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# NYT: Sperner's lemma defeats the rental harmony problem

Welcome to another Mathologer video. Today is about
a math story that made it into The New York Times
something that's really very rare.
It's about a familiar problem. A couple
of friends,
Ashwin, Bret and Chad want to rent an
apartment. Now the rooms are quite
different and the friends have different
preferences, different ideas about what's
worth what. Is there a way to split the
rent and assign the rooms to the
friends so that everybody is happy?
It's really quite a tricky problem. Quite
recently it was discovered that a very
famous and very pretty result in
abstract math can be brought to bear on
this and many other related fair
division problems. So my mission today is
to sort all this out for you as best as I
can. Let me know in the comments how
I did. My favorite math channel is
Grant Sanderson's 3blue1Brown. Now this
week Grant and I have conspired to bring
you videos about two very different
fair division problems. If you like my video
or even if you don't, check out 3Blue1Brown,
Grant really does some wonderful work there.
Alright let's start with this
mathematical results, it's called Sperner's
lemma and it's due to the mathematician
Emanuel Sperner who published it in
1928. It's been awhile. In maths this pretty result
comes up in all sorts of
different contexts. So if you are in the know you
might recognize some of these. Okay so
here we go, take a triangle, any triangle
and triangulate it, that means chop it
into a network of little triangles like
this, ... or like that
We'll actually use a super-nice
triangulation of an equilateral triangle
for illustration. You all know this one, right?
Important: everything I'll say will work
for all other other triangulations, all other
triangles.
Color the corners with three different
colors, red, green and blue.
On the bottom edge randomly color every
vertex green or blue like this. On the
left edge randomly color every vertex
red or blue, for example like this.
The vertices on the right side are colored
red and green. Color the vertices in the middle
randomly, no restrictions apart from them having to be
red, green and blue.
Maybe something like this. Now,
there are infinitely many ways to
triangulate and infinitely many valid
ways to color the vertices of
triangulations and yet we can be
absolutely, absolutely certain, and that's
really what Sperner's lemma is about, that the
corners of at least one of the little
triangles in there feature all three
colors. For example here's one. Now there
can be more than one. For example, here's
another one, and there's another one. But
we can always be sure that there is at least one
of them. So how do you prove something
like this.
Well, there's a really really nice proof.
Think of this big triangle here as
a house with the little triangles being
the rooms. Every room is bounded by
three edges.
Let's say an edge is a door if it's two
ends are colored blue and green like that
one here.
Now there's are lots of doors in here, so
let's just highlight all of them. Here
they are. Important: instead of making
the blue-green edges the doors we could
also have chosen the red-green edges or
the red-blue edges as door. That would have
given different sets of doors but the
following argument stays the same for all
these choices. Here are three
important insights about the doors in our
house.
Let's start with a question: What are the
possible numbers of doors in a room.
Zero doors is definitely possible like
this. Exactly one is also possible and if
this is the one door here
what can we say about the color on top?
Well it cannot be blue or green because
that would give a second door like this, or
like that and so this means there's got to
be a red on top right? And that means
that the rooms that have exactly one
door are exactly the special 3-colored ones.
OK so that's pretty good already. Now as
we've already seen two-door rooms are
possible.
They have either two blue vertices and
one green, like this, or two green and one
blue. But then once a room has two
doors it's going to be one of these two
types which means that three doors is
impossible.
So to summarize, a room has either
zero, one or two doors and the
ones with exactly one door are exactly the
special 3-colored ones. Next, because of the
restriction of what colors can go on the
sides of the house it's clear that all
doors on sides have to be along the
bottom. We can also say something about
the number of doors at the bottom
namely there will always be an odd
number of doors at the bottom, so 1, 3, 5, 7
etc. In this case we've got three.
Why always an odd number of doors at the bottom?
Well I leave this for you to figure out
in the comments, actually pretty easy.
Now we're ready for the proof that
there will always be at least one of
those special 3-colored rooms in our
house. Let's enter the house through one
of the doors at the bottom, here we go.
Either the room we're in has exactly one
door or exactly two doors. If it has
exactly one door
the room is special and we're done, we
found a special room. Otherwise let's step into
the next room via the second door. Again
either we are in a special room and
we're done or we can keep going and as
we continue like this will never enter
the same room twice. But then since the
only finitely many rooms our trip must
end either in a special room in which
case we're done or we leave the house
through a second door at the bottom. So
far we've walked through two doors at the bottom ,
right, but since there's an odd number of
doors at the bottom there's at least one
more left over,
at the bottom, that we have not passed yet.
So let's reenter the house through one door
like this, at the bottom. Again we'll never
enter a room that we've already visited
and so the second trip will also either
end in a special room or another door at
the bottom. So, going like this, going like
that. Every time we end up outside
we've used up an even number of doors at
the bottom and since there's an odd
number of doors at the bottom, we can
always find a door to re-enter at this
stage. Now obviously this cannot go on
forever and therefore we eventually must
end up in one of those special
3-colored rooms no matter how hard we try
to avoid this. Isn't this a wonderful
proof. Now there's a bit more: you can
actually show that there's not only just
one special room but an odd number of
them. So, for example, in this case we've
got three so maybe also see whether you
can extend our proof in the comments
to show this. Now, how can all this be used to
achieve rental harmony. Let's say the rent
is 500 dollars,
Well it's obviously not New York City :) So
then there are lots of different ways to
split up the rent, for example,
\$100+ \$200 + \$200 or this one here or,
crazy but or well it's just one rich guy
paying for his friends. Now we turn the
big triangle into a mathematical space
of rent divisions.
