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In
the
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last lecture we introduced you to the map
method of Boolean simplification. we drew
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a Karnaugh Map which is a graphical representation
of a truth table filled this graph with 1s
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corresponding to the cells for which whose
min terms had a output true and 0s of course
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we didn’t mark 0s to avoid cluttering of
the map. Wherever we did not mark 1 in the
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map the entry was a 0 which was purposely
left out so that the map doesn’t get cluttered.
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The objective is to identify groups of 1s
as large as possible of course satisfying
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the adjacency rule and remove or knock off
as many variables as possible so that by repeatedly
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doing this we can get a simpler representation
of a given Boolean function. So we look at
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a few more examples today some special cases
and all that.
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We will take an example of a map like this,
four variables I will call these variables
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ABCD, want to say this map as 1s in the following
cells, this is the pattern of 1s. That means
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this is a function this expresses the sum
of the following min terms, this is min term
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number 2 so you will have to sort out from
here 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15,
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so we take (2, 6, 7, 9, 13, 15) this is the
map.
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Now the idea is to group the adjacent ones
with as large groups as possible keeping in
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mind the group should be as large as possible.
If you have a smaller group which can be totally
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submerged into a larger group we should not
consider the smaller group and we have to
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make sure that all 1s are covered and we do
not mind one being covered more than once
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these are the rules. Make as large groups
as possible keeping in the mind the adjacency
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rule, make sure all 1s are covered, do not
worry about combining a given 1 more than
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once. So in this map groups of two 1s are
only possible 1 2 3 4 so I can say this because
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this one has to be covered any way, this is
standing out separately.
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The most efficient way would probably be not
covering anything more than once. You can
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cover 1 more than once if it can result in
a smaller group. If something is not going
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to result in a smaller group then there is
no point of covering 1 more than once. So
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in this case there are only two ones this
is going to be a group, this is going to be
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a group, this is going to be a group all groups
are having two 1s so there is no point in
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unnecessarily doing this and doing this. So
this is how I will write this three groups
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of two 1s each and what would this be in terms
of the product expression? F would be sum
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of these products 1 2 3 this would be AC barD
plus then this will be BCD plus, this would
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be A bar C D bar
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Now instead of doing this way suppose I marked
it slightly differently. That means suppose
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I did it like this it is obviously not the
most efficient way you can see that because
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there is a small map with few 1s but in general
when you have a large map with large possible
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groups you may miss out and may do things
like this. I brought this up to explain some
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of the definitions we gave in the last lecture.
I talked about implicant, the prime implicant
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and an essential prime implicant. An implicant
is any group of 1s and a prime implicant is
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the largest possible group of 1s in the given
group of course that cannot be an essential
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prime implicant, a prime implicant is a group
which cannot become part of another implicant.
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An essential prime implicant is one which
I said if you remember has at least an entry
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of one not covered in any other prime implicant.
So this way here this is the essential prime
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implicant I will call this 1 2 3, this is
1, 1 is an essential prime implicant because
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this one cannot be covered otherwise, 3 is
an essential prime implicant because this
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cannot be covered otherwise.
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Now is 2 an essential prime implicant or not
is the question? Do you think 2 is an essential
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prime implicant or not? The way I draw it
is an essential prime implicant because these
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two 1s cannot be covered anyway. But when
I draw this map there are two ways of combining
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these two 1s. This is an essential prime implicant,
this is an essential prime implicant, is this
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a prime implicant? What is a prime implicant?
The largest group of 1s possible within that,
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it’s a largest group of 1s possible in that
group. So all the four here I call this 1
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2 3 4, here all the four are prime implicants,
here all the three are prime implicants but
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one and three are clearly essential prime
implicant, 2 is not an essential prime implicant
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it is a non essential prime implicant for
the simple reason I can cover these two 1s
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in two other ways. So in this case again 1
is an essential prime implicant, 4 is an essential
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prime implicant so 2 is a non EPI.
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But the reason that this one could have been
combined with this I could have it small it
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is an essential prime implicant. I could have
covered this to make another prime implicant.
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So this is the only way in which I can make
prime implican otherwise that one gets left
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out only then it is an essential prime implicant.
