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Today
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we will continue with the simplification of
Boolean functions or logic functions. The
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last lecture we talked about Boolean algebra
being a set of formulae or identities to be
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used to simplify a given logic expression
without changing the Boolean relationship.
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That is for a given set of input conditions
an output is defined as true or false so when
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we make a simplification that should not change.
Whatever simplifications you make the output
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should still be true for all the input combinations
for which it is supposed to be true or it
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would still be false for all input combinations
for which the output is supposed to be false.
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Now we will continue this and use a graphical
method. As I mentioned earlier it is not a
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new theory or anything. It is a systematic
procedure actually. If you want to call it
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as a procedure at best it is a procedure,
systematized for easy handling or when the
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function gets bigger and bigger so that it
can be automate.
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There are other automated reduction techniques
other than graph method. We will not see them
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in this course. But these are all having the
same type of concept. Identify wherever possible
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the combining of the terms. What did you do
yesterday? In order to reduce or simplify
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the function if you can reduce the number
of terms it is good and in each term if you
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can reduce the number of variables it is good.
So we will use the same method to see wherever
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possible we can combine the terms into a smaller
term. Two terms can be combined into a smaller
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term then there is a saving in hardware. The
same concept will be applied in graph method
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and later on there are methods called as computerized
methods, implicant methods and all that. We
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will not see in those things because conceptually
they are same, they are merely procedures.
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If you know how to do it conceptually you
can learn the procedure any time depending
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on your requirement.
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So we will start with the map method of Boolean
simplification today. It can also be called
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a graphical method but usually it is called
a map method. What we have to do is to map
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the truth table on a graph. Look at the truth
table there are 1s and 0s in the output and
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we want to combine all these 1s together to
give the minimum possible solution. Now we
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will map it on a graph and see whether there
is a possibility of combining those that’s
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all we are going to do.
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We are going to repeatedly use the same concept
of Boolean algebra again and again. There
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is nothing like we are going to use a different
set of equations or identities, it is the
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same set of equations or identities we are
going to use but in a systematic way.
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So let us take the example; what is a map?
A map is nothing but all the variables are
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present in a map in true form and complement
form. In other words we will have a value
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corresponding to each input combination that
is possible. Let us take an example of 3 variables.
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Let us say the variables are A, B, C, what
are the eight combinations of A B C? They
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are A bar B bar C bar; A bar B bar C etc finally
ABC. That means there are eight possible values.
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So we will now draw a graph with eight cells.
This is how it is going to look like. You
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can draw it horizontally or vertically doesn’t
matter.
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I need eight cells in this map I call this
a 3 variable map, eight cells corresponding
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to this A bar B bar C bar; A bar B bar C;
A bar B C bar; A bar B C etc will be represented
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here. I will say on this vertical axis I will
let A vary the true complement value, true
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value. this 0 here means A bar this cell will
have a value, the value of this cell A is
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A bar, value of A for this cell is A bar and
value for this cell and for the horizontal
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axis we will let B and C vary there are four
possibilities of B and C varying such as B
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bar C bar, B bar C, B C bar, BC so that can
be represented by 0 0, 0 1, 1 0, 1 1.
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When I say 0 0 here for BC, A0 B0 C0 correspond
to A bar B bar C bar that’s all. Here this
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will be A B bar C bar you don’t have to
write it, it is the first time I am doing
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so I am writing the corresponding values in
the input combinations, input combination
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I am writing in the cell but the cell is supposed
to have that combination. Once you are familiar
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with this you don’t have to write it. Then
the next will be 0 1 for B and C so this will
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be A bar B bar C, this will be A B bar C.
So what will be the next value for BC? It
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is 1 0 or 1 1 but if we put 1 0 what will
happen is I have to write A bar B C bar. But
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I would like to have this adjacency rule.
Remember, this type of identity I am used
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to, repeatedly to combine to knock of a variable.
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For example, if we have A bar B bar C OR A
bar B C then I can take this as A bar C and
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B or B bar and knock it of and write it as
A bar C. that means I should only let one
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of these two variables change. Between B and
C I should let only one of those variables
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change from cell to cell. From one cell to
the next cell either in the horizontal direction
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or in the vertical direction if you let only
one variable change and the other variable
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keeps its value then I will be able to combine
it with a previous cell.
