So, the methodology for creating flip classroom is what we talked about in the session as well as in the lab. So all that what we have written as guidelines that is actually the methodology and if you want more information on that if you look at the references slide if you go to the links that are there in the references slide you will get more detailed information about how to create flip classroom videos and how to use the flip classroom videos in your class PSG college of technology. What is the percentage of marks allotted for each levels that is a recall level understand level in question in question purpose. So in fact there is I am glad you brought up this query because there is a formal way of systematic way of doing this and it is actually called a blueprint where what a teacher does is takes all the topics that need to be that need to have questions in the exam and also looks at all the levels and the teacher has to decide how much percent and how many marks to allocate for each topic as well as for each level. So this process starts like that however it is really up to the teacher to decide what percentage of marks to allocate and the teacher should decide based on the learning objectives again because it depends on the depth at which the teacher expects the students to learn the topic and the depth at which the teacher has done the instruction on the topic. So a lot of these decisions in fact it is the teacher who has to make but having said that what is recommended or what does not exist if you try to Google for something like this you will not find any answer because there is no single correct answer but what is recommended is that you actually have a mix over different levels and especially in the engineering and sciences disciplines a lot of the discipline involves doing and problem solving. So you should at least have some questions and some problems at the apply and analyze level. What happens currently in many university question papers is that a lot of questions are at the recall and understand level. So what we are recommending is do go beyond that go to the apply and analyze level and if you are able to do create level in the exam if it is hard to do that then do create level questions as part of the lab exams. Madam like after the teaching method or the blackboard teaching appears to be best because the students are able to follow what the teacher is able to present there in the classroom. So like I said during the flipped instruction class if your goal is information transmission and if you feel that blackboard instruction works best for information transmission then sure that is fine but if your goal is for the students to be able to assimilate the knowledge for them to be able to do something to be able to apply the knowledge then you do need to implement these active learning strategies. So it depends upon what your goal really is. Sir how this flipped classroom videos will work out for theory courses because if there is no problem solving sessions it will be very difficult to give problems or discussions on theory papers. So if a theory course has higher level cognitive objectives if it has learning objectives at analyze level or if it has learning objectives at create level for example you know even if you take a topic or it may have some proofs which have to be done or it may have to have it may have pros and cons debates of various topics. So even in a theory course all of this is possible to execute. So if you Google for flipped classroom activities you will find lot of courses in architecture and even in English humanity social sciences many such topics there are people who are conducting flipped classrooms in that. So it is possible to do so for theory courses in engineering also again it depends upon what your objective is. This is regarding the flipped classroom is it possible to cover the entire syllabus otherwise if it is going to be only for 10 to 20 minutes and there is always every day at least 25 percent of the students will be coming without seeing the video course. And what are we going to do for that and if it is only for 10 to 20 minutes is it possible for us to show the video in the classroom itself where all the students are present and then we can have the active classroom and we can conduct the activities. So this is a good question so there are two parts one is what to do if the students come to the classroom without watching the video. So different instructors do different things some instructors take a quick quiz at the beginning of the classroom to ensure that they have watched the video before coming. Some others actually do summarize whatever they have learnt whatever was supposed to be there in the video before they start. According to the question of can we have students watch the video in the classroom itself before we start see that is possible but it may not achieve the entire purpose because the students have not spent time with the material before coming to the class. So the desirable thing is that the students should watch it before coming to class do a small quiz in the beginning of the class to ensure that they know the topic at the recall understand level and then do the further learning in the class itself. So if you look at the examples that we have given in the slide so if you see the one on arrays what is done is just the simple definition and a simple demonstration example is there in the flip classroom part but in the actual classroom they are extending that example all the way to writing a sorting algorithm on an array. So this again is something which the instructor might feel that I need to tell them how to write a program for sorting but if we use the flip idea of creating a think-pair share activity it can turn out that they will discover most of it on their own and the instructor simply at the end has to compare it with their own program that they have written. So in such mechanisms the so there is no fear of not being able to complete the syllabus because the flip classroom video is short. Now whether to do the flip classroom video in class or not you will have to just try it you know you will have to do it in your course you will have to see whether students are watching how many students are watching see whether the incentives work or not and different strategies will have to be used. Good evening so we just have a related question being discussed. So do these cognitive levels relate to the undergraduate or post do they differ for undergraduate and postgraduate courses in the sense that if you have a first year student or a final year student or a postgraduate course almost every course is expected to have their learning objectives for all the orders of cognitive levels. Yes every course would have you know so an undergraduate or an early course may have a lot of learning objectives which are at the recall understand and apply level and some analyze and some create level. The higher courses you know if the second course on the same topic or you know an advanced course you may not have any learning objectives at the recall or the understand level and you may focus completely on create level where we are talking about designing of systems or you know identifying pros and cons between various approaches. But in all courses it is desirable to have a mix of learning objectives so even in a basic course it is desirable to have some at least some topics or some objectives at the create level because that is what keeps the students engaged. I want to add one point to this. If you recall today morning's assignment you all took one topic it was just a single topic from a single course and you wrote questions in all six levels. So one reason we gave that assignment was to actually help you realize that that is indeed possible and if you think a little hard you will see that you will be able to write even evaluate and create level questions for introductory courses which are not outside the syllabus and they are not something difficult to do but they are still interesting and challenging enough. So you can have a few of such questions and few of those objectives that students can do as homework or as assignment or a project or a group project but it is desirable to have a range in most courses and not say that keep only the lower levels for the lower courses. Related to concept maps, C maps which we are designing, how could the C maps be benefited to the students or are they meant for teachers only? In fact the concept maps are definitely recommended for students as a learning aid. There are a number of ways you can use this so for example once you create it as a teacher for your syllabus and for your course plan one straightforward way you can use it is in fact to help the students realize where they are in the course as a course goes on and also to show these interconnections that we discussed yesterday. So if you are in chapter 5 and when you are beginning the class on chapter 5 you can show the C map and see how the concepts of chapter 5 relate to chapter 2. So that is one thing you can do. The other way you can use C maps is actually teach this technique to students and help them or make them