Boolean Algebra

KARNAUGH MAP METHODS TO SOLVING
SOLVING THE BOOLEAN FUNCTIONS
Ms.S.Swathi Sundari, M.Sc.,M.Phil.,
IIIB.ScMathematics
Boolean Algebra(15UMTC63)
1

MINIMIZATION OF BOOLEANEXPRESSIONS
• The minimization will result in reduction
of the number of gates (resulting from
less number of terms) and the number
of inputs per gate (resulting from less
number of variables per term)
• The minimization will reduce cost,
efficiency and power
consumption.
• y(x+x`)=y.1=y
• y+xx`=y+0=y
2

MINIMUM SOP AND POS
SOP - SUM OF THE PRODUCTS AND POS – PRODUCT OF SUMS
• f= (xyz +x`yz+ xy`z+ …..) is called sum of
products.
The + is sum operator which is an OR gate.
The product such as xy is an AND gate for
the two inputs x and y.
• The minimum sum of products (MSOP) of a
function, f, is a SOP representation of f that
contains the fewest number of product terms
3

EXAMPLE
• Minimize the following Boolean
function using
sum of products (SOP):
• f(a,b,c,d) = m(3,7,11,12,13,14,15)
3
abcd
0011 a`b`cd
7 0111 a`bcd
11 1011 ab`cd
12 1100 abc`d`
13 1101 abc`d
14 1110 abcd`
15 1111 abcd
4

EXAMPLE
f(a,b,c,d) = m(3,7,11,12,13,14,15)
= a`b`cd + a`bcd + ab`cd + abc`d`+
abc`d + abcd` + abcd
= cd(a`b` + a`b + ab`) + ab(c`d` +
c`d + cd` + cd )
= cd(a`[b` + b] + ab`) + ab(c`[d` + d]
+ c[d` + d])
= cd(a`[1] + ab`) + ab(c`[1] + c[1])
= ab+ab`cd + a`cd
= ab+cd(ab` + a`)
= ab+ cd(a + a`)(a`+b`)
5

MINIMUM PRODUCT OF SUMS (MPOS)
• The minimum product of sums (MPOS)
of a function, f, is a POS representation
of f that contains the fewest number of
sum terms and the fewest number of
literals of any POS representation of f.
• The zeros are considered exactly the
same as ones in the case of sum
of product (SOP)
6

EXAMP
LE
f(a,b,c,d) = M(0,1,2,4,5,6,8,9,10)
= m(3,7,11,12,13,14,15)
=
[(a+b+c+d)(a+b+c+d`)(a+b`+c`+d`)
(a`+b+c`+d`)(a`+b`+c+
d)(a`+b`+c+ d`)
(a`+b`+c`+d)(a`+b`+c`+d`)] 7

KARNAUGH MAPS (K-MAPS)
•Karnaugh maps -- A tool for
representing Boolean functions of
up to six variables.
•K-maps are tables of rows and
columns with entries represent 1`s
or 0`s of SOP and POS
representations. 8

• An n-variable K-map has 2n cells
with each cell corresponding to an
n-variable truth table value.
• K-map cells are labeled with the
corresponding truth-table row.
• K-map cells are arranged such
that adjacent cells correspond to
truth rows that differ in only one
bit position (logical adjacency).
9

•If mi is a minterm of f, then place a 1
in cell i of the K-map.
•If Mi is a maxterm of f, then place a
0 in cell i.
•If di is a don’t care of f, then place a
d or x in cell i. 10

EXAMP
LES
• Two variable K-map
f(A,B)=m(0,1,3)=A`B`+A`B+AB
1 0
1 1
A 0 1
B01
11

THREE VARIABLE MAP
• f(A,B,C) = m(0,3,5)=
A`B`C`+A`BC+AB`C
A`B`
0 0
A`B
0 1
A B
1 1
A B`
1 0
1 C`
0 A`B`C `
1
A
C
`1BC
1
AB`C
12

MAXTERM EXAMPLE
0 0 0
0 0
A`B` A`B AB AB`
C`
C
(A+B) (A+B`) (A`+B`) (A`+B)
C
C`
f(A,B,C) = M(1,2,4,6,7)
=(A+B+C`)(A+B`+C)(A`+B+C) )(A`+B`+C) (A`+B`+C`)
Note that the complements are (0,3,5) which
are the minterms of the previous example13

FOUR VARIABLE EXAMPLE
(a) Minterm form.
(b) Maxtermform.
f(a,b,Q,G) = m(0,3,5,7,10,11,12,13,14,15) = M(1,2,4,6,8,9)
14

