 आस्टनामलातिकम स्दूँन्त, अँगी वासीम् मिक्राम. इस ellege 12th lecture in a series of 45 lectures on digital logic design कैसे आप चलजा अच्छे होंगे? किषते दो लोगचा से लेक्छे फीं लिख कर रहें काड्रार्फ मैप्स में, तोस्ब कि आपताा है, अचो़ के नहीं शीमने खापते है, you while use the equation of the rational way of simplifying Boolean expressions. क्यों टुछ त्यों क्यों, और नृन रोट् मन्वल की तरीके से, solve करनां, मुशकल होट आच्वाजादु अस्में। Today we are inside, मैद्ट्ट, to simplify boolean expressions. But before that, let us revise the contents which we studied in the last lecture. In the last lecture, we talked about representing a product of some expression in the form of k-maps, rather mapping product of some expressions to k-maps. We talked about representing standard product of some expressions, how will you map? Basically, the standard product of some expression, you will have maximums in it. So, whatever maximums are directly in the k-map, the cells representing those maximums, you will put zeros there. The rest of the maximums that are not present, they would be represented by ones. Then we talked about mapping non-standard product of some expressions to a k-map, k-map, k-map. Again, what we have to do in that? Let us suppose you have an expression a plus d-bar into b plus c-bar. So, this is a non-standard product of some expression. How will we map in k-map? Basically, the first sum term, the second sum term, wherever the values are available in the cells, you have to put zeros there. The rest of the cells will be one again. Then we talked about simplification of product of some expressions using the k-map. Well, simplification is a standard method. When you map the entire information in the k-map, then you have to form groups. Groups, you could have group of 8, you could have group of 16, 4, 2 and so on. You start with the largest group. Group, basically you have to make a group of zeros because product of some expression is not talking about. So, groups of zeros will be formed in this. One more thing we talked about. If you have a kind of map, it is fully mapped, that is all cells either have ones or zeros, you can easily extract expressions from that. If you group one, the expression giving you is basically sum of product terms. If you group zeros, then you will get product of some forms. So, basically you can easily convert. You have a kind of map, product of some expression or sum of product expression by grouping ones or zeros. We talked about five variables, kind of maps. Five variables, kind of maps, basically we said are composed of 32 cells. So, 32 cells, kind of map is difficult to make. And it is difficult to represent or group. So, a five variable kind of map is represented in the form of two, four variable maps. So, a two dimensional map is being made. And there you will group basically two dimensional grouping. Again, I mean defining groups in a five variable kind of map is a bit confusing. You might not be able to recognize the appropriate patterns. So, we mentioned in the last lecture that if the expression goes beyond four variables, five variables, six variables, another method is going to be used which we will study today. We also talked about functions having multiple outputs. Till now, whatever functions we talked about we said they have only a single output. Multiple inputs are possible. Output is always one of the examples we have done. Well, practically you are going to have circuits which have multiple outputs and of course multiple inputs. Multiple output circuits how to simplify the circuit? So, basically in the last lecture we talked about BCD to seven segment converter circuit. So, basically the input to the circuit is four bit BCD. The output is a seven segment display or the circuit has seven pins. So, the pins are basically connected to a seven segment display A, B, C, D, E, F, G. Segments So, the circuit turns on the appropriate segments to display the appropriate digits. Let us say you have to display zero. So, you would provide BCD zero at the input. Appropriate segments would be turned on and of course you would display the digit zero. So, you had a circuit with seven outputs. उसको आपने कैसे expression के फाँ में कैसे define करने. Basically you would have seven expressions. हमने बात कीते to table की to table would have sixteen possible combinations because you have a four input four bit input to the circuit. Since you have seven outputs therefore you would have seven different to tables or you could have a single to table representing each of the seven segments. We are going to start today's lecture with another example which has multiple outputs. उसको आपने कैसे then of course we would be talking about the other method which is used to simplify Boolean expressions. Let us talk about a comparator circuit. A comparator circuit basically compares two values. In digital logic, in any system you need to compare several values. So you might need to know if X is equal to Y if it is greater than Y or less than Y. In digital logic maybe we need circuits which have to compare two values and then provide you with the answer are the two values equal is the first value less than the other or otherwise. What type of circuit is it? Basically how many outputs should