use it for their own learning. So you can have students create concept maps for a specific topic and in fact assess them to see how well they have understood the topic. So getting students to draw their own concept maps is again one technique you can use. First of all thanks for this learning episode it is really very informative and looking forward to implementing my teaching in forthcoming classes. There are two concerns right now coming in my mind that it is not very always possible and to motivate the student to actually go and do the homework and watch the video and all. So the next day if they come in the class and they find that a huge chunk has not actually watched the video and then discussing about the video will make this huge chunk off the hook and they will not understand what is going on. So in that case become very difficult and I don't know how to understand the situation. And the next thing is that in asking the questions I found that students mostly appreciate if the questions are just the very basic level either maybe the understanding level or recall level. Any question beyond this level it makes them that they say that it is very difficult. So how to tackle this situation actually how to make them understand that actually these are also important and they should go ahead with that. Okay so there are two parts to the question. For the first part I have already answered that for another query is that we have to try various mechanisms. What works for my university may not work for your university students. So you have to try with your own students as to how to incentivize them to watch the video before coming to class. In the worst case like the other person had suggested maybe you play if it's a 10 minute video you play that 10 minute of the video in the first 10 minutes of your class. Okay so that is the first part of or you conduct a quiz and give some one mark for every such quiz that also serves as the attendance and you can say that I will take the best 10 out of 15 quizzes something like that. So you have to try various strategies to ensure that they look at the videos and once they figure out that looking at the videos and coming to class actually helps them you will find that students start doing that more and more. Coming to the second point of students being interested only in simple questions see that is where sometimes as teachers we don't give adequate credit to our students. See students often also enjoy challenge. So the only disconnect that happens for them is that they need a situation in which they can implement this challenge. So if you give them a real life context like what we saw for the Pythagoras theorem example many of you came up with questions like okay an airplane is approaching what is the distance or there is A to B there are two places and that's where you have to find out. So if you give problems which are with whose context that the student can identify then they will actually want to solve these challenging problems rather than to only solve simple problems. So again this is something which you just need to keep trying. Just a little clarification actually sir that when I last time I taught one course called Heat Transfer and it was with the coordination with the QEE program with one professor from IT Madras because actually so in that case inadvertently in the first exam internal exam I said the question in such a manner that they could easily avoid the entire numerical and write on the theory and then they can still write the 100 percent. So they scored very well. So next time I was very much careful that they will go partly this and that way. So next time the scoring was very low. So it was a very difficult situation for me that how to make them realize and also make them sure that they score good. So I don't know how to do that. What you need to do here the question is about how do you get students motivated to answer or to think about more challenging questions. Your view I mean you're right in the sense if you spring one of these challenging questions straight in the exam they may not be motivated to do it. So what we've been talking about the last three days is bring in real life context bring in challenging questions into the classroom right from the beginning. So once students see that this course in the material in fact is useful for addressing real life problems either through videos or bring in some industry connections connected to what students relate to Facebook or there you know find out what actually they relate to and do problems which connect to their interests in class itself and my problem I don't necessarily mean a numerical problem but connect the theoretical abstract concepts of heat transfer to something that they see regularly. Once they get used to it in class itself then they won't be so surprised when they see it in an exam but you can't directly give something very hard in an exam and not do it in class. Yeah go ahead. We are happy to attend this three days training. We thank to Mumbai IIT for giving the wonderful training to us. We follow it. So here the first question is we framing the assessment of the any subject shall I include the all the level of the blooms takes and I mean. So it's not necessary that in a single question paper you have equal weightage for all levels that's not what is recommended that's not for anybody else to say. Again what you have to do as a teacher before you decide the assessment pattern is go back to your objectives and see what weight how many learning objectives do you have at the apply level in which topics are there and how many learning objectives do you have at the analyze and create levels. So based on what you expect the students to do at the end of the course keep your assessment pattern consistent with it. So you have to know what you want students to do and the assessment should reflect that. And the next question is how will you plan the times for the one hour class. So I guess what you mean is what will you do in a one hour class. So that is where your lesson plan comes into play you know if you have decided that you are going to do a flip classroom then you have to decide what video they are going to watch and what activities that they are going to do in the class. So for example today's flip classroom session was like a one hour 15 minute class ok but what all did I do I did some telling in the beginning and then there was an activity in fact we started off with an activity we started off the scenario we started off with a debate which got all of you thinking about which way of teaching is better for what purpose and so on. So that is an activity and then there was some 10 minutes of telling where I was talking about what are the limitations of traditional classes. And then there was another activity where you were to create something for a flip classroom think pair share that we did. And finally there was another activity where you had to come up with some activities for a given video. So there were three activities with some telling interspersed and if you do that so even though it was you know 2 p.m. activity just after lunch and so on it did keep most of you engaged with the content. So that is typically the formula that you do one activity to begin with let it go on for 10 15 minutes do some telling for 10 15 minutes do another activity and then again do some telling. So that model is what you would typically do in a one hour classroom. One broad guideline that's recommended is that you don't do the same activity for more than 20 minutes. So just keep this broad guideline in mind and every 15 to 20 minutes or even every 10 minutes just do something different. Yeah Vishwakarma college please go ahead. My question is can we apply a sea mapping chart to every sub course say for example for power plant engineering where we can start any power plant from the beginning. So there is no mapping I think so can we able to apply for any subject such type of sea mapping. There is no rule that says that you can't or there are only these subjects for which you can apply because the sea map is nothing but a visual representation of the topics and the connections. So if you have the topics and whatever connections you have you represent it visually instead of writing it out as a list that that's really the essence of a sea map. Yeah Don Bosco go ahead. In an institution where attendance is very much important students have normally this attitude that even if they are assigned some homework they don't do it at home thinking that everything will be discussed again in the class. Now when students have this kind of attitude how to tackle this actually supposing we are asking them to watch some videos or do some reading still they think that everything will be everything will be discussed in the class again then what is the use of doing it at home. They have this kind of attitude and by just by listening or what is done in the class they are able to follow and they are satisfied with that because they are very confident that all all the things are going to be discussed in the classroom itself. Are students expecting that you will discuss everything in class