 In this video, we provide the solution to question number nine, which given the graph illustrated right here, we're trying to find a Hamilton path among the six options that are provided. Now, well, with Hamilton paths, we don't have the same nice criteria that we do for like the Euler paths, like with Euler's theorem, but there are some things to look for, right? When it comes to, I mean, one worst case scenario, we just try each and every one of these six options and see what happens there. That's not a possibility. But another thing that does point out to me is this right here, there's a bridge between F to B, like so. And because it's a bridge, it means the vertices associated to it are bottlenecks. And to form a Hamilton path, you want to try to avoid bottlenecks until absolutely necessary. So what I'm seeing here is that because the only way that F connects to the rest of them is by this edge right here, once you cross this edge, you can't go back. And since there's only one vertex over here, this tells me that you either have to start at F or you have to end at F for a Hamilton path. Because there's no other way to get to F, to visit it, to solve the traveling salesman problem, unless you either start at F or end at F. So I get rid of some options here. So choice A starts with F, so that might work. Choice B does not, it starts with A, ends with G, so that doesn't work. This one starts with F, so maybe, this one starts with F, maybe. This one starts with G, ends with E, so that can't be choice E. This one starts with C, it ends with G, so that one can't be it. So I can throw out three of them. The other ones start with F, so that seems promising. Also, be aware that there is one, two, three, four, five, six, seven. There are seven vertices on this graph. A Hamilton path will hit every single vertex exactly once, never repeats. So my path has to involve exactly seven vertices. So that also helps me out here. It's like, notice, one, two, three, four, five, six, seven. That's a possibility, one, two, three, four, five, six, seven. That's a possibility, I think I drew the wrong number of lines, but I counted it right. One, two, three, four, five, six, seven. But this one, this one looks too long. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's too big, right? So I can't be choice D, so I have it down to two options just based upon those observations right there, okay? But again, worst case scenario, you just try each and every one of the six. So let's try A and C what happens there. So if you do A, you're gonna go from F to B to A to D to E to C to G. That actually looks like a Hamilton path right there. We visit every vertex once and only once. I'm good with that. So it looks like the correct answer is A, but just for the sake of completeness, let's look at choice C and see why that one didn't work. You go from F to B to E to D to G to A to C. If you follow that, it looks like, when you look at this, you use every vertex exactly once, so it's tempting to think it's a Hamilton path. But you're utilizing edges that don't exist in the graph, so that is not a Hamilton path. So in fact, A is the correct answer. It's the only Hamilton path listed amongst these options.