AIML Module 2 Important Questions | ONE SHOT VIDEO FOR EXAM | VTU 5th sem| AIML #21cs54 #aiml #vtu

in this video we'll be discussing the important topics of AIML subject second module that is informed search strategies and ml introduction so if you have not watched my previous video I've explained all the important topics module wise so as an extension to that I'll be discussing the answers here other the important topics so in second module in informed search strategies we have GFS that is greedy best first search so greedy best first search is it displays the path that appears to be best at that particular moment so right so the the actual optimal path may not be the uh the same which which which the result gives in Greedy best for search but then at that particular Point whichever it finds to be the best would be selected okay the order the order or the basis on which the selection is done is the herotic function here so a node is considered for expansion based on evaluation function that is f of N and constructed as the uh cost estimate so that other node with lowest so that the cost estimate so that the other node with lowest cost is expand first so here this is pretty important f ofn is equal to H of n h ofn is the cheapest cost or the estimated cost from the node and to the gold goal node so as I said greedy best for search may not give the optimal solution but then uh yeah the solution has some sort of value so looking at the algorithm here looking at the algorithm here function best first search we enter the node so initialize open to the starting node so firstly we take the open and then initialize it to the starting node if open is empty then return failure if open is empty you can't go further right return failure and stop if not minimum value of H of n from node n placed in the closed list this is the second step this was the first step and this was the Second Step adding values in the closed list and then we expand in and generate the accessor so from n we move forward considering the path cost okay minimum value of path cost then expand and yeah this is done if n is equal to goal State then return closed solution once we reach the closed State whatever is there in once we read the solution whatever is there in the closed state would be given as the output then else continue visiting all successor and obtain closed so once you visit all the successes once you reach the end point then you can give the closed U list closed as an output here the time complexity is order of B power M okay then moving on to the a star search so a star search is a combination of UCS and greedy BFS so here we make use of herotic function that is H of N and cost to reach n node from State n so previously we had used F of n is equal to H of n for greedy Bas first search rate here we'll make use of f of n is equal to G of n plus h of n so here G of n would give us the cost to reach the node and then h of n would be the cost to get from no to the goal okay so f ofn is estimated cost of cheapest solution through n so compared to greedy best for search AAR search expands less search re and provides optimal solution faster so looking at the algorithm so function a star search and starting node you the starting node is present in the open list and then uh and then if open is empty then return failure and stop if not the smallest value of evaluation from f ofn is equal to G of n plus h of n the sum of this would give F of n value right the smallest among this would be selected if n is the goal goal node then return success and stop otherwise you look into this part that is to expand in and generate and general all successor and generate all successor nodes and put in closed so if n is already open and closed then it should be attached to back pointer which reflects lowest G of n value once you reach once the closed list is completed with all the result with the path follow path cost and the path followed we generate the return list so the time complexity here is order of B to the power D let's look at next we have the eight puzzle problem using hero stics so we had discussed eight puzzle problem in the first module right we had defined the state goal State problem uh the gold test and the transition sequence and all so here we'll try to give one possible solution or one possible possible way to address this problem using heroist function so here we would be making we would be using the Manhattan or the priority function that is also considered as a distance so this this is the initial State and we need to reach the goal state that is this empty TI should come in the middle empty TI should come in the end okay so this is the uh the goal state which we have to reach so from the initial State there are three possibilities that is either two can move here eight can move here or seven can move here I'll write it accordingly so 158 if two moves here firstly I'll write three Space 2 and then 4 6 and 7 now I'll consider that uh 7 has moved up okay so I'll write 158 327 and 46 in blank here next I'll consider that 8 has moved down so I'll write one five space and then here I have 3 to8 and 4 67 now I'll calculate the Manhattan distance okay so Manhattan distance says that let's let's write all the notes Here 1 2 3 4 5 6 7 8 eight uh tiles are there here right I'll consider this as M1 you consider this as M2 and this is M3 the distances so now for calculating the