yes but then students are also looking at what kind of questions you're asking in the exam so the moment you start having these apply level and higher level questions in the exam and you start saying that okay these questions you will be able to do only if you watch the video and come to the class for the active learning activities you will find that students are automatically beginning to do the homework before they come to the class the second point is that often students don't watch the homework because they don't get the point of watching that they don't get we have never communicated to them why they should do that particular activity so why should you learn about Pythagoras theorem did anybody tell before they taught you that a square plus b square is equal to c square so if we communicate the why they should learn a particular topic well enough then they will have the motivation to watch the video and come and in the worst case you can always use a stick you can always say that you give a surprise quiz and you say that okay this quiz is going to be of five arcs total weightage so if the price does not work then the stick will work one of the two things you have to use and that is where you have to exercise the creativity as an instructor to keep changing your pattern and ensure that they are engaged with your class yeah yeah actually we are introducing this team for the first time there are students so this is to be propagated to all the faculty that is who are handling the subject for the particular class otherwise students may feel that a single person is doing this kind of activity there will be any good response from the students side okay so there are mixed responses for this what happens is that sometimes initially there will be resistance from the students okay so even in our classes initially we do get resistance because the students are used to coming to the class and going to sleep whereas here you are forcing them to use their brain and you know work and at the end of the class they often leave with a feeling of you know tiredness because they have actually done some work in the class so initially there will be some resistance from the students but eventually or very soon they realize that they are not having to do any work outside of class they realize that all their work all their understanding all the effort that they need to put in order to get good grades in the exam is all happening in the class itself so at that point they flip over and they start appreciating so if you look at if you do this for a course and you look at your course evaluation you will find that many students actually appreciate these techniques and the moment that happens other faculty colleagues start getting buy-in into such techniques so that is the threshold point I also have another suggestion especially that I see I see that there are some 20 or 30 of you here the more number of people that the students see the more number of classes they see this sort of techniques being used in it will be easier for them and it will be easier for you you are right that if you're the only class where they see these things there will be more resistance especially in the beginning but let's say there are say three of you from one department all three of you agree that in the next course you'll do these kind of activities and slowly try to get more faculty from your department even one at a time one new faculty at a time trying these techniques every semester means there's one more new person who tries this so if the 20 of you who are there try these techniques next semester students are likely to see it in multiple classes and their buy-in will also increase and one last point about this is that don't be too ambitious to begin with so just because we have got exposure to peer instruction and debate and think and share and flip classroom and all these techniques don't attempt to try all of them at once in set a let's say in a month itself or so because then what happens is it's possible that students will get over well so try one technique let students get used to that technique you also get used to the technique then try the next technique and so on so that is a easy way of gradually getting into introducing active learning instead of trying to do all of it at once and then students getting over well okay so let's see if there are questions on the chat okay so the question is when we evaluate the class based on measurable learning outcomes how many percentage students are expected to give the outcome okay so that is actually hard to answer because it can be that if all the students in your class are highly motivated all of them will be able to show that outcome but generally a thumb rule a very broad thumb rule is called an 80 80 rule if 80% of your students can achieve 80% of what you expect them to achieve then you are doing a good job as a teacher so that's a very good yardstick to follow 80% of the students being able to show 80% of the outcomes for what you want here is another question which I think all of us commonly face this is on chat that in many colleges there is a mix of students and the mix in terms of their prior knowledge their interest and motivation and so on so how can we come up the question says how can we come up with a strategy to motivate and bring our average students to match with the excellent so I think I'll ask this question slightly differently it's not how do we get our average ones to match with the excellent but in if you have if you take a learning outcome based approach you're actually not looking at relative understanding but you're you keep absolute standards and average students improve and the excellent students actually go even higher so if you many of these active learning strategies have shown that it's not only students who are low achievers but also the high achievers everybody goes up a little more than if they're passively listening so that's one way you can think about these strategies that keep some absolute criteria and standards and make sure that there's a minimum bar which everybody has to attain and do strategies accordingly but also keep some challenging strategies so that people can go higher. Okay so now coming to what we expect many of you have thanked us for a good three day workshop so what we want to remind you is that it's not a three day workshop it is an eight week workshop and only three days of those eight weeks are done so for the next six weeks we are going to meet online and then we will meet again for three days in the seventh week and then the final week once again we will do online so what can you expect in the next six weeks. What you can expect is you will have assignments posted on Monday or Tuesday which will be about half a day's worth of work for you and that assignment will be due at the end of that week okay maybe on a Saturday or a Sunday depending upon when you find time you will have to do that assignment and that is what will happen for the next six weeks so one of the things is what are these assignments so some assignments will be where we will post videos it's sort of like our flip classroom videos so we will post some videos that you have to watch on different strategies like think per share peer instruction and you will have to come up with such questions for use in your course for the next semester so all of these activities are geared towards giving you more and more material not giving you more and more material but towards assisting you to create more and more material that you can use for your upcoming course so there will be so another week we will have activities of find visualizations find videos of your course or for topics of your course and then create those activities so once you have this after the six weeks what you will find if you participate diligently what you will find is you would have created a lot of lesson plans for many of the classes for your next semester's course and that is our goal so if that happens then to an 8080 ratio our course objectives will be sort of met in order to help you with these assignments we will do two things one is that we will respond to moodle queries so you can always post queries on moodle we will respond to them and the second thing that we will do is we will have office hours so this notion of office hours is that we will be available on every Saturday on moodle through the chat forum of moodle live so we will not be live on video we will not be live on video but we will be live logged in for exactly one hour on Saturday from from 11 a.m. to 12 p.m. okay so we will be logged in from 11 a.m. for one hour on Saturdays and you can post your queries on the chat moodle and we will reply almost in real time to those queries so that's like what is considered as office hours where you get to meet us at that time every week so these are the two things that we will do in that in the next online phase so I'll not say more about what to expect because that you will see anyway you will not be able to absorb all of that right now and you will see as the things progress week by week you will see all of them there's actually one question on chat which I would