the circuit have? Basically it should have three outputs. One output should be set to one if both the numbers are equal. The second output should be set to one if let us say number A is greater than B and the third output would be set to one if number A is less than B. So three outputs one output indicates if the two numbers are equal the second output which indicates if number A is greater than B and the third output which indicates if number A is less than B. How many inputs will there be in the circuit? Well it depends on the two numbers. Let us suppose you have two numbers A and B and both are two bit numbers. So how many bits should be input? Basically A is two bits B is two bits so you have a four bit input to the circuit and the output is three bits. Let us consider the function table of the comparator circuit. There are four bit inputs so the function table will have four columns of course representing the two numbers each is of two bits. How many input combinations will there be? Basically four bit input will have 16 combinations. Output is basically you have three outputs. So there will be three outputs in the function table or you can make three different function tables. One function table will have three different outputs. The output would represent let us say the number A greater than B. So you would look at the input combinations wherever you find number A to be larger than B you would do what? With that output column A greater than B basically you would write a one there. Let us talk about the second column where you have A equals to B the output represents the equality between two numbers A equals to B. So again you would look at all the 16 input combinations wherever you find both the numbers to be equal the output is set to one. Similarly you have the third output which represents A less than B again you would again look at all the 16 possible input combinations where A is less than B of course you mark the corresponding output with a one. So you have a function table which represents the operator function. After this what you have to do basically you need to write Boolean expressions. After Boolean expressions you will of course make a circuit but the work to make a circuit we will discuss later. Let us first of all define the expressions. How many kind of maps do you need to use and how many variables do you need to use the second kind of map which you need to simplify the expression has to be a four variable kind of map how many you need basically three kind of map each is of four variables because you have three outputs so one kind of map would be used to simplify the expression for the numbers A greater than B the second kind of map would be used to simplify the expression to B and the third kind of map would be used to represent the Boolean expression or rather simplify the Boolean expression which represents the relationship A is less than B. So let us have a look at the comparator circuit let us have a look at the function table and the kind of maps. Let us first have look at the function table for the relationship A greater than B output A greater than B represents the output of the circuit output logic line representing A greater than B for sixteen combinations of the two bit numbers A and B the input numbers A and B are represented by A1 A0 and B1 B0 columns of the function table output A greater than B line is set to 1 when A is greater than B thus for input combinations of A and B that is 0 1 0 0 1 0 0 0 1 0 0 1 1 1 0 0 1 1 0 1 and 1 1 1 0 the output A greater than B is 1 the function table for the output A equals to B represents the output of the circuit logic line representing A equals to B for sixteen combinations of the A and B the input numbers A and B are represented by A1 A0 and B1 B0 columns of the function table output A equals to B line is set to 1 when A equals to B thus for input combinations of A and B that is 0 0 0 0 0 1 0 1 1 0 1 0 and 1 1 1 1 the output A equals to B is set to 1 let us have a look at the function table for the relationship A is equal is less than B the function table for the output A less than B represents the output of the circuit logic line representing A less than B for sixteen combinations of the two bit numbers A and B the four variables are represented as A1 A0 B1 B0 output A less than B line is set to 1 when A is less than B thus for input combinations of A and B that is 0 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 0 1 1 1 and 1 0 1 1 the output A less than B is set to 1 let us look at the simplification of the output A is greater than B simplifying expression for A greater than B through a four variable Kmap yields A1 B1 bar plus A0 B1 bar B0 bar plus A1 A0 B0 bar note that the left most column has binary values 0 0 0 0 1 1 1 0 representing A1 and A0 variables the top row has binary values 0 0 0 1 1 1 and 1 0 representing variables B1 and B0 the number B simplifying expression for the relationship A equals to B is very simple it again requires a four variable kind of map it yields the Boolean expression A1 bar A0 bar B1 bar B0 bar plus A1 bar A0 B1 bar B0 plus A1 A0 B1 B0 plus A1 A0 bar B1 B0 bar the kind of map for a relationship A less than B is again a four variable kind of map simplifying the Boolean