We are not talking about the most efficient
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way here remember that I am only talking about
definitions. Just because it is an essential
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prime implicant it does not mean that it is
the most efficient way of doing it. so these
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are non essential prime implicant, it is a
non essential prime implicant because this
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one could have been combined with this to
get another term, this is two in this case,
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this term could have been combined with this
term to get another term 3 in this case so
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this is a non essential prime implicant but
the same argument 1 and 4 are EPI’s and
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2 and 3 are non essential prime implicants.
Having drawn in this case F is equal to 1
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or 2 or 3. In this case F is 1 or 2 or 3 or
4 and all are two terms, two 1s all are two
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variables, a three variable term there are
only three terms in this case, four terms
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in this case and it is very obvious that this
is a less efficient simplification compared
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to this.
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But the fact remains that there are some non
essential prime implicants that has to come
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to the final expression in order to complete
the expression. This is a very clear case
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of a non essential prime implicant inefficiently
used. In this case a non essential prime implicant
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has been used efficiently because it was very
clear and obvious apparent just looking into
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this map.
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Sometimes there may be two non essential prime
implicants, each of them can lead to a final
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solution which is correct and both of them
may have the same simplicity or complexity
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in which case you have to choose between 1
and 2 or one or the other of the essential
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prime implicant in which case the choice is
not unique but in this case the choice is
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almost unique. I would not choose this against
this if my aim is to reduce the logic which
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is again you have been proclaiming I will
not go for this. But I will give you another
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example where I could have combined in two
ways both of them resulting in a same simplified
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expression in terms of number of variables
and number of terms involving a non essential
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prime implicant that has been an interesting
case. So we will take another example.
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For a change I will call it WXYZ and this
is my map. So what is the map now? F is sigma
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M (0 2 3 5 7 8 10 11 14 15) these are the
min terms for which the output is true for
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all the min term outputs. I just wrote the
map and then wrote the min term. Normally
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it is the other way. Normally you are given
the min terms for which the output is 1 from
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the truth table or from the word description
of the problem. Either a word description
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of the problem or a truth table is given to
you which will tell you what are the min terms
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for which the output is 1 and automatically
the other outputs are 0. But in this case
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just to illustrate a point in terms of implicants,
prime implicant, essential prime implicants,
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non essential prime implicants and so on I
am just taking maps to give you examples so
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I have to first draw the map and then tell
you what are the min terms which are true
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in this map. Now what are the different ways
of combining them? We have to again efficiently
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do it.
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You have to make smallest possible number
of prime implicants and make a combination
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which is the minimum possible combination.
So now this one, this one, this one, this
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one together form a map group a prime implicant
because they roll over top to bottom and left
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to right so I will call this first prime implicant.
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I said the other day the adjacency, this cell
is adjacent to this cell as well as to this
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cell, this cell is adjacent to this cell as
well as to this cell, this is adjacent to
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this as well as this and so forth. Hence all
the four cells are adjacent to each other,
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it is very similar to something like this
or something like this. and this is in essential
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prime implicant because this one would have
been left out otherwise or this one for that
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matter because I could have combined this
some other way but not this, I could have
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combine this but not this. So because of these
two 1s we present this becomes an essential
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prime implicant and essential prime implicant
has to necessarily find its place in the final
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expression.
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The debate is only about how many non essential
prime implicants to be included to keep the
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function minimum and at the same time not
omitting any one in the table that is the
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only thing will have to be concerned with.
now this one again has to be combined there
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is no other way I could combine this one except
this way otherwise this one gets left out
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completely
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so I will call this 2, one this is 2 now I
have two options, I can combine these four
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into an essential prime implicant. In fact
I should combine this way because this one
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would be left out otherwise of course I can
always combine these two, these two are combining
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is not a solution because a prime implicant
is one which absorbs smaller implicants so
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you put this one is not a solution because
when I have four larger prime implicant I
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cannot put this one as an implicant I have
to cover this. So this is an essential prime
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omplicant for the simple fact this one would
not be covered otherwise so this will be number
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3. Thus the only one that is left out is one.
In order to complete my simplification I have
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an uncovered one and that one has to be included
so this I can do it in two ways I can do it
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in this way I will call this 4 and these four
1s are adjacent to each other because this
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is adjacent to this and so forth or I could
have combine these two 1s and these two 1s
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so either this way or this way
these two 1s.
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Now this one was what was left out remember,
this min term number 3 was left out I had
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to include it in order to include it in a
largest possible group of 1s I can do it this
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way or I can do it this way and this one I
will call 5. Unlike in this case where we
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had an option which is obviously less efficient.