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In this case when I say 0 1 and 1 0 B changes
from 0 to 1, C changes from 1 to 0 both the
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variables B and C change at the same time
that means it is not possible to combine them
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effectively. So the technique is then not
to write here for 1 0 for BC write here as
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1 1. That means I will let B change from 0
to 1 and C remains and I let C remain as 1.
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That means this will be A bar BC and this
will be ABC. And finally the only other value
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which you have not used is 1 0 that means
B is 1, C is 0, A bar B C bar A bar A B C
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bar.
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The advantage of this representation is between
two adjacent cells either in this direction
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or in this direction only one of the variables
changes from 0 to 1 or 1 to 0 the other variables
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remains same that means I can use this formula
repeatedly. This formula identity A OR A bar
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is 1 can be used if two adjacent cells are
same and both of them are one I can combine
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them to knock of the variable which varies
from this to this and retain the other two
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variables which do not vary from this cell
to this cell.
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The same thing applies here also. here A only
changes, B value C value remains same in all
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these cases and this adjacency work also backwards
round its a wrapping it is something like
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a circular symmetry here so between this cell
and this cell A bar B B bar C bar C bar that
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means B only varies, B is 0 here 1 here C
is 0 0, A is 0 in both cases. And if I had
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four columns or four rows in the vertical
direction then also I can apply rotational
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symmetry if I had two more rows this and this
will be adjacent.
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So in this map I will call each of them as
a cell
and the cells in the map which are adjacent
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to each other in horizontal direction or vertical
direction will differ in only one variable
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and all other variables will be same these
are called adjacent cells, differ in the value
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of only one variable.
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Adjacency applies from end to end also that
is what I am trying to say. I am trying to
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say that adjacency applies from end to end
as if we are folding it around wrapping it
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around in a circular way. So what does it
mean now? If I have values of output is 1
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or true for this case and for this case that
means the truth table will have a 1 in this
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cell as well as in this cell when I have a
1 and 1 in two adjacent cells of the truth
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table I can knock of the variable which is
different from these two cells and have a
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simplified expression, that is what we did.
For example, yesterday we said we had A bar
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B bar C bar AND A bar B bar C both of them
are one in my truth table and I have written
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it as
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so that is what we are going to use in a graphical
method.
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We are going to use the identity A plus A
bar is equal to 1 repeatedly in the graphical
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method till we can identify the entire group
of 1s till we can do it no more. Hence it
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is a more systematic way of doing it and you
more or less certain at the end to be exhausted
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all the possibilities because it is a graphical
representation. So let us map our original
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function which you have been talking about
in the last couple of classes. This is how
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you write the map you always write the map
like this with all the cells and mark with
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0s and 1s on the left hand side for this variable
and 0s and 1s on the top for these variables
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then only the map is complete. This map is
also called as Karnaugh Map. Karnaugh is the
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guy who probably invented this, it is named
after him, they simply call it K’ Map. You
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will see this K’ Map mentioned in text books
or Karnaugh Maps.
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Let us apply that function in the truth table
that we had originally for F is equal to A
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plus B C bar. Which are the entries of the
truth table? It was 1 in the output. You remember
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ABC this was the truth table, this was 1 BC
bar. This was the original truth table we
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have been talking about in the last few classes,
the examples.
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So there is a 1 corresponding to this row
and all these four rows the output and for
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all other rows the output is 0. So the output
was 0 for the first row, second row and the
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fourth row, 1 for third row, fifth sixth seventh
and eighth rows but we don’t call it 1 to
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8 we always call from 0 to 7, the reason is
very simple, with 3 variables I can have eight
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values but if 0 has been included as one of
the values I can go only from 0 to 7 or eight
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values.
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So when you have a number representation 0
is an essential thing I can’t have a number
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representation where I cannot represent 0.
So with the binary number one with one digit
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or one bit of binary even though there are
two values possible the one of them happens
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to be 0 or the other one happens to be 1 so
I can only represent a maximum of 1 using
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a binary bit.
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We have two binary bits or two bits I can
have four values but the first value is 0
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so it is 0 1 2 3. 0 is 0 0, 1 is 0 1, 2 is
1 0, 3 is 1 1, I can’t represent four using
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two bits. Since it is going to be binary and
we always have to have 0 as the one of the
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values in digital design or digital system
or digital representation wherever digital
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things are involved we always start with 0
and not 1. So the first entry is always a
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0 entry so that way I will call this zeroth
row, first row, second row, third row, fourth
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row, fifth row, sixth row, seventh row and
we also call these variations or combinations
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of the inputs each of the input combinations
of a truth table are called min terms. So
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first min term is min term 0, second min term
is min term 1 so there is a symbol for the
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min term a small ‘m’.
