You probably know we transmit all our messages on the internet in 1s and 0s and it's kind of
like a morse code, imagine beeps going around the world, these are our 1s and 0s. What
happens if something goes wrong? Right, when your transmitting a message a 1 turns into a 0 or
0 turns into a 1 which means the other person receives the wrong message. Well how can we fix
this? Well we're going to fix this by turning your message into code words - that's what they're called,
code words. And these code words are a bit more distinct from each other and the idea is kind of like
the phonetic alphabet? Have you ever tried spelling out like a word or name using the phonetic
alphabet? That's that Alfa, Bravo, Charlie thing? The reason you do that is because when you're
spelling out a word the letters kind of sound a bit similar, they can get confused. The letter B
might be confused with the letter d because they sound similar; but then Bravo is less likely to be
confused for Delta in the phonetic alphabet, so it helps you transmit messages without mistakes. So
let's say we're trying to transmit something like a coin toss, right? Heads or tails. When I get a head
I'm going to transmit a 0, and if I get a tail I transmit a 1. But when I'm transmitting the
result a mistake might happen. So just to be sure I'm gonna transmit 000. If I get a tail
I transmit 111. Then it's less likely that I'm gonna get a mistake here because if a mistake
happens - for example if I got something like 100 at the other end you know something's
gone wrong because I'm expecting one of these answers. So something's gone wrong. Now I don't know
which one it is, but I would say it's more likely to be heads because there's only one mistake for
it to be heads. If it was tails that would be two mistakes so I might interpret that as heads.
In the same way if I received 010 or 001 I would say those are heads;
if I received 011 or 101 or 110 I think they're more likely to be tails. So that means that
this list on the left gives me all the messages that I'm going to reinterpret as heads, and the
list on the right gives me all the messages I reinterpret as tails. So it helps me fix mistakes.
Well we can actually think of this geometrically: think of these 0s and 1s as coordinates,
coordinates in three-dimensional space. Let's say this one down here is 000 and then this
one would be 100. This would be 010 and if I go up 001. And let's just
fill in the coordinates we've got here: 011, 110, 101 and up here this
is 111. So all the possible messages become points in three dimensional space. Now imagine we
have a sphere centred on those code words. So let's have a sphere centred at 000
and another sphere centred at 111. And let's say the sphere has radius 1. That means
that these messages that are closest to heads are all contained within that sphere; so a message that
is contained within a sphere is interpreted as the code word that is at the centre of the sphere. Now
it would be nice if we could fit more code words in, that means we could transmit more information,
not just heads and tails - we could transmit more things. So the more code words we can fit into
that space, right, the more things we can send. I might use instead as my code words 000
again, let's use 011, 101 and 110. So I've got four code words, they're
packed in a bit more densely; but the problem with that is, if I use those as code words, if I have a
mistake, if I had then 111 which would be a mistake, it's actually quite close to two code
words so I don't know which one to pick. Although I can tell there's a mistake that's happened I don't
know how to reinterpret the mistake. Now that is a more dense packing though so that is nice;
in fact if you extended that packing there forever you get hexagonal packing which is the densest
packing possible in three dimensions. But let's try another way of sending four code words. So
I'm rolling a tetrahedral die, I'm trying to send the result, and I'm gonna either get 0, 1, 2
or 3. 0 is going to be transmitted as 00000 - so five digits. I'm gonna have
1 be 10011. 2 I'll transmit as 11100, and 3 I will transmit as
01111. So they are five digits long, they are coordinates in five dimensional space.
So these would be our code words as points in five dimensional space and then imagine them surrounded
by a five dimensional sphere as well. So if you get a mistake then - let's say I get something like
11111 it would be closest to 3, that would be reinterpreted as 3. Actually I'm
going to kind of draw this out just to show you what it kind of looks like, it's going to be
2-dimensional of course. So these are going to be our code words, so let's say this is 0 here, this
is a point in 5-dimensional space surrounded by a sphere. That's a 1, that would be code word 2,
and this would be code word 3. This is only a 2-dimensional slice of what I've done there
but that's not the densest packing possible. Let's see what is the densest packing possible. First of
all what's the density of this packing? Let's just treat it as circles here. If this extended forever
we would have a packing of circles. Let's just create a unit that I can repeat over and over
again and then that will be the full packing and if I can work out the density of this unit
I've made then I've worked out the density of the packing. So I've made a little square here, and if
I just repeat that square over and over that means I get the whole packing. Well that square contains
one circle. So the area of that circle, it's pi r squared isn't it? Let's say it's radius 1 so
I'm just going to say the area and then is pi. And what is the area of the square that I've created?