We now an option in which one is essential
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prime implicant, II is essential prime implicant,
III is essential prime implicant because of
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these individual things, IV and V are non
essential prime implicants because essential
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prime implicant is 1 in which at least one
unique 1 is covered in that group no other
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way it should be able to cover that one that
is not the case with IV and V because this
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one can cover this way or this way so everyone
of this 1 can be covered and all these other
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three 1s can be already taken care off. In
this case all these have been taken care off
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except this one this has been taken care of,
this has been taken care off, everything other
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than this has been taken care off so this
one can be covered either this way or this
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way so when IV and V are two options of covering
this min term represented by this III then
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it is not an essential prime implicant, IV
and V are non essential prime implicants.
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Therefore I II III are EPI’s and IV and
V are non EPIs. But it is enough one of IV
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or V is included because four and five both
cover this one and this is the only one which
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has uncovered. After considering I II and
III the only entry in the truth table which
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was not covered in our simplification was
this entry which will be covered either by
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this or this. Both IV and V can cover this
one so where is the need for both of them
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to be present and the choice is not easy because
both are of same complexity, this is four
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ones this also is four ones. If I is 4 1s
the other is 8 1s so the complexity is different
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so you know what to choose. If both have equal
complexity where is anything to choose from?
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So now I can write the final expression as
I OR II OR III OR IV this is one option or
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this OR is English OR not the operator OR
I OR II OR III OR IV, this explains to you
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a non essential prime implicants and its role
in the final simplification. So I wanted to
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bring about the definitions of implicant,
prime implicant, essential prime implicant
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and non essential prime implicant and the
role of non essential prime implicant in simplification.
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It is not that they are trivial they can be
ignored they are essential sometimes so you
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may have to include. But how many of them
you will include and what are the non essential
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prime implicants you will include depends
on the overall goal of minimum sum of products.
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I don’t want to re write this now, you know
each of these. We can complete this, I will
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have I II III IV but I II III IV are variables
of or product terms with variables ABCD or
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in this case what WXYZ. So I II III IV and
V are each product terms of variables WXYZ
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you write it and finally get the simplified
minimum sum of product expression.
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Now we can have a four variable map. What
is meant by this? Let us do this for example.
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we started with simple gates and we drew the
simple structure using gates and wrote down
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the expression and then said that write a
truth table for that expression and we wrote
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the truth table found out the truth table
had many more entries than the original circuit
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had so we tried to see why and we saw that
the simplification of the truth table resulted
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in a smaller circuit that we started and we
said simplification is possible using either
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Boolean algebra which you saw or by mapping
method which you have seen so let us revisit
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that and see now. We will do the reverse now.
This is the sum of product expression you
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have given. a system of four variables WXYZ
having an output F circuit of system has the
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following Boolean functionality as a functional
relationship between input and output given
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by this Boolean equation not identity Boolean
expression this Boolean expression gives the
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relationship between F on one hand of the
output and WXYZ on the other hand as inputs.
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What will it look like we will see because
we have simplified this? For that I need to
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explain this 1 2 3 4 5. I have no reason to
choose this or this so we will go with this.
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So let us quickly write what is 1. 1 is these
four 1s, what are the expression terms of
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WXYZ? This is X bar Z bar so I is X bar Z
bar, II is these two W bar XZ, III is this
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which is WY and IV I am going to take instead
of V so IV would be YZ. So my expression is
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done X bar Z bar
W bar XZ WY YZ these are gates with inputs
X and Z of course inverted WXZ where W is
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inverted W and Y, Y and Z these are four AND
gates, outputs of these four AND gates tied
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together by an OR gate that is, this is X
again, W, Z second term, third term is WY
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and
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fourth term is YZ, Z has to come from here
which is here.
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So if you don’t want to lose track of these
inputs
you can write it one more time here, YZ, WY,
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this is ZW bar X, XW bar Z, this is W bar
X bar Z bar all of them combined in an OR
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gate with four inputs. This is the function
represented by this circuit whose Karnaugh
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Map is this and whose functional description
is this.
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Of course I can always write the truth table.
