 Welcome to the session minimization techniques using Kmap. At the end of this session students will be able to describe different logical functions on Kmap and also solve different logical functions using Kmap. So these are the contents of video lecture. Simplification of logical functions makes digital circuits simpler and faster. But simplification of logical functions using Boolean algebra laws and rules is time-consuming and error-prone. So to overcome above problem in 1953 Morris Karnoff, an engineer at Bell Laboratories invented a graphical way of visualizing and then simplifying logical functions. Therefore this graphical representation of logical functions now known as Karnoff map or simply Kmap. Kmap is a matrix consisting of rows and columns which forms squares known as cells. Each square or cell represents the output values of a logical function. The output values placed in each cell are derived from the min terms or max terms of a logical function. All the cells of Kmap are arranged in a gray code sequence where only one bit changes at a time. Karnoff maps or Kmaps are mostly used to simplify logical problems with 2, 3 or 4 variables. Now let us see Kmap for two variables. For two input variables Kmap consists of 2 raised to 2 cells that is 4 cells as shown in below diagram. So to know how many cells are there for that we have formula cell is equal to 2 raised to n where n is a number of variables. So you can see this is the structure for two input variable where we have considered a and b are two variables where a may have value 0 or 1 and b also having value 0 or 1. So you can see the structure here where the cell represents 0 0 0 1 1 0 1 1 where these are cell number 0 1 2 and 3. Now the next Kmap will show you min terms for each cell. So here is a bar b bar a bar b a b bar a b bar a b. And for max term you can see here a plus b a plus b bar a bar plus b a bar plus b bar. So this is how min term and max term can be represented using two variable Kmap. For three input variable the Kmap having 2 raised to 3 8 cells as shown in below diagram. So here a variable a is taken at a row side and b and c are taken at column side. So a may having value 0 or 1 and b and c having value 0 0 0 1 then 1 1 and 1 0 because here we are following gray code. So total 8 cells are there. For four input variables a Kmap consist of 2 raised to 4 is equal to 16 cells as shown in below diagram. So for 4 variable 2 variables are taken at row side and 2 variable c d are taken at column side. Only standard or canonical form of logical function can be represented on Kmap. So for example here is the function f 1 of a b c equal to a bar b bar c bar plus a b c bar plus a bar b c bar is a standard swap. A standard swap is a swap where in all the terms all the variables are present. If it is not then it is not a standard or a canonical. Now let us have a next equation function 2 of a b c equal to a plus b plus c into a plus b bar plus c bar into a bar plus b plus c bar. This is what a standard pos where again in all these terms all the variables are present. So this is a standard pos or canonical pos product of some. Now let us see how to represent a standard swap on a Kmap. A logical function in a standard swap form can be represented on Kmap by just entering once in the sales of Kmap corresponding to each minterm in the equation. And let us have one example f 1 a b c is equal to a bar b bar c bar which represents 0 a b c bar which represents 4 plus a bar b c bar which represents 2. So in swap form this equation can be written as summation m 0 to 4. So to represent this equation on Kmap we require 3 input variable Kmap as we have 3 variables here and just by entering once into respective cell 0 to 4 you can represent swap equation. Now let us see how to represent a standard pos on Kmap. A logical function in a standard pos form can be represented on Kmap by just entering zeros in the sales of Kmap corresponding to each max term in the equation. So here is the example f 2 of a b c equal to a plus b plus c into a plus b bar plus c bar plus into a bar plus b bar plus c which is a standard pos and can be represented in a max term as pi capital M 036. So we have to enter zeros into cell number 036. So this is how we can represent a pos on Kmap. Now let us see how to simplify a logical function using Kmap. The simplification is based on the principle of combining terms that is min terms or max terms in adjacent cells. Two or more cells are called adjacent cells if they differ in only one variable. So let us see how to make a pair which is nothing but two adjacent cells. So by making a pair one variable can be eliminated. For example here we have two input variable Kmap where 1 1 is present in cell number 1 2 3. So here we can make a pair of cell number 1 and 3 because here only variable A is changing its value 0 to 1 but B is constant throughout this column and the next pair can be made by making a cell number 2 3 a pair. So here variable A is remaining same throughout cell number 2 and 3 but B is changing its value. So in Kmap the variable whose value is changed is cancelled. So by making this pair variable B is get cancelled so only A remains whose value is 1. So in this equation Y is equal to A and B. So for this pair B remains same having value 1 but A changes its value so A is cancelled. So in this way a variable is get cancelled whose value is changed. Now let us see how to make a quad which means by making a group of four adjacent ones. By making a quad two variable can be eliminated different ways of making a group of four adjacent ones that is quad or as below. So first group is making a horizontal ones. So here you can see it is a four input Kmap where you can see all ones are present in a row. So this is a picture as horizontal ones. So you can make up a group of four adjacent ones then you can eliminate two variables. So here by solving this you can get A bar B. So as A is having value 0 and B is having value 1 so this will give you A bar B. Now next is vertical ones. So when all ones are present in a column it seems as a vertical. So in this way you can eliminate two variables. All A and B's are get cancelled only C and D having the same values throughout the all these cells. So C is having value 1 and D is having value 1. So this equation gives you C D. Now let us see forming a square. So you can form square like this so that will give you A bar C. Now one of the form is left most and right most ones. So here you can see left most and right most ones are made in a group called as a quad and it will give you B D bar. Then this is a picture where you can make a group of upper one and lower one making a quad. So here it is get reduced to B bar C and one more grouping is done where all corner ones are present. So here it gives you B bar D bar. Now let us see how to make octet which means making a group of eight adjacent ones. So by making octet three variable can be eliminated. Different ways of making group of eight adjacent one are horizontal ones. So again here if all the ones are present in eight adjacent ones so you can make a group of horizontal ones. So it will give you one variable by limiting three variables. So here you can see A is changing its value and all C and D's are changing value but only B is having same value one one so this gives you B. By making group of four vertical ones use you D. Now let us see how to make a group where left most and right most ones are present. So in this way you can make octet and it gives you D bar. Now one of the method where all ones are present at upper most and lower most so it can form an octet. So in this example it gives you B bar. Now let us have one example where we have a function f of w x y z and these are cell numbers. By putting once into these cells as this equation is solved we can make an octet and two pair. So octet will eliminate three variables so only one variable is present and pair will eliminate only one variable so there are three variables each in these terms. So always you have to form a bigger group that is you have to give the priority to the octet then to the quad and into the pair because octet eliminate three variables then quad eliminates two variables and pair eliminates only one variable. Thank you.