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Already there is K’ Map here so this is
called min term 0, min term 1, min term 2,
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min term 3, min term 4, min term 5 so min
term 0 to 7 are the eight min terms possible
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for with 3 inputs and when it comes to max
terms we talked about max terms also yesterday
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we use a capital ‘M’, a capital ‘M’
for max small ‘m’ for min. so I don’t
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even have to write this truth table. Of course
this one is what we should get finally this
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is what we started with. But this is the truth
table which we considered and said that this
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is same as this. We first do a circuit for
this then you said that this circuit can represent
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a truth table of this type and then we proved
yesterday using Boolean algebra that this
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table can be represented to this function.
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Now my objective is to reduce this table again
to this using Karnaugh Map. Yesterday we did
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it using Boolean algebra today we will do
it using Karnaugh Map. So I don’t need this
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any more unless at the end to verify whether
the result is right or wrong. This is what
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I should get. We will see whether we get that.
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So I have the truth table and I have to represent
it in Karnaugh Map, map it in a graphical
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representation in a K’ Map and you don’t
even need this truth table. Somebody is going
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to give you the truth table and then you are
going to map it So why waste time because
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you know what the truth table is all about
all possible input combinations and the combinations
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for the input for which the output is 1 and
this information is called the truth table.
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You list all the input combinations possibilities
and also list those combinations for which
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the output is 1.
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Therefore instead of saying all these in such
a big way can I not simply say this function
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F has output 1 for m2 m3 m4 m5 m6 m7 because
you know what is m0 for 3 variables it has
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to be A bar B bar C bar or P bar Q bar R bar
or whatever X bar Y bar Z bar whatever. It
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can be X Y Z, P Q R, alpha beta gamma, A B
C, doesn’t matter. but as long as there
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are three variables, only these eight combinations
are possible so what is the point in writing
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everything in a big table and say this is
1, this is 1 and this is 1 instead I am saying
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this F is equal to min term 1 m2 min term
m4 min term m5 min term m6 min term m7. For
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these min terms output is true or 1.
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Since it is a sum of product expression if
you remember each of these product term I
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represent it by a sigma. So we now say my
function is I will remove this now the given
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function is F is equal to sigma m standing
min terms instead of even having to say m
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m m repeatedly try to minimize having to repeatedly
do the same thing over and over again out
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of laziness or whatever you want to call it
out you want to say stationery or just more
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or less boring just don’t have to keep on
removing the same thing over and over again
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so what I am saying is sigma m means the sum
of the products and what are the product terms
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involved or the min terms? It is 2, 4, 5,
6, 7. So if I give you the function like this
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F is equal to sigma m2 four five six seven
you know the truth table and you know the
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entries in the truth table for which the outputs
are 1 and the others are 0.
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So my job is to map this into the Karnaugh
Map represent into this Karnaugh Map this
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truth table so I don’t need to give you
this so from here it should be able to directly
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come to this you would directly be able to
get this.
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Now, in order to help you in the beginning,
later on you may drop this, you can write
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these min term numbers in the cells. This
is min term number 0, 1, 2, 3, 4, 5 so that
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you won’t make a mistake in the beginning
and then later on you know all that by experience.
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These are the min terms my job is to map or
represent on this map the min terms for which
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the output is 1 or true these are 2 4 5 6
7. So this is the K map of the function given
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as this which can also be expressed as this
or can be expressed as sum of products as
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A bar B C bar plus A B bar C bar plus plus
plus five terms we wrote yesterday.
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So this is the Karnaugh Map and this is the
function how do you simplify it? Deduce it
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so that fewer literals are used and fewer
terms are used in final implementation in
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the final representation. that is where adjacency
rule comes in. you know the two adjacent cells
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differ in only one variable so you find a
one in one variable one cell and one in the
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adjacent variable cell, adjacency works left
to right, top to bottom, end to end not diagonally.
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Adjacency does not work diagonally you have
to remember that. There is no adjacency diagonally.