expression yields A1 bar B1 plus A1 bar A0 bar B0 plus A0 bar B1 B0 we just looked at the function tables of the comparator circuit the Garnoff maps basically circuits of functions which have multiple outputs they are solved in a manner which is similar to solving circuits functions which have a single output the only difference between functions having multiple outputs and functions having a single output is that as much as your outputs are as much as your expressions are as much as your kind of map okay let us now talk about the alternate method for simplifying expressions which have more than four variables we talked that if five variables are solved in kind of map then it becomes difficult to solve you need two maps two dimensional then when you write ones or zeros it becomes difficult to understand the patterns of the group you may miss something so a better method is the Quen-McLuskey method so Quen-McLuskey method is used to solve expressions with more than four variables five, six, seven, eight the Quen-McLuskey method is based on a computer program basically you will not make any kind of map you will not do any mapping you will not see any pattern the Quen-McLuskey method is based on a computer program so a program will write you would run it on a computer you would provide the min terms of the expression and of course the program would come up with the solution the minimum product terms how do you do the work basically if you say that there is an expression which has a lot of product terms you have to simplify it what will be the method the kind of map you are using basically you will see every two terms there are some common variables which you can remove if you cannot do it then you will compare it with another term basically you would be comparing all terms with all the remaining terms so basically you would be doing an exhaustive search the complete search you will see all of them so the Quen-McLuskey method is based on this exhaustive search all the variables are searched to find the minimum possible terms we will see in detail how it works but let me just describe you the main features of this method let us take an example you have a four variable so in left column you are representing the variables A in the upper row you are representing the CD variables let us suppose the two rows below you have written all the ones you can say solve it what is the term basically a group of eight cells so the expression simplifies to simply A the variable A okay if you cannot make this cell so the method is basically let us suppose the first column the one written in the first row we will compare it the one written in the same column we will compare both the ones so if we combine these two then which variable will come out basically A will be there B will come out CD is also common so if these two in the first column the two ones basically B will come out A,C,D similarly if you in the next column the two ones below which variable will come out basically again B will come out similarly if you are in the third column and in the fourth column the one in both the rows we will compare them again in all of them after comparing you can see in which form basically you have four cells of two ones okay, four cells each have two ones next step is basically the two adjacent cells we will compare them so in the first column two cells are made and in the second column we will compare them what will happen basically A remains there D variable is removed so A stays there C stays there D variable is removed similarly if you compare the two cells in the third column and the fourth column again D variable will be removed now you are left with two groups of four cells okay, if you compare these two groups of four cells what will happen basically C variable is removed you end up with A variable originally we talked about if you have two rows in which all the ones are the two rows of four cells then there is a group of eight cells which is equal to A and in such an exhaustive search basically what you have done you are comparing each cell because of the map the cells in which one is written so the Quinn-McLuskey method also uses this technique since you are not doing it manually you have a computer program doing this search so it does not matter because you have so many variables so let us have a look at Quinn-McLuskey method but before that before I describe the method let me show you two examples of four variable Carnoff maps although we can solve an expression of four variables using four variable Carnoff map but still even with four variable expressions it is sometimes difficult to see appropriate patterns so if you use a four variable Carnoff map you might not come out with the simplest form of the expression so let us have a look at two examples of four variable Carnoff maps let us have a look at a four variable Carnoff map six groups of four cells each are formed the six groups form A B A C bar A D B C C D and B bar D out of these six groups three groups are redundant and therefore they are introducing three extra product terms which are not required can you spot the groups basically the essential terms that are required are A C bar B C and B bar D so the term A B is redundant it is not required