M1 distance compare the M1 U M1 box with the gold state so check if one and one are in are in the same state or not one and one are in the same position so write zero here now check two two is here and two is here so in order to move the two from here to this position from here to this position I need to perform two shifts right first shift would be like this and the second shift would be like this so I'll write 0 + 2 here for three three is here and it has to come here so 1 2 and three so I'll write for three I'll get three and then for four four is here and four is here so one shift if I shift up I'll get four here I'll write one so that is one shift is required and for five also one shift is required five is here it has to come down I'll get five for six also one shift is two shifts are required like this go here and then go here so I'll get two shifts for six and for seven also two shifts 1 and two for eight I would require eight is here right 1 2 and three 1 2 and 3 three shifts are required for eight so if I add this 4 I'll get similarly calculating the Manhattan distance for M2 for one one is in the same state two I need to shift two by up one and then three three is here I need to shift 1 two and three so three shifts four one go up once four five go down once six go up once 7 um 1 2 and 3 3 for 7 and for 88 should come here right so 1 1 2 and 3 so this I'll get 14 and when I calculate M3 I get 0 + 1 + 3 + 1 + 1 + 2 + 2 + 2 I get 12 so this M3 has the least Manhattan distance so from M3 M3 I can further split so now there are two possibilities either five can come here or eight can come here so this keeps on repeating until I find this particular goal state so that was about eight puzzle problem using hero stics uh this is not that important but in case they ask we should be capable to calculate the Manhattan distance at least next we have the greedy best first search so in this so in this they have given the they have given the graph here and then here the Hof and values are specified so as we have seen greedy best for search let's write the formula that is f of n is equal to H of n so in order to perform the searching we need to consider the F of n the H of n value herotic values okay so whichever node has h of n as zero would be the end or goal State okay and wherever we have S would be the start state so s is the start State here and G is the end I need to reach G here so in order order to reach from s to G I need to consider only the H of n values I need not consider whatever is written here okay so that is one point to remember about greedy best first search so I'll start from s s has h of n value 13 then from s to a for a I have how much for a I have 12 and then for B I have for B I have four here I'll write B here so here 13 + 4 and 13 + 12 so 13 + 4 would be 17 right so I'll write 17 here here 13 okay now from B from B I have two options further E and F right so for e i i I'll check for E for E it is8 so 17 + 8 and for uh f it is 2 so I would consider F right so for f it would be 17 + 2 18 19 and from F I have two options further from F to I for I it is 9 so 19 + 9 would be there and for G it is zero so I can reach G directly and G is the end state which I had to reach right so I have reached G the end state so now the path cost would be 30 so from s to b b to F and F to G so if I write s + B + f + G I would get 19 19 + 0 here right it is 19 only and considering the path distance in order to write the path distance I will make use of these values whichever are specified here so S2 B I have 2 2 plus B2 f is 3 and F2 G is also 3 3 + 3 is 6 so the path distance would be six and the path cost would be9 19 so this was about uh greedy best first search now we'll look into the next topic so now we'll be looking into the a star search algorithm so here the graph is given and as well as the Heros values are also specified here so for a star search we know F of n is equal to H of n plus G of n that is we need to consider this cost as well as the H of n values so let's start from s s is the start State and G is the goal which I need to reach so from s to a from s to a I have 1 here and S to G I have 10 here so so now calculating the F of n from s to a I have this G of n is 1 and H of n s to a so calculating the H of N I would get a for a it is three right so three I'll get four here so S2 a I get 4 and for S2 G I get 10 + G is for G it is zero so I get 10 so I'll select this path okay now next step from s to a I have reached and from a I have two options either to go to b or to C A to B is 2 and a to c is 1 now I'll calculate s to c s a c for s c i have 1 + 1 that is 2 plus the H of n value for C is 2 that is I'll have four here and on this side I have sa ab 2 + 1 is 3 sa B I have 3 plus for B it is 4 h of n value is 4 4 5 6 7 is the value so I'll consider this path as it is the lowest so next from S I moved on to a and from a I went to C here one here one and now from C I have two options either to go to G or to go to D so from C to D it is three here and from C to G it is four path cost is four now I'll calculate from s to D from s to s a i have reached right now I'll calculate from sa CD okay s d sa a c d okay so here 1 + 1 2 2 + 3 is 5 + 4 d h of n value is 6 got 11 and for s a c and G I get 1 + 1 2 2 + 4 is 6 and for g h of n value is zero so I get 6 so I can conclude include that this path is the optimal path s a c g s a c