 Hi, I'm Zor, welcome to Inezor Education. So we continue talking about something which I call math plus and problems. So basically it's problems which are not exactly typical problems presented to let's say high school or teenagers, first college year, something like this. Now I have divided all these problems into categories like logic, algebra, geometry, etc. So today is something which I call logic 01, which means it's the first lecture about logical problems and I will present three problems here. So obviously the best way is to not to listen to my solution to these problems but to solve yourself. And to do this you can either read the notes for this lecture, Unizor.com, all the notes are there. So you go to basically a course called math plus plus problems and find the topic called logic and this lecture will be among them. So you can stop video after I present the problem before solving it or you can read the notes for this lecture and the notes contain problem and solution. So again you read only problem and then think about this problem just by yourself. So that's the best way to approach all these problems and definitely recommend it because that's the purpose of the whole course actually, to develop your analytical mind, your logic, your creativity. So that's why we need to solve these problems. And I perfectly understand that these problems will never occur in your life but being prepared for all kinds of unusual problems your mind will be dealing with real life problems much much better. So three logical problems. Problem number one, let's say we have a matrix of rows and columns, m rows and columns. And in this matrix we place numbers. So in this case I suggest numbers, one, two, three, four. Then what we do? Then we do something which is called calculating minimax or maximine. So first we do row by row and we find the minimum, in this case one or two minimum is one, three or four minimum is three. And then we do maximum of these minimums and maximum is three. Now we do the same thing column wise but now we start with maximum and then go to minimum. So this is called maxi, mean. So first minimum in each row and then maximum among them. And this will be maximum first which is three here, two and four, four. And now we do the maximum, so sorry, minimum. So this is minimax. In this particular case they happen to be equal to each other. But that's not typical. Here is another example. Instead of this we do one, three, four, two. Minimum, one, one and three and three, four and two minimum, two. And maximum among them would be two. Here we will have first maximum which is one and four maximum, four, three and two maximum, three. And we have minimum out of these maximums which is three. So now as we see minimax is greater than maximine. Now so what's the problem? The problem is to prove that maximine would be less than or equal to minimax. So equal or less, in this case less, in the previous case it was equal. Now how can we prove it? Okay, so that's the proper point to stop to pause the video and do it yourself. And now I will continue. So let's mark maximine is this one and minimax three is this one. Okay, in this case, in this particular case both maximine which is two and minimax which is three are in the same color. But now let's remember how did we obtain this? We do by column and we have the maximum among columns. Maximum is first. Maximum is always first in this case, right? Which means is greater. So this would be greater than this. So if minimax and maximine are in the same column then minimax would always be greater because we start from choosing the maximum. So we have chosen this one and that is greater than this obviously. Now what if it happened to be a different case when they are in the same row? Well, if they are in the same row, now let's just think about how we choose the maximine. We have the row and we have the minimum. So again, maximine would be less because it's a minimum than minimax. And finally the third case when they are not in the same row and not in the same column. Let's say it's this and this. Well, in this case we will do something like find something in between. So let's assume that our... So this would be my minimax and this would be my maximine. Not in the same column but in the same row. What do we do? Well, let's choose something which has the same row with one and the same column with another. Obviously it's always possible because it's matrix, right? So we have this row and we have this column and we have something on the crossing of these things. Okay, now let's think about this particular number. So if this is minimax, so it's maximum among... No, this one. If this one is the column, now in the column we always choose for the minimax. We always choose the maximum. So this one is greater than this one. Then let's talk about the row. In the row we always choose for minimax the maximum. So this one is greater than this one. So this is greater than this, this is greater than this. And that's why this is greater. Greater or equal, sorry. It would be greater or equal to this one. Well, that's it. So we have three cases when they are in the same row. Then we use the principle of minimum because we are choosing the minimum. Or the same column we are using the principle of choosing the maximum. Or if they are not in the same thing which is the intermediary number which is the same column with one and the same row with another. And basically do twice the same logic. That's it, that's the first problem. Next. Next problem is solving the crime. That's basically like Sherlock Holmes did. Let's assume that there was a crime committed and there are three different people X, Y and Z who are interrogated. And basically they are asked the question, did you did it? Or if you don't, who did it? Okay, here are the answers. A, I did not. Or B, Y did not. So the Y says X did not, Z did it. What does Z say? I did not, X did it. Okay, these are answers. And it's not sufficient yet to tell who is the criminal, who committed the crime based on these answers. But there is an additional condition. In the process of further interrogation, they found out that one of them told truth, both questions were answered