the term A D is again not required it can be removed and the term C D it is again not required so only three terms are essential let us look at the second Carnoff map in the four variable K map five groups of four cells and two cells are formed the five groups form the five terms A B C bar A C D A bar B C A bar C bar D and B D now when groups are formed you start with group having maximum number of ones so you would start performing group of four now in this particular case group of four is basically redundant it is not required the essential terms that are required are A B C bar A C D A bar B C and A bar C bar D we just saw two examples of four variable Carnoff maps and we saw that it is possible to make mistakes not recognize proper patterns even with four variable Carnoff maps so five variable Carnoff maps it will have more problems okay let us start with the quen maklasky method quen maklasky method is basically a two step method basically you would perform an exhaustive search what happens in an exhaustive search you are comparing all the minterms as I mentioned in the example you have four variable Carnoff maps in which there are two rows in which all the ones are written so first you compare two cells then the two cells then four then you are comparing all of them similarly in quen maklasky method in the first step you do an exhaustive search this search will be repeated again and again it is possible that you compare a term twice or thrice the exhaustive search starts stops rather when no more comparisons can be done by comparison what do I mean when you cannot remove a variable in the example you were doing you were removing the variables one by one so when you are comparing in exhaustive search you try to remove a variable so when you are not able to remove a variable that means the search has stopped so this completes the first step after that what you have to do the second step is quen maklasky method in that you have the minimum terms which you call prime implicants so in the second step the prime implicants you have to take the minimum possible product terms which express the Boolean expression in the simplest form so we are going to have a look at two examples and we are going to follow through all these steps so let us have a look at the first example consider the function defined in canonical some form as sigma a b c d 1 3 6 7 8 9 11 12 13 14 and 15 the min terms along with variables a b c and d are written in a tabular form the numbers 1 3 6 7 8 9 11 12 13 14 and 15 of course represent the min terms of the expression of the sum of product expression each min term is represented in terms of its binary value as can be seen in the table we refer to this table as table 1 the table of min terms is reorganized in terms of groups of min terms having 0 1 2 3 and 4 1s thus min terms 1 and 8 have only single 1s the min terms 3 6 9 and 12 have 2 1s each the min terms 7 11 13 and 14 have 3 1s each and the min term 15 has 4 1s an extra column is added to the table of min terms which indicates which min terms can be compared together to eliminate a variable this table is referred to as table 2 all pairs of min terms which can be compared together to eliminate a variable are marked as used when comparing min terms the rule is to compare each min term in one group with each min term in the other group thus in the in this example min terms 1 and 8 in group having single 1s are compared with each of the 4 min terms 3 6 9 and 12 in the group having min terms of 2 1s each similarly each of the 4 min terms 3 6 9 and 12 are compared with each of the min terms in the next group having 3 1s that is min terms 7 11 13 and 14 lastly min terms 7 11 13 and 14 are compared with the min term 15 in the last group having all 1s or 4 1s the results of the comparisons between 2 min terms are represented in a separate table referred to as table 3 the first column lists the min terms that have been compared together to eliminate common variables so terms 1 and 3 form a single term eliminating variable c forming the product term a bar b bar d the comparison terms 1 and 3 are marked as used in table 2 similarly terms 1 and 9 form a single term eliminating variable a forming the product term b bar c bar d both these terms are marked as used in table 2 similarly terms 8 and 9 eliminate variable d terms 8 and 12 eliminate variable b terms 3 and 7 eliminate variable b and so on all these terms which eliminate variables are marked as used in table 2 as a result of comparison a total of 16 3 variable product terms are formed eliminating a single variable from each term all the 16 terms are represented in table 3 the exhaustive search for finding prime implicants has not completed an unmarked term represents a prime implicant comparison of terms and elimination of single variable continues with terms in table 3 all terms that combine to eliminate a variable are represented in table 4 thus terms 1 3 and terms 9 11 in table 3 form the product term b bar d eliminating variable 9 whilst comparing terms in table 3 a pair of terms which are different in a single variable are used the terms 1 3 and 9 11 are different in a single variable a only all terms in table 3 which form a simpler product term eliminating a single