call that vertex J. There's a probability that's 
the one we're after. probably not the probability  
[11:53] that it's not I or J equals X to the 2 now because 
we got two of them and the key thing here now is  
[12:02] that X squ is going to be smaller than X because 
X is less than one so if we were to pick another  
[12:07] vertex at random K and then check is it not I J 
or K we're now up to X cubed it's even smaller and  
[12:14] if we had n different vertices that we've looked 
at the probability that we've not found the one  
[12:22] we're after yet is x to the n. And as n approaches 
infinitely many, x to the n is going to approach  
[12:32] zero. So the probability after we've checked 
infinitely many other vertices outside of a and  
[12:38] b that we haven't found the one that we're after 
that satisfies the extension property is zero.  
[12:44] So the probability that vertex does exist is one. 
It's almost certain, which is subtly different to  
[12:53] guaranteed. I don't want to get into probability 
like measure theory stuff. There's always like  
[12:59] some malicious graph like what if there were 
no edges at all? I don't care about that. The  
[13:03] point is if you've got infinitely many vertices 
that are joined together randomly, it is almost  
[13:09] certain with probability one that the extension 
property holds. And we haven't picked a value for  
[13:15] x that's anything between zero and one. So this 
logic holds for coin flips. It holds for rolling  
[13:21] a dice. As long as there's a nonzero chance you've 
linked any given pair of vertices together, you  
[13:27] will have the extension property almost certainly 
with probability one. And who can ask for more  
[13:34] than that? The point is all random graphs converge 
on the rattle graph with the extension property.  
[13:44] We now know that all our graphs have the extension 
property. And we'll deal with proving that they're  
[13:49] all the same graph in a moment. It's easier to 
first of all prove that the extension property  
[13:54] implies all other graphs are subgraphs. So as 
an example here, I have a nice friendly small  
[14:02] graph. I have the diamond graph and I've 
numbered the vertices 1 to four. Again,  
[14:08] that's arbitrary. You can number them however you 
want. We're going to prove that this indeed any  
[14:13] other graph is somewhere in the radar. And we do 
that by building it up one vertex at a time. So  
[14:20] the lowest numbered vertex one, just pick a vertex 
that's going to be our corresponding number one.  
[14:27] Then to fill in the second vertex, we work out 
what we need it connected to. So we've already got  
[14:34] one. So we do want it connected to one. So we're 
going to call that category A. And we don't want  
[14:42] it connected to well, we haven't got anything else 
yet. So strictly speaking, B would be a null, some  
[14:48] kind of null set over here somewhere. Uh that is 
just completely empty. We don't care. And we know  
[14:54] by the extension property, we can definitely find 
a vertex that's connected to everything in A and  
[14:58] nothing in B. So now we've got two. Now three has 
to be connected to both of the previous ones we've  
[15:06] already got. So now we redefine A to not just be 
one but to be one and two at the same time. And  
[15:15] B is still the empty category. And we know by the 
extension property we can definitely find a vertex  
[15:23] three add it to the graph and then we just keep 
going. Each time we want to add one more vertex  
[15:29] we define our categories A and B to force that 
verticy to exist. And we know by the extension  
[15:36] property it will. So four four has to be connected 
to two and three but not one. Give me a second.
[15:51] All right. I've defined three and four to be in 
A and one to be in B. So we know by the extension  
[15:57] property we can find a vertex that's exactly 
the same as what we want for our four links  
[16:02] to both of these but not that one. And then we 
just keep going for each new vertex. Redefine A  
[16:07] and B. It guarantees we can find the next vertex 
down. Fill it in. Repeat over and over and over.  
[16:14] As long as we have the extension property, we 
can build up any graph we want of any size of  
[16:20] our choosing. Finally, I have two infinite graphs 
that are generated however you want. Coin flips,  
[16:27] binary, your choice. The point is they both have 
the extension property. And we're going to prove  
[16:32] they're exactly the same graph. And we're going 
to do that by gradually putting each of them into  
[16:37] a region within the graph. And we'll keep the 
two regions identical until they cover all of  
[16:44] both of the graphs, showing that the whole graphs 
are identical. Okay, just bear with me. Here's  
[16:49] how we're going to do it. First of all, we'll use 
the lowest numbered vertex in each of the graphs.  
