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Pune Vidyarthi Griha’s
COLLEGE OF ENGINEERING, NASIK - 4
“Minimization techniques”
By
Prof. Anand N. Gharu
Assistant Professor
Computer Department
Combinational Logic Circuits
Introduction Standard representation of canonical forms (SOP &
POS), Maxterm and Minterm , Conversion between SOP and POS forms
K-map reduction techniques upto 4 variables (SOP & POS form), Design of Half Adder, Full Adder, Half Subtractor & Full Subtractor using k-Map
Code Converter using K-map: Gray to Binary Binary to Gray Code Converter (upto 4 bit)
IC 7447 as BCD to 7- Segment decoder driver IC 7483 as Adder & Subtractor, 1 Digit BCD Adder Block Schematic of ALU IC 74181 IC 74381
8/29/2017 2
Standard Representation
Any logical expression can be expressed in the
following two forms:
Sum of Product (SOP) Form
Product of Sum (POS) Form
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SOP Form
For Example, logical expression given is;
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Sum
Product
. . .Y A B B C AC
POS Form
For Example, logical expression given is;
8/29/2017 5
Sum
Product
( ).( ).( )Y A B B C A C
Standard or Canonical SOP & POS Forms
We can say that a logic expression is said to be in the standard (or canonical) SOP or POS form if each product term (for SOP) and sum term (for POS) consists of all the literals in their complemented or uncomplemented form.
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Standard SOP
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Each product term consists all the literals
Y ABC ABC ABC
Standard POS
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Each sum term consists all the literals
( ).( ).( )Y A B C A B C A B C
Sr. No. Expression Type
1 Non Standard SOP
2 Standard SOP
3 Standard POS
4 Non Standard POS
Examples
8/29/2017 9
Y AB ABC ABC
Y AB AB AB
( ).( ).( )Y A B A B A B
( ).( )Y A B A B C
Conversion of SOP form to Standard SOP
Procedure: 1. Write down all the terms. 2. If one or more variables are missing in any
product term, expand the term by multiplying it with the sum of each one of the missing variable and its complement .
3. Drop out the redundant terms
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Example 1
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Convert given expression into its standard SOP form
Missing literal is A
Missing literal is B
Missing literal is C
Term formed by ORing of missing literal & its complement
Y AB AC BC
Y AB AC BC
.( ) .( ) .( )Y AB C C AC B B BC A A
8/29/2017 12
Example 1 Continue….
Standard SOP form Each product term consists all the literals
.( ) .( ) .( )Y AB C C AC B B BC A A
Y ABC ABC ABC ABC ABC ABC
Y ABC ABC ABC ABC ABC ABC
Y ABC ABC ABC ABC
Conversion of POS form to Standard POS
Procedure: 1. Write down all the terms. 2. If one or more variables are missing in any
sum term, expand the term by adding the products of each one of the missing variable and its complement .
3. Drop out the redundant terms
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Example 2
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Convert given expression into its standard SOP form
Missing literal is A
Missing literal is B
Missing literal is C
Term formed by ANDing of missing literal & its complement
( ).( ) ( )Y A B A C B C
( ).( ).( )Y A B CC A C BB B C AA
( ).( ) ( )Y A B A C B C
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Example 2 Continue….
Standard POS form Each sum term consists all the literals
( ).( ).( )Y A B CC A C BB B C AA
( )( ).( )( ).( )( )Y A B C A B C A B C A B C A B C A B C
( )( )( )( )Y A B C A B C A B C A B C
( )( )( )( )Y A B C A B C A B C A B C
Concept of Minterm and Maxterm
Minterm: Each individual term in the standard SOP form is called as “Minterm”.
Maxterm: Each individual term in the standard
POS form is called as “Maxterm”.
