Hello again, everyone! Last time, I set you a challenge– a challenge to write a sequence of numbers using numbers from 0 to 1, so that the first two numbers occupy different halves, and the first three numbers then occupy different thirds, and the first four numbers occupy different fourths, and so on. I set it up as a puzzle, building houses on a strip of land, so that every time a new person comes along, you have to re-share the land, so that everyone gets a plot of equal length. And I asked you, how many houses can you build? Can you do this indefinitely, or is there some sort of limit to it? Now, if you haven’t tried this already, please do! You can do this with a pen and paper, or if you haven’t seen this, we did make a computer game out of it. There’s a link in the description to this. So you just sort of point and click and build your houses as you go along. Do try this out now if you haven’t already, because I am going to give the solution. You see, you can’t do this indefinitely. There is a limit to how many houses you can build. And that limit is 17. Now, that’s really odd! 17 houses! So, for a start, it’s weird that you can’t do it indefinitely. That’s not obvious. There is a limit. The limit is 17, though. Why 17? Why not 18, or something else? Now, I am going to show you a solution, with 17 houses on it, which I’m going to do later. But first, I want to show you a proof– a proof that it’s not 18 houses. Now, to do that, we’re going to have to consider a number of steps with the land split up into different sized plots. I’m going to start by splitting the land into 9 equally-sized plots, and then I’m going to consider the tenths and the elevenths and the twelfths, and so on. And first, I’m going to consider the point that occupies the fifth plot after division by 9. Let’s call that ‘x.’ So x is somewhere between 4/9 and 5/9. But let’s say x is equal to 0.46. But if x is 0.46, then I know that 0.46 occupies the fifth plot after division by 10. And it also occupies the sixth plot after division by 11. And the sixth plot after division by 12. That’s all automatic. Now let’s consider the point that occupies the fourth plot after division by 9. Let’s call that ‘y,’ and we’ll try and work out the value of y. Now, y might occupy the fourth or fifth plot after division by 10. But it can’t occupy the fifth plot because that belongs to x, So y must occupy the fourth plot. That means y must be less than 4/10, or 0.4. Looking at division by 11, y crosses a border again, so it might occupy the fourth plot, or the fifth plot. If y occupies the fourth plot, that means the fifth plot is empty. That’s OK. That just means the empty plot must be occupied by the 11th point. But that’s a problem when you look at division by 12, because the eleventh point must either share a plot with either x or y, and that’s not allowed. So y cannot occupy the fourth plot after division by 11. It must occupy the fifth plot after division by 11. And that means we’ve worked out that y must be between 4/11 and 4/10. In other words, y is between 0.364 and 0.4. Now let’s add a new point. We’ll call that zed, and that occupies the fourth plot after division by 12. And we’re now going to go through the same process considering what happens after division by 13, 14 and 15. And we’ll discover that y occupies the sixth plot after division by 15, and zed occupies the fifth plot after division by 15. Continuing on again, we now see that y occupies the seventh plot after division by 16, and zed will occupy the fifth plot. But that means the sixth plot is going to be empty. So that must be filled with the 16th point. Now consider the next division. So y would occupy the seventh plot, after division by 17, and zed will occupy the fifth plot, and the new point occupies the sixth plot. But like before, this is a problem, this means the 16th point will share a plot with either zed or y, when we divide by 18, and that can’t happen. So we’ve found that we have failed at division by 18. But this was all working perfectly fine up to and including division by 17. This is what happens when I choose x to be 0.46. But what if I chose a different point? Well, the full proof shows that you will always get stuck for any value of x. And remember, x was defined to be the point in the fifth plot after division by 9. In fact, I chose this value of x because we could get as far as division by 17. Other choices for x actually fail a lot sooner. So the proof shows that it fails at division by 18. So a sequence of 18 points does not exist. That’s impossible. But the same method may also be applied to construct solutions, including solutions up to and including 17 points. For example, here is one solution where the sixth point is 0.46– that was x in my proof. And zed and y are exactly where I predicted they would be. The thing to remember is it’s not just the values of the solution that are important, but it’s the order, as well. For example, we can’t just divide the line into, say, 100 equal parts, and then put a point in each part. No–the first two points have to be in different halves. The first three points have to be in different thirds, and so on. And it’s this restriction that gives us this limit. The problem is related to the Farey sequence of order 18. That’s the ordered sequence of all fractions with denominators less than, or equal to, 18. And the problem is actually called “The Irregularity of Distributions,” or “The 18 Point Problem.” Now, this proof actually goes back to the 1960’s. It’s very functional, though. It does show that you can’t construct an 18th point, and that you can construct a 17th point. So there are solutions with 17. But on a deeper level, it doesn’t really tell you why it’s that number. And if you’re thinking that, I agree. It’s really curious; it’s really odd. I don’t think it’s a very well-known fact, which is why I wanted to share it with you. Finally, I just want to give some thanks to Christian Perfect for making the game for us to use. And, well–if you have been, thanks for watching!