K-maps

WEEK 4
Simplification: K-maps
1
More on BA: The Consensus Theorem
Given a pair of terms for which a variable appears in one term, and its
complement in the other, then the consensus term is formed by ANDing the
original terms together, leaving out the selected variable and its complement.
Example.
The consensus of X.Y and X'.Z is Y.Z
The consensus of X.Y.Z and Y'.Z.W' is (X.Z).(Z.W')
check.
X.Y + X'.Z + Y.Z = X.Y + X'.Z
or dual form as below
(X + Y).(X' + Z).(Y + Z) = (X + Y).(X' + Z)
Page 2
The Consensus Theorem
XY  X ' Z  YZ  XY  X ' Z
Consensus Theorem
Proof :
XY  X ' Z  YZ  XY  X ' Z  ( X  X ' )YZ
 ( XY  XYZ )  ( X ' Z  X ' YZ )
 XY (1  Z )  X ' Z (1  Y )  XY  X ' Z
Example:
consensus
a' b'ac  bc'b' c  ab  a' b'ac  bc'
consensus
Dual form of consensus theorem
Example:
( X  Y )( X 'Z )(Y  Z )  ( X  Y )( X 'Z )
(a  b  c' )(a  b  d ' )(b  c  d ' )  (a  b  c' )(b  c  d ' )
Page 3
Multiplying Out and Factoring
Expressions
To obtain a sum-of-product form  Multiplying out using distributive laws
X (Y  Z )  XY  XZ
( X  Y )(Z  X )  XZ  XX  YZ  YX  X  YZ
PROOF: see next slide
567
( X  Y )( X '  Z )  XZ  X ' Y
Theorem for multiplying out:
144244
3
If X  0, (3 - 3) reduces toY(1  Z)  0  1*Y or Y  Y.
If X  1, (3 - 3) reduces to(1  Y)Z  Z  0 *Y or Z  Z.


As the equation is valid for both x=0 and x=1, it is always valid
Theorem for factoring:
567
A
A4
B2
'C  ( A  C )( A' B)
1
4
3
(3-3)
Multiplying Out and Factoring
Expressions
To obtain a product-of-sum form  Factoring using distributive laws
567
A
A4
B2
'C  ( A  C )( A' B)
1
4
3
Theorem for factoring:
Example of factoring:
AC  A' BD' A' BE  A' C ' DE
 AC  A' ( BD' BE  C ' DE )
XZ X '
Y
X
(X+Y)(X’+Z)
Y
 ( A  BD' BE  C ' DE )( A' C )
 [ A  C ' DE  B( D' E )]( A' C )
(XX+YZ)
X
Y Z
(X+Z)(Y+X)
 ( A  B  C ' DE )( A  C ' DE  D' E )( A' C )
 ( A  B  C ' )( A  B  D)( A  B  E )( A  D' E )( A' C )
(3-5)
Page 5
Proof
(X+Y)(X+Z)=X(X+Z)+Y(X+Z)
= XX+XZ+YX+YZ
=X+XZ+XY+YZ
=X(1+Z+Y)+YZ
= X+YZ
All from Nand
Can build AND, OR, NOT from NAND gates:
NAND(X,X) = NOT (X)
A A Nand
00 1
11 0
AND from NAND gates:
AND(X,Y) = NAND followed by NOT (X)
OR from NAND gates:
Use Demorgan’s:
NAND(A’,B’) = (A’  B’)’ =A+B
How to invert inputs?
Page 7
Gate Delay


Due to the physics of chip electronics, when one or more
input changes causes the output to change, the output
change does not occur instantaneously.
Gate delay (tG ) is the time difference between an input
change(s) and the resulting output change:
Input
Output
1
0
tG
1
0
0
tG = 0.3 ns
tG
0.5
1
1.5
Time (ns)
Page 8
Gate-Level Minimization Via Karnaugh Maps
(K-Maps)
SOP and POS representations using
minterms and maxtems are far from
minimal in terms of the #gates and #gate
inputs used.
 Boolean algebra helped, but it is hard to
automate, not very systematic, etc.
 For small problems (<=5 variables), we can
use a Karnaugh Map (KMap) to help us
figure out how to get a more minimal
representation for a logic function.

Page 9
What is a K-Map?
It’s an alternative representation to a truth
table and is good for about 5 variables.
 Whereas a truth table is tabular, a K-Map is a
grid of squares; each square corresponds to
one row of the truth table.