truthfully. Another person answered both questions falsely, lied in both questions. And third suspect said one truthful statement and one lie. Using these additional conditions, it's sufficient to find out who actually committed the crime. Now, there are probably a few different approaches. One of the approaches would be, for example, okay, let's assume X did it. Okay, let's check if it corresponds to all the conditions in the problem. Well, let's see. This would be false, right? If X really did it. Y did not. And that would be true. X did not, that would be false. Z did it, that would be false. I did not do, that would be true. And X did it, that would be also true. So as we see, all our conditions are satisfied. So if we assume that X is the person committed to crime, all these questions were answered according to basically conditions of the problem. Well, that's one approach. And we already found basically the answer. X did it. Now, if we start, not with X, but for some chance we start with Z, for example, what would be in this particular case? Well, let's do it with Z. So this is if we assume X. But if we assume Z, I did not. That would be true. Y did not, that would be true. X did not, that would be true. Z did it, that would be also true. And now we already have a contradiction because we have two people who said both truths. And we have only one person who said twice truths. And other person said twice lie. And the third person, one lie and one truth. So this is already out. Now if we assume Y, I did not, that would be true. Y did not, that would be false. X did not, that would be true. Z did it, that would be false. So again, we have contradiction because we have two people who said one truth and one false statement. So only assume is only assumption that X did it, gives us basically complete solution which corresponds to all the conditions. Or alternatively we can do it in a different way the way how I have presented it in the notes for this lecture. That's slightly different. I assume basically that X is the person who said twice truths. And then did some other logic. And we also came into exactly the same conclusion that X did it. So that's basically the second problem. Very easy. All you have to do is basically go with case by case and do some logical comparison. Okay, the third problem is also relatively easy. You just have to kind of think about it a little bit. But what's interesting, when I was working I was in charge basically of a group of programmers and I was always when I was interviewing somebody else for the group, I was always asking this question. Just to check how person thinks, how quickly he reacts, et cetera. Okay, so here is the problem. We have a person who works and after work he likes to go to a restaurant. He has two favorite restaurants. So one restaurant, here is where it works. This is restaurant one and this is restaurant two. And there is a train. Well, I live in New York so let's consider it's a subway train. And what the person does, he goes down to the platform and whatever train comes first, this direction or that direction he takes and goes to this restaurant or that restaurant. Okay, now, a little bit more details about this. Let's say that he is ending his work relatively randomly from five to six p.m. at the end of the work. But it's not really like an exact time when he finished the work. Let's also assume that train goes to each direction with interval of five minutes. Now, at the end of the year, let's say, he counted how many times he was dining in this or that restaurant and it appeared that in this restaurant he was four times more often than in this restaurant, approximately. Now, again, from the statistical standpoint, from the standpoint of probability, he is randomly ending his work within an hour. Hour is a lot. Now, during the hour like 20 trains goes to each direction, right? Five minutes, not 20, 60 divided by five, what is it, 12 trains. So, in theory, just thinking about probabilities, he's supposed to basically be more or less even in both cases. Approximately. But now they have such a drastic difference. One restaurant was attended four times more than another. So, my question was, how come it be? Okay, here is the solution. It's all in the schedule of the train. Let's assume that train to this direction comes at five, five, four, five, five, 10, five, 15, et cetera, every five minutes. Train in this direction comes at five, zero, one, five, zero, six, five, 11, five, 16, et cetera. So, what happens? If he comes from five to 501, he takes this train. But if he comes from 501 to 505, he takes this train. So, the interval of time between this and this is one minute, but between this and this, four minutes. Well, obviously that's the reason why he attended this four times greater. Same thing is here. If he arrives between 505 and 506, he goes to this train. But if it's from 506 to 510, which is four minutes interval, he goes to this, again, one and four. So, that's the reason why he attended that restaurant greater, four times more than this one. And by the way, even if this number is 10 or 100, doesn't matter what number, we can always adjust our schedule in such a way that interval of time to take this train would be 10 or 100 times less than interval for that train. It all depends on how shifted relative to each other these trains are. Okay, so, what can I say? That number one, most likely these problems would not be part of the regular course of mathematics in regular schools. So, that's why I called it math plus, basically, and problems. But at the same time, I would like to say that it's very important to be familiar with these problems because they are not really mathematical. They're kind of go a little bit towards thinking and not checking the theoretical material which you have learned on the lesson, but just to force you to think about something. And that's the beauty of mathematics in general and this particular kind of mathematical problems in particular, they are preparing you to think to find solution for some problem which is completely unrelated to mass or anything but it requires thinking and making the right decision. That's it, thank you very much and good luck.