like to respond to because it is relevant I think to a lot of us and this question is what to do if we have large classrooms and are there some tips so I want to just give a very brief answer to this that a number of techniques that we have discussed in these three days are valid even in large classrooms and especially the the polling and peer instruction which is going to be the topic of the very first online assignment so you've seen you've actually been the students when we do polling but how do you write your own questions and the pedagogical technique here is called peer instruction how do you implement it in class all that we'll do in in the first online week where we'll post a video and we'll also create worksheets and where you can write your own questions so peer instruction as well as that strategies that you that we discussed for use with visualizations all of these are absolutely implementable in large classrooms and here we have tried it in classrooms as large as 100 to even 300 so we've it's completely valid it works exactly how to do it will be there in the online phase so finally the other point is that we mentioned it in the first day so I'll just repeat it that there is a conference called technology for education so this is an IEEE conference for technology for education this is held in India every year and some of us are on the steering committee of that conference so what would be good for you is one to attend the conference and two to write a paper on some technique that you have tried in your class so let's say you created a flip classroom and you tried that in your classroom okay so what you could do is so we have created some templates for creating for writing papers based on such experiments so you could take those templates and write a paper on what happened in your classroom what did students think about it what did you think about it there are various ways of measuring effectiveness so one of them could be from the perception of students one of them could be from actual what happened in the exam did the students marks improve from last year questions like that so you could perform experiments of that sort and write papers to for this conference called technology for education and we would be happy to see you there at the conference so this year's conference is in Amrita University in Colomb from December 18th to 21st the only thing is that the deadline for paper submission is just about a month from now it will be sometime in middle of July we'll post the call for papers and the website will post it on Moodle so if there is a technique that you can quickly do and implement in class in case it's possible do think of writing a paper for it otherwise at least you can attend it this year and use these techniques over the next semester and do something for next year also but we'll be posting a link to the T4 I think that brings us to the end of these three days thank you


Ukrainian Defence Forces receive new drones capable of striking deep into Russia. The Ukrainian Armed Forces have received the first batch of new unmanned aerial vehicles AQ-400 Skythe. These UAVs are capable of carrying out strikes hundreds of kilometres deep into Russia report defence blog. Terminal autonomy, formerly known as the startup One-Way Aerospace, has announced the successful completion of procurements aimed at supporting the combat deployment of the AQ-400 Skythe system in the Ukrainian Armed Forces. The AQ-400 Skythe is a long range unmanned aerial vehicle with a range of up to 750 kilometres designed for takeoff from short runways or catapults. The vehicle is capable of carrying a payload of up to 32kg which can be increased to 70kg by reducing the flight range. It stands out among the competitors for its quick assembly, mass production capabilities and economic efficiency making it a key type of weaponry in modern warfare. The cost of the drone starts at $15,000. Terminal autonomy has strategically organised the production of kamikaze drones through a network of smaller enterprises and engineering firms with experience in manufacturing similar platforms. Although the company procures certain components outside Ukraine, primarily in Germany and the United Kingdom, the project's priority is to maximise the localisation of production within Ukraine. In the initial stage, the company plans to produce 100 units of drones per month. Gradually, this figure is expected to increase to 500 units. With the production of hundreds of drones per month, these specifications allow for regular strikes on objects within the territory of the Russian Federation.


In this lesson we're going to be learning about the Kosiski test, which is one method that will learn how to use to break the vision air cipher. Now this test is named after Friedrich Wilhelm Kosiski. You can see he was born here and lived mostly during the 1800s. And this work was first published in a 95 page book on cryptography that he wrote here in the original German, but translated into English, meaning the secret writing and the art of deciphering. So we've mentioned before that Charles Babbage was actually the first person really to break the vision air cipher, but it was Kosiski was the first person to have a general attack that would work for really any messages that we were to intercept. And this method relied on the analysis of gaps between repeated fragments in the cipher text, meaning the distance between repetitions in your cipher text. So let's let's look at a sample cipher text here. And if you were to pause your screen and look at this, you might see some repetitions in the cipher text. Actually, why don't you try and do that by hand real quick? Hit that pause button. And then we'll show you some of the repetitions in three, two, one. Here's some right there. So this fragment Z V R A O, we see shows up not once, not twice, not three times, but actually five times in this short paragraph. And in fact, it's not the only fragment that repeats. Here's another set of repeated cipher text and B D B M V S shows up also five times in the cipher text. So we'll have to figure out first, why are these repetitions even occurring? Is it by chance? Or is there something else at play? So let's look at a smaller one here. Let's let's start with some plain text. Let's start with a small message.  And notice the word plane shows up twice in this message. We're going to pick a random small keyword here is going to use the word milk. So this would work for any four letter keyword. And if we were to write out our plain text and create our key stream on top and encrypt the message, you'd see we do get some repetitions in the cipher text due to a couple of things. First, there's a repetition in the plain text. We saw that that word plane showed up twice. But also we can see that above that repeated plain text word, our key stream happened to line up the exact same way K M I L K both times we started with the word plane. Let's think about why that might be the case. Between the repetitions in the plain text, there are eight characters. So if you started the first P and the first word plane and count over this eight more characters until you get to the P and the second plane. And since our keyword has a length of four, it just happens to line up really nicely. So we have K M I L K M I L, and then K again, starting the sequence again. Well, that is that actually doesn't have to happen with the four letter keyword. If we use the different word this time hospital line up our plain text with the key stream encrypt. And again, under the word plane, we get the exact same cipher text ETT NP. And again, it's because our key stream lined up in the exact same way over the word plane. This time it was P I T I L, there's eight letters in the word hospital. And because again, there's eight letters from the first P and the word plane to the P and the second word plane, that eight letters has happened to repeat perfectly right on top. Now this won't always happen. Here's a five letter keyword water. And when we set up our plain text, key stream and cipher text, we don't get a repetition. And again, because the length of the word water is five characters in it. But we started with an E over the P in the first word plane, we had an A over the P in the second, it didn't line up correctly. There's eight characters between the repeated letters in the plain text, but five characters in the keyword. It didn't really stay in sync, we got off sync by the time we got to letter P. So we didn't get a repeat in the cipher text. So you might have already drawn this conclusion, but it appears that if the length of the keyword, which we can call n, can evenly divide the distance between the repeated plain text, which we can call D, the cipher text will also repeat and in fact, must also repeat. So if the distance between the repeated plain text and cipher text is large, however, there could be a lot of different keys that could divide length D a lot of different lengths, which would still leave a lot of options for the true key size. So it'll be helpful for us if we could have multiple repeats, like we saw in the example, in the plain text and cipher text, which