and G is the optimal path so this was about AAR search algorithm so this is one part of second module that is the informed search strategies now we'll look into the second part that is ml so ml here again we have um several theory topics uh which would be asked for 6 to 10 marks so let's look into them one by one firstly understanding the definition of ml ml is a sub branch of AI we know that so the textbook definition says that Arthur Samuel stated that ml is a field of study that gives computers to learn without being explicitly programmed so one textbook definition and then the the normal definition is converting logical and reasoning from experts knowledge into set of rules and programs in order to construct or model the expert system so this process is comes under machine learning so here ml model is explicit description of patterns within the data so here the pattern could be either mathematical equations or the relational diagrams like trees or graphs or The Logical if else rules or different clusters okay so this was about ml model now we'll look into the deep the relation of ml with other fields okay so firstly we have uh ai ai is the broad spectrum under which ML and deep learning come okay so AI is the broader FEI field focused on intelligent agents we have seen what agents is in the first module capable of perceiving perceiving their environment and making decisions so depending on the uh the the input is taken as sensor and depending on the input certain action is performed through the actuator that is what agent refers to so it it comes under the AI so ml is the subfield of AI that extracts patterns and regularities from data to make predictions so this was about ml then we have the Deep learning deep learning is further a sub branch of ml that uses neural networks to model complex patterns in relationship of data here neural networks refers to the interconnected layer of neural of neurons like how we have neurons in our brains uh the Deep learning is based on that particular Theory so interconnected layers of neurons to process the data and apply activation function to generate the output we'll be seeing this in depth in the fifth module so this was about the basic ml intro now we'll look into the types of ml uh now we'll look into the labelled data and unlabelled data so the data set in which the each example is paired with corresponding label or output so whatever the input whatever you have the data is defined properly or labeled properly it it falls under label data so in data set of images each one is associated with label that indicates object or a category present in the image so out of thousand images we can classify them as cat dog and car so that is what you're labeling the data it is used in supervised learning then coming to unlabelled data it is data set in which each example is not paired with corresponding data output labels so that is opposite to this so in a data set of text documents the content of each document is provided but there are no explicit labels indicating the sentiment of document so this is used in unsupervised learning so this was about labeled and unlabeled having the brief knowledge of labeled and unlabeled we'll look into the type of ml so firstly we have there are four types majorly supervised unsupervised semi-supervised and reinforcement um looking into supervised firstly we have in supervised a supervisor component provides label data so that model is constructed and generate and generates a test data so there are two stages here that one is the training and another one is the testing So based on the label data a training uh a training model would be developed and and then a test data would be given as the input and further the output or the results would be generated so in supervised learning there are of two types one is the classification another one is the regression so classification is to predict discrete labels from set of finite labels so the model learns from labeled data and is test on unseen data so once you train the model after training the model you classify them and then you give an input of unseen data to assign the labels to it so it is used in recognition disease diagnosis and all now um let's look into then the example here so here I have the label data so these are considered as independent variables so there are boxes and humans so through the classif classification algorithm the classification algorithm could be decision tree or Nave base or artificial neural networks and many others so through this algorithm we generate a model after generating a model you give a input the that is the the new test data and based on this test data it will be labeled as a human or a box so that's what is done here simply the simple definition of it next we have the regression model here so it predicts continuous variables so it creates a regression Tree in order to generate a regression tree we make use of regression model so it predicts continuous variable test variable variables like price and temperature model learns from a fit of lines fit of line or function that is through linear regression so predicting product sales based on weekly weekly data or monthly data these come under the continuous variables right so we make use of regression model here so algorithm used here is linear regression polinomial regression and so on now looking into the unsupervised part so an UNS