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So anywhere I find in a 1 top to bottom or
side to side I will try to combine these two
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and find out which is the variable which is
appearing in one cell as a complement and
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the next cell as a true. In one cell it will
appear as B bar and in the next cell it appears
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as B then B goes because of the A plus A bar
is equal to 1 or B plus B bar is equal to
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1 relationship. That is easy to identify for
example these two cells A appears as 0 here,
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1 here. When you
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take these two cells and connect it like this,
this is equal to, later on you will be read
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00:26:55,230 --> 00:27:02,230
it directly from the map. Now the first time
I am going to write it as this is A bar B
192
00:27:02,490 --> 00:27:09,280
C bar this corresponds to B, this corresponds
to C bar and this corresponds to A bar. So
193
00:27:09,280 --> 00:27:16,280
you write it as A bar B B C bar and this corresponds
to A B C bar. These two together can be written
194
00:27:23,640 --> 00:27:25,090
like this.
195
00:27:25,090 --> 00:27:32,090
What is common between these two? B is common
and C bar is common so B C bar is common so
196
00:27:37,480 --> 00:27:44,480
take BC bar using that distributive property
say A plus A bar is 1 and since A plus A bar
197
00:27:50,320 --> 00:27:57,320
is 1 it is a known identity this is same as
this. So these two cells can be merged or
198
00:28:03,350 --> 00:28:10,350
combined into a single term of BC bar. Each
of the cells have three terms three variables
199
00:28:13,840 --> 00:28:20,650
I have knocked of one of the variables so
two 3 variable terms have been reduced to
200
00:28:20,650 --> 00:28:24,240
one 2 variable terms that is the simplification
procedure.
201
00:28:24,240 --> 00:28:29,850
There is nothing very great about it, all
you have done is identified the adjacent cells
202
00:28:29,850 --> 00:28:36,150
where 1s are marked and have found out which
of the variables was (constan….28:35) these
203
00:28:36,150 --> 00:28:43,150
two cells which of the variables change between
these two cells and the changing variable
204
00:28:43,920 --> 00:28:50,570
is removed, that is all. Hence you don’t
need all these algebra because you know B
205
00:28:50,570 --> 00:28:56,860
is constant between these two, C bar is constant
between these two, A goes from 0 to 1 so since
206
00:28:56,860 --> 00:29:03,860
A goes from 0 to 1 you can write this without
A with B and C bar. So you don’t need this
207
00:29:04,490 --> 00:29:09,140
algebra I just wrote it for your convenience.
208
00:29:09,140 --> 00:29:16,140
Now let us extend this argument to these two
and these two. These two will have A common,
209
00:29:18,430 --> 00:29:25,430
B common, C varying so I can write this as
AB. Between these two A is common, B bar is
210
00:29:26,960 --> 00:29:33,960
common, C changes so these two if I write
together it will be A B bar because A is common
211
00:29:35,870 --> 00:29:40,400
between these two B bar is common between
these two and C changes from here to here
212
00:29:40,400 --> 00:29:47,400
which has to be knocked off. and between these
two A is common, B is common in these two,
213
00:29:49,610 --> 00:29:56,610
C changes from here to here so C gets knocked
off so this is AB. But again these two cells
214
00:29:59,040 --> 00:30:05,440
and these two cells adjacent here it is B
bar, here it is B, A remains constant throughout
215
00:30:05,440 --> 00:30:12,200
so instead of doing it in two steps and instead
of grouping two 1s each time and then combining
216
00:30:12,200 --> 00:30:19,200
them again with one more step I can group
also four 1s at the same time. These two can
217
00:30:19,440 --> 00:30:26,440
be combined to as the one group in which it
is A B bar OR AB which is equal to A. I don’t
218
00:30:39,890 --> 00:30:44,570
have to do it this way step by step but I
just showed it to you how it works. But you
219
00:30:44,570 --> 00:30:49,930
don’t have to do it this way.
220
00:30:49,930 --> 00:30:56,930
All you have to do is see that A is constant
throughout these four cells across the four
221
00:30:58,340 --> 00:31:05,340
cells A is common, B changes from 0 to 1 across,
C changes from 0 to 1 1 to 0 so both these
222
00:31:11,780 --> 00:31:17,280
variables B and C or neither of these variables
B and C remain constant throughout so they
223
00:31:17,280 --> 00:31:21,050
both get knocked off. So it is not necessary
to knock off only one variable, you can knock
224
00:31:21,050 --> 00:31:24,690
off two variables, you can knock off three
variables and in order to knock off three
225
00:31:24,690 --> 00:31:28,840
variables you need eight 1s in a group.