The truth table is only an expansion of this,
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I will have to put WXYZ all the four combinations
0 0 0 0 to 1 1 1 1 and then put here finally
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1 0. So it requires one 3 input AND gate,
one 4 input OR gate, three 2 input AND gates
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and three inverters, this
is inventory. You have to ask for these components
in order to build a circuit and you can now
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associate with WXYZ some signals as we did
in the example previously about some system
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which is to represent a particular condition
WXYZ may be four parameters we are monitoring
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and on these incidence of parameters in this
particular way will trigger an event at the
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output of this system that could be an example
of the use of this.
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So you go for a definition of the word description
of the truth table and the Boolean relationship
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mapping and then simplification and then drawing
the circuit and then of course you have to
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get the components, build it and test it.
Now the variables XYZ we have used three or
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four WXYZ and IV or V you may have used, VI
we may use, seven we may use and digital systems
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today are very complex. A system like a micro
processor, a system like a controller, a system
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like a automotive system or a fuel injection
or even very complex systems like missiles
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I already told you missiles and aircrafts
and things like that so I don’t you can
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work with three variables four variables and
three variables and four variables classroom
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exercises.
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What will happen if variables are more than
four? We have already represented a two dimensional
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now it has to go for three dimensional. For
a five variables what do we do? So let us
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say we have a
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five variables let us call these variables
ABCDE.
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I can only represent sixteen cells in a four
variable map two dimensional and five variables
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will have how many min terms? It will have
32, 2 power 5 so we need to have one variable.
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The variables we start listing them in a truth
table will become starting with 0 0 0 0 0
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0 0 0 0 1 and finally it will become this
is min term 0, min term 1, min term 31 or
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00:34:42,929 --> 00:34:49,919
I told you that you will have to start always
with 0 min term and go on till the last min
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00:34:49,919 --> 00:34:56,329
term which is 2 power n minus 1, 5 is 2 power
5 is 32 minus 1 is equal to 31 so 0 to 31
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00:34:56,329 --> 00:35:03,329
will be min terms. So one way to do this would
be draw two maps
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considering only four of these five variables
each. You leave the A out because first sixteen
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00:35:30,999 --> 00:35:37,999
entries of the truth table m15 would be 0
1 1 1 1, m16 would be 1 0 0 0 0 so these entries
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00:35:46,609 --> 00:35:53,609
will have A is equal to 1, these entries will
have A is equal to 0. First sixteen entries
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00:35:55,969 --> 00:36:01,229
of the truth table will have A is equal to
0 and second sixteen will have A is equal
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00:36:01,229 --> 00:36:08,229
to 1. So I will draw this map with variables
BCDE always remembering A is equal to 0 in
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this map. Likewise I will do BCDE one more
time always remembering A is equal to 1 in
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00:36:41,299 --> 00:36:44,019
this map.
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00:36:44,019 --> 00:36:51,019
Now the min terms would be m0 1 2 3 4 5 6
7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
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00:37:07,499 --> 00:37:14,499
22 23 24 25 26 27 28 29 30 31 so whichever
min term is one output you map it map in the
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00:37:23,079 --> 00:37:27,229
corresponding cell either here or here if
it is the min term corresponding to terms
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00:37:27,229 --> 00:37:34,229
between 16 and 31 the 1 will up here so it
will go in this part of the map. So you can
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map it the way you would map the four variables
map graphically but while reading you have
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to be careful. The adjacency works between
these two maps. The one here and the one here
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are considered adjacent because the only variable
different from these two maps this is B is
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equal to 0 C is equal to 0 D is equal to 0
E is equal to 0 and this is also B is equal
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00:38:01,650 --> 00:38:06,650
to 0 C is equal to 0 D is equal to 0 E is
equal to 0. If I have 1 here and a 1 here
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the variable A is 0 here, variable A is 1
here so these two cells are adjacent, the
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cell number 0 and 16 are adjacent.
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00:38:18,450 --> 00:38:25,450
Therefore the corresponding cells in this
map and this map are adjacent that property
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has to be built in while reading the map that’s
all. So I map the entire variable so we will
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use the same arbitrary thing, you will do
this 0 2 and this for a change I will put
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00:38:52,729 --> 00:38:59,729
here. Now these four 1s which you called prime
implicant one essential prime implicant one
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00:39:21,309 --> 00:39:28,309
here will appear here also in both the maps.
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00:39:29,299 --> 00:39:36,299
So this I will call 1 including this, these
two together are 1. Now these four again I
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don’t know what we called it here we called
it III but don’t worry about it, this and
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00:39:53,150 --> 00:40:00,150
this are again are common between these two
maps the same cells so I will call it II this
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00:40:00,410 --> 00:40:07,410
and this. Now what is left out?
