 Hi, I'm Zor. Welcome to Unisor Education. So we will solve a couple of problems more. It's all part of the course which I called Maths and Problems, Maths Plus and Problems presented on Unisor.com. Now, this is basically a continuation of the course called Maths for Teens on the same website. So I do suggest you obviously to be familiar with the main course. These are just problems which are in some way not exactly standard but still very interesting to basically it helps to develop your thinking, your analytical abilities, your other qualities, very important like logic, for example, or creativity. All right, so these problems are not exactly typical as I was saying. They require certain maybe unorthodox approach. And some of them are really very easy as soon as you just get very important like the first the right thought how to approach this problem. Some of them are maybe a little bit more time-consuming, let's say. So today I have two problems. One of them is quick. You just have to like come up with something very simple but that's basically the whole problem is to get that one little thing which really gives you the solution. Another problem, the second one which I'm going to present today, is a little bit more time-consuming but well, it's still important to be creative, to be persistent in your approach to solving problems. Okay, so the problem number one, you have two piles of stones. One pile of stone and another pile of stone. Exactly the same number of stones are in each pile. And you have two players. So the game is, on each move a player takes as many stones from one pile as he wants, basically. Then the second player does exactly the same, takes as many stones as he wants from one particular pile, one of these two. Now the winner is the person who takes the last stones from whatever the pile remains. So here's the question. You have two players, player number one, player number two. So first player number one starts, second player number two starts. Is there a strategy for player number two to always win? Well, obviously the answer is yes and our purpose is to develop that strategy. Okay, now it's a good thing to pause the video if you did not really solve this problem before and think about it. Now, here is the solution. It's very important that you think about this problem before basically listening to this solution which I am suggesting. Either you will come up with solution yourself, which is the best of the best of the best or if you don't, it's still important to think about this because the whole process of thinking and changing what kind of strategies should they choose, it's still very useful process. It's a training for your brain. Okay, now here is the strategy. For example, you have n stones here and n stones here. Now player number one takes n stones and we don't know actually how many he will take, but from the one particular pile of stones. Maybe all of them, maybe all capital N. So maybe lowercase n is equal to capital and there is nothing left in this. So as a result the pile number one contains n minus n and pile number two still n. So this is the strategy now for number two. He takes also n, but from the second. This is from the first pile and he takes from the second to equalize number of stones. So as a result, you have n minus n and n minus n. Again, equal. So we started with equal and number two makes it equal again. So no matter how many stones is taken by the first player, the second player takes exactly the same number of stones from another pile, equalizing the number of stones. Why is it a winning strategy? Well, if lowercase n is equal to capital, so he will take everything from here, then the number two will take everything from here, taking the less stones and therefore winning. If it's less than capital N, if lowercase is less than, then there is something left in this pile and after the second player makes his move, the second pile has exactly the same. And we are returning to the same situation as before with lesser number of stones. Well, I didn't tell it in the beginning that to take some stones is mandatory. So gradually, little by little, these piles of stones are getting lower and lower in number of stones and eventually there will be no more than one, obviously, in which case the first player would be obliged to take that one and then the second player will take from another pile exactly the same and that would be the last one. So equalizing the number of stones is the strategy. If it was equal, it should be equal after the second player makes its move and he will always win by doing that. Now, in the notes for this lecture, and every lecture has notes with solutions in this case, etc., I stated the problem slightly differently. I did not specify that the number of stones is equal to each other. Now, what to do in this case? Well, in this case there is no winning strategy, except if I will allow for the first player to either take some stones or skip his turn and let the second player to start the game. Well, in which case if the number of stones is equal, then the first player should actually yield the right for the first move to the second player, in which case the first player always be the winner. So if we will add another rule that there is one particular case in the very very beginning of the game when the first player can actually yield the right to the first move to the second player, then it's the first player who will get the winning strategy. So either if number of stones is not equal in the beginning, then the first player equalizes that and then he becomes actually the master of the game because he will always equalize. Or if the number of stones is equal, then the first player should yield the right to the second player, actually making the second player the first, and the first player makes himself the second, in which case basically the same as I have just explained here. So whatever it is, this is a little bit more simple case than the one which I put in the notes for this lecture. But in any case, the idea is exactly the same. Equalizing the number of stones will give you the win. Okay, so that's my first problem and again it's very kind of short and easy and the only