[16:56] Doesn't matter how you number them. Just assign a 
whole number to every single vertex and put vertex  
[17:01] one and vertex one in a region. Doesn't matter 
which because they're not joined to anything  
[17:05] else in the region. We're now on the first graph 
going to move in the second lowest numbered vertex  
[17:13] which may or may not be connected to the first 
one. In this case it is and we have to do the  
[17:17] same thing on this side. So using the extension 
property we find another vertex that is connected  
[17:22] to the first one may not be the second one and we 
move that in. We then pick the next lowest number  
[17:27] vertex outside the region in this graph. Check 
how it's connected to all the ones already in  
[17:33] the region. Move it in. And at the same time, 
use the extension property to find a matching  
[17:39] one over here and move it in as well. And then 
find the next lowest on this side. Move it in.  
[17:43] Find the matching one. Move it in. And then we 
go backwards and forwards. And because each time,  
[17:48] whichever one we move in on one, we can find a 
matching one using the extension property in the  
[17:53] other one. And because we're going backwards and 
forwards, we're not leaving any unmatched. We will  
[17:59] gradually match up all the vertices in both of 
these infinite graphs into pairs where each pair  
[18:06] are identical vertices. And that's it. That's 
we can do that and we can prove both graphs are  
[18:15] identical. Now again, this does in involve not 
getting too stressed out about infinity and  
[18:22] these kind of matching things up. But if you've 
done stuff involving, you know, Hilbert's hotel  
[18:26] and and matching stuff with countably infinite 
sets, you're already you you'll recognize this  
[18:32] this kind of matching approach. But it does work. 
As long as you're systematic, you can prove these  
[18:38] are exactly the same graph. All infinite random 
graphs are the radraph. There's only one. So  
[18:46] that is the rando graph. And isn't it satisfying 
that we actually went through the explanation,  
[18:52] the logic, the proof, and we took what were 
counterintuitive weird properties and eventually  
[18:58] made them well almost obvious. In fact, the longer 
you think about it, the more obvious they get. And  
[19:03] the whole extension property is just kind of 
codifying obviousness. I mean, it's my theory.  
[19:10] Any sufficiently well-explained mathematics is 
indistinguishable from being obvious. for some  
[19:18] in some cases a lot of explanation. But the point 
is understanding things is its own joy. And if you  
[19:24] enjoy learning things, oh, here comes a segue for 
you cuz this video is sponsored by Jane Street and  
[19:31] they have opened their internships in their Hong 
Kong office. Yes, they do also have internships  
[19:37] available in their New York and London offices, 
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[19:42] enthusiastic individuals to join the team in Hong 
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[20:20] you what they know. And when you're not working, 
you get to enjoy Jane Street's fantastic office in  
[20:24] Hong Kong, join in the social events, and watch 
the guest speakers, which this year included  
[20:30] me. I try to go out to Hong Kong as often as I 
can. I was there to see the most recent fleet,  
[20:36] actually the most recent two groups of interns in 
Hong Kong. I gave a talk. I got to hang out. Uh,  
[20:42] we ate pizza. We talked about maths. We had a 
great Well, I had a great time. I think everyone  
[20:46] else had a great time. Anyway, I always enjoy 
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[21:36] Please do check out the details or pass them on to 
someone who you think would benefit from them. And  
[21:40] huge thanks to Jane Street for sponsoring this 
video. Well, that's it. Thank you so much for  
[21:45] watching the video. But before we go, for one last 
time, here comes the whiteboard again. Like magic.  
[21:50] How's that even happening? Huh? It was me. That 
was me the whole time. Yeah. Yeah. And while we're  
[21:56] here, if you have watched right to the end of this 
video, I mean, thanks. It's been a wonderful year.  
[22:02] Next year, calculating pie on the moon. Yeah. 
And before the year is over, you can still sign  
[22:08] up for a digital Christmas card from me. Yes, 
it's too late for us to post a physical card.  
[22:15] If you do sign up, uh we will email one to you. 
So that's it from me and me. Thanks for watching.