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These four are left out I could have combined
this way or this way doesn’t matter I will
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00:40:16,329 --> 00:40:23,329
combine this way but this is not common to
this map 1 1 1 1 here unique to A is equal
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00:40:27,289 --> 00:40:34,289
to 0 OR not to A is equal to 1. So I am going
to call this III in which I will have to make
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00:40:34,799 --> 00:40:41,799
A is equal to 0 in that. This one is unique
to this map but not to this map. So this is
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00:40:45,459 --> 00:40:52,459
III and these two 1s I will call it IV because
this is not here, this one is unique not to
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00:40:58,410 --> 00:40:59,880
the other map so V.
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So there are five prime implicants all are
essential I made sure that is simple because
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it is an example, with five variables I wanted
to drive home the point of how to simplify
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a five variable map so I made them essential
prime implicants so you need not worry about
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00:41:21,380 --> 00:41:28,380
non essential prime implicants or proper choice
of that so we have five EPI’s 1 2 3 4 and
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00:41:32,079 --> 00:41:39,079
5, actually writing the minimum sum of product
expression based on this is called reading
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00:41:46,829 --> 00:41:53,829
the map. Somehow they call this reading the
map so we should be able to read the map now,
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00:41:54,199 --> 00:41:56,440
they use the term reading the map.
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00:41:56,440 --> 00:42:01,059
After putting all these 1s and grouping them
identifying them in groups you are writing
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00:42:01,059 --> 00:42:05,789
the final expression and that operation is
called reading the map. So now I have to read
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00:42:05,789 --> 00:42:12,789
the map to get 1 2 3 4 5 and then put it together
as one expression of F in terms of ABCD and
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00:42:14,249 --> 00:42:21,249
E. already we have said that I was here this
is C prime E prime and since this is common
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00:42:30,630 --> 00:42:37,630
between A is equal to 0 A is equal to 1 so
1 is C prime E prime or C bar E bar want to
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00:42:40,579 --> 00:42:47,579
call it C prime E prime C bar E bar or NOT
C AND NOT E whatever, II is again common between
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00:42:49,449 --> 00:42:56,449
these two maps that would be BD then comes
the third map third EPI and the third EPI
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00:43:02,979 --> 00:43:09,979
is this between these four cells which is
DE. But this is not available here so I will
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00:43:11,989 --> 00:43:16,680
have to include the information and it is
only available in the A bar part of the map.
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00:43:16,680 --> 00:43:20,709
I will make sure that you know it is in A
bar part of the map and not A part of the
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00:43:20,709 --> 00:43:27,709
map so I have to say A bar and B so III is
A bar DE. And IV is again only this essential
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00:43:37,709 --> 00:43:44,709
prime implicant which is B bar C B bar CE
but you have to add A bar to that so it is
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00:43:48,959 --> 00:43:55,959
A bar B bar CE but it is found in the A bar
part of the map so it is A bar. And finally
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00:44:03,699 --> 00:44:10,699
this fifth prime implicant this part A is
equal to 1 part of the map which is BCE with
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00:44:18,839 --> 00:44:23,209
A.
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00:44:23,209 --> 00:44:30,209
So now my final expression if you want we
can write it as F is equal to C bar E bar
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BD A bar DE A bar B bar CE and you can draw
your gate structure or whatever it is. This
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00:45:23,650 --> 00:45:30,650
is how you read a five variable map identify
the common cells. Of course some people draw
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00:45:31,089 --> 00:45:38,089
it side by side, some people draw it together
left and right and so on. It is not a mathematical
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00:45:41,900 --> 00:45:47,199
procedure or anything it is only for convenience
whichever way you feel comfortable with you
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can do it.
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00:45:48,410 --> 00:45:55,410
Now I can extend this to six variables. if
I want six variables I want four maps each
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00:45:56,680 --> 00:46:03,680
of sixteen cells I will have to say six variable
map A bar B bar, A bar B, A B bar, and AB.