226
00:31:28,840 --> 00:31:35,820
Therefore if you have a more number of 1s
in a group adjacent cells it is better for
227
00:31:35,820 --> 00:31:42,060
us because we are going to simplify better,
our reduction is going to be more. That means
228
00:31:42,060 --> 00:31:48,710
without going through all these algebra all
this background theory if I give you this
229
00:31:48,710 --> 00:31:55,710
same map now all I said was 1 1 1 1 1 I will
simply do it combine this and combine this
230
00:32:08,820 --> 00:32:15,820
this combination I will write it as A this
combination I will write it as BC bar so my
231
00:32:17,570 --> 00:32:24,570
final function is the OR combination of this
or this BC bar. This is the function we started
232
00:32:35,160 --> 00:32:42,160
with that means K’ Map can also give you
the simplification that the Boolean algebra
233
00:32:44,740 --> 00:32:50,950
gives. But here you know that I have exhausted
all 1s, how do you know it is done? I have
234
00:32:50,950 --> 00:32:55,440
exhausted grouping all 1s no individual 1
stays out of course if it stays out you have
235
00:32:55,440 --> 00:33:02,440
to write it separately and we have also always
considered the largest possible groups of
236
00:33:02,510 --> 00:33:03,010
1.
237
00:33:03,010 --> 00:33:09,420
There are four 1s possible plus here there
is a group of two 1s possible so whereever
238
00:33:09,420 --> 00:33:14,020
possible you have taken the largest possible
group in that particular group so there is
239
00:33:14,020 --> 00:33:19,810
no question of doing it better than this.
So this way you are more confident of your
240
00:33:19,810 --> 00:33:26,810
final output compared to Boolean algebra where
you might have or might not have used an identity
241
00:33:27,080 --> 00:33:32,760
which should have been used. there lies the
difference between a Karnaugh Map method and
242
00:33:32,760 --> 00:33:37,340
the Boolean algebra but the Boolean algebra
is what we have used finally A plus A bar
243
00:33:37,340 --> 00:33:43,270
is 1 is Boolean algebra. It is not as if some
new technique I will not say new theory it
244
00:33:43,270 --> 00:33:50,200
is a technique, the procedure may be different
but not the theory. And now you can extend
245
00:33:50,200 --> 00:33:57,200
this to four variables. Four variables will
have sixteen cells, two variables here two
246
00:34:01,360 --> 00:34:08,360
variables vertically and two variables horizontally
and you can do like that.
247
00:34:13,300 --> 00:34:20,300
We will take an example of four variables.
Simplify minimize using K’ Map the function
248
00:34:37,690 --> 00:34:44,690
F I can now give the variables. For a change
we will go from ABC P, Q, R, S, function of
249
00:34:53,869 --> 00:34:59,700
four variables P, Q, R, S which is true for
the following min terms. I am going to write
250
00:34:59,700 --> 00:35:06,700
arbitrarily I am not having any particular
function in mind. When you put sigma m that
251
00:35:10,529 --> 00:35:16,039
is the following min terms are having 1 as
the output or true as the output and rest
252
00:35:16,039 --> 00:35:19,329
of the min terms are having 0 as the output
that’s what we write. This is a sum of the
253
00:35:19,329 --> 00:35:22,509
product representation in min term form.
254
00:35:22,509 --> 00:35:29,509
So let us say 0 2 3 7 11 13 14 15 and so on.
It is an arbitrary example. No specific system
255
00:35:42,529 --> 00:35:49,059
I have in mind no specific circuit I have
in mind. so I have a function of PQRS four
256
00:35:49,059 --> 00:35:55,690
variables and if I write a truth table I will
have sixteen rows and to this sixteen rows
257
00:35:55,690 --> 00:36:02,690
min terms min term number 0, min term number
2, 3, 7, 11, 13, 14, 15 I have true outputs
258
00:36:05,920 --> 00:36:12,920
or one output and other min terms namely 1,
4, 5, 6, 8, 9, 0, 12 I have 0 or false as
259
00:36:19,079 --> 00:36:26,079
outputs. For such a function I wanted to do
a Karnaugh Map and minimize it, reading it
260
00:36:28,420 --> 00:36:35,420
by combining the ones as effectively as possible
write the minimum possible expression minimum
261
00:36:37,029 --> 00:36:44,029
sum of product MSP not sum of product expression
SOP is sum of product but this is non minimum.