 The “co-ordinates” of a square depend on the input
pattern.
 The “content” of the square is the output value.

Adjacent squares share a very important
property; namely that only one variable
values changes.
Page 10
2-Variable K-Maps

Consider a logic
function with two
inputs.
 The Canoncial SOP using minterms
is:
x1
x2
0
1
0
1
0
1
1
1
m0
m1
m2
m3
 The Minimum SOP using product
terms is:
Page 11
3-Variable K-Maps

Consider a logic function with three inputs.
yz
x2x3
x
x1
0
1
00
01
11
10
0
1
0
0
1
1
0
1
f=(x2)’(x3)+x1x2
 The Canoncial SOP using minterms would require 4 minterms:
 Note: we always number such that only one variable changes between adjacent
columns or rows (Gray codes).
 Note: columns (and rows) wrap around from left-to-right (top-to-bottom).
Page 12
K Map with 3 Variables
yz
x


3 Variable, F = f(x,y,z);
Given F3 = ∑(3,4,6,7)

Minimize the function using K map
Function minimization
00
11
10
1
0
1
01
1
1
1



Find maximum size groups that cover all 1s in the map
(Comment – a group should not be a subset of other group)
4 cell group  2 literals can be removed
2 cell group  1 literal can be removed
Guidelines for logic synthesis (SOP)

Fewer groups  fewer AND gates, and fewer inputs to the OR gate

Fewer literals (larger group)  fewer inputs to an AND gate
Synthesis Objective: Fewest # of gates and # of inputs
13
K Map with 3 Variables


3 Variable, F = f(x,y,z);
Given F2 = ∑(2,3,4,5)


yz
x
Represent it on the K map
minimize the function
yz
x
00
01
0
11
10
1
1
00
01
11
10
0
m0
m1
m3
m2
1
m4
m5
m7
m6
00
01
11
10
yz
x
1
1
1
0
x’y’z’ ’x’y’z
x’yz
x’yz’
1
xy’z’
xyz
xyz’
xy’z
14
More Examples of 3-Variable KMaps

Examplex1x21:
x3
00
01
11
10
0
1
0
0
1
1
1
1
1
1
m0
m1
m2
m3
m6
m7
x1x2
x3
00
01
11
10
0
1
0
0
1
1
1
1
1
1
m0
m4
m2
m1
m5
m6
m3
m4
m7
m5
 Example 2:
x1x2
x3
0
1
x1x2
x3
00
0
m0
1
m1
01
1
m2
1
m3
11
1
m6
1
m7
00
01
11
10
0
0
1
1
1
1
1
1
1
0
10
1
m0
m2
m6
m4
m4
0
m1
m3
m7
m5
m5
Page 15
Comments on K-Maps
To properly implement a function using KMaps, we must make sure to select enough
rectangles (product terms) to cover all of
the squares for which the function outputs
a “1”.
 Rectangles in the K-Map (product terms) are
always powers of 2.

 Minterms
are rectangles of area 1 (1x1).
 Product terms are rectangles of area 2
(1x2,2x1), area 4 (2x2,1x4,4x1), area 8, area 16,
etc…
Page 16
4-Variable K-Maps

Consider a function with 4-inputs (truth table
not shown).
x1x2
x3x4
00
01
11
x1x2
x3x4
00
1
m0
1
m1
1
m3
01
1
m4
1
m5
1
m7
11
0
m12
0
m13
0
m15
00
01
11
10
00
1
1
0
1
01
1
1
0
1
11
1
1
0
1
10
1
1
0
0
10
m0
1
m2
1
m6
0
m14
m12
m8
1
m8
m1
m5
m13
m9
1
m9
m3
m7
m15
m11
1
m11
m2
10
m4
m6
m14
m10
0
m10
 Enclosing larger squares are better, as it implies fewer literals and product terms.
Page 17
Another Example of 4-Variable KMaps

Sometimes there are multiple
representations for a function that are
equally as good in terms of the number of
literals and product terms required.
x1x2
x3x4
00
01
11
10
00
1
m0
1
m1
0
m3
0
m2
01
1
m4
1
m5
0
m7
0
m6
11
1
m12
1
m13
1
m15
1
m14
10
x1x2
x3x4
00
01
11
10
00
1
1
1
0
01
1
1
1
0
11
0
0
1
1
10
0
0
1
1
0
m8
m0
m4
m12
m8
0
m9
m1
m5
m13
m9
1
m11
m3
m7
m15
m11
1
m10
m2
E&CE 223 Digital Circuits and Systems (A. Kennings)
m6
m14
m10
Page 18
Watch Out for Corners!