will help us narrow down the choices for n. So if one gap had a length of 10 between it and the next gap had a length of seven between it, maybe that'll help us figure out something about the key length, more than if we just had one gap to identify. So let's look back at our example, cipher text here, and we'll just highlighted all of those repetitions. So if we look at the ZVRAO repeats, the first one of those is at index 102, so we kind of counted those there for you. The next one would appear at index 134. So the distance between those character Zs would be the difference between the indexes. So 32, they're 32 characters apart. The next occurrence ZVRAO is at index 390. So we could do 390 minus 134 to get another difference between the repeated fragments in the cipher text. And we could do it again at 402. And again at 426. So these five kind of occurrences of the same fragment, we can get four values for that D, the distance between the repeated cipher text. And now we know that the key, the length of the key must divide all four of those distances. We could even do better. We could go to the green words, the MBDB, MVS. And we can see that the first occurrence is at 143, the next at 155, the next at 307, and then 411, 439. We can get four more distances. And since the key is the same for this entire message, we know that the length of that keyword must divide all four of those as well. So we've got eight differences between repeated fragments in the cipher text. And we know that the length of the key must divide all eight of those numbers. Let's see if that's helpful for us to figure out what the true value for the key length is. Now, before we do this, there's a big assumption here. The assumption is that these repeated fragments of cipher text is only caused because of the repeated values in the plain text and our keywords getting kind of back in sync due to their length. In theory, it could be just by chance alone that we're getting these repetitions, maybe we have some keyword and it lines up just right to create a repetition otherwise wouldn't have been there. But if we have a long repeat makes six or seven characters like we've seen, and you get multiple repeats of the same fragment, the chance of this repetition in your cipher text just being by just pure luck is very, very small. So we're going to make this assumption, but just realize it could be wrong in very specific situations. Alright, so let's take a look. We've organized these indices, so there's our 10 indices, five for the first fragment, five for the second. And then there's the corresponding values of their differences, which we called the values for D. So those differences are 32, 256, 12, 24, and then 12, 152, 104, and 28. Now remember the value n, the length of the keyword, must divide all of those. And we have a nice handy algorithm that helps us find divisors and in fact a very specific divisor, the greatest common divisor, which is actually really handy. You probably, if you were to find multiple divisors for all of these values here, you could probably assume that your encryptor was using the largest value of those because that would reduce the most secure message as we've learned about in other lessons. So let's find the greatest common divisor of all of those numbers. And it happens to be four. It's actually not a very large divisor at all. So that is going to be our most likely length of the keyword, since it is the largest value that divides all eight of those numbers. But it could be other ones. There are other values that would divide those eight numbers. They're the numbers that also divide four. So one or two. Now, we know a key value of one or a key value with a length of one in vision air is really just a Caesar cipher. So that's probably not what was used here. And if a key value of two, a length of two, not very secure. So again, it's a pretty another good assumption, especially when the value is this small, that the greatest common divisor of all the values of D is the most likely candidate for the length of your keyword. Now, now that we know how long the keyword is, how does that even help us? Let's take a look at our cipher text again here. I've just shown the first line of it. I'm going to color code each kind of grouping of every fourth character. So color code them here, the green ones were all then in ciphered using the same letter from the keyword, assuming we guessed the length of the keyword correct, that j q c j q l and so on. We're all in ciphered using the same letter from the keyword. The pink letters a s w b and so on are all in ciphered using the same letter of the keyword, the orange ones, the gray ones, likewise, that we don't know those letters yet, but let's group them together. So all of the green letters there are all of the letters from the cipher text that were in ciphered using the same letter of the keyword, the pink ones, likewise, and so on down the line. Now, we've seen in other lessons that if all of those letters were in ciphered using the same letter of the keyword, that means they were all added to their numerical representations were all added to with the same value, which means that they should follow the same distribution of the Caesar cipher, meaning any other sort of mono alphabetic cipher. And in our case just been shifted left or right from the English standard distribution. So let's actually take a look at the bar charts for each of these four groups. Here they are color coded. And we can see starting from left to right, that we have some telltale spikes that we saw earlier when we were doing the crypt analysis of Caesar. I can see that this J is the most likely keyword for the green group because I see the five letters down the way at the end, we have a larger spike, so that's probably the a e spike right there, meaning that if a got shifted to J, that means that the first letter in the keyword is most likely J. The second and the pink here, we can see kind of the O and the S are the most likely a and the E from the English that have been shifted. I don't see any other kind of spacings like that with the double spikes. Likewise, with the orange group K seems to be where a got shifted to so K is probably the most likely letter in the keyword in the third position. And see if you can guess on the last one. E. And it actually works out pretty nicely here that these four letters form an in English word joke. And now if we think that's the most likely four letters of that make up our keyword, let's actually go ahead and decipher the ciphertext of the joke as the keyword. And if we guessed it right, are able to deduce that correctly, we should see some English pop out.  Now in this case, it does happen to work out pretty nicely that we get a joke out of this about a mathematician, a physicist and an engineer that you can read. But we can also see that there's a lot of repeated plain text in this message, all of these numbers like three, five, seven, nine, the word math shows up, engineer shows up, there's a lot of repetition in this plain text. So there's probably no surprise that we are four letter keyword happened to line up over those repeated words in the same position several times. But it turns out that English is also very repetitive in general, that it didn't have to be just this joke with a lot of repeated words, any length message of sufficient size, we'll have some repetitions in it, maybe it's the word the or four, or two, these common words that we use quite a bit. And regardless of the length of your keyword, they're going to show up fairly frequently as repetitions in the ciphertext as well. So that's it for the Kosiski test, we can see it's it's somewhat manual, but we could automate this fairly well with Python, we just need to find a way to find repeated fragments in our ciphertext, calculate the difference between those repetitions, and then compute the greatest common divisor of those different delta values. Once we've got that, we can then use our bar charts to make pretty good guesses at what those letters were in our keyword.


What I have been doing for this meeting, particularly, is a special issue for the Journal of Orthopedic Research. As the editor-in-chief, I can put together different papers that we've published over the last few years into a virtual issue. Every year, when Dr. Huard and the people here at SPRI have this meeting, it's good to have everyone who's here and interested in regenerative medicine actually read the original literature and read what the scientists have to say about this topic. As you well know, we have a lot of information out there in the public that's not necessarily correct. We're all scientists here, and we really want people to understand what is really the scientific basis for stem cells, or the scientific basis for PRP, or for synthetic drugs, or any of the things that we're talking about here. And they can actually read what the scientists say about it. First of all, it's just great to see everyone in person. Interacting in all these different ways, whether it's at lunch or whether it's on a break or over coffee, just talking about science, and everybody really enjoys that.