thing which was necessary is to basically come up with idea of equalization of the number of stones as the winning strategy. That's the idea. All this about unequal stone number of stones in the beginning and yielding that that's just details. Okay, the second problem is slightly different in the way how well it's kind of solved and in the way how you really have to consume certain amount of time to think about different cases because this is the problem where you really have to consider different cases and that's where the logic comes. So here is the problem. You have six points. They can be in space then can be on the plane but what's important is that no three points are lying on the same straight line. So I will put them into a circle basically but it doesn't really matter. In this case we don't have any three points lying on the same line. That's what's important. Okay now any subset of three points out of six three out of six we will call it a triplet. It's just probability because every time to say subset of three points out of six points that's too much. Okay triplet is any group of three points out of this six. Now next thing is we are connecting these points with segments for instance this or this or this. It doesn't really matter. Now there are certain triplets in this particular case which have at least one connection. For example one two three it has two connections. One two three it has one connection this one. One two three these do not have any connections these three points right. So we have plain triplets and we have I will call them good triplets. They have one or two or even three connections but three connections I will obviously call a triangle. So if I will add this particular segment then these three points are obviously making this triplet makes a triangle. All right so we have triangle we have good triplets like this this and this because there is one connection and we have just plain triplets which are not connected at all. Three points not connected at all. Okay now what if I will tell you and this is the given that all triplets are good. So this is the given that all triplets are good which means every group of three points has at least one connection or two or three. If three it's a triangle but all of them are good. There are no triplets like this one by actually the initial given condition. There are no triplets without any segments. So that's given so what's to prove? Prove that there is at least one triangle. I cannot satisfy my condition that all triplets are good which means they have at least one connection without making a triangle. Well let's just try as a kind of an illustration what if for example I will connect them this way. Now are all triplets good no this one these three points are not connected and these three points are not connected okay. So now again it makes sense for you to pause the video and to start thinking how can we prove that there are only there is always at least one triangle if we want if we want all of the triplets to be good then we need we have to have at least one triangle. Okay now in the notes for this lecture I consider all different cases and they have nice pictures much nicer than this one obviously just to clarify my points. So I will try to basically do more or less the same first of all the approach. Now the approach which I have chosen maybe there are some other better ones and by the way if you have you know better solution by all means send it to me and I'll publish into the website with proper references. Okay so my approach is the following let's count how many triplets in theory exist and then let's count how many triplets we are making by connecting points. If in the process of connecting points I will have less than the number without making a triangle well it means that obviously we need to make a triangle. So I cannot avoid triangle and still make all my triplets good here is basically the approach. Well the first part is simple how many triplets are from the six points well that's number of combination from six by three which is six factorial divided by six minus three factorial and three factorial which is 20 so there are 20 different triplets that's simple now we have to count how many triplets we are making now before that I have one very simple theorem to prove auxiliary theorem lemma it's called if you have a point connected to three other points if then there is a triangle why because these three points make a triplet which must be good by the given condition of the of the problem which means there is some connection either this one or this one or this one but in any case if we have this one we have this triangle if we have this connection between these three points we have this triangle and if we have this we have this triangle so in any case if you have a point with three connections it means you have a triangle basically very next step in in in logic so if you want to avoid triangles by basically making our triplets good by connecting the points you have to avoid three segments from the same point but that would lead to a triangle anyway and basically what I will just show that if you are trying to avoid three segments from the same point you will not be able to make 20 triplets covered you will make only like less 18 or 16 17 15 whatever and that's what I'm going to prove if I will be able to prove it that avoiding triangles and avoiding three segments from the same point if I will be able to prove that by avoiding this situation I will not be able to reach 20 that means that I will have to really have either triangle explicitly built or I will have three lines three segments coming from the same point which next logical step would lead to a triangle anyway so that's basically the plan so my problem is how to prove that by avoiding three segments from the same line and avoiding explicit triangles how can I prove that I will not reach 20 well let's just count here is how we will count for example I do not have yet any connection at all and I make the first segment connecting how many good triplets I am making in this particular case well obviously four because this triplet now is good because it has one connection this triplet is good because it