supervisor that is Learning Without label uh without the label data that is unlabelled data the model identifies patterns and relationships in the data so firstly we have clustering so clustering is group of objects group object based on the attributes in into disjoint clusters so you took the unlabeled data which comprised boxes and humans and then through the clustering algorithms it was further classified or divided or clustered into two parts that is one would uh one which has humans of humans that is similarities match here and then on this side we have clustered to where all the boxes are kept so this was about clustering then we have U semi supervised so in semi-supervised we have uh it is combination of labeled and unlabeled so here um for here the unlabeled and labeled data is taken so application is pseudo labeling is done oftenly used to combine labeled and unlabeled data set for the learning then you have the reinforcement that is Agent interacts with environment receiving rewards and penalties based on its action so uh each and every time the um the each and every time the agent generates an action a reward would be awarded reward would be given to it so goal is to maximize the cumulative rewards over the time so if the rewards are more over the time that means um whatever the process or the operation was expected to be done has has been successfully completed quick in a certain time is what it refers so it is an approach which involves trial and error learning from the actions and their consequences so uh it is used in game game game playing and Robotics and all then we have they can ask the applications of ml it could be used in image processing data science medicine natural language processing telecommunication banking business gar gaming and many others then we have the challenges of ml so if the problems if the problems are not proposed clearly that is ill posed problems it could lead to Pro it could lead to challenges then we have huge data when the data set is of large quantity it againe creates a problem then the complexity of algorithms the algorithms being used here are very complex then we have the bias variance that is underfitting or overfitting condition can also happen here next we have the ml process so firstly understand the business then understand the data then we perform the pre-processing then we have the modeling and then the model is evaluated and then once evaluation is done your deployment uh the deployment of model is done here let's Gove on to the next topic here we have uh from the introduction to data topic we have big data so big data is large volume of data so the characteristics of Big Data it comprises of six vs so the first first one is volume that is volume is stored in pabes or exabytes then we have the Velocity that is fast arrival of speed then we have the vacity that is confirming confirming of facts and believability then the variety here the data could be graph or video or audio map anything then the validity that is accuracy of data the data which you have obtained need to be accurate enough to perform the further operations and you have the values here that is value of data which is extracted further uh the types of data we have structured data that is when the data is organized properly it it falls under structure data here we have the record data which is stored in the Matrix here the rows are called as records and columns are called as Fields then you have the graphical data an example to it would be the web pages then we have the data Matrix in which the numeric attributes or the vectors or points or the dimensional space could be stored then we have the unstructured data so in unstructured data we have video games images programs blog text documents or 80% of the organiz organizational data falls under the unstructured data then moving on we have the semi-structured data here we have XML data RSS feed and hierarchial Records then the last topic by big data analytics so big data analytics assist business organizations to take the decisions depending on the analysis which they have performed data analysis is activity that takes the data and generates useful insights and the information required for the companies so types of uh data analytics we have a descriptive that is here we are describing the entire process or whatever the problem is then we have the diagnostic so in this uh a solution to the problem would be addressed then we have the predictive so depending on the data and the analytics uh the further Trends could be uh predicted here and then the prescribe so depending on the predictions or the diagnosis we can say that if the company is in loss or profit certain prescriptive measures could be given given so that is what prescriptive data analysis analytics refers to moving on we have the binning techniques so this is the problem consider following set the set comprises of all these values apply various binning techniques so here we have three binning techniques first one is smoothening by means so uh the next one is we have smoothing by bin medium that is to place the value here and then we have smoothening by the bin boundaries here that is we replace the closest bin boundary here so starting let's write the solution first so by equal frequency bin method data can be distributed across the bin let's assume that the bin size is three Okay so bin 1 could be written as 12 