 Hi there. Let's take a look at the rendezvous problem. It's a logical dilemma that was first devised by Steve Alpern in 1976. Here's how it goes. Let's say two people want to meet up in a park. We'll call them A and B. Unfortunately for them, this is a big park and they can't see each other. We'll model this park as a grid with nine squares. Each person can only see within their own square. If they want to meet up, they'll have to get into the same square. And since there's no cell service in this park, neither person can communicate to the other person what square they're in. To model how the people can move, let's say that on each round, person A and person B can move to any square on the grid. There are no restrictions on movement, meaning both players can go to any square each move. Each person also has the choice to stay put for a round and not move at all. Once they both get into the same square, they win and can get together. So the question is, what is the optimal strategy that A and B should use to determine where to go on each move? Before we take a look at that, here are a few other points about this problem. A and B are both very smart guys, so we'll assume that they both know the best strategy to use. Perhaps they both watched this video or read the papers on the subject. This problem gets a lot more tricky if one player is using an optimal strategy, but the other isn't. The only way A and B could act differently is through randomness, since any other way would require communication between the two. Each player sees the grid differently, so what A might see in the top left, B might see in the bottom center. This point isn't very relevant to the strategies I'll talk about today, however. Alrighty, keeping all that in mind, what are some possible solutions? The symbolist would be for each player to move to a random square each round. Here are a few example games. On average, the number of rounds this will take is the same as the number of squares, so if we have nine squares in the park, it'll take an average of nine rounds. But can we do better? In a paper written in the year 1990, E.J. Anderson and R.R. Weber found a better way. They weren't sure if it was the best way, but it was better than picking out random. Here's how it works. On the first round, both players start on a random square. The rest of the game is divided into 8 round chunks. The chunk length, 8, is one less than the number of squares, 9. For each of the 8 round chunks, there are two possible options a player can take. The first option is to just stay still for 8 rounds. The second option is to take a random route through every other square on the grid, which takes 8 rounds. A random 27.53% of the time, a player takes the first option of staying still. The other random 72.47% of the time, the player will just visit every other square. These numbers were determined to be the best through some math that I won't get into in this video, but you can read the paper linked in the description for the full details. Note that these numbers change depending on the number of squares in the grid. Then, both players repeat either staying still or moving until they meet each other. The advantage of this strategy is that it maximizes the chance that one player will remain still while the other visits all the squares. If one player stays still while the other moves, then the two players are guaranteed to meet since the moving player will visit all the squares, including the one the still player is on. So, is this the best strategy? Probably. No better strategy has been found, but there isn't any proof that this strategy is the best when there are 9 squares. However, it's been conjectured that this strategy is the best for all grids with more than two squares. It has been proven to be the best when there are just three squares, though, linked to the proof in the description. And for two square games, just moving around randomly is best. In fact, for the optimal strategy I just described, it would recommend moving 50% of the time and staying still 50% of the time, which is just the same as moving around randomly. Thanks for watching. I've linked some papers on the rendezvous problem in the description.