What this means is that every one of the
points inside it will stand for exactly
one of these possible divisions. OK
let me explain.Take a point on the inside
of this equilateral triangle here and
add up the distances to the three sides.
Now and this is really quite wonderful
this distance sum is always the same in
equilateral triangles. Red plus blue plus
green is constant. This is called Viviani's
theorem. So this distance sum is the same
as that one here, it's the same as that
one and it is the same as, well let's
make red and green vanish by moving the
point to the corner.
So what this means is that our constant
is really the height of the triangle.
Now let's superimpose our
triangulation. Using this special
triangulation we can illustrate all its
nicely. So the big triangle is five
little triangles high: 1, 2, 3, 4, 5
So let's say the height is 5 units.
Let's move the points. As you can see red is one
unit long, blue is two units long and
green two. And 1+2+2=5.
Works. Try another one: red is
zero, blue is 2 and green is 3. Works!
In this way every point in the big triangle
corresponds to exactly one of the
possible ways to write 5 at the sum
of three non-negative integers or,
if we scale everything by 100, the
vertices of our triangulation correspond
to the different ways of writing \$500
as sums of multiples of \$100.
For example, here we've got \$0 + \$200 +\$300.
Or, moving to a different vertex and
maybe one more, like that.
OK, now we somehow want to use Sperner's
lemma to choose how we should split up
the rent and who among the friends
should move into which room and the friends
should end up happy with all this.
To do this we first label the vertices
with the friends, like so. Now the letters
are distributed in a regular pattern
such that all three letters occur at the
vertices of every one of the little
triangles, like this little triangle
there has ABC around it. That little
triangle has ABC around it and really you can
check, every other little triangle has ABC
around it.
Now let's look at one of the possible sums.
\$0 + \$400 + \$100.
The corresponding vertex is labeled B, so let's
ask Bret which room he'd go for if the
rent was split up like this.
Well the red room costs nothing
and so it's reasonable to assume that
Bret will go for the red room
In fact,
we'll assume that everybody who's offered
a free room will go for it.
OK, Bret goes for the red room, so we
color this vertex red. Next vertex. The
new vertex is labeled C so it's up to
Chad to choose. Since red still costs
nothing Chad also goes for this room.
So in this way all the vertices on the left
will all end up red. For the same reason
the bottom edge will end up blue and the
right edge will end up green. Now in a
corner we've got two free rooms and in
this case
Ashwin will choose either one of them.
You get the same sort of choices in the
other corners. Now we fill in the rest by
going to the middle, and in the middle
there's no 0 here, so Bret can choose
whichever of the rooms he likes best
given this rent division, say he goes
for red. Now we continue to fill in the rest
and it looks like we may be able to somehow
apply Sperner's lemma with all those
colors. And we do if the choices in the corners lead to
three different colors,
for example like this. If this happens then
we have a coloring to which
Sperner's lemma applies, that it, we have a
little 3-colored triangle, there we go :)
How does this help with our problem?
Well the way the corners are labeled
ABC tells us that Ashwin should take
the green room,
Bret the red and Chad the blue. What
about the division of the rent.
Well the divisions corresponding to the
vertices of the little triangle are very
close, especially if we use a fine
triangulation like this one here.
Just in case you cannot see properly, the arrows
point at the tiny special triangle that I've
magnified up here. The three corners of this
this triangle correspond to
\$120 + \$150 + \$230 dollars, ... that one
here, and finally that one here. So
the differences are in the
ten dollar range and every friend was
happy with this assignment of rooms there
for one of these divisions, and so
picking any of these three divisions
randomly should work for all three friends.
Now if some unhappiness persists finer
triangulations will get the divisions to
differ in arbitrarily small amounts. Now it's
possible to use the same sort of ideas
to come up with all sorts of other types
of fair divisions. I've linked in the paper by
Francis Su on which all this is based, as
well as to link to the New York Times
article which contains a nice apt to
explore as well as a website that allows
you to actually make real-life decisions
based on this circle of ideas.
A couple afterthoughts: all this
also works if you're dealing with less
or more than three rooms you just have to
use lower or higher-dimensional versions of
Sperner's lemma. I''m not going to go into
details here.
You may also have gotten the impression
that all this is impractical because to
color in a triangulation like this you
have to ask the friends to make as many
decisions as they are vertices, in this
case it is more than a thousand. However,
remember, when we looked for the special
little triangles we only visited a very
small part of the large triangle, so the
idea is to color in the triangulation
"on demand" as we walk around it.
This will cut down on the number of
decisions, and using some other
refinements it's possible to reduce the
number of decisions to something
manageable as in the online app.
One more thing:
Remember, at this point of the story,
choices at the corners either lead to
three different colors, like this. Then
Sperner's lemma applies directly. However,
you can also end up with this sort of
coloring here which is slightly
different from a Sperner coloring because
the corners of the big triangle don't
feature all three colors. Can you explain
how things have to be adjusted to save
the day here? Again let's have some fun
discussing this in the comments.
And that's it for today. Now I'll go and
have some rest but you should now go and
explore a bit what other miracles
mathematicians work using Sperner's lemma.
For example, if you understood everything
in this video you are actually on the brink
of understanding the nifty proof of the
famous Brouwer's fixed-point theorem that I've
linked in in the comments. So have fun
with that, too :)