has one connection this triplet is good and this triplet is good so by connecting two previously unconnected points it doesn't matter how what what here doesn't really matter as long as these two points are naked not connected to anything then connecting them makes four additional good triplets okay that's one case what if I have two points and one of them has already been connected to something and I'm making a new connection this one how many new triangles new triplets sorry are becoming good in this particular case I could not count this one because this one was already good because we had this connection so I have to count only this this and this so I will have three new good triangles new that's what's important three to three new triangles and finally there is final case if my point points which I would like to connect already have connections before that both so I had first case no connection second case one connection and the third case three connection okay how many new triplets become good in this particular case if I will connect them again I cannot count this one because it already was containing two connecting points it was already good and I cannot count these this triplet because again this already was there so it was already so how many new ones but obviously two one and two so I have either four or three or two new uh good triplets by connecting two points okay that's fine so let's start connecting now we should not have situations when from one point I have three connections because that leads to triangle and that would finish basically the problem so I have only two from each point two connections from each point you're coming to it and you're coming out and then no more than two on this one no more than two on this one etc now if they do not close a loop let's say one two three four five if I do not close this at all does it make sense or not it doesn't make sense it makes sense to close why because I will create new good triangles I will not diminish I will only increase maybe the same but at least it not diminish so connecting the points always make sense I will always I might increase number of good triangles without basically breaking the rules of not having more than three points from one no more than three segments from one point so we can consider that these six points make at least one or two loops etc no two loops doesn't make any sense actually because if it's a loop then it's a three right six divided by two loops I cannot make it and that already triangle so this situation doesn't make any sense obviously we should not connect this way what we can do is we can arrange a loop of all six of them and we can count how many good triangles are in this case we can arrange a loop of five of them leaving the six one naked and we can arrange a loop of four of them having two obviously connected as the third case so we have only three cases basically we cannot have a loop of three because it's already triangle so we are we're trying to prove that we can do it without making triangles right so let's try and we will fail obviously without making triangles we can't do it all right let's count about let's count how many new triangles uh new sorry new good triplets are created if we will do this well let's start from the beginning connect the first pair nothing yet is connected yet we have four as I said right new uh triplets which is this this this and this now we will next connect it how many this one we have already count it as I said so there are only three of them which is this this and this same thing with this same thing with this same thing with this and the last one when both already are connected this segment and this segment already exists we add this one and we have only two how many uh 12 18 in this case again the first one gives you four then the next one three three three and two now we cannot connect this one to any of these because it's already if we connect this one let's say to to these two now this triangle already had this so it's not new so basically everything is already counted so how many nine and six fifteen now in this case again let's start four three three two now this is not connected at all to any of these so whenever these connected it's four again so what's total six and six twelve sixteen again in all cases we will have less than 20 which means that if we do not create triangles ourselves and we do not have three segments from each point from at least one point we cannot make all triangles good we can what we can make maximum 18 triangles good and two remains basically uncovered in this particular case it's this one and this one they do not have connected points between themselves and in this case as well so exactly two by the way 18 plus these two will be 20 now in this case we also can count all the five triangles which we have not covered basically in any in any case for example in this case what we have not covered this one this one then this one anyway we have not covered all of them we can basically count how many we still can do but we will have to add the third segment from some points which would which would lead us to triangle so without breaking the rule of not creating triangles and not creating three segments from the same point we cannot really cover all the good all the triangles we cannot make them all good so basically this is a solution so you're logically divided the whole situation into cases and decide every case separately in this case just a little bit of combinatoric you have that to count how many triplets you can make out of six points but that's just details which you have to by the way the the topic of combinatorics is very important and very interesting actually it does give you a very good push towards developing your analytical and logical thinking and there is a chapter in the course math for teens and the unisor.com which has a very detailed explanation of what kind of different combinatorical aspects exist and some problems very important problems with dice with with card deck etc so I do recommend you to read this chapter it contains lots of different lectures about different aspects of combinatorics okay that's it for today thank you very much and good luck