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The concept of minterm and max term allows us to introduce a very convenient shorthand notation to express logic functions
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Minterms & Maxterms for 3 variable/literal logic function
8/29/2017 Amit Nevase 18
Variables Minterms Maxterms
A B C mi Mi
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
1ABC m
2ABC m
3ABC m
4ABC m
5ABC m
6ABC m
7ABC m
4A B C M
3A B C M
2A B C M
1A B C M
0A B C M 0ABC m
5A B C M
6A B C M
7A B C M
Minterms and maxterms
Each minterm is represented by mi where i=0,1,2,3,…….,2n-1
Each maxterm is represented by Mi where i=0,1,2,3,…….,2n-1
If ‘n’ number of variables forms the function, then number of minterms or maxterms will be 2n • i.e. for 3 variables function f(A,B,C), the number of
minterms or maxterms are 23=8
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Minterms & Maxterms for 2 variable/literal logic function
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Variables Minterms Maxterms
A B mi Mi
0 0
0 1
1 0
1 1
0AB m
1AB m
2AB m
3AB m 3A B M
2A B M
1A B M
0A B M
Representation of Logical expression using minterm
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m7 m3 m4 m5
Logical Expression
Corresponding minterms
where denotes sum of products
OR
Y ABC ABC ABC ABC
7 3 4 5Y m m m m
(3,4,5,7)Y m
( , , ) (3,4,5,7)Y f A B C m
Representation of Logical expression using maxterm
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M2 M0 M6
Logical Expression
Corresponding maxterms
where denotes product of sum
OR
( ).( ).( )Y A B C A B C A B C
2 0 6. .Y M M M
(0,2,6)Y M
( , , ) (0,2,6)Y f A B C M
Conversion from SOP to POS & Vice versa
The relationship between the expressions using minters and maxterms is complementary.
We can exploit this complementary
relationship to write the expressions in terms of maxterms if the expression in terms of minterms is known and vice versa
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Conversion from SOP to POS & Vice versa
For example, if a SOP expression for 4 variable is given by,
Then we can get the equivalent POS
expression using the complementary relationship as follows,
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(0,1,3,5,6,7,11,12,15)Y m
(2,4,8,9,10,13,14)Y M
Examples
1. Convert the given expression into standard form
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2. Convert the given expression into standard form
Y A BC ABC
( ).( )Y A B A C
Combinational Logic Circuits
Introduction Standard representation of canonical forms (SOP & POS),
Maxterm and Minterm , Conversion between SOP and POS forms
K-map reduction techniques upto 4 variables (SOP & POS form), Design of Half Adder, Full Adder, Half Subtractor & Full Subtractor using k-Map
Code Converter using K-map: Gray to Binary, Binary to Gray Code Converter (upto 4 bit)
IC 7447 as BCD to 7- Segment decoder driver IC 7483 as Adder & Subtractor, 1 Digit BCD Adder Block Schematic of ALU IC 74181 IC 74381
8/29/2017 26
Karnaugh Map (K-map)
In the algebraic method of simplification, we need to write lengthy equations, find the common terms, manipulate the expressions etc., so it is time consuming work.
Thus “K-map” is another simplification
technique to reduce the Boolean equation.
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It overcomes all the disadvantages of algebraic simplification techniques.
The information contained in a truth table or
available in the SOP or POS form is represented on K-map.
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Karnaugh Map (K-map)
K-map Structure - 2 Variable A & B are variables or inputs 0 & 1 are values of A & B 2 variable k-map consists of 4 boxes i.e. 22=4
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Karnaugh Map (K-map)
A
B 0 1
0
1
K-map Structure - 2 Variable Inside 4 boxes we have enter values of Y i.e.
