 In this video, we provide the solution to question number 13 for practice exam number one for math 1030, in which case we're given the graph K4. It's a weighted, complete graph, and we need to solve the associated traveling salesman problem using the brute force algorithm. Now because it's K4, we can start all of our vertices at A, or all of our circuits at A. And so it turns out there's only six possibilities, which isn't so bad. But honestly, each of the possibilities of reverse gives a different possibility. So there's really only three options that we have to worry about. So this one, brute force isn't so bad. We just have three options we have to do. So one option is just to go around the whole thing, A, B, C, D, A, like so. So we get A, B, C, D, and A. Of course, if you reverse that process, so I went around clockwise, you could also go counterclockwise. That's a second option. I'm not gonna list that one though. So that would cost 48 plus 32 plus 18 plus 22, which is then $120. So that's the number to be right now. Another option is we could go from A to B, then we go to D, then we go to C, then we go to A like so. So we could go from A to B to D to C to A. You could also do that backwards if you wanted to. So you could go from A to C to D to B to A. That's the same thing there. If you do that one, you're gonna get a cost of 48 to go to B, then you're gonna get a cost of 20 to go to D, then 18 to go to C, and then 28 to go back to A. That's gonna cost 114, so that one's a little bit better. So the last option is instead of going to B first, we could go to C first, so you go from A to C. And then from here, we don't wanna do like this one backwards, because if you did A to C to D to B to A, that's the same thing backwards there. So we actually wanna go from, we wanna go from A to C to B to D to A. So that's the one we're gonna be looking for right there. So if you go from A to C to B to D to A, or going backwards gives you the same thing here. In this case, you get A to C, which is $28. You go from C to B, which is $32. You could go from B to D, which is $20, and you can go from D to A, which is $22, like so. If you add those up, you end up with a total of $102, and so that actually is the optimal tour on this complete graph, and so that would give us the final answer. The tour that I've currently drawn on the board is the optimal tour, and it'll cost $102 to transverse that one.