Hi, I'm Zor. Welcome to Unizor Education. We continue the course of advanced mathematics for teenagers presented on Unizor.com. I do suggest you to watch this lecture from this website because it contains notes. Well, in this particular case, notes are basically the problems which we are going to solve today and answers and I would definitely encourage you to try to solve these problems just by yourself. Check with the answer if you are correct and that would be probably even a better schooling for you than to just watch this lecture. Anyway, I will present these problems right now. I will try to solve them. I will try not to make any arithmetic mistakes, which I know I can do. Well, basically, let's just start solving problems. It's about cones. I have how many? One, two, three, four, five problems. So let's just go one by one. All right. So the first problem, well, it might have been actually presented as a theoretical material. I did not really mention in previous lectures the concept of a truncated cone. So let's do it now. So let's consider you have a cone and then you truncated it by a plane, which is parallel to the base. Now, this is the radius. This is the overall height. Now, let's consider you have a plane parallel to the base or perpendicular to the x-axis and it cuts this circle. Now, the fact that this is a circle is obvious because this plane is parallel to this plane, which means that if you take any two points here and here, now, these two radiuses are the same, right? So it's very easy to prove that these are also the same by, let's say, cutting the plane through the axis through this point, then it will actually intersect this circle at this point. And from the parallelism of these planes, follows the parallelism of these two lines obviously, right? Because it's two parallel planes intersecting by one line. So it looks like these guys are proportional to the altitudes, right? And the same thing exactly with any other two points obtained by another plane. So that's why we have the proportionality between these and this also like altitudes. And since these are equal to each other, these two will be also equal to each other because they are proportional to the same ratio of the altitudes. Okay, so that's simple, that's a side issue. So every plane parallel to the base cuts a circle. Now, if I will replace this cone with a truncated cone, now the obvious problem is to evaluate its volume in terms of now two radiuses, right? r1 and r2 and the height, which is h. So we need the formula, basically formula of the volume in terms of r1, r2 and h. All right, let's do it. Obviously, we will calculate the volume of this truncated cone as a difference between volume of the big cone and the volume of the small cone on the top, right? So what we don't know about these two things is this piece called h, right? So this is h from the apex to the top of the smaller circle, the smaller base. Well, but we can very easily determine this. Now, consider the proportionality between these triangles and obviously h, lowercase h to r2 equals big altitude to r1. Big altitude is h plus h to r1, right? From here, we can determine lowercase h in terms of r1, r2 and h. And then having all these parameters at hand, we will just calculate the volume of the truncated cone as a difference between two volumes. So from here, h r1 equals h r2 plus lowercase h r2. From here, h is equal to, this goes to this. So h r2 divided by r1 minus r2, right? I just solved this linear equation, transfer h r2 to the left by subtracting h r2 from both sides. And I will have h times r1 minus r2. That's why I divide. Okay, fine. Now, knowing this, we can calculate the big volume. Okay, the volume 1, let's say. It's equal to volume of the whole cone. It's one third pi r1 square times the height, which is h plus lowercase h. So it's h plus h r2 divided by r1 minus r2. So that's my top pyramid, the big pyramid. Sorry, cone, not pyramid, cone. Now, this small cone has a volume 1, so I have to subtract. One third pi r2 square, that's the area of the base. So one third area of the base, and the height is lowercase h, which is this one. h r2 r1 minus r2 equals. So we really should somehow simplify this thing. Well, it's one third pi r1 square times, well, I don't really need this parenthesis. If I will use the common denominator, it will be h r1 minus h r2 and plus h r2. So it will be h r1 divided by r1 minus r2 minus one third pi r2 square h r2 divided by r1 minus r2. Okay. Now, I don't need this anymore. So what could we do here? Well, obviously we should one third pi h divided by r1 minus r2 and open parenthesis. r1 square and r1 would be r1 cube minus r2 cube. r2 cube. Everything else is outside of the parenthesis. One third pi h and r1 minus r2. So this is almost the end of it. You can simplify by dividing r1 cube minus r2 cube as we all know this is equal to r1 square plus r1 r2 plus r2 square. If you don't remember it, you can always try multiply this by this and you will get this r1 cube minus r1 square r2 and then there will be plus r2 square r1 and then there will be r2 square with a plus. So whatever is necessary will cancel out and that will be the remaining. And this is the formula. This is the answer. So in terms of r1, r2 and h, this is the volume of the truncated comb. Number two. Okay, side surface. Okay, so if you have a comb and you cut it, you cut it by one of its generatrix which goes along the side surface, all right, connecting an apex with one particular point on a circle which is the base. You can roll it out into a sector. We're already talking about this. It will be a sector. So this is r and this is h. Now the side, the lengths of this line along the side the generatrix, let's say, would be l. And obviously l2 is equal to r2 plus h2 from the Pythagorean theorem. Now if you will roll it out on a flat surface, then the radius of this sector would be l, right? Because you are using all these points on the circle, they are becoming this arc. And the arc length would be 2 pi r, where r is the radius, right? So if this is the cone, this is the result of its rolling out after the cut along the generatrix. Now what's known about this is the area of this thing is s and this angle is 120 degrees. That's given. r and l and h, whatever else, it's not given. And we have to determine the value. Okay, first of all, let's just think about it. How many parameters define the cone? Well, two parameters, let's say r and h. Everything else can be derived from it. For two parameters, we need two equations, right? So, and these two equations are given basically. There is one information, there is an angle of a 120 degree, and there is another information about the area of the side surface. So basically, we somehow should represent the angle and the surface in terms of r and h, and we will get two equations with these two unknowns, and then we will solve them for r and h, and then we will determine the value. All right? All right, so what can we say about this? Well, first of all, we have a very easy thing here. It's 120 degrees. It's one-third of the full circle of 360, right? Now the full circle of the radius l has the length 2 pi l. At the same time, it's three times greater than this. So it's three times 2 pi r. So what follows is that l is equal to 3r. Okay? That's important. Now the surface area, the side surface area is s. That's given. Now, do you remember how the side surface area is represented in terms of h and r or l or whatever? Well, let me just