14 and 19 bin 2 can be written as 22 24 and 26 B1 B2 and B3 then we have B3 as 28 30 here we have 28 also 28 31 and 34 so these are the V values by by the first method I can write bin one as by smoothening this I can get the value as 15 15 15 and for this considering the mean here for 12 14 and 19 I get 15 so I I wrote 15 directly and for 22 24 and 26 I get 24 then for bin 3 I get 28 31 and 34 31 31 and 31 using the bin boundaries method so I can get for 12 14 and 19 you can get 12 12 and 19 that is for bin 1 and for bin 2 I can write 24 it was 24 22 24 26 so I can write 22 22 and 26 and Bin 3 could be 28 34 and 34 because this is the closest one right and here this is the closest one so that's how bin boundaries is done so from here on this is the end of theory topic now we have only the problems here in the second module so the first one we have is so this is a set of data the values are given here we need to find mean median mode range standard deviation and the variance here so in order to find a mean I I I hope it is very easy sum of all these divided by the number of values present here so we can get 17.5 directly in order to calculate the median that is the middle value of the entire data I'll consider n + 1 by 2 okay so here n value is 6 so 6 + 1 by 2 is 7 7 divided by 2 is 3.5 so value of 3.5 fth item would be considered 3.5 item is not present here it would be in between three and four right so I'll consider 3 and fourth item and divide it by 2 so 15 + 20 15 + 20 divided by 2 I get 17.5 one major rule before solving all these is to arrange the data in ascending order okay so assuming that the data is here I've taken the data in assum ascending order only that's why I have directly calculated it if not first step would be to do this then we have mode mode is the um the number or the data which has occurred most number of times it has high frequency so here um there is no data which is repeated twice tce or n number of times then coming to range range is the maximum versus the minimum I get 30 - 5 is 25 then comes the standard deviation in order to calculate the sigma uh we can make use of calculator so that is I'll I'll show you how to do it mode in mode we select um statistics that is three and in statistics we select one 1 minus wi and then here we enter the data 5 10 15 20 and 25 and 30 so 5 10 15 20 25 and 30 the data is entered once the data is entered press uh AC now press shift and press start 1 now here select four where and then select four again SX so you get 9.35 41 so even if you calculate it using the normal the uh the traditional formula you get the same uh 9.35 so Sigma is equal to summation I = 1 to n x - x² by nus1 so this is the formula for standard deviation instead of substituting the value here directly calculate it using the calcium then for variance it is the square of standard deviation so 9.35 Square I get 8 87.5 then we have the five point summary Five Point summary comprises of the median and then the inter quartile range inter quartile range in that I have q1 and Q3 so 1 2 and three then I have the semi quartile range and then the box plot so drawing all these formulating all these three would come under five point summary so the data would be given I need to firstly find the mean here that is the um sorry I need to find the median here that is the middle value middle value is considered as Q2 Then followed by that I we'll form two sections here right in this section whichever is the middle value would be the q1 and in this section whichever is the middle value would be the Q3 q1 and Q3 is done finding the range now that is Q3 - you inter quartile range I get three that is 7 - 4 I get three then I have semi inter quartile range that is 3x2 whatever I've got here by two I get 1.5 now I need to represent it using the box plot so from 8 to two I have the data so put dots there and at four q1 starts and at 7 Q3 ends so put lines here and then at five there is Q2 so put a box for from four five and 7 so this was about box plot then skewness formula need to be remembered uh this may not be asked I'm not sure about it uh it was not asked in internals so then we have curtois also so curtois also this is the formula then we have covariance and correlation so in order to find the correlation we need to find the covariance first so formula of covariance is 1 by n summation I equal 1 to n x IUS e e to the E of X into y IUS e of Y so here e of x is the mean of X and E of Y is the mean of Y so X and Y values would be given find the mean of them and then um write this formula so again you can make use of caly here so write an equation write the standard equation that is here x i minus E of x e of X is constant that is 13.5 whereas XI value would vary X1 would be 10 X2 would be 12 X3 would be 14 and X4 would be 18 similarly 10 -3.5 into 40 - 44 uh plus again 12 - 13.5 into 48 - 44 it will continue till n value reaches till it reaches four instead you can make use of caly you can write the standard equation that is um that is x minus here I have e of x value is 13.5 so 13.5 into again I'll write Yus Yus I'll write 44 here I'll solve this so first I'll enter the x value that is 10 and the Y value that is 40 so I get 14 write 14 here first okay 1X 4 You Keep It Outside Inside the bracket write 14 here first next I have 12 and 18 12 and 48 sorry I write 48 I get- 6 so- 6 is entered here then again 14 and 56 so I get 6 