262
00:36:44,140 --> 00:36:50,119
So it is a canonical sum of products which
is the standard truth table representation
263
00:36:50,119 --> 00:36:55,819
then the sum of product is anything which
is having some product relationship and minimum
264
00:36:55,819 --> 00:36:59,420
sum of product is what you cannot reduce further
as a sum of product expression.
265
00:36:59,420 --> 00:37:06,420
Let us do the Karnaugh Map, we will have four
variables P, Q, R, S sixteen cells must be
266
00:37:17,109 --> 00:37:24,109
there, this is the K’ Map of the 4 variable
0 0, 0 1 remember 1 1 and then 1 0 because
267
00:37:43,900 --> 00:37:50,900
the same argument we had earlier for BC and
the horizontal scale we will now apply to
268
00:37:51,150 --> 00:37:52,710
PQ the vertical scale.
269
00:37:52,710 --> 00:37:57,700
I want only one variable to change from here
to here so only one variable changes. If I
270
00:37:57,700 --> 00:38:01,640
had put 1 0 both the variables would have
changed I don’t want that situation then
271
00:38:01,640 --> 00:38:06,250
they are no more adjacent. Two cells are defined
to be adjacent only if they differ in one
272
00:38:06,250 --> 00:38:12,589
variable and if they differ in more than one
variable then they are not adjacent according
273
00:38:12,589 --> 00:38:16,940
to the adjacency relationship. Even though
physically you put them together they don’t
274
00:38:16,940 --> 00:38:23,940
become adjacent. So this is min term 0 1 2
3 min term 4 5 6 7 min term 8 will start here
275
00:38:34,650 --> 00:38:41,650
because there is 1 1 that comes later than
1 0 so 1 0 will have a min term 8 9 10 11
276
00:38:42,269 --> 00:38:49,269
12 13 14 15. this is the 4 variable Karnaugh
Map which is standard whether variables are
277
00:38:51,390 --> 00:38:58,390
A, B, C, D or P, Q, R, S or X, Y, Z, W or
whatever alpha, beta, gamma, delta.
278
00:38:59,799 --> 00:39:06,799
Now what are the min terms for which the output
is true or 1 those have to be mapped on to
279
00:39:10,119 --> 00:39:17,119
this graph? These are 0 2 3 7 11 13 14 15.
Later on when you start doing a few more of
280
00:39:34,069 --> 00:39:39,009
these exercises you will drop these min term
numbers so this will become less cluttered
281
00:39:39,009 --> 00:39:44,970
from a clearer map. Now I put it purposely
for you to get used to this so you can remove
282
00:39:44,970 --> 00:39:47,970
all these and later on you will not see those.
283
00:39:47,970 --> 00:39:54,970
Now we have to combine them try to group them
as many 1s as possible a in a given group
284
00:39:57,480 --> 00:40:03,680
but they should all be adjacent. I can straight
away see a group of four 1s 1, 2, 3 etc. You
285
00:40:03,680 --> 00:40:08,670
always start with the largest possible 1s
largest possible group and then come down,
286
00:40:08,670 --> 00:40:13,630
don’t start with the smallest number because
sometimes the smaller group may get submerged
287
00:40:13,630 --> 00:40:17,329
into a larger group later on so start with
a larger group.
288
00:40:17,329 --> 00:40:24,329
The idea is to combine all the 1s but you
can combine a 1 more than once. remember the
289
00:40:25,400 --> 00:40:31,539
Boolean algebra, yesterday when we tried to
simplify the Boolean algebra we used the same
290
00:40:31,539 --> 00:40:38,539
time twice because A AND A is A, A OR A is
A so I can use the same term twice no problem
291
00:40:39,059 --> 00:40:45,319
but I should not leave out anything everything
should be included but it is alright to include
292
00:40:45,319 --> 00:40:51,390
a particular cell which is asserted as one
more than once doesn’t harm us. In fact
293
00:40:51,390 --> 00:40:53,650
it will help us sometime.