output
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Karnaugh Map (K-map)
A
B 0 1
0
1
A
B 0 1
0
1
K-map & its associated minterms
AB
AB
AB
AB
0m 1m
3m2m
B
B
AA
Relationship between Truth Table & K-map
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Karnaugh Map (K-map)
A
B 0 1
0
1
A B Y
0 0 0
0 1 1
1 0 0
1 1 1
0 0
1 1
B
A 0 1
0
1
0 1
1 0
B
B
AA
B B
A
A
K-map Structure - 3 Variable A, B & C are variables or inputs 3 variable k-map consists of 8 boxes i.e. 23=8
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Karnaugh Map (K-map)
AB
C
0
1
BC
A 00
0
1
01 11 10
A BC 0 1
00
01
11
10
3 Variable K-map & its associated product terms
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Karnaugh Map (K-map)
AB
C
0
1
BC
A 00
0
1
01 11 10
A BC 0 1
00
01
11
10
00 01 11 10
ABC ABC ABC ABC
ABC ABC ABC ABC
ABC
ABC
ABC ABC ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
3 Variable K-map & its associated minterms
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Karnaugh Map (K-map)
AB
C
0
1
BC
A 00
0
1
01 11 10
A BC 0 1
00
01
11
10
00 01 11 10
0m
4m 5m
1m
6m
2m3m
7m
5m
4m6m
7m
0m 2m
1m 3m0m 4m
5m1m
6m2m
3m 7m
8/29/2017 35
Karnaugh Map (K-map)
AB
CD
00
00
01
01 11 10
11
10
CD
AB
00
00
01
01 11 10
11
10
K-map Structure - 4 Variable A, B, C & D are variables or inputs 4 variable k-map consists of 16 boxes i.e. 24=16
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Karnaugh Map (K-map)
AB
CD
00
00
01
01 11 10
11
10
CD
AB
00
00
01
01 11 10
11
10
4 Variable K-map and its associated product terms
ABCD ABCD ABCD ABCD ABCD
ABCD ABCD ABCD ABCD
ABCD ABCD ABCD ABCD
ABCD ABCD ABCD ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
8/29/2017 37
Karnaugh Map (K-map)
m0 m4 m12 m8
m1 m5 m13 m9
m3 m7 m15 m11
m2 m6 m11 m10
AB
CD
00
00
01
01 11 10
11
10
m0 m1 m3 m2
m4 m5 m7 m6
m12 m13 m15 m14
m8 m9 m11 m10
CD
AB
00
00
01
01 11 10
11
10
4 Variable K-map and its associated minterms
Representation of Standard SOP form expression on K-map
For example, SOP equation is given as
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The given expression is in the standard SOP form. Each term represents a minterm. We have to enter ‘1’ in the boxes corresponding to each
minterm as below
1 1 0 0
1 0 1 1
BC
A 00
0
1
01 11 10
Y ABC ABC ABC ABC ABC
ABC ABC
ABC
ABC
ABC
BC BC BC BC
A
A
Simplification of K-map
Once we plot the logic function or truth table on K-map, we have to use the grouping technique for simplifying the logic function.
Grouping means the combining the terms in adjacent cells.
The grouping of either 1’s or 0’s results in the simplification of boolean expression.
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If we group the adjacent 1’s then the result of simplification is SOP form
If we group the adjacent 0’s then the result of simplification is POS form
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Simplification of K-map
Grouping
While grouping, we should group most number of 1’s.
The grouping follows the binary rule i.e we can group 1,2,4,8,16,32,…..…number of 1’s.
