 Let's talk about
the card game SET.
To play, you start with a
deck of cards, each of which
has a certain number of
shapes and different colors
and shadings.
You deal out 12 cards
and start looking
for a set, a collection
of three cards that
have either all the same
or all different patterns.
Now, once you deal
out those 12 cards,
it's possible that there
might not be a set among them.
When that happens, you just
deal out three more cards,
and in some cases, there
still might not be a set,
so you can add three more cards.
And this begs the
question, what is
the maximum number
of cards you can deal
that might not contain a set?
[MUSIC PLAYING]
Before I tell you
the answer, let
me start by clarifying
how the game works.
You have a total of 81 cards,
and each card displays an image
with four features,
shape, color, number,
and shading, each of which
can vary in three ways.
For example, this card has
three green striped diamonds,
this card has one
purple solid squiggle,
and this card has two red
outline ovals, and so on.
To start the game, you
deal 12 cards face up,
and again, the goal is
to collect the most sets.
A set is a collection
of three cards where
either each feature on
the cards are all the same
or each feature on the
cards are all different.
For example, this is a set,
because all three cards have
the same color, same number
of shapes, same shape,
and all three cards
have different shading.
This is also a set.
All three cards have different
colors, different shapes,
different numbers of shapes,
and different shading.
And here are some
other examples of sets.
Now, if your 12 cards
don't have a set,
then you can keep dishing
out cards three at a time
until a set is present.
So how many cars does this take?
In other words, what
is the maximum number
of cards you can deal that
might not contain a set?
It turns out the answer
is 20, and the proof
is pretty involved, but
to whet your appetite
for the kind of math
behind the scenes, let's
ask this exact same question
but for a lower dimensional
version of the game.
That is, let's suppose
that the cards only
have two instead
of four features,
like number and shading.
In other words, pick a color,
like green, and a shape,
like diamond, so only the
numbers and shadings will vary.
That means our deck now
consists of nine cards.
And we can ask
the same question,
which I will pose as this
week's challenge problem.
Out of these nine cards,
what is the maximum number
you can deal that might
not contain a set?
You can approach the
problem in any way
you like, but in the
next few minutes,
let me try to steer you
in a geometric direction.
It's a small mathematical
detour, but at the end,
we'll see how it circles right
back to the challenge problem.
OK, here's the detour.
If you watched our video,
"How to Divide by Zero,"
then you're familiar
with the quotient set
Z mod nZ For example,
if n equals 5,
then Z mod 5Z is the set of five
equivalence classes, 0, 1, 2,
3, 4.
Or if n equals 3, then Z mod
3Z is the set 0, 1, and 2.
They're equivalence classes,
but for the rest of the episode,
let's just think
of them as numbers.
And you know that we can do
arithmetic with these numbers.
For instance, here's
the multiplication table
for Z mod 3Z times Z mod 3Z.
0 times anything is always 0.
1 times 1 is 1, and 2 times 2 is
4, which is 1 mod 3 and so on.
OK.
Now I want to show
you something else.
Let's replace each
of the numbers
by their original factors.
For example, this 1
is really 1 times 1,
so let's replace it with
the coordinate 1 comma 1.
Similarly, this 2
is really 1 times 2,
so let's replace it by
the coordinate 1, 2.
Now do this for all the points.
For each coordinate,
the first number
is the x, or
horizontal component,
and the second number is the
y, or vertical, component.
So we end up with a
three-by-three grid.
This grid is called the
Cartesian product or Z mod 3Z
cross Z mod 3Z,
because it's just
like the usual Cartesian, or
x-y plane that you're used to.
So really, you can think
of the Cartesian plane
as a kind of multiplication
table for real numbers,
except instead of considering
pairs of real numbers,
we are now considering
pairs of modulo 3 numbers.
And that's why we get a
plane comprised of only nine
and not infinitely many points.
Now, what's cool is that
we can still do geometry
with these nine points.
In particular, we can
still make sense of lines
in this three-by-three grid.
Remember, a line in the
usual Cartesian plane
is a collection of points, x,
y that satisfy the equation
y equals mx plus b, where m, the
slope, and b, the y-intercept,
are real numbers.
In the exact same way, a line
in the grid of nine points
is a collection of mod
3 numbers, x and y,
that also satisfy the equation
y equals mx plus b, where now, m
and b are numbers in Z mod 3Z.
For example, these
three points comprise
the line y equals 2, since the
y component of each coordinate
is 2.
So here, m is 0, and b is 2.
And this diagonal line has
the quation y equals x.
So m is 1 and b is 0, because
the x and y components
of all three
coordinates are equal.