294
00:40:53,650 --> 00:41:00,650
So first group of 1s is this. I will call
this 1 because there is no space here I will
295
00:41:09,569 --> 00:41:16,569
write later on. then we have these two 1s,
these two 1s, these two 1s and so on so there
296
00:41:20,049 --> 00:41:27,049
are four 1s available here and all of them
are to be combined I can’t see this one
297
00:41:27,200 --> 00:41:32,230
being combined anyway other than this because
all other adjacent cells are blank so this
298
00:41:32,230 --> 00:41:39,230
has to be combined this way so is the case
here. here though I don’t have to combine
299
00:41:45,579 --> 00:41:52,579
in this way because this cell is adjacent
to this cell because this is 0 0 0 0 this
300
00:41:53,079 --> 00:42:00,079
is 0 0 1 0, R changes from 0 to 1 between
these two cells 0 and 2 R changes from 0 to
301
00:42:03,009 --> 00:42:08,099
1 whereas PQ and S are remaining as 0 0 0.
302
00:42:08,099 --> 00:42:13,499
So cell number 0 and cell number 2 are adjacent
even though they are physically not adjacent.
303
00:42:13,499 --> 00:42:17,319
That is what I am saying physical adjacency
is different from logical adjacency we are
304
00:42:17,319 --> 00:42:22,349
talking about here. We are talking about logical
adjacency. So that way I don’t have to mix
305
00:42:22,349 --> 00:42:25,519
this here since this is going to be combined
this is going to be combined, this is not
306
00:42:25,519 --> 00:42:32,519
being combined anyway so I will combine these
two so I will call this 2, I will call this
307
00:42:37,359 --> 00:42:41,450
3 there are four terms.
308
00:42:41,450 --> 00:42:48,450
There are now four terms of groups of 1s and
each of them I have to write a sum of product
309
00:42:51,109 --> 00:42:58,109
expression then my final output F is sum of
all these product expressions SOP minimum
310
00:42:58,259 --> 00:43:05,259
sum of product. So what is 1? 1 is this, variable
P AND Q change all over so you have to write
311
00:43:09,859 --> 00:43:16,859
only this R AND S are 1 so RS. So 1 corresponds
to RS, 1 is RS, 2 would be, PQ are 1 1 they
312
00:43:27,859 --> 00:43:34,609
remain as 1 1 here and here so they have to
remain in the expression, here S is 1, here
313
00:43:34,609 --> 00:43:39,329
also S is 1 where R is 0 here R is 1 here
so R changes from 0 to 1 between these two
314
00:43:39,329 --> 00:43:46,329
cells so R gets knocked off. R gets knocked
off from this cell to this cell whereas S
315
00:43:46,490 --> 00:43:53,490
remains as 1 so 2 is nothing but PQS. For
3 it is the same argument PQ R remains 1 S
316
00:44:08,390 --> 00:44:15,390
changes from 1 to 0 S gets knocked off R remains
so 3 is PQR.
317
00:44:22,549 --> 00:44:29,549
Finally the fourth one is PQR 0 0, R is 0
here R is 1 here R is 0 here R gets knocked
318
00:44:31,400 --> 00:44:38,400
off here S remains 0 so it is P bar Q bar
S bar. That means my final expression is this
319
00:44:51,609 --> 00:44:58,609
RS plus PQS plus PQR plus P bar Q bar S bar.
Each of these terms 1 2 3 4 which are product
320
00:45:24,809 --> 00:45:31,809
terms which combine to become sum of product
later on is called a prime implicant.
321
00:45:50,529 --> 00:45:57,529
An implicant is any group of 1s. Suppose I
write a term I won’t do that, supposing
322
00:46:00,859 --> 00:46:07,859
I did between these two ones that means it
is RS 1 1 P 1. So, if I write P RS P RS will
323
00:46:11,230 --> 00:46:18,220
mean these two cells cell number 11 and 15.
Cell number 11 and 15 if I combine together
324
00:46:18,220 --> 00:46:24,710
I will write it as PRS. I will not do it because
I am covering a larger group of 1s four 1s
325
00:46:24,710 --> 00:46:27,670
so there is no need to do this. But suppose
I did it it’s called an implicant, it’s
326
00:46:27,670 --> 00:46:29,470
not a prime implicant.
327
00:46:29,470 --> 00:46:36,470
A prime implicant is a largest possible group
of 1s in that particular context for that
328
00:46:37,049 --> 00:46:44,049
group, you cannot find a larger group. An
implicant gets submerged into a prime implicant.
329
00:46:44,910 --> 00:46:51,519
A prime implicant is a largest possible group
of 1s that you can find for a given group
330
00:46:51,519 --> 00:46:56,640
out of all the prime implicants.