variable are marked as used in table 3 in table 4 there are 6 product terms of 2 variables each if the terms in table 4 are compared none of them form pairs to eliminate a variable thus all the 6 terms are marked as not used the exhaustive search for prime implicants has been completed the prime implicants are the product terms which are represented by the terms in table 4 in the second step of quen maklasky method the essential and minimal prime implicants are found the prime implicants found in the first step are listed in left most column of the table the table is referred to as table 5 so the 6 prime implicant terms found in table 5 all the original terms are listed in the top row so the original terms were 1, 3, 6, 7, 8, 9, 11, 12, 13, 14 and 15 in each cell an x is marked indicating that the prime implicant listed in the left column covers the minterm mentioned in the table 5 all the original covers the minterm mentioned in the top row thus the prime implicant b bar d covers the minterm 1, 3, 9 and 11 if you write an expression representing the minterm 1, 3, 9 and 11 all these four terms would have the product term b bar d the table 5 can be directly implemented from table 4 circles are marked in cells having x which represent minterm covered by only a single prime implicant thus the minterm 1, 6 and 8 are covered by only the prime implicant b bar d, a c bar and b c respectively so they have to be included in the final expression these three implicants in fact are the three essential prime implicants that cover all the minterms the simplified expression therefore has the terms b bar d a c bar and b c the prime implicants c d, a d and a b are redundant product terms which are not required for example the product term c d covers minterms 3, 7, 11 and 15 the four minterms are covered by the essential prime implicants b bar d and b c the example which we just saw was in fact based on the four variable Carnoff map which we saw earlier in which there were 6 terms we told that there are 3 redundant and 3 are being used exactly the same minterms we used and we applied the Quince McLeod's method what we got the answer the same essential terms so let's summarize this how we did this we reorganized in the form of groups in every group we had minterms which have single ones two ones, three ones and four ones then what we did we compared all the minterms in one group with all the minterms in the next group so we continued and by comparing this we in fact were removing a single variable so this step it didn't complete the search until we were not able to remove any more variables so in the end what we had left we had six product terms which we called prime implicants if we compare them you won't see any common variable which you can remove this completes the first step of Quince McLeod's method in the second step again the prime implicants which we got we wrote all the minterms the prime implicants in the column then we saw the prime implicants we removed which minterms we are covering whatever minterms we are covering we marked x there now in the table you would have some columns where you only have a single x what is this representing this is representing that only one prime implicant which is covering one minterms and no prime implicant is covering that minterms in this particular case that particular prime implicant has to be an essential prime implicant it has to be included in the expression as we saw three essential prime implicants they have to come in that expression in addition to them they don't need prime implicants because the three essential prime implicants cover all the minterms let me remind you that the method we did is actually not going to solve the table basically it will be a program this is an algorithm or a method so basically it would be a programmed based method let us look at the second example you will remember the second kind of map there were five groups four cells and four or two cells we had decided that the big group is redundant so four small groups were needed so let us use that same example let us find the simplest expression using the quaint-mecholoski method so similarly we will write all the minterms we will reorganize them we will compare them we will get prime implicants we will make a table without any essential prime implicants we will select them so let us have a look at the two steps each minterm is represented in terms of its binary value we refer to this table as table 1 the table of minterms is reorganized in terms of groups of minterms having 0, 1, 2, 3 and 4 ones thus minterms 1 has a single one the minterms 5, 6 and 12 have two ones each the minterms 7, 11 and 13 have 3 ones each and the minterm 15 has 4 ones an extra column is added to the table of minterms which indicates which minterms can be compared together to eliminate a variable the table is referred to as table 2 all pairs of minterms which can be compared together to eliminate a variable are marked as used when comparing minterms the rule is to compare each minterms in one group with each minterm in the other group thus in this example minterm 1 in group having single ones is compared with each of the three minterms 5, 6 and 12 in the group having minterms of two ones each similarly