Remember these are circles with radius 1 so it's just 2 x 2 so that's gonna have an area of
4. And the density is going to be the area of the circle divided by the area of the square that
contains the circle - pi over 4 - which is equal to 78.5%. So that's the
density there; and that's not the best density, we can do better than that, you might be able to see
we can do better than that. A better one would be this, we have the circles here and then we can
fill that kind of dimple there so we've got this kind of packing that would go on forever. I'm just
going to split it up with using this parallelogram that is containing one complete circle, just like
before. And it will have an area of pi and now we're going to look at the- I'll just call it a box,
that parallelogram. What is the area of that? It has a base of 2 multiplied by the height - with a bit
of Pythagoras that's going to be square root of 3. So the box has an area of 2 times
the square root of 3. And then the density: we divide one by the other, which is equal to
90.7%. Much better, much more dense. So if you can pack in these code words more densely you're
going to be able to fit more code words into your space. So that is the best density for circles,
um let's have a look at other dimensions. Because we're going to have to use higher dimensions for
this idea, let's have a look what the best possible packing is in all the dimensions that we know of.
So in dimension 2 is what we've just done, pi over 2 root 3, the 90.7%. Proven
to be the best possible by a Fejes Tóth in the 1940s. Dimension 3 is pi over 3 root 2
which is 74.1% which was proven to be the best possible by Thomas Hales in the 1990s. The
best we've found in dimension 4, pi squared over 16. That's 61.7%. In dimension 5 pi squared
over 15 root 2, 46.5%. In dimension 6 the best possible is pi cubed over
48 root 3 which is 37.3%. For dimension 2 we've absolutely
proved it; dimension 3 we've absolutely proved it. The ones I've shown you so far, best known, 7 is pi
cubed over 105 which is 29.5%. And now in dimension 8, the best known is pi to the 4
over 384, the density of that is 25.4%. In 2016 this was proven
to be the best possible by Maryna Viazowska. Not that long ago. And that's because in dimension 8
there is a particular structure - it's called E8 it has a name - where the gaps become big enough that
you can start to fit in extra spheres into those gaps and then the structure becomes rigid again,
and then you get this properly dense packing but it's also been proven to be the densest possible;
including regular and irregular packings. When they did that they very quickly then showed that you
can do the same thing in dimension 24. And in dimension 24 they showed that the best packing
is pi to the 12 over 12 factorial, 0.19%. And that structure is a regular structure called
the Leech lattice. And John Leech who came up with this Leech lattice in in the 1960s was actually
studying error correcting codes when he discovered this structure. But it was Maryna and others that
showed that this structure was also the densest possible in dimension 24. - (Brady: In dimension 24 - that's)
(that's not very dense is it?) - It's not dense, it's not very dense. However we do use dimension 24, the
Leech lattice, or E8 in dimension 8 we do use those structures in error correcting codes that we use
today to transmit our messages. Yeah, so every message is transmitted with this kind of idea.
So I mean I was talking about rolling a die or just sending heads and tails but yes,
if you're sending a message on the internet, so you're sending an email, so your the letters of
the message uh gets turned into a number - which is not that complicated, A's 1, B's 2, Z is 26.
Then you transmit that number. Now you could just transmit that as a binary number if you
want to, which is kind of how ASCII works, or you can use this bit of a code word
idea, have a bit of extra redundancy in there, just so that mistakes don't happen along the way.
So this was the third part in a trilogy of sphere videos and they were brought to you
by Brilliant. Brilliant is an excellent site, it's full of puzzles and quizzes all about
science and mathematics, computer science, all designed to really help improve the way
you think. But one of my favourite things on the site are their problems of the week, have a look
at these, these are just for one week. You can see there are beginners, intermediate, advanced.
You can try and answer them but you can also go down below to a discussion in the community
all about the problems, it's really great. Go to brilliant.org/numberphile to have
a look at the site. There's loads of free stuff but if you use that URL /Numberphile
you can also get a 20% discount on their premium service which unlocks all the good stuff. I'll
also include a link in the description to the problems of the week so you can check them out.
Our thanks to Brilliant for supporting this trilogy of videos all about spheres.