We cannot group 3,5,7,………number of 1’s Pair: A group of two adjacent 1’s is called as
Pair Quad: A group of four adjacent 1’s is called as
Quad Octet: A group of eight adjacent 1’s is called as
Octet 8/29/2017 41
Grouping of Two Adjacent 1’s : Pair
A pair eliminates 1 variable
8/29/2017 42
0 0 1 1
0 0 0 0
BC
A 00
0
1
01 11 10
BC BC BC BC
A
A
ABCABC
Y ABC ABC
( )Y AB C C
( 1)Y AB C C
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0 0 0 0
1 0 0 1
BC
A 00
0
1
01 11 10
0 1 0 0
0 1 0 0
BC
A 00
0
1
01 11 10
1 1
1 0
B
A 0
0
1
1
0 1 1 1
0 0 1 0
BC
A 00
0
1
01 11 10
Grouping of Two Adjacent 1’s : Pair
BC BC BC BC
A
A
BC BC BC BC
A
A
B B
A
A
BC BC BC BC
A
A
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0 1 0 0
0 0 0 0
0 0 0 0
0 1 0 0
CD
AB 00 01 11 10
00
01
11
10
Grouping of Two Adjacent 1’s : Pair
CD CD CD CD
AB
AB
AB
AB
8/29/2017 45
Possible Grouping of Four Adjacent 1’s : Quad
0 0 0 0
0 0 0 0
0 0 0 0
1 1 1 1
CD
AB 00 01 11 10
00
01
11
10
0 1 0 0
0 1 0 0
0 1 0 0
0 1 0 0
CD
AB 00 01 11 10
00
01
11
10
A Quad eliminates 2 variable
CD CD CD CD
AB
AB
AB
AB
CD CD CD CD
AB
AB
AB
AB
8/29/2017 46
Possible Grouping of Four Adjacent 1’s : Quad
0 0 0 0
1 1 0 0
1 1 0 0
0 0 0 0
CD
AB 00 01 11 10
00
01
11
10
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
CD
AB 00 01 11 10
00
01
11
10
A Quad eliminates 2 variable
CD CD CD CD
AB
AB
AB
AB
CD CD CD CD
AB
AB
AB
AB
8/29/2017 47
Possible Grouping of Four Adjacent 1’s : Quad
1 0 0 1
0 0 0 0
0 0 0 0
1 0 0 1
CD
AB 00 01 11 10
00
01
11
10
0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0
CD
AB 00 01 11 10
00
01
11
10
A Quad eliminates 2 variable
CD CD CD CD
AB
AB
AB
AB
CD CD CD CD
AB
AB
AB
AB
8/29/2017 48
Possible Grouping of Four Adjacent 1’s : Quad
0 0 0 0
0 1 1 1
0 1 1 1
0 0 0 0
CD
AB 00 01 11 10
00
01
11
10
0 0 0 0
0 1 1 0
0 1 1 0
0 1 1 0
CD
AB 00 01 11 10
00
01
11
10
A Quad eliminates 2 variable
CD CD CD CD
AB
AB
AB
AB
CD CD CD CD
AB
AB
AB
AB
8/29/2017 49
Possible Grouping of Eight Adjacent 1’s : Octet
0 0 0 0
0 0 0 0
1 1 1 1
1 1 1 1
CD
AB 00 01 11 10
00
01
11
10
0 1 1 0
0 1 1 0
0 1 1 0
0 1 1 0
CD
AB 00 01 11 10
00
01
11
10
A Octet eliminates 3 variable
CD CD CD CD
AB
AB
AB
AB
CD CD CD CD
AB
AB
AB
AB
8/29/2017 50
Possible Grouping of Eight Adjacent 1’s : Octet
1 1 1 1
0 0 0 0
0 0 0 0
1 1 1 1
CD
AB 00 01 11 10
00
01
11
10
1 0 0 1
1 0 0 1
1 0 0 1
1 0 0 1
CD
AB 00 01 11 10
00
01
11
10
A Octet eliminates 3 variable
CD CD CD CD
AB
AB
AB
AB
CD CD CD CD
AB
AB
AB
AB
Rules for K-map simplification
1. Groups may not include any cell containing a zero.
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0
1
A
B 0 1
0
1
0
1 1
A
B 0 1
0
1
Not Accepted Accepted
B
B
AA
B
B
AA
Rules for K-map simplification
2. Groups may be horizontal or vertical, but may not be diagonal
8/29/2017 52
0 1
1 0
A
B 0 1
0
1
0 1
1 1
A
B 0 1
0
1
Not Accepted Accepted
B
B
AA
B
B
AA
Rules for K-map simplification
3. Groups must contain 1,2,4,8 or in general 2n cells
8/29/2017 53
1 1
0 1
A
B 0 1
0
1
1 1
0 1
A
B 0 1
0
1
Not Accepted Accepted
0 1 1 1
0 0 0 0
BC
A 00
0
1
01 11 10
0 1 1 1
0 0 0 0
BC
A 00
0
1
01 11 10
B
B
AA
B
B
AA
BC BC BC BC
A
A
BC BC BC BC
A
A
Rules for K-map simplification