each of the three minterms 5, 6 and 12 are compared with each of the three minterms in the next group having minterms 7, 11 and 13 lastly each of the three minterms 7, 11 and 13 are compared with the minterm 15 in the last group having all ones or four ones the results of the comparisons between two minterms are represented in a separate table referred to as table 3 the first column lists the minterms that have been compared together common variables the second column shows the variable in terms of its binary value so terms 1 and 5 form a single term eliminating variable b forming the product term a bar c bar d variables a, b, c and d have binary values 8, 4, 2 and 1 respectively the comparison terms 1 and 5 are marked as used in table 2 similarly terms 5 and 7 form a single term eliminating variable c forming the product term a bar b, d both these terms are marked as used in table 2 terms 5 and 13 eliminate variable a terms 6 and 7 eliminate variable d terms 12 and 13 eliminate variable so on as a result of comparison a total of 8, 3 variable product terms are formed eliminating a single variable from each term all the 8 terms are represented in table 3 the exhaustive search for finding prime implicants has not completed an unmarked term represents a prime implicant terms 5, 7 and 13, 15 compare to form a product term b, d eliminating variable a the terms 5, 7 and 13, 15 are marked as used in table 3 similarly terms 5, 13 and 7, 15 compare to form an identical product term b, d eliminating variable a both the terms 5, 13 and 7, 15 are marked as used in table 3 to speed up the comparison process terms having same missing or removed variables are compared however the comparison should eliminate only a single variable thus in table 3 terms 1, 5 and terms 11, 15 have their b variable eliminated considering that 1, 5 represents the product term a power c bar d and terms 115 represent the product term a, c, d cannot be compared as 2 variables are different terms 5, 7 and 13, 15 can be compared as in both the product terms the variables c is missing and by comparing the 2 product terms removes variable a no more comparisons of terms and elimination of variables takes place thus the prime implicants have been found there are 4 prime implicants in table 3 and another prime implicant in table 4 the 5 prime implicants are represented by the product terms a bar c bar d a bar b c a b c bar a c d and b d in the second step of quen maklasky method the essential and minimal prime implicants are found the prime implicants found in the first step are listed in the left most column of the table referred to as table 5 all the original terms are listed in the top row that is 1, 5, 6, 7, 11, 12, 13 and 15 the prime implicants are listed as a bar c bar d a bar b c a b c bar a c d and b d in each cell an x is marked indicating that the prime implicant listed in the left column covers the mentum mentioned in the top row thus the prime implicant a c bar d covers the mentums 1 and 5 in other words mentums 1 and 5 all have the product terms a bar c bar d the table 5 can be directly implemented from table 3 and 4 circles are marked in cells having x which represents mentums covered by only a single prime implicant thus the mentums 1, 6, 11 and 12 are covered by the prime implicants a bar c bar d a bar b c a c d and a b c bar respectively these 4 implicants in fact are the 4 essential prime implicants that cover all the mentums the simplified expression therefore has the terms a bar c bar d a bar b c a c d and a b c bar the prime implicant b d is a redundant product term as the mentums covered by the b d product term are covered by the essential prime implicants in the example which we saw when Mcplus's method was used again we had 5 prime implicants 4 of them were essential 1 was redundant even in the map we saw the group of 4 cells was basically redundant now the method which I have just described basically the method was representation slightly different in the first method we had a column where we had binary values of all the terms in this particular case we had the term removed or the variable removed basically the same programming so perhaps you could implement the Quen-McLeusky method as a programming exercise then you would be able to solve circuits having multiple variables more than 5 variables let's apply this Quen-McLeusky method to write some expressions which have multiple variables basically more than 4 variables comparator circuit we discussed basically 4 variables why 4 because we had 2 numbers a and b each number was of 2 bits so the inputs were basically 4 bits so 4 variables now let us suppose we need to design a comparator circuit which again compares 2 numbers but the numbers are 4 bits each or 3 bits each if there are 2 numbers of 3 bits how many inputs will be there basically 6 bit inputs if the comparator you are designing it compares 2 numbers each number is of 4 bits how many variables are required 8 variables will come 8 variables or 6 variables so expression, boolean expression how will it come if we will use the Karnaugh map then you need either a 6 variable Karnaugh map or a 8 variable Karnaugh map as you know 6 variable Karnaugh map 8 variable Karnaugh maps would derive very difficult