remind you. I will use exactly the same thing. I will cut it and open it up, right? So I know that the whole circle area is pi l square, right? Now, if this arc has the length 2 pi r and the whole circle has the length 2 pi l, then what I can say is that whatever part of the whole circle, whole circumference this 2 pi r takes, which is 2 pi r divided by 2 pi l. That's the ratio of this area relative to the area of the whole circle. So s divided by pi l square. So what do we have from here? l square l. We have the s equals pi r l, right? Pi r. So this is the formula which we have derived a long time ago. And instead of l, we can always have square root of h square plus r square. But this is the formula. So let's just write it here. s equals to pi r l equals to pi r square root r square plus h square. So we will use it. And this is our second equation because s is given, right? So s is given, which means this is given. And this is another equation. So we have two equations. Now I think it would be easier if I will also have l square here. Do we need l square or not? Actually we probably don't need l square. I'll just use directly this formula. Okay, so s which is given is equal to pi r l. And l is 3r. So it's 3 pi r square. Now from this we can determine r. r is equal to s divided by 3p, 3pi square root, right? s divided by 3pi is equal to r square. So r is equal to this. So we got r. Now from this we can actually define h. Let's put it this way. l square which is equal to h square plus r square is equal to square of this 9r square. Okay, so h square is equal to 8r square. And h is equal to r square root of 8 or 2r square root of 2, right? Because 8 is 4 times 2, square root of 4 is 2, and square root of 2 is remaining. So I have h and I know r here so I can find out what's my h exactly. h is equal to 2 square root of 2 times r. r is this. Square root of s divided by square root of 3pi. So I have r and I have h. All I need is what? The volume, right? Right, so let's just volume is equal to, now it's just, you know, pure technicality. And as I was saying, I always 1 third pi r square which is s divided by 3pi times h which is 2 square root of 2 square root of 3pi and square root of s, right? What else? We can simplify it a little bit, basically, I know. We probably should multiply it by square root of 3pi so we don't have radicals in the, so we will have what? Pi and pi are reducing so we will have s square root of s, 2 square root of 2, square root of 3pi. I multiply by 3pi and in the denominator I will have, therefore, just 3pi times 3 times 3, it's 27pi. Something like this, right? Is this right? 2s, yeah, that's what it is. You can combine square root of 2, it's square root of 3, so it's 2 square root of 6pi s square root of s divided by 27pi. Oh, that's not a very pleasant formula but anyway, we did the problem. We did solve it. That's the answer. All right, 2 down, 3 to go. Okay, this is a very interesting problem actually. Here is what's given. So, given site surface s and what also is given is the following. Consider this as one of the generatrix, so a point connecting the apex with a single point in the base along the straight line. Now, if I will have this plane all AB and I will draw a perpendicular to AB within this plane, that would be distance d. That's also given. d is given. Now, obviously, if you take any other line like this on the surface and connect it with this one and draw a perpendicular, it will be the same d, right? Because these triangles AB prime O and AB O are obviously congruent because they're all right triangles with the common catatris and other catatris, other catatris are radiuses of the circle, right? So, they are right triangles with equal catatris. So, that's why the perpendicular, the altitude to hypotenuse would be always the same, or this one, for instance. So, these are all perpendicular from the center to generate resist. All right, that's given. Well, question is the value. All right. So, I actually first solved this problem like directly without thinking much. And that's what I'm going to do right now, right? So, let's assume that this is R and this is H. So, what I would like to do is, in terms of R and H, I will put the surface area. Now, I know it's pi R L, which is pi R square root of, all right, don't have enough space here. So, here equation is S is equal to pi R square root of R square plus H square, right? Pi R L, where L is this particular diagonal, hypotenuse of this triangle. Now, at the same time, if you consider just this right triangle, H, R, L, where L is this, this is D, how to determine D? Well, obviously, the area of this triangle is R times H, well, one half of this, and it's equal to one half of L times D, right? This is base and this is altitude, or this is base and this is altitude. So, from here, D is equal to R H divided by L. So, that's my second equation. R H divided by square root of R square plus H square. Now, I have to find, from these two equations, I can find R and H, right? And calculate my value. So, I did that. How can we resolve it easier? Well, we can say something like this. R H is equal to D square root of R square plus H square, and S divided by pi R is equal to square root of R square plus H square, right? So, we can substitute this to this. We will have R H equals D S divided by pi R. So, pi R square H, pi R square H is equal to D times S, right? That's interesting. This is basically almost the formula for the volume. It's triple. So, one third of this, pi R square H, which is the volume of the cone, is equal to S times D divided by three. It's too simple, right? Now, and then I was thinking, why this is such a simple formula? It looks like there are complicated calculations, but the formula seems to be very simple. Strange, right? And then, as I was thinking, I came up with a perfect explanation. Let me just draw this better. Okay, here is my cone. Okay, let's take two planes which are connecting two lines, two generators. And consider, this is C, and this is B, this is O. Consider A, B, C, and O in front of it. What is it? It's a triangular pyramid, right? Now, if I will consider A, B, C as a plane, this pyramid is basically almost like a sector of the cone. So, the plane A, B, C, the triangle A, B, C, the base of this pyramid is very much close to the surface of the cone, right? Now, if my points B and C are closer to each other, then this line will almost be like a circle, right? The closer B and C together, the closer segment B, C would be an arc B, C, right? So, that's how we were calculating something like an area of the circle, if you remember. We doubled the number of sides of these inscribed polygons, and the more vertices we have placed on the circle, the closer perimeter would be to a circumference, right? Now, here we have exactly the same situation, but it's in a three-dimensional thing. The triangle A, B, C would be closer and closer to the surface of the cone. So, now, if I will divide my circle into many small pieces, the volume would be divided into small pyramids like these ones, right? And the volume of any individual pyramid would be what? Area of A, B, C, A, B, C, right? If O is apex, so I need the area times the distance from O to this triangle A, B, C, right? Which is like called H, whatever, divided by three. So, one-third of the area of the base, which is a triangle A, B, C, times the altitude, which is from O to some whatever point B. Now, if I will add them all together, all these pyramids, areas of A, B, C