so six is entered here next I have 18 and 32 18 and 3 32 so I get - 54 - 54 is entered so perform the addition of all these you get the final value correlation covariance once the covariance value is calculated you can find the correlation that is co Varian of XY by Sigma X and sigma y we have previously seen how to find the standard deviation in the previous question right similarly make right make use of the features available in caly you can easily calculate Sigma X and sigma Y and then divided you get the correlation value so this was about correlation and co-variance in case if they ask us the visualization part of the data we can they can give bar chart or P chart or histogram Dot Plot box plot stem and leaf chart or line chart the data would be given here two variables would be given X and Y so constructing a bar plot one would be taken as X and the other one would be taken as Y and depending on the value we plot it we make different separate boxes and then for p chart we calculate the sum and then we formulate for each one that is for 4.2 we find the sum of all this firstly we get 60 right so 4.2 by 60 into 100 that is 7% so out of 100% of a pie chart 7% would be allotted to the 4.2 when age is one the weight 4.2 okay then we have histogram that is continuation of uh the difference between bar plot and histogram is here you create separate boxes whereas here it is continued okay then box we have seen it in the previous question how to calculate the box plot and the Dot Plot also it's similar create Dots here on the points and then put different dots or lines here and then coming to the um sixth one we have is stem and leaf so suppose the values are 4.2 4.5 4.7 and all these so in order to calculate the stem and leaf plot before decimal whatever is that 4 4 4 5 6 6 7 7 7 8 so 4 5 6 7 8 are common I'll write them in stem part after that whatever is there that will fall under the Leaf part so here I have 4.2 4.5 and 4.7 so it will be 257 here so that's how we write for stem uh like that's how we write for stem plot and leaf plot and for line chart it is the simple you for this extension of this only you connect the dots you get the line chart next we have the gsan elimination method so we need to solve the given variables using gosan elimination method our first step would be to write the a is to B Matrix so this would comprise of a values would be on whatever is there on the lhes so 2 1 and 4 all the coefficients of x y and zed 2 1 and 4 here and then 4 11 and minus 1 here and then 8 - 3 and 2 here and on B part whatever is there on the rhs would be written so 12 33 and 20 so my main aim here is to convert the upper ular Matrix to get this Matrix in the form of upper triangular Matrix that is this part should be converted to zero that is my main Moto if you look in YouTube there are they complicated but then uh understanding it in a pretty simple way you need to make these three values zero that's it once these three values are zero I'll explain you the next step okay so first I I'll get this value as zero so this is Row one this is row two and this is Row three so so I what I'll do row2 I'll perform an operation row2 needs to be updated such that R2 - 2 * r 1 so when I multiply 2 here 2 here 2 here 4 - 2 into 2 is 0 so here I'll get zero so I'll write the updated Matrix here after performing the operation I'll get upper part would be the same and here I would get 4 - 2 into 2 that is 0 so I'll write Zer here next here I have 11 so 11 - 2 into 1 so 11 - 2 into 1 I get 9 so I'll write 9 here 12 that is uh 9 so I'll write 9 here again now I'll write this part as it is because I've performed operation only on R2 so I'll write this as it is 8 - 3 2 and 20 here now my next operation would be R3 writing R3 is R3 is equal to R3 - 4 * R1 so writing the updated Matrix now above it will be the same 2 1 4 12 here also it will be the same 0 9 - 9 and 9 here here it would be 8 - 4 * of 2 that is 0 so I'll write 0 here here - 3 - 4 * of 1 I would get - 7 then I get 14 -4 here and then on this side here here I have 20 - 4 into 12 I get- 28 here so these are the values I've got so my aim was to get this particular these particular values are Z 0 0 is obtained now here Min - 7 I need to convert it to 0o so for this what I'll do I'll write R3 as 9 * R3 - 7 times R 2 so if I write the Matrix I'll get 2 1 4 and 12 as it is here also 0 9 - 9 and 9 here as it is here 0 0 would be there here I'll get zero that's I'll explain you 9 into - 7 - 7 into R2 value is - 9 so 9 into - 7 and 9 into - 7 I get 0 and here I get 9 into -14 - - 9 into 9 that is - 189 and so once I've obtained this all these three the upper triangular Matrix the this part is zero my next step would be to consider the last row consider the last row we get - 189 is equal to - 189 here uh this row was for Zed right this was for y and this was for X so I'll write z here so I get Z is equal to 1 if I can if I cancel these two - 189 - 189 now I'll consider the second row I get 0x + 9 y - 9 Z is = 9 here right now Z value I know it is 1 so I'll write one here so send this 9 that side so Y is equal to 18 by 9 9 1's 9 2 Y is equal to 2 I get two here now considering the first row I get 2x + 1 y + 4 Z is = 12 uh y value I know it is 2 so 2 here and four Z value I know it is 1 so 1 here 2 + 4 is 6 so 12 - 6 is 6 and here I have 2X so X is equal to 3 I have obtained the required values that is z