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00:46:33,900 --> 00:46:40,900
So min terms 0 to 16, 16 to 31, 30 to 47,
48 to 64 and then you have to see adjacency
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see the adjacency between these adjacency
between these, adjacency between all the four
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00:46:53,579 --> 00:46:58,799
and all that and you can go on writing the
expression and have unlimited fun all these
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00:46:58,799 --> 00:47:05,799
checker games you play. But there is a limitation,
why I am saying all these is because there
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00:47:06,130 --> 00:47:11,459
is a limit beyond which after six how will
you do? Of course you can always find a method
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of doing it and then it becomes inefficient.
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We started with the map method because of
which simplicity an efficiency then we are
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00:47:19,809 --> 00:47:26,099
reaching a point where the simplicity is lost
it would become an ordeal so beyond six variable,
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00:47:26,099 --> 00:47:30,979
five variables is most comfortable, four is
good we would like to have a problem with
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00:47:30,979 --> 00:47:37,979
four in the exam may be five manage, six is
okay but seven and above it becomes cumbersome.
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00:47:40,160 --> 00:47:47,160
After all this is a procedure what we have
done is adjacency. Find 1s which are adjacent
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00:47:47,170 --> 00:47:54,170
to each other when you represent in a graphical
way. So it is always possible to write a computer
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00:47:54,779 --> 00:48:01,779
program to find out the adjacent 1s in a truth
table and keep on repeatedly doing it. The
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00:48:06,049 --> 00:48:10,979
advantage or the problem whichever way you
see it in the computers what you have to do
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00:48:10,979 --> 00:48:16,079
is step by step. Combine two 1s at a time
not more than two 1s at a time then it is
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00:48:16,079 --> 00:48:22,479
too much to handle for computers. Put two
1s at a time and then two 1s at a time will
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00:48:22,479 --> 00:48:26,719
become 1 1 then you say put two more so keep
on giving it in steps you write a program
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00:48:26,719 --> 00:48:31,339
algorithm they call it. So, algorithm is a
step by step program. You write an algorithm
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00:48:31,339 --> 00:48:38,339
for a program to systematically find out all
the 1s in a truth table merge them to the
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00:48:41,459 --> 00:48:43,890
best possible way and repeatedly do this.
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00:48:43,890 --> 00:48:50,890
Sometimes there may be several terms in which
some redundancy may be there. The essential
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00:48:52,339 --> 00:48:55,699
prime implicant and non essential prime implicant
that the computer cannot resolve you know.
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00:48:55,699 --> 00:49:00,189
After all the computer is what you program
computer, it cannot become more intelligent
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00:49:00,189 --> 00:49:02,920
than you are because you are the one who is
feeding in the program in the computer.
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00:49:02,920 --> 00:49:09,920
In a computer the program is limited by your
intelligence cannot be more than your intelligence.
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So finally we may have intervened in some
place and say we will list all essential prime
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00:49:13,630 --> 00:49:17,559
implicants and non essential prime implicants
and make a choice of the right combination
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00:49:17,559 --> 00:49:22,239
that we have to do. Such a method is called
a tabulation method as against the map method.
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00:49:22,239 --> 00:49:27,499
We will not do that and I thought I should
tell you. So the number of variables becomes
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00:49:27,499 --> 00:49:34,499
extremely large so you go for what is known
as the tabulation method. These are computer
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00:49:34,829 --> 00:49:41,829
based methods. Of course you can also do it
by hand to understand the algorithm. Like
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00:49:44,930 --> 00:49:51,930
the same example of 31 cells or even 61 cells
you can take for fun, 64 cells 0 to 63 or
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00:49:53,299 --> 00:49:59,420
even 31, 0 to 31. Do it the way the computer
will do it. Combine two two at a time and
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00:49:59,420 --> 00:50:02,410
do it like that just to get an understanding
of this method.
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00:50:02,410 --> 00:50:06,719
I am not going to teach this in this class
but it is available in many books, many books
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00:50:06,719 --> 00:50:13,719
talk about this method called tabulation method.
Some people call it prime implicant method.
315
00:50:15,089 --> 00:50:21,209
Some books call it prime implicant method;
some books call it graphical method a tabulation
316
00:50:21,209 --> 00:50:26,880
method. You can do that to understand the
concept behind this but these are all computer
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00:50:26,880 --> 00:50:30,759
programs very easy to understand once you
know how it works. So you can do a simple
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00:50:30,759 --> 00:50:35,109
example of the same example we do in the class
try to do it using this method so you will
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00:50:35,109 --> 00:50:42,109
have an understanding of this program. We
will stop here for today.
320