4. Each group should be as large as possible
8/29/2017 54
Not Accepted Accepted
1 1 1 1
0 0 1 1
BC
A 00
0
1
01 11 10
1 1 1 1
0 0 1 1
BC
A 00
0
1
01 11 10 BC BC BC BC
A
A
BC BC BC BC
A
A
Rules for K-map simplification
5. Each cell containing a one must be in at least one group
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0 0 0 1
0 0 1 0
BC
A 00
0
1
01 11 10
BC BC BC BC
A
A
Rules for K-map simplification
6. Groups may be overlap
8/29/2017 56
1 1 1 1
0 0 1 1
BC
A 00
0
1
01 11 10
BC BC BC BC
A
A
Rules for K-map simplification
7. Groups may wrap around the table. The leftmost cell in
a row may be grouped with rightmost cell and the top cell in a column may be grouped with bottom cell
8/29/2017 57
1 0 0 1
1 0 0 1
BC
A 00
0
1
01 11 10 1 1 1 1
0 0 0 0
0 0 0 0
1 1 1 1
CD
AB 00 01 11 10
00
01
11
10
BC BC BC BC
A
A
CD CD CD CD
AB
AB
AB
AB
Rules for K-map simplification
8. There should be as few groups as possible, as long as this does not contradict any of the previous rules.
8/29/2017 58
Not Accepted Accepted
1 1 1 1
0 0 1 1
BC
A 00
0
1
01 11 10
1 1 1 1
0 0 1 1
BC
A 00
0
1
01 11 10 BC BC BC BC
A
A
BC BC BC BC
A
A
9. A pair eliminates one variable. 10. A Quad eliminates two variables. 11. A octet eliminates three variables
8/29/2017 59
Rules for K-map simplification
Example 1
For the given K-map write simplified Boolean expression
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0 1 1 1
0 0 1 0
AB
C 00
0
1
01 11 10
AB AB AB AB
C
C
Example 1 continue…..
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0 1 1 1
0 0 1 0
AB
C 00
0
1
01 11 10
Simplified Boolean expression
AB AB AB AB
C
C
AC
ABBC
Y BC AB AC
Example 2
For the given K-map write simplified Boolean expression
8/29/2017 62
1 1 0 1
1 0 0 1
AB
C 00
0
1
01 11 10
AB AB AB AB
C
C
Example 2 continue…..
8/29/2017 63
1 1 0 1
1 0 0 1
AB
C 00
0
1
01 11 10
Simplified Boolean expression
AB AB AB AB
C
C
AC B
Y B AC
Example 3
A logical expression in the standard SOP form is as follows; Minimize it with using the K-map technique
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Y ABC ABC ABC ABC
Example 3 continue……
8/29/2017 65
1 0 1 1
0 1 0 0
BC
A 00
0
1
01 11 10
Simplified Boolean expression
Y ABC ABC ABC ABC
BC BC BC BC
A
A
ABC
AB
AC
Y AC AB ABC
Example 4
A logical expression representing a logic circuit is; Draw the K-map and find the minimized logical expression
8/29/2017 66
(0,1,2,5,13,15)Y m
Example 4 continue…..
8/29/2017 67
1 1 0 1
0 1 0 0
0 1 1 0
0 0 0 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14 Simplified Boolean expression
(0,1,2,5,13,15)Y m
CD CD CD CD
AB
AB
AB
AB
ABD
ABDACD
Y ABD ACD ABD
Example 5
Minimize the following Boolean expression using K-map ;
8/29/2017 68
( , , , ) (1,3,5,9,11,13)f A B C D m
Example 5 continue…..
8/29/2017 69
0 1 1 0
0 1 0 0
0 1 0 0
0 1 1 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14 Simplified Boolean expression
CD CD CD CD
AB
AB
AB
AB
CDBD
f BD CD
( , , , ) (1,3,5,9,11,13)f A B C D m
( )f D B C
Example 6
Minimize the following Boolean expression using K-map ;
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( , , , ) (4,5,8,9,11,12,13,15)f A B C D m
Example 6 continue…..
