 Hi, I'm Zor. Welcome to Unisor education. We continue course which is called Mass Plus and Problems presented on Unisor.com where, well, as I'm sure you know, we solve problems. That's basically the whole purpose of this particular course. Now there is a prerequisite course which is called Mass for Teens on the same website and that contains basically some theoretical material based on which we are solving these problems. In addition, I just have to say that the problems which we are talking about in this course are not exactly the I would say classic problems presented in regular schools. They are more towards Olympic kind of problems, which means they require certain creativity, certain analytical thinking. Well, thinking, it requires thinking and what's very important every problem which is presented on this website in this course it has a video part and it has textual part. Textual part is like a text book basically where the problem is presented. In some cases, I present the solution as well, sometimes known in the textual part. Sometimes I have a suggestion how to approach this particular problem. So what I suggest you to do is if you're watching the lecture first, then just pause after I present the problem and try to solve it yourself. If you go to a textual part, just read it and basically again, try not to look at the solution if there is one, try to solve it yourself. That's very, very important. Even if you don't solve it, it's just a thinking process. Your kind of attempts to solve the problem is also very important. So the purpose of this particular course is to force you to think and that's the most important part. Okay. Back to the problem. Today, we will talk about logical problems and this particular lecture is number six in the category of logic. There are other categories like algebra, geometry, etc. in this same course, mass plus in problems. Okay. So I will present, it's basically one problem, but I'll kind of present it in three different incarnations. Okay. The first problem is we have five towns on the map. One, two, three, four, five. Now, I know that among any group of four of these towns, there is one connected to three others, any group of four. This group of four, this group of four, this group of four, any group of four has one, let's call it connector, connector town, which is connecting to all others with some kind of roads. So that's a known fact. Now, what's necessary to prove is that there is one town connected to all four other towns on this map. Okay. So again, any group of four has one connector who has roads to three others, but we have to prove that there is one connected to all four of these other towns. All right. So again, that's the proper time to pause the video, read the conditions of this problem, or you just can be satisfied with whatever I just said. The problem is presented to you and try to solve it. Okay. Now it's solved basically by kind of considering different cases. It's a logic, right? So let's first take any group of four. For example, this group, this group of four doesn't really matter which one. And I know that there is some town connected to three others. Well, let's assume it's this one. So I have these three roads. Again, it doesn't matter which town I choose. Now I just call them. Let's call this one A, B, C and D. Okay. Okay. So I just use the condition of the problem. I took one particular group. I have one problem is telling me as given that one is connected. I call it A and these are connections. Okay. Let's consider a different group. This group, B, C, D and E. Now, again, it's a group of four and it's given that it's supposed to have one particular town connecting to all three others. Okay. Here is a very simple case. What is this particular town connecting to three others is one of these. It doesn't matter which one. Let's say this one. It's connected to three others like this. Then this becomes connecting, connected to all four because it's already connected to A. So it's connected to these three and it used to connect to A. So everything is fine. We found basically, we found exactly the town which is connected to all four. Now, if it's instead of B, if it's a C or D, it's exactly the same thing because if it's connected to all these, to this one, it's already connected. Non-trivial case is when it's not this one. So it's not B, not C, not D. If it's E in this group which is connected to these three. Now, we still don't have any town connected to four others. All right. But we still have different groups of four to consider. Different groups of four. So let's consider this group, A, B, C and E. There are basically two different extra connections which we can build. Now, which of these can be the connector? Well, if it's A and connected to B and C, it's already connected. Now, if it's already connected to E, problem solved. It's A who connects to all four of these. Now, if it's B, then it's not obvious yet because, again, there is one still town not connected to B or A. I mean, it's connected to A and E, but it's not connected to B. So none of these can actually be the solution. A is not connected to E. B is not connected to D. Well, D is not connected to A obviously. So it's still not solved the problem. Okay. But we do use this particular connection for the future and let's consider yet another group of four. This group. A, C, D and E. Again, if A is connected, if A is a connector, which means it needs a connection to E, the problem solved. A is the main guy who is connected to everybody. If E is connected to A, it's the same thing basically. It's one road. And the only thing which remains is this one. So in this group, it's C supposed to be or D is supposed to be connected to between them. The missing link. So if these four needs to be connected using one particular connector, in all cases, whatever we consider, if it's A, if it's E, if it's C or if it's D, in any case, we have to add one road, either this one or this one, which will make some city connected to all four. In this case, it's C which is connected to all four. If instead of this, we draw that from A to E, it will be A or E which is connected to all four. So in any case, we are inevitably coming to a situation when one of the cities, one of the towns, will be connected by four roads to all four other members of this group. So that's basically it. We construct it. We kind of use the condition of the problem that out of any group of four, there is one connector which is connected to three others. We use this in this particular case three times for this, for this, or four times actually, for this group, then for this group, then for this group, and then for that group. And by applying the same condition up to four times, we actually came to a necessity of some town to be connected to all four. It's inevitable. That's it for this first problem. Now, the second problem is, let's just increase the number of towns to six. So let me just get rid of all the roads. Let's add the six here, E and F. So I will start with exactly the same case as before. So I will take the group of four, any group of four. Let's say this one, A, B, C, D, and I know that A is connected to all of them. I'll do it this way. Now, let's take another group of four. Let's say this one, A, C, D, and E. Well, again, there are different variations. I will consider the variations A which is connected to E. So in this group, A, C, D, E, A is the connector. Now, there are other cases. And other cases, I actually suggest you to do it, but in any case, it's not really necessary right now. And here is why. Let's consider a group of five, A, B, C, D, and E. According to whatever I have already proven before, that