331
00:46:56,640 --> 00:47:02,940
Any prime implicant which is essential for
the final representation, I told you two things
332
00:47:02,940 --> 00:47:07,670
you have to make sure all 1s should be covered
irrespective of the fact that 1s can be covered
333
00:47:07,670 --> 00:47:12,180
more than once. For example, this one has
been covered 3 times this was covered along
334
00:47:12,180 --> 00:47:18,119
with this, this was covered along with this
so that is okay. But in each time we have
335
00:47:18,119 --> 00:47:25,119
included at least one 1 at least one cell
which has not been combined earlier. When
336
00:47:25,599 --> 00:47:32,599
I wrote this number 2 term this one has been
included new, when I wrote three terms I included
337
00:47:33,779 --> 00:47:40,779
this one new.
338
00:47:41,680 --> 00:47:48,680
I defined an implicant, I defined a prime
implicant and now I am defining an essential
339
00:48:05,900 --> 00:48:12,900
prime implicant. In an essential prime implicant
there will be at least one 1 or one cell which
340
00:48:18,910 --> 00:48:25,910
has not been covered in any other prime implicant.
there should at least be one cell or one true
341
00:48:28,980 --> 00:48:35,980
value which has not been included in any other
group so such a prime implicant is called
342
00:48:36,380 --> 00:48:42,759
essential prime implicant in this case all
the four are essential prime implicants because
343
00:48:42,759 --> 00:48:46,079
this one and this one cannot be covered and
these three 1s cannot be covered other than
344
00:48:46,079 --> 00:48:51,089
this group, this one will not be covered other
than this, this one will not be covered other
345
00:48:51,089 --> 00:48:56,619
than, these two 1s will not be covered in
any other way. So here all the four prime
346
00:48:56,619 --> 00:49:03,049
implicants are essential prime implicants.
Sometimes occasionally we will get non essential
347
00:49:03,049 --> 00:49:06,849
prime implicant.
348
00:49:06,849 --> 00:49:13,849
Supposing you have different ways of combining
a 1 in two groups you can combine this way
349
00:49:19,809 --> 00:49:25,029
or that way both of them will lead to the
same result or similar results. Such a group
350
00:49:25,029 --> 00:49:30,359
is called non essential prime implicant. That
means there is a more than one way of combining
351
00:49:30,359 --> 00:49:37,359
a 1 which is not covered otherwise as a essential
prime implicant. But one of them is essential
352
00:49:38,109 --> 00:49:40,960
otherwise that 1 will be left out.
353
00:49:40,960 --> 00:49:46,619
Suppose I have two different ways of combining
a particular 1 I have exhausted all 1s in
354
00:49:46,619 --> 00:49:53,249
a map, all entries in a cell for which the
output is true I have exhausted except 1 there
355
00:49:53,249 --> 00:50:00,249
is a 1 hanging out which I have not covered
yet in any essential prime implicant or any
356
00:50:00,460 --> 00:50:05,749
prime implicant. Now I want to combine this
1 and I can find several possibilities or
357
00:50:05,749 --> 00:50:12,749
at least two possibilities but one of them
may be required the other may not be, if I
358
00:50:14,680 --> 00:50:18,900
take this the other one is not required if
I take the other one this is not required.
359
00:50:18,900 --> 00:50:21,619
So this is a non essential prime implicant.
prime implicant is important it has to be
360
00:50:21,619 --> 00:50:27,940
covered but each of those two prime implicant
which I will get by combining this one with
361
00:50:27,940 --> 00:50:32,769
two different ways are called non essential
prime implicants so it is not necessary to
362
00:50:32,769 --> 00:50:36,920
include all non essential prime implicants
for the final output but it is only necessary
363
00:50:36,920 --> 00:50:43,259
to include as many non essential prime implicants
as required to cover all the ones in the map.
364
00:50:43,259 --> 00:50:47,410
Whereas essential prime implicants all of
them should be covered in your final output.
365
00:50:47,410 --> 00:50:51,599
We will take an example of non essential prime
implicant, how a particular one can be covered
366
00:50:51,599 --> 00:50:58,599
in different ways and how only one way may
be required and all other ways may be redundant
367
00:50:58,609 --> 00:51:03,009
that will tell you more about a non essential
prime implicant. We will see it in the next
368
00:51:03,009 --> 00:51:03,349
lecture.