value y value and x value so this was about gos elimination so for the main aim is to convert these three the lower part as zero suppose here uh they have given three equations if they have given two equations then my Matrix would be of the form x y z w and here 1 and two some coefficient in this case you need to convert this as zero this Z as zero so that's about goian elimination method now we'll look into the Lu decomposition method so for this Lu refers to uh a is equal to we need to write it in this form the output in this form l u where L is the lower triangular Matrix and U is the upper triangular Matrix so from the given uh Matrix we need to convert it into that particular form so starting this is the Matrix they have given I'll multiply it with the identity Matrix so 1 0 0 0 1 0 and 0 0 1 so this is the identity Matrix right so my aim is to apply Matrix op operations and reduce these two as lower triangular Matrix and upper triangular Matrix lower triangular Matrix means this part should be zero that is what they have said in the uh that's what we have seen in the previous question right lower triangular Matrix upper triangular Matrix means this part instead of zero there has to be some value in it so we need to perform operations such that both the changes happen simultaneously so selection of the operations is pretty important here let me start from um the R2 operation R2 you can write it as R2 - 2 * are 1 so here I I'll get 2 1 3 as it is in place of 4 I get 4 - 2 * of 2 I get 0 and then here 3 - 2 * of 1 3 - 2 is 1 so I'll write 1 here and then 10 - 2 * of 3 3 2 6 6 - 10 10 - 6 is 4 so I'll write four here so R3 would be as it is 2 4 and 17 I am done with the this part now I'll look into the identity part in order to change this value 1 0 I'll write it here and in order to change this value I'll write this is this is I'll consider it as l21 okay so for l21 operation I'll take a Sorry r21 by R11 r21 here is um r21 a21 or r21 anything is 4 by 2 right 4 by 2 so 4 by 2 is 2 so from Zer I've changed the value to 2 right then I can write 1 and 0 as it is and Below also I'll write 0 0 1 as it is my aim was to get here uh zeros and get here nonzeros so we are done with the first variable we have two more left in the second operation I'll write R3 as R3 - 2x 2 R1 so writing for this I'll write the first one as it is second row also as it is for the third row I'll write R3 is 2 here 2 minus um 2x 2 is 1 and R1 is 2 here so I'll get zero here next I have 4 - 2x2 of uh 4 - 2x2 of 1 that is three here I get three here and then I have 17 minus 3 I get 14 16 17 14 here so I am done with changing the value of second uh number here and on this side I have 1 0 0 2 1 0 I need to change the value of this this part this is a l31 okay so for l31 I can write it is a31 r31 by R11 r31 is r31 is 2 and then R11 is also two so I get one here so I got one here zero as it is and one as it is so I have changed this this this and this now these two are remaining so I'll perform next operation I'll write R3 is equal to is equal to R3 minus 3 * 3 * R1 so here I'll write here I'll get R3 is 0 so 0 - 3 * 0 I get 0 here so I write 2 1 3 0 1 4 and here 0 next I have 3 here so 3 - 3 * 1 again 0 here and then here I have 14 - 3 * 4 here 14 - 3 * 4 two here so I'm done with changing these values I have obtained the lower triangular Matrix and on this side I need to get upper triangular Matrix so I'll write 1 0 0 2 1 0 here one this part I need to change this part this is a L3 in order to write value for this I'll consider l32 that is I'll consider R32 by r 22 R32 value is 3 R22 value is 1 so I get three here so I'll write three here and here I'll write uh one as it is so my aim was to get this as lower triangular Matrix and this as upper triangular Matrix I've obtained both so they can ask they can ask these questions these type of questions so it is easy if they have given only four values here 2 is to2 Matrix then it will be pretty easy to solve next topic we have is principal compound component analysis in principal component analysis we try to reduce the dimension so the higher Dimension data to the lower dimensional data the reduction is done here in order to perform this we make use of co-variance Egan values and Egan Vector so this is the question they have given 4 8 13 and 7 the values are given here we found the mean first and then we identify the number of features here here there are X and Y two values are there right so two features and number of samples given for each is 1 2 3 4 four values are given so I compute this table firstly which comprises of x y x - xar y - Y Bar x - x bar squ y - Y Bar Square x - x bar into y - Y Bar I perform all the operations find out the values then I draw this Matrix which comprises of X and Y this side X and Y this side and here I'll be Computing the co-variance so covariance of XX coari of XY coari of YX and co-variance of y y by the obtained values I'll be finding all these and then fill the value in The covariance Matrix now from here I'll be finding the EG Vector e values that is determinant of s minus Lambda i s is the value which I had obtained here I wrote it minus Lambda * I I is the identity Matrix I have written so next I'll multiply this so I get 14 14 14 and - 11 here - 11 and 23 here this Lambda would be multiplied inside so I get Lambda 0 0 Lambda now perform the subtraction operation so 14 - Lambda minus - 0 - 11 - 0 23 - Lambda so solving this I get Lambda 1 and Lambda 2 values so solving the quadratic equation through Cal caly I can easily get the Lambda 1 and Lambda 2 values so once these Ean values are obtained for each Lambda I need to perform two operation two two times substitutions so first for Lambda 1 I'll consider U1 and U2 here so I'll substitute the Lambda value then I found the result multiply it with u so I get one equation right right so this equation I I send U U2 to the other side okay so if I send the U2 to the other side you get I get um I take it as T here and then I get U1 and U2 values here and this was for Lambda 1 and then I normalize it so this is the normalization procedure then I move on to the Lambda 2 considering the Lambda 2 so again perform the same operation now I step five is to derive the new data set so P1 P11 P1 p13 and p14 so all the values which I previously obtained are up updated here in this so in The Matrix I need in the graph I'll consider the new points and mark them as E1 E2 E3 E4 E1 E2 sorry and then from E1 E2 I find the X Y X and Y values so this was about U this was about the principal PCA topic I would suggest that if you get this question just skip it because it's pretty lengthy and time consuming so this was about PCA next we have now we'll be looking into the probability distributions so under continuous probability distribution we have three types so the first one is normal distribution so normal distribution has the Bell curved shape graph right so we have studied this in our fourth Sim so the same formula Z is equal to x - mu by Sigma here instead of Z they have taken is yeah Sigma Square they have taken x - mu s by 2 Sigma Square so this is the formula for normal distribution in rectangular distribution we have p Al to capital x equal to small X where X if it ranges between A and B then 1 - B by a else it is zero then in exponential we know the exponential formula Lambda e x into e minus Lambda X this formula we have studied in four zero coming to the discrete probability distribution part we have binomial that is B of X is equal to n n CX P of X into Q n- X and then for poison we have e^ minus Lambda Lambda X into here Lambda is the mean mu P then we have burnol so burnol formula we know Q is equal to 1 - P if K is equal to 0 if p k is equal to 1 then it's just P then we have the hypothesis topic so hypothesis is assumption about the outcome or existing belief that researcher wants to test so in hypothesis we have null hypothesis and alternating hypothesis null hypothesis is h0 and alternative hypothesis is H1 so in null hypothesis we have the existing beliefs so so in null hypothesis uh we have existing Bel Bel so there's no difference uh like after performing the hypothesis also in alternative hypothesis the researcher claims whatever he wants to he or she wants to establish so there is an effect or difference in this hypothesis then we have types of hypothesis test in that we have parametric and the non-parametric ones parametric ones are the based on parameters like we we can make use of mean or standard deviation we assume that the the data follows a specific distribution so on the basis of distribution this test would be performed then we have the non-parametric test in this we don't have any specific distribution so it is independent of that then we have the statistical test here so then we have the statistical test so we perform various operations statistical operations in order to test and we have the P value here so U like assuming that the null hypothesis is is true we perform operations here then we have error in hypothesis resting so we know that we have studied this in fourth s right so type 1 error and type two error type one error is false the hypothesis is false and on the other side it is positive here it is false and negative so then we have confidence levels so confidence level is so below certain point if the probability comes then that is not accepted that's what we have seen in confidence level right then we have different types of hypothesis test here Zed hypothesis test when used when the data is large and then the T hypothesis test which is used to compare smaller data and then K Square hypothesis test is observed frequency and estimated frequency values so this part could be asked in the theory in the theoretical topics they can ask us to Define these and even here also they can give us simple values and expect us to answer but probably the these won't be asked because you have big chunk of problems here serious ones so those would be asked so I have tried covering all the topics of fourth module of all the topics of second module whichever is important so share it with your friends if you have any doubts do let me know in the comments and stay tuned to my channel I'll be uploading third module fourth module and fifth module also so thank you

Channel: VTU padhai

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Keywords: 21CS54, aiml important questions, vtu aiml important questions, vtu aiml, aiml module 1, aiml classification, kmeans classification, how to pass aiml exam, aiml vtu important questions, vtu 5th sem aiml, aiml imp questions, aiml passing package, 21cs54, aiml, vtupadhai, vtu aiml important notes, gaussian elimination, hypothesis testing, pca problem

Id: kvhI7_vygpU

Length: 52min 38sec (3158 seconds)

Published: Thu Apr 11 2024