Introduction to Machine Learning Unit-4 Notes for II-II Mechanical Engineering
DLD Lecture No 15 Prime and Essential Implicants, Five Variable Map.pptx
1. Digital Logic and Design
Lecture No 15 : Prime and Essential Prime Implicants
Five Variable Karnaugh Map
1
2. Another Example
• Four adjacent corners can be combined to form the two
literal term x’z’.
• Four adjacent squares can be combined to form the
two literal term x’y.
• The remaining 1 is combined with a single adjacent 1 to
obtain the three literal term w’y’z’.
• F = x’z’ + x’y + w’y’z’ 2
3. Another Example
F=A’BC’ +A’CD’+ABC+AB’C’D’+ABC’+AB’C
F=BC ’+CD ’+ AC+ AD ’
0
AB
1 1
0
00 01
00
01
CD
0
0 1
1
11 10
1
1 0
1
11
10
1
1 1
1
3
5. Working With Maxterms
At times, we may be required to work with maxterms.
The previous process actually worked with minterms. Remember that
the numbers used for minterms are the opposites of the numbers
used for maxterms:
F(w, x, y, z) = ∑(0, 1, 2, 8, 9, 10, 11), uses minterms
F(w, x, y, z) = ∏(3, 4, 5, 6, 7, 12, 13, 14, 15), uses maxterms
If you are given minterms, fill in 1’s for the minterms and then fill the
remaining cells with 0’s
If you are given maxterms, fill in 0’s for the maxterms and then fill the
remaining cells with 1’s
For SOP simplification, solve the map for the 1’s
For POS simplification, solve the map for the 0’s to get complemented
function. Taking the complement of this complemented function we
obtain function in POS form
5
6. Example 3-5
F(w,x,y,z) =
(0,1,2,4,5,6,8,9,12,13,14)
Sol:
1 is marked in each minterm
that represents the function
Find the possible adjacent
squares and mark them with
rectangles
We combine eight adjacent
squares to get a single literal
term y’
The top two 1’s on the right
are combined with the top two
1,son the left to give the term
w’z’
We combine the single square
left on right with three adjecent
squares that are already used
to give the term xz’
The logical sum of these three
terms gives:
F = y’ + w’z’+xz’
Correction in the book:
Add 1 in the square
6
7. Example 3-6
F = A’B’C’+B’CD’+A’BCD’+AB’C’
Sol:
Each of three literal term in map is
represented by two squares and four
literal term in map is represented by
one square
We combine the 1’s in the four corners
to give the term B’D’
The two left hand 1’s in the top row are
combined with two 1’s in the bottom
row to give the term B’C’
The remaining 1’s may be combined in
the two-square area to give the term
A’CD’
The logical sum of these three terms
gives:
F = B’D’ + B’C’+ A’CD’
7
8. Overview of Karnaugh Map
• Karnaugh Map?
• Made up of squares
• Each square represent one minterm
• The variables in squares change in gray code
• Each variable covers an area in the squares
• Grouping of Squares Rules?
Every cell containing a 1 must be included at least once.
The largest possible “power of 2 rectangle” must be enclosed.
The 1’s must be enclosed in the smallest possible number of rectangles
Mapping of Functions into the Karnaugh Map
Function expressed in sum of products or sum of minterms
Function expressed in product of sums or product of maxterms?
Convert it to sum of minterms or sum of products form
8
10. Five-Variable Map
A five-variable map holds
thirty-two minterms for five
variables.
We use two four variable map
with one of the variables
distinguishing between the
two.
Each square in the first map is
adjacent to the corresponding
square in the second map (i.e.
4 and 20 are adjacent). It is
just like placing one map on
the top of the other.
10
12. 5-Variable Map Patterns
The number of adjacent squares that may be combined always
represent a number that is a power of 2 such as 1, 2, 4, 8, 16,
and 32.
One square represents one minterm with five literals.
Two adjacent squares represents a term of four literals.
Four adjacent squares represents a term of three literals.
Eight adjacent squares represents a term of two literals.
Sixteen adjacent squares represents a term of one literal.
Thirty-two adjacent squares represents the entire map and produces
a function that is always equal to 1.
12
16. Product of Sums Minimization
By definition, all the squares in a map that are not
marked with a 1 represent the complement of the
function.
If we mark the empty squares with 0s and then combine the
zeros into valid adjacent squares, we obtain a simplified
expression of the complement of the function i.e., F’
The complement of F’ [as (F’)’ = F] by DeMorgan’s theorem
(by taking the dual and complementing each literal, section 2-
4), gives us the product of sums form
16
17. POS Minimization Example
• F’ = w’x + yz + xz + xy
• F = (F’)’
• =(w’x + yz + xz + xy)’ = (w + x’)(y’ + z’)(x’ + z’)(x’ + y’)
0
0
0
0
0
0
0
0
0
w’x
xz yz
xy
17
18. Example 3-8
F = (0,1,2,5,8,9,10)
Simplify the function in
Sum Of Products (SOP)
Product Of Sums (POS)
Sol:
The squares marked with 1’s
represents minterms and are
combined to form simplified
function in sum of products
(SOP). F=B’D’+B’C’+A’C’D
If the squares marked with 0’s
are are combined we obtain the
simplified complemented
function F’=AB+CD+BD’
Applying DeMorgan’s theorem
we obtain the simplified function
in product of sum form (POS)
F=(A’+B’)(C’+D’)(B’+D)
18
23. Working With Maxterms
At times, we may be required to work with maxterms.
The previous process actually worked with minterms. Remember that
the numbers used for minterms are the opposites of the numbers used
for maxterms:
F(w, x, y, z) = ∑(0, 1, 2, 8, 9, 10, 11), uses minterms
F(w, x, y, z) = ∏(3, 4, 5, 6, 7, 12, 13, 14, 15), uses maxterms
If you are given minterms, fill in 1’s for the minterms and then fill the
remaining cells with 0’s
If you are given maxterms, fill in 0’s for the maxterms and then fill the
remaining cells with 1’s
For SOP simplification, solve the map for the 1’s
For POS simplification, solve the map for the 0’s to get complemented
function. Taking the complement of this complemented function we
obtain function in POS form
23
How to get my simplified form in Product of Sums?
If I combine 0’s What type of expression I will get?
If I take complement of Sum of Products, what form I get?
It is AND-OR implementation
It is a OR-AND implementation
Sometime we assume that complement inputs are available
It is a two level implementation
What is the advantage of two level implementation?
Are the two functions same?
How to find that they are same?
Let us plot their truth table, if we get same output they are same