And this line has equation
y equals x plus 1.
And this takes a
little extra thought.
It doesn't look like a
line, so is it really?
Well, the answer is yes if
each of the three points
satisfies the equation
y equals x plus 1.
So do they?
Absolutely.
Look at the point 1, 2.
It's y component, 2, equals
it's x component, 1 plus 1.
And that's exactly what it means
to say 1, 2 lies on the line y
equals x plus 1.
And the same goes for the
other two points, therefore,
these three points
do form a line.
So why doesn't it look like it?
Well, I claim it does.
Notice what happens if we were
to expand our grid a little.
Modulo 3, the point 0,
1 is equivalent to 3, 1.
Similarly, 1, 2 is equivalent
to 4, 2, and voila,
there's our line.
The idea is that once we
reduce mod 3, lines simply
wrap around the edges.
And this is exactly
what happens when
forming a quotient like
in the Pac-Man universe
or a flat torus that Gabe talked
about in his episode "Telling
Time on a Torus."
OK.
This is all very nice, but
it's not just a cute fact.
This is exactly what we need
to solve the challenge problem.
Let's recall what it is.
Out of these nine cards,
what is the maximum number
you can deal that might
not contain a set?
OK.
How does that
relate to geometry?
Remember that the shading of the
cards comes in three varieties,
solid, striped, and outlined.
Now, think of each option
as an element in Z mod 3Z.
For example, let's assign
0 to solid, 1 to striped,
and 2 to outlined.
Likewise, there are three
different numbers of diamonds,
one, two, and three.
So let's assign 0 to three,
1 to two, and 2 to one.
And now, we can encode
features of the cards
via pairs of numbers, a, b.
The first component
represents a shading value,
and the second component
represents a number value.
For example, this card
corresponds to 1, 1
because it has two
striped diamonds.
And this card
corresponds to 0, 2
because it has
one solid diamond.
Do this for each card and we
recover our coordinate plane
with nine points.
Now, let's think back
to what a set is.
A set is a triple
of cards, where
the shading and number of shapes
is either the same or different
on all three cards.
For example this forms
a set, because the cards
have the same number but
each has a different shading.
Moreover, this set
corresponds to a line, namely,
y equals two.
Interestingly enough, the x
components all add up to zero,
and the y components
all add up to zero.
As another example,
these three cards
also form a set,
because the cards
have different shading
and different numbers.
Moreover, this set corresponds
to a line, namely, y equals x.
And interestingly enough, the x
components all add up to zero.
And the y components
all add up to zero.
And here's one last example,
these three cards form a set,
because again, their numbers
and shadings are all different.
Moreover, this set also
corresponds to a line, namely,
y equals x plus 1.
And interestingly enough, the
x components all add up to zero
and so do the y components.
OK.
So it seems like sets correspond
to lines in the Z mod 3Z grid,
and here's a hint, it's true.
They do.
You can prove it.
Now, can you use that
to solve the challenge?
Here it is again.
Among the nine cards,
what is the maximum number
of them that may
not contain a set?
Can you rephrase this question
in an equivalent but geometric
way and then answer
it using the hint?
Once you've got your
solution, send it to me
in an email at
pbsinfiniteseries@gmail.com
with the subject
line "SET Challenge."
but don't just send me a number.
I want a number and a proof.
Then, we'll select
a random winner
and send you a T-shirt
from PBS Digital Studios.
Good luck, and
see you next time.
Hey, everyone.
Just responding to some comments
from last week's episode
on Fermat's Last Theorem.
First of all, I'm
glad you enjoyed it.
Abstract algebra
and number theory
are each beautiful
subjects in their own right
but combining them makes
for some really great math.
Secondly, as Matheus
and many of you
pointed out, in this
second definition of prime,
an integer is prime if whenever
it divides a product of two
numbers, then it divides
at least one, not exactly
one of the numbers.
So Deoxal, I hope you
didn't lose too much sleep.
That was a slip of
the tongue on my part.
So good catch.
Also, some of you who are
more familiar with the nuances
of ring theory, like
Nicholas and tpog,
pointed out that my
statement around nine minutes
in, that prime and irreducible
are the same if and only
if your ring is a
UFD is true when
your ring as an atomic or
a factorization domain.
So yeah, absolutely.
You need factorization
before you can even talk
about unique factorization.
Good catch.
And don't worry if you're not
familiar with these terms.
What folks are pointing
out is that rings
come in tons of
different flavors,
and in the last episode, we were
dealing with really nice rings.
But ring theory has a lot
of nuances to explore.
If you're interested
in diving deeper,
I put some great references in
the description of that video.
See you next time.