8/29/2017 71
0 0 0 0
1 1 0 0
1 1 1 0
1 1 1 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14 Simplified Boolean expression
CD CD CD CD
AB
AB
AB
AB
ADAC
f BC AC AD
( , , , ) (4,5,8,9,11,12,13,15)f A B C D m
BC
Example 7
Minimize the following Boolean expression using K-map ;
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2( , , , ) (0,1,2,3,11,12,14,15)f A B C D m
Example 7 continue…..
8/29/2017 73
1 1 1 1
0 0 0 0
1 0 1 1
0 0 1 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14 Simplified Boolean expression
CD CD CD CD
AB
AB
AB
AB
ACDABD
2f AB ABD ACD
AB
2( , , , ) (0,1,2,3,11,12,14,15)f A B C D m
Example 8
Solve the following expression with K-maps; 1. 2.
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1( , , ) (0,1,3,4,5)f A B C m
2( , , ) (0,1,2,3,6,7)f A B C m
8/29/2017 75
Example 8 continue……
1 1 1 0
1 1 0 0
BC
A 00
0
1
01 11 10
0 1 3 2
4 5 7 6
1 1 1 1
0 0 1 1
BC
A 00
0
1
01 11 10
0 1 3 2
4 5 7 6
Simplified Boolean expression
Simplified Boolean expression
1( , , ) (0,1,3,4,5)f A B C m 2( , , ) (0,1,2,3,6,7)f A B C m
BC BC BC BC
A
A
BC BC BC BC
A
A
AC
BBA
2f A B 1f AC B
Example 9
Simplify ;
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( , , , ) (0,1,4,5,7,8,9,12,13,15)f A B C D m
Example 9 continue…..
8/29/2017 77
1 1 0 0
1 1 1 0
1 1 1 0
1 1 0 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14 Simplified Boolean expression
CD CD CD CD
AB
AB
AB
AB
BDC
f C BD
( , , , ) (0,1,4,5,7,8,9,12,13,15)f A B C D m
Example 10
Solve the following expression with K-maps; 1. 2.
8/29/2017 78
1( , , , ) (0,1,3,4,5,7)f A B C D m
2( , , ) (0,1,3,4,5,7)f A B C m
8/29/2017 79
Example 10 continue……
1 1 1 0
1 1 1 0
BC
A 00
0
1
01 11 10
0 1 3 2
4 5 7 6
Simplified Boolean expression
Simplified Boolean expression
1 1 0 0
1 1 1 0
0 0 0 0
0 0 0 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14
BC BC BC BC
A
A
CB
2f B C
1f AC AD
1( , , , ) (0,1,3,4,5,7)f A B C D m 2( , , ) (0,1,3,4,5,7)f A B C m
CD CD CD CD
AB
AB
AD
AB
AB
AC
K-map and don’t care conditions
For SOP form we enter 1’s corresponding to the combinations of input variables which produce a high output and we enter 0’s in the remaining cells of the K-map.
For POS form we enter 0’s corresponding to the combinations of input variables which produce a high output and we enter 1’s in the remaining cells of the K-map.
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But it is not always true that the cells not containing 1’s (in SOP) will contain 0’s, because some combinations of input variable do not occur.
Also for some functions the outputs corresponding to certain combinations of input variables do not matter.
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K-map and don’t care conditions
In such situations we have a freedom to assume a 0 or 1 as output for each of these combinations.
These conditions are known as the “Don’t Care Conditions” and in the K-map it is represented as ‘X’, in the corresponding cell.
The don’t care conditions may be assumed to be 0 or 1 as per the need for simplification
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K-map and don’t care conditions
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K-map and don’t care conditions - Example
Simplify ;
( , , , ) (1,3,7,11,15) (0,2,5)f A B C D m d
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X 1 1 X
0 X 1 0
0 0 1 0
0 0 1 0
CD
AB 00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
8 9 11 10
12 13 15 14 Simplified Boolean expression
K-map and don’t care conditions - Example
CD CD CD CD
AB
AB
AB
AB
CDAB
f CD AB AD
( , , , ) (1,3,7,11,15) (0,2,5)f A B C D m d
AD
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