The document discusses Karnaugh maps and their use in simplifying Boolean functions. Karnaugh maps are a pictorial representation of truth tables used to reduce logic functions with up to 6 variables. The maps arrange minterms in a grid based on the number of variables. Rules for grouping 1s in the map include no zeros, no diagonals, groups as powers of 2, and obtaining the fewest number of groups. Don't care conditions and overlapping groups can further simplify the resulting logic expression.

1. DIGITALCOMPUTER FUNDAMENTALS

2. G.Vinothini M.sc.,M.phil.,
Department of Information Techonology,
Bon Secours College for Women,
Thanjavur.

3.  It originated from the “Map Method” proposed
by veitch also called the “Veitch Diagram” and
then modified by Karnaugh
Karnaugh map is introduced by telecom
engineer Mauruce Karnaugh in 1953 at Bell
labs.
It is a pictorial from of the truth table and
could handle up to 6 variables.
It is used to reduce or simplify a Boolean
function.

4. The map is made up of a table of every possible SOP
using the Number of Variables that are being used.
If 2 variables are used then a 2×2 map is used.
If 3 variables are used then a 4×2 map is used.
If 4 variables are used then a 4×4 map is used.
If 5 variables are used then a 8×4 map is used.

5. KARNAUGH MAP WITH 2 VARIABLES
 The karnaugh map with two variables is show in below
figure. There are four minterms for two variables; hence the
map consists of four squares, one for each minterms.
 The below map show the relationship between the squares
and the two variables.
 The two values 0 and 1 marked for each row and each
column designate the values of variables A and B
respectively.
m0 m1
m2 m3
B
0
B
1
A 0 AB AB
A 1 AB AB

6. KARNAUGH MAP WITH 3 VARIABLE
There are 8 minterms for three binary variables.
Therefore, the Karnaugh map consists of eight squares.
The map is show the relationship between the squares
and the three variables.
It may be noted that the minterms are not arranged in a
binary sequence, but in a sequence similar to the gray
code.
m0 m1 m3 M2
m4 m5 m7 m6

7. KARNAUGH MAP WITH 4 VARIABLES
The karnaugh map for four variables has 16 minterms
are shown. The map redrawn to show the relationship
between the four variables.
The rows and columns are numbered in a reflected –
code sequence, with only one digit changing value
between two adjacent rows and columns.
The minterm corresponding to each squares can be
obtained from the concatenation of the row number
with the column number.

8. RULES OF KARNAUGH MAP
No zeros allowed.
No diagonals.
Only power of 2 Number of cells in each group.
Group should be as large as possible.
Every one must be in at least one group.
Overlapping allowed.
Wrap arround allowed.
Fewest Number of groups possible.

10. Let us consider the following Boolean equation
Y=A.B+AB=AB+AB
This equation implies that A is AND with B, A
is ANDed with B, and AB is ORed with AB. Logic
circuits to use when variables and their complements
are available from other circuits. If complements are
not available we have to use an inverter .

11. NANDAND NOR IMPLEMENTATION
NAND IMPLEMENTATION NOR IMPLEMENTATION
The implementation of
Boolean function with NAND
gates requires that the
function be simplified in the
sum-of-products form.
Consider the Boolean
function
F=AB+CD+E
The implementation of
Boolean function with NOR
gates requires that the
function be simplified in the
product-of-sums form.
Consider the product-of-
sums expression
F=(A+B)(C+D)E

13. DON’T CARE CONDITION
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.

14. OVERLAPPING GROUPS
We can use the same 1 more than once. In the
words, the same 1 can be common to two or more
groups. In figure the 1 representing the fundamental
products ABCD is part of the pair as well as part of the
octet.
Y=A+BCD
CD CD CD CD
AB 0 0 0 0
AB 0 1 0 0
AB 1 1 1 1
AB 1 1 1 1

15. We can also roll and overlap to get the
largest groups that can be found . The octet and
pair have a Boolean equation of y=C+BCD.
We can do better by rolling and overlapping.
The Boolean equation now becomes y=C+BD
EXAMPLE
CD CD CD CD
AB 1 1 0 0
AB 1 1 0 1
AB 1 1 0 1
AB 1 1 0 0

16. ELIMINATING REDUNDANT GROUPS
After encircling groups, there
is one more thing we should do
before writing the simplified
Boolean equation, for example
eliminate any group whose 1’s are
completely overlapped by other
groups.