[MUSIC PLAYING] You're throwing a party, a huge party, with a thousand bottles of wine.

But there's a problem-- exactly one bottle is poisoned.

Turns out, this is the set up for a classic math problem, and the solution will come from an unexpected place-- binary numbers.

10 00:00:27,270 --> 00:00:29,130 First, I'm going to give you the setup for this wine-filled puzzle, and then you can try to solve it.

Like I said, you have a thousand bottles of wine and exactly one of them is poisoned before the start of your epic party.

Fearing that your guests might suffer from sobriety, you want to save as many bottles of wine as possible, but you really don't want to kill anyone, so you only want to serve bottles that you're confident aren't poisoned.

Fortunately, you have 10 pesky rats at your disposal and no qualms about poisoning them.

You know you're going to have to test the wine on the rats to narrow down which bottle is deadly, but how?

The party starts in one hour, which by some sort of contrived math problem coincidence is exactly the same amount of time for the poison to take effect and kill any rat that drank it.

This means that you only have one chance to run your rat poison experiment.

It wouldn't work to feed some of the wine to the rats, wait, see if anyone dies, then feed them more wine, wait again, and so on.

By then, all of your guests have left because your boring party has no wine.

Because this is a math problem and not a real life problem, we can make some funny assumptions.

Like the rats are capable of drinking a lot of wine.

If you want to give one rat a drop from every bottle, go for it.

Except, that's not a very helpful math strategy because that rat will definitely die and won't provide you with any information about what bottle contained the poison.

So that's the entire setup of the puzzle.

But how do you solve it?

Well, here's my first suggestion.

Number the bottles 1 to 1,000.

Feed the first rat wines number one to 100.

Feed the second rat wines number 101 to 200.

The third rat gets wines 201 to 300, and so on.

Each of the 10 rats samples from a hundred bottles of wine.

This way, if the seventh rat dies, you know that the poisoned bottle is numbered between 601 and 700.

Using this strategy, you can confidently narrow it down to 100 possible bottles of wine.

So you throw away 100, but you save 900.

Not bad.

Can you think of a better strategy than this one?

Can you save more than 900 bottles of wine?

Is there a way to figure out exactly which bottle is poisoned saving 999 other bottles?

If you want to try and solve this problem with no help from me, pause the video now, and come back when you're done.

If not, stay tuned for the unexpected solution.

Amazingly, you can figure out exactly which bottle was poisoned, but the method is tricky and involves binary numbers.

Here's a crash course in binary.

Normally, we write numbers in base 10.

This means that the digits represent increasing powers of 10-- 1, 10, 100, 1,000, and so on.

But in base 2, or binary, the digits represent increasing powers of 2-- 1, 2, 4 8, 16, and so on.

Now, instead of having a tens place or hundreds place, we have a 16th place, or a 64th place.

In base 10, you can put the digits 0 through 9 in each place.

When you write a number like 2508, the digit 8 represents 8 ones, the 0 represents 0 tens, the five is 5 hundreds, and the 2 is 2 thousands.

But in base 2, you can only put a 0 or 1 in each place.

You can think of it as 1 indicating the presence of that digit, and the 0 indicating its absence.

So the number 53 is 1 1 0 1 0 1 because 32 plus 16 plus 4 plus 1 is 53.

Here's how you represent the numbers 1 through 20 in binary.

Cool, right?

But back to what's important.

How is all this binary number stuff helping us save wine?

Basically, by giving us a way to encode each bottle.

Similar to the last strategy, label each bottle of wine with the number one through 1,000, but this time underneath their base 10 representation, also label them in binary.

Now, set up the rats in a row.

These rats will represent the digits of a binary number.

So the right most rat represents the ones.

Then the next one over is the twos, then the fours, and so on.

The left most rat represents the 512th place.

Take each bottle of wine, and feed it to those rats which represent a one in the binary number on the bottle's label.

That's kind of a mouthful, but here's some examples.

Bottle number 4 is 1 0 0 in binary.

So feed one drop of bottle 4 to the third rat from the right.

Bottle number 295 is represented in binary as 1 0 0 1 0 0 1 1 1, so feed it to these corresponding rats.

Bottle 1000 is represented in binary as 1 1 1 1 1 0 1 0 0 0, so feed it to these corresponding rats.

After you've given a few drops of wine to the appropriate rats, wait an hour.

Now, how can you tell exactly which bottle is poisoned?

This is the part that's so amazing.

You can just read the number for the poisonous bottle from the rats.

You have a line of rats, some dead and some alive.

Just imagine all the dead rats are ones, and all the alive rats are zeros.

This gives a 10 digit string of zeros and ones.

Specifically, it's the binary number for the poisonous bottle.

So if the sequence of dead and alive rats looks like this, you know that the poisonous bottle is 1 1 0 1 0 1, or as it's more commonly referred to in base 10, number 53, and the party is saved.

Here's a follow up question for you, if you only had 100 bottles of wine and one was poisoned, how many rats would you need to precisely determine which bottle it is?

What about if you had a million bottles of wine?

If you know the answer, let me know in the comments.

On the next episode of Infinite Series, I'll shout out to some of the folks who got it right.

We had some great comments in the hyper-spheres video, our first episode, and I wanted to share two of them with you.

First, remember that when we nestle a sphere inside a cube, the ratio of the sphere's volume to the cube's gets smaller as the dimension gets bigger.

Mike Guitar suggests a neat way to visualize where all that extra volume comes from.

A square has four corners, whereas a 3D cube has eight and a 4D cube has 16 and an ND cube has two to the N. The sphere nestled inside the cube will never reach the corners, and as the dimension goes up the cube gets more and more corners, so the circle gets smaller and smaller relative to the cube.

David de Kloet had a great comment about the portion of the video where we talk about placing 2 to the N spheres inside a cube and then constructing a central sphere which bursts out of the box after dimension 9.

Talking about a cube with side length 1, he says, this sounds super exotic, but after thinking about it, I realized that the math is actually trivial with Pythagoras' theorem.

The 2 to the N balls all have radius 1/4, but the distance from their center to the center of the cube is square root of N over four, so the center ball has radius square root of N over 4 minus a 1/4.

When this becomes greater than a half, the center ball grows out of the cube.

And when n equals 9, this is exactly equal.

This is definitely a mathematician's use of the word trivial, but he's right.

It's kind of a fun exercise in geometry, if you know that the Pythagorean theorem holds in higher dimensions.

If you have a cube in N dimensions where the length of each side is 1, then the distance between opposite corners is the square root of N. Try to work out the details for yourself.

[MUSIC PLAYING]