SIMPLIFICATIONOFBOOLEANFUNCTIONS
USING K-MAPS
• K-map cells that are physically
adjacent are also logically adjacent.
Also, cells on an edge of a K-map are
logically adjacent to cells on the
opposite edge of the map.
• If two logically adjacent cells both
contain logical 1s, the two cells can
be combined to eliminate the variable
15

USING K-MAPS
•This is equivalent to the
algebraic operation,
aP + aP =P where P is a
product term not containing
a or a.
•A group of cells can be
16

USING K-MAPS
• Always combine as many cells in a group
as possible. This will result in the fewest
number of literals in the term that
represents the group.
• Make as few groupings as possible to
cover all minterms. This will result in
the fewest product terms.
• Always begin with the largest group,
which means if you can find eight
members group is better than two four
17

EXAMPLE
SIMPLIFY F= A`BC`+ A B C`+ A B C USING;
(a) Sum of minterms. (b)
Maxterms.
• Each cell of an n-variable K-map has n
logically adjacent cells. C
AB
10
0 2 6 4
1 3 7 5
1
B
0
0C
C
AB
00 11 10 00 01 11
0 2 6 4
1 3 7 5
0
1C
A A
B
AB
BC
01
1 1
1 0
0 0
0
a-
b-
F`= B`+ A`C
f(A,B,C) = AB + BC
f(A,B,C) = B(A + C)
F = B(A+C`)
18

EXAMPLE
SIMPLIFY
CD
AB
00 01 11
0 4
1
12 8
1
1 5
1
13
1
9
3
1
7
1
15
1
11
2
1
6 14 10
1
00
01
11
10
B
(a)
D
C
AB
10 CD 00 01 11 10
00
01
11
10
B
(b)
D
A A
0 4
1
12 8
1
1 5
1
13
1
9
3
1
7
1
15
1
11
2
1
6 14 10
1
C
CD
AB
00 01 11 10
0 4
1
12 8
1
1 5
1
13
1
9
3
1
7
1
15
1
11
2
1
6 14 10
1
00
01
11
10
B
(c)
D
A
C
CD
AB
00 01 11 10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
00
01
11
10
D
A
C
B
(d)
1 1
1 1 1
1
1 1
1
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
19

EXAMPLE MULTIPLESELECTIONS
CD
AB
00 01 11 10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
00
01
11
D
A
C
B
(a)
CD
AB
00 01 11 10
0 4
1
12 8
1
1 5 13 9
1 1
3
1
7
1
15
1
11
2
1
6 14 10
1
00
01
11
10
B
(b)
D
A
C
1 1
1 1
1 1
10 1 1
1
CD
AB
00 01 11 10
0 4
1
12 8
1
1 5 13 9
1 1
3
1
7
1
15
1
11
2
1
6 14 10
1
00
01
11
10
B
(c)
D
A
C
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
c produces less terms than a
20

COMPLETE SIMPLIFICATION PROCESS
21
1.Construct the K map and place 1s and 0s in the squares
according to the truth table.
2.Group the isolated 1s which are not adjacent to any other 1s.
(single loops)
3.Group any pair which contains a 1 adjacent to only one other 1.
(double loops)
4.Group any octet even if it contains one or more 1s that have
already been grouped.
5.Group any quad that contains one or more 1s that have not
already been grouped, making sure to use the minimum number
of groups.
6.Group any pairs necessary to include any 1s that have not yet
been grouped, making sure to use the minimum number of
groups.
7.Form the OR sum of all the terms generated by each group.

• Minterms that may produce either 0 or 1 for the function.
• They are marked with an ´ in the K-map.
• This happens, for example, when we don’t input certain minterms to
the Boolean function.
• These don’t-care conditions can be used to provide further
simplification of the algebraic expression.
(Example) F = A`B`C`+A`BC` +ABC` d=A`B`C
+A`BC +AB`C
F = A` + BC`
DON’T-CARE
CONDITION
22

23
 “Don’t care” conditions should be changed to either 0 or 1
to produce K-map looping that yields the simplest
expression.
More “Don’t Care” examples

SUMMARY
 Compared to the algebraic method, the K-map process
is a more orderly process requiring fewer steps and
always producing a minimum expression.
 The minimum expression in generally is NOT unique.
 For the circuits with large numbers of inputs (larger
than four), other more complex techniques are used.
 SOP and POS –useful forms of Boolean equations.
24

Boolean Algebra

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