together, A, B, C, and A, C, D, etc., etc., they will be more or less like a side surface of the cone. And the H, the distance from center to this plane would be very much, as soon as my B and C are closer, it will be very much like the distance from this center O to the generator, which goes somewhere here. So, the more vertices we will put on this circle, the closer the total volume of all these pyramids will be to the volume of the cone. The sum of all bases, sum of all areas of the bases will be closer to the side surface of the cone. And the altitude of each pyramid will be very much like this distance from O to a generator on the side of the cone. And that's why, as soon as I will go to a limit with the number of divisions going to infinity and sum them all together, the total area would be S and this H would be D. And that's why we have this particular simple answer to this problem. Okay. But I have came up with this only after I saw that the formula is very simple. I didn't really think about this, I just saw the first directive without any. Okay. Side surface is twice as big as base. Okay. So, again, we have a cone and side surface, which is pi RL is twice as big as pi R square, right? Now, this is H, this is R, and this is L. And obviously, L is equal to square root of R square plus H. It's just more convenient to use side surface in this format. All right. So, I've got that. Side surface is twice as big as base. Now, central section through the main axis. So, if I will cut this cone with a plane going through the main axis, through the main altitude, which connects the apex with the center of the base. It will cut a triangle, obviously, right? So, within this triangle, the area of this triangle, let's call it Bc. Area of B ABC is equal to S. Find the volume. Okay. All right. Let's find the volume. Well, in terms of R and H, the area of ABC obviously is what? This is a diameter, right? Since we cut the plane through the point O, this is a straight line. So, it's BC is equal to 2R. So, the base is equal to R of this triangle. The height is H. So, 2R times H divided by 2. That's the area of the triangle. And it's equal to S. So, we have two equations. One and two. All we need is, again, it's technicality right now, to find out R and H. Well, L is equal to this. All right. So, let's just do it. Let's do it a little bit simpler. So, from here, I can reduce by pi R and I will have what? That L is equal to 2R. So, square root of R square plus H square is equal to 2R. That's one equation. And the second equation is R times H is equal to S. That's my system of two equations with two unknowns, right? Okay. Let's square this one. We will get R square plus H square is equal to 4R square. H square is equal to R square times 3. H is equal to R square root of 3. We can substitute it into this. So, we will have R square square root of 3, right? Equals S. So, R is equal to square root of S divided by square root of the fourth degree of 3, right? From here. R is equal to square root of S divided by square root of square root of 3, which is fourth degree. Okay. I've got R and H here. And H is equal to R times square root of 3, which is square root of S, square root of 3 divided by square root of fourth root of 3. Now, what is square root of 3 divided by fourth root of 3? This will be square root of square root, which is fourth of 3. Square root of, again, fourth root of 3 times fourth root of 3 is square root of 3, right? So, that's why you divide this by this. You will get this. So, I've got both. But I can do actually, well, I'll simplify it later. All right. So, now all we need to do is substitute it into the formula for the value. 1 third pi R square H equals to 1 third pi. R square is this, which is S divided by square root of 3, right? Fourth root of 3 double, I mean square would be square root times H. And H is square root of S, fourth root of 3. Well, that's the answer. And if you don't want to have this, let me check. Well, you probably should multiply it by square root of 3 to get rid of the radical in the denominator. So, in denominator we will have 3 and 3, which is 9. So, on the top we will have pi S square root of S. Now, what is this? This is square root of 4, fourth root of 3, sorry, times fourth root of 9, right? Square root of 3 is fourth root of 9. Same thing. Divided by 9, 3 times 3, which is pi S square root of S, fourth root of 27, 3 times 9 divided by 9. So, that's the answer. All right. Fourth down, one to go. I hope you're not scared with all these fourth degree roots and stuff like this. Well, that's what it is. All right. Here's an interesting problem. Okay, you have a cone altitude, and you divide the altitude into n equal parts. So, somewhere here you will have a 1, a 2, etc. a k minus 1, a k, and the center of the base would be a nth, right? So, we have n minus 1 from 1 to n, from a 1 to a n minus 1 division points, which divide into n pieces. All right. Now, let's cut with the planes through each point. So, you have n truncated combs. Well, actually n minus 1 truncated, and the top one is not truncated. So, the question is, if you will take the cone number k, what is its volume? In case you know the volume of the whole cone, and you know the number of divisions. All right. Okay. Now, this is actually a very simple problem. It sounds complicated, but it's simple. So, let's have the plane representation of this. Let's cut the whole thing by a plane, which goes through the central axis from a 0 to a n, and it will cut the whole cone in section, which looks like this, right? This is a k. This is a k minus 1. This is a n. This is a 0. Now, obviously, let's call this one r, which is equal to r n. This would be r k minus 1. This is r k. Now, let's talk about these heights. This one would be h n. This one would be h k. This one would be h k minus 1. Now, obviously, you have the similarity of all this, right? So, how can we determine what is the value of r k in terms of r, let's say? Well, very simply, since the distance from a 0 to a k is k times small part, and the small part is actually equal to h divided by capital N, right, where h is the whole height. So, the height, which is h k, is equal to h divided by N times k. Same thing, radius r k, because it's all proportional, right? So, it's r divided by N times k. Now, since you know this, you can find out this truncated cone as a difference between the k's cone and k minus 1's cone, right? So, the volume of the truncated part would be the 1 third pi r k square times h k, right? Minus 1 third pi r k minus 1 square times h k minus 1 equals 1 third pi open parenthesis. r k is k nth of the r, so it's k nth of the r square times h k, which is k nth of the h minus 1 third pi. Now, instead of k, I should put k minus 1 r divided by N square and k minus 1 h N equals. So, what do we have now? Well, we can 1 third pi r square h out of the parenthesis, and actually N also we can put out of the parenthesis, N square and N, N cube. That's what it is. So, let me just write it more accurately. So, it would be pi r square h divided by 3 N cube.  What would be inside? Well, k cube minus k minus 1 cube. That's what it is. Now, this is the volume which is given to us. So, the whole thing is equal to v divided by N cube times k cube minus k minus 1 cube. That's the answer. That's it. I do suggest if you didn't do it before, why don't you go through the same problems again and try to solve them by yourself. Check the answers. Answers are on the web page. And good luck. Thank you.