work so basically you would directly apply Quinn-McLuskey method if you have written a program then of course you would just define the min terms for the 3 bit comparator circuit and for the 4 bit comparator circuit of course the Quinn-McLuskey method would come up with the minimum product terms which represent the circuit the comparator circuit let us actually do an example to implement a boolean expression the example is of an odd prime number checker we have done an example an odd prime number between 0 to 9 we have to detect it we have made a boolean expression let us suppose we need to do the same function that is detect an odd prime number but the number range has increased basically from 0 to 31 let us say how do you represent numbers between 0 to 31 basically you need 5 bits so the input to the circuit would be a 5 bit binary number representing of course the decimal numbers 0 to 31 now how do you come up with an expression representing the function of this particular circuit if you use the Karnaugh map how many variables will be the Karnaugh map basically 5 variables it is a difficult task there are chances of making a mistake so what do you do you would express the function for that odd prime number generator by the canonical sum form ABCDE there are 5 variables the min terms will represent the odd numbers the odd prime numbers the min terms are 1, 3, 5, 7, 11, 13, 17 and so on basically these min terms are representing odd prime numbers okay now this is the expression you have represented in the canonical sum form how to solve it basically we have to apply the method of quenmeclus basically what we are doing you organize all these min terms in a tabular form then what we have to do we have to reorganize them in terms of min terms having single 1 and so on when you have organized them then what you have to do you compare min terms having single 1 with min terms in the next group which have 2 1s you will try to remove the single variable similarly min terms having 2 1s would be compared with min terms in the next group having 3 1s you will try to remove a common variable ultimately you would be comparing all these terms and removing a single variable till now your exhaustive search the first step was not completed what we will do again you would try to compare these min terms and try to remove the second variable if it is possible ultimately you would reach a stage where you cannot remove any more variables that means the first step has completed the exhaustive search has completed next what we will do basically this first step would give you the prime implicants from prime implicants if you want to know minimal implicants or minimal product terms which represent the simplest expression what we will do next step we will go again the prime implicants we will see them and the mentums you will write then you would find out which of these prime implicants cover which of the min terms some of them will be essential some of them will be redundant you will find out the essential min terms and basically that would give you the expression so as I have said before this sari jeez ko karnaf map se karna ke li bada muskul hoga aap ko correct expression bhi nahi milega so in fact we would look at the actual steps we would look at the actual expression in the next lecture at the end I would once again like to remind you that karnaf map based method does not work beyond 4 variables 5 variables hoga to usme ghalti ke imkaan hai kyunke you would not be able to recognize certain patterns jise aap ke paas paps jo important product terms hain jo essential product terms hain wo nahi ya aap ko redundant product terms bhi chme include kare hain even with the 4 variable karnaf map apne jasit do examples mein dekhah different redundant terms bhi chme aasakthi hain usme ya aap ek redundant term include kar sakta hain so kwain maklaskhi method is an appropriate method for solving expressions which have more than 4 variables secondly when you have circuits which have a large number of inputs more variables 5, 7, 8, 10 then of course karnaf map to bilkoli fail ho jayega usme the example which we were doing odd prime number jo checker hai humne baat ki thi ke 5 variables tak jaara because we are trying to find that prime number within the decimal range 0 to 31 agar main aapko kaun ke 0 se 100 tak maloom karnaf kya ho ga of course the number of variables would increase to kwain maklaskhi method he use karnaf badega in fact there are other methods similar to kwain maklaskhi jo program based hain so you have to write a program you have to indicate the main terms or wo aap ko calculate karke result deyega I mentioned that kwain maklaskhi method is a program based method a computer mein program likhainge jo solve karega kisthana jo humne tables banaya tha yeh soru aapko sumjhane ke liye tha tabler form mein nahi ho ga usme so just take it as a challenge you have learn programming so write a program which implements kwain maklaskhi method ao usse solve karne ke koshish kar hain so I hope ke jab hum next lecture mein miltein to aapne yeh program we can find expressions for very large complicated circuits so we will meet again in the next lecture either say discussion start karenge khuda hafiz and aslam aleykum