字幕列表 影片播放 列印英文字幕 - You are Mondrian's worst art critic 100 years ago. You tell him he can no longer create art where two rectangles have the same dimension. That's just wrong. You're going to say, okay, I am going to score each of your pieces of art. We're going to insist that all of the rectangles have got different dimensions. And your score is going to be equal to your largest area of rectangle minus your smallest area of rectangle. Your smallest score possible is what you want to go for. The other condition is you have to tile the whole canvas. So let's start out. Brady, how big do you want this first rectangle to be? - [Brady] Three by two. - [Gordon] Three by two, okay. There's a three by two. So we can no longer make a two by three or three by two anywhere else. But I can make a six by one? You can make a six by one. In fact, why don't we do that? There we go. There is a six by one. What else would you like in here? - [Brady] Should we make that little square, two by two? - [Gordon] Two by two? Yeah. If you were to stop right now, this is area 20. So if you were to stop right now, your score would be 20. That's your biggest minus your smallest, four. Your score would be 16 and that is not very good. You want low and 16 is not low enough. So what we could we do? We could split this into an 8 here and a 12. We could do that. And so now our score is 12 minus 4 is equal to 8. You can tell Mondrian you are scoring this an eight. You can score five. So let me show you this one. Your largest rectangle is eight, minus your smallest, it is three. So that score is five, and that is the best possible score that you can get for a six by six Mondrian art puzzle. Of course, you have to ask yourself, what about a seven by seven? What about an eight by eight? What about an n by n? What is the optimal? Ed Pegg has happened to prove solutions. He's proven that we have these optimal solutions, beautiful results and great for algebra class as well to come up with general solutions. So for example, for an odd square, let's say that you are dealing with a square 101 by 101. Well, at least you know for sure that you can split this nearly in two. So this is a rectangle 50 by 101 And this rectangle is 51 times 101. And you can see right away that your score for any odd square, You can see that it's just the edge length. So you can see can guaranteed that you can at least do 101. You can actually do a lot better than 101. But you can at least do 101. An unsolved problem here is when this number ever goes to 0? You wpi;d think, well, this has got to increase. But it doesn't always increase. For example, 17 by 17, the score for that is only 8. And the score for 16 by 16, 15 by 15, I think they're larger. So this does not continue to increase. It's a very complicated function. And it is not clear that this, for example, doesn't go to zero. - [Brady] Presumably called the Holy Grail is some kind of generalization for an n by n square. The solution will always be n cubed minus four plus four. - [Gordon] Oh, that would be a joy again! But good luck. (laughs) Good luck. Dream on. - [Brady] You can follow it all the way through.