out of five, there is one which is connected to all four of them, right? So basically we have proven this already just before. So if I have a situation when any four has a connector, then I have logically derived that in a group of five, also there is one which is connected to all four of them. So let's just do it differently. Let's just consider a group of five, A, B, C, D, and E. According to the previously proven results, I know that there is one which is connected to four others. So I call it A and the rest I will call B, C, D, and E. So that would be kind of easier approach. And it's kind of inductive. From five, we go to six. Given, okay. Now let's consider these five. Again, I'll use exactly the same results which have been just proven. That out of group of five, well, considering that any four has a connector, I have proven that there is one which is connected to all others. Now, if in this group of five, the one which is connected to four others is this, this, this, or this, we have basically solved the problem. Because if any one of them, let's say E, is connected to D and C and B and F, then E becomes connected to everybody because already connection to A is established. Same thing will be with D. If D is connected to E, C, B and F, D becomes the one which is connected. So in any case, we got the town which is connected to, in this case, five others. So these are trivial results. What if it's F in this group which is connected to four others? Then situation becomes a little bit more difficult, right? Because right now we have, for example, A and F are not connected by direct road. That's okay. That's okay because here is what we can say right now. Let's consider this group of four in this case. By conditions of the problem, we have one of them which is connected to all others, right? So it's either E connected to all others or D or C or B. But any one of them already have the roads to A and F. So if any one of these becomes connected to all others, let's say it's D. One, two and three. Then D becomes the focal point. And D has connections to all other five cities. One, two, three, four, five. Okay? Same thing will be if we choose C connected to all other members of these four, B, C, D and E. So we basically proven that out of six towns, if again any subgroup of four has the connector connected, connecting to three others, then there is one connected to all of those guys. Okay, fine. That's good result. But it looks like we can actually extend it to seven or eight or et cetera. We will just do exactly the same thing. We will use the previous result to prove the next one. So in this particular case, I'm thinking that by induction, we can prove that for any number of towns, 100,000, whatever, condition is that out of any four, there is one connector which connects to three others. If any group of four has such a connector, then there is one town connected to all of them with direct roads. Okay, let's try to prove it. So first of all, how the method of induction actually works. First, you present it for some number like n is equal to one or n is equal to two or whatever, doesn't matter. You establish that your statement which you are trying to state is true. Well, in our case, we need at least four towns because the condition of the problem is about four. So for four towns, there is one by definition of this particular problem. That's how I define the problem. From any group of four, so if there is all together only four towns, then my statement is true because it's given. Well, actually, I have also proven it for five, even for six. Okay, fine. Now, let's assume that for all k is equal to one, two, et cetera, and my theorem has been proven. I will try based on this, prove it for k is equal to n plus one. This is the method of induction. So if I have proven it for four and five and six, then using this, I can say if I will prove this, then I can say that, okay, for seven, it will be the same thing. Now I have proven it for seven, which means I can prove it for eight, et cetera. So let's start with this. Let's say our n is equal to, well, in this case, six, and I have already proven for six, right? I can start from five, actually, it doesn't matter. Okay, so what I will do is I will assume that my n plus one cities are here, n minus, minus one, one, and one. So it's n plus two, right? n minus one, plus one, plus one. Okay, so these are my terms. So first I choose this group is n towns, and I know that out of n towns, I have proven that my statement is true, which means there is one connected to others. So I call it a one, and these are connections. Okay, now I will apply exactly the same logic as I did for six. Let's consider this group, which is a two, a three, a four, a, and this will be n plus first. Okay, so let's take this group from a two to a n plus one. It's also n towns, and out of n towns, there is one according to my inductive proposition, there is one connected to all. Again, the trivial case is if one of these is connected to all of these groups, because since it's already connected to a one, we already have a winner. So the one from these, which is connected to all of these, would be also connected to a one, and that's the problem solved. We have found the town which is connected to all others. Non-trivial case is this one. So if out of this group, it's this one, a n plus first, which is connected to all others, we still have unsatisfied condition, because a one and a two, sorry, and a n plus one are not connected by a direct road. But let's continue our logic. Let's consider this group. Now, this group has even less elements. It's not n, it's n minus one. But we have proven it for all up to n, which means for n minus one also proven. So from these, we already know that there is some town connected to all others. Now, choose any one of them. If this one is connected to all others, it becomes the one which is connected to all others, because a one and a n plus one already connected to this one. So even in this case, when out of this group it's this one which is connected, we still can find going further in our logical assumptions that there is one among these which is connected to all. So that's basically the end of it. So again, we have given as a condition of the problem that out of four we have one always connected to all three others. Then we have proven actually that for five it's true as well. And then we have proven it for six as well. But then we have proven this inductively that for actually any number of towns, we use the already proven results to prove this one. So if it's proven up to n, then it's true for n plus one. And we did prove it for four, it's given for five, we have proven for six, we have proven. So we have satisfied this and using this inductive transformation, so to speak, we are saying, okay, since we have proven it up to six, it's true for seven. Since we have proven up to seven, it's true for eight, et cetera. So that's what inductive approaches. By the way, if you want, there are many problems on induction in the course which is prerequisite for this one, which is mass before teens. Okay, so I do suggest you to read the problem again in the textual part of this lecture. So you go to Unisor.com, choose the course, mass plus and problems. And then the category is logic and this is logic 06. So in this part I basically present the problem. And then I do have some explanations for certain cases. I don't remember if I put a complete proof for n plus one. Maybe I did, but you don't read it. Again, you try to restore this proof yourself in your memory from whatever you have just listened to. Okay, so that's it for today. Thank you very much and good luck.