Important to make sure we see all rectangles,
especially around the edges and corners.
x1x2
x3x4
x1x2
x3x4
00
00
1
m0
01
0
m4
11
0
m12
10
m8
0
0
0
0
11
0
0
0
0
10
m3
1
m2
m5
m7
0
m6
m13
m15
0
m14
01
11
10
00
1
0
0
1
01
0
0
0
0
11
0
0
0
0
10
1
m0
m1
m3
m4
m5
m7
m12
m13
m15
m8
m9
m11
0
m6
0
1
m2
00
01
11
10
00
1
0
0
1
01
0
0
0
0
11
0
0
0
0
10
1
0
0
1
1
01
m1
00
x1x2
x3x4
m14
m10
m9
m0
m4
m12
m8
m11
m1
m5
m13
m9
1
m10
m3
m2
m7
m6
m15
m14
m11
m10
Page 19
5-Variable K-Maps
x1x2

Slightly harder to visualize, but still do-able by hand.
x3x4
00
01
11
10
00
x1x2
x3x4
00
01
11
10
x1x2
x3x4
00
01
11
10
11
00
1
m0
0
m8
0
m24
1
m16
00
1
m1
0
m9
0
m25
x5=0
01
1
m17
x1x2
10
x3x4
00
01
0
1
1
0
01
0
1
1
0
11
0
1
1
0
11
0
1
1
1
10
0
1
0
0
10
0
1
0
1
m2
m6
m10
m14
m26
m30
m18
m22
m3
m7
m11
m15
m27
m31
00
01
11
10
x5=1
01
m19
11
m23
10
m4
m12
m28
x5=0
m20
m5
m13
m29
m21
x5=1
A BCDE
0
K-maps
f1
f=A’f1+Af2
1
K-maps
f2
Page 20
Summary of Sum-Of-Products
Minimization Using K-Maps


Find product terms with as few literals as possible
(by finding larger rectangles in the K-Map covering
“1”s).
Select as few product terms as possible, so that all
of the “1”s are covered.
 It is okay if a “1” is covered by more than one
product term.
 It is possible that there are multiple solutions
that are equally as good.
Page 21
Product-Of-Sums Minimization
Using K-Maps

More or less the same thing as Sum-OfProducts minimization, except…
 We
try to surround the “0”s in the K-Map, and…
 We AND together the resulting sum terms.

An example with four inputs:
Another way is to find SOP of f’ and
Then use DeMorgan to obtain f in POS
x1x2
x3x4
00
01
11
10
00
1
1
1
0
01
1
1
1
0
11
0
0
1
1
10
0
0
1
1
f '  ( x3' .x1 .x2' )  ( x3 .x1' )
f  ( x3  x  x2 ).( x 3  x1 )
'
1
'
Page 22
Don’t Cares
There are possible input combinations where it doesn’t
matter what the output is; for example, for we know that
input is never possible.
Example, in a 4 bit BCD, inputs from 1010 to 1111 are not in
the set of BCD numbers. But we still need 4 bits to
represnta BCD.

 These
situations are called “don’t cares”.
 Rather than using a “1” or a “0”, we can use an
“X” to mark a don’t care.


Why are don’t cares important and useful???
Well, if we are looking for a minimum representation using
as few literals and gates as possible, we can use the don’t
cares however we want; i.e.,
 We can set them to 0 or 1, however it helps us.
Page 23
Example of Don’t Cares

Consider the following four input function where there are three input patterns for
which the output can be either 0 or 1, and we don’t care which.
x1x2
x3x4
00
01
11
10
00
X
1
1
X
01
0
X
1
0
11
0
0
1
0
10
0
0
1
0

For the smallest SOP implementation, it appears good to let one of the don’t cares
be a 1, while the other 2 are 0.

Question: Is there another equally as good implementation that uses the don’t
cares differently?
Page 24
Logical vs. Algebraic Equivalence

When we have don’t cares, we can get equally good expressions that are logically
correct (i.e., we get output of 1 when it must be 1, and 0 when it must be 0).

However, depending on how we use the don’t cares, multiple expressions will not
equal each other algebraically.
x1x2
x3x4

00
01
11
10
00
X
1
1
X
01
0
X
1
11
0
0
10
0
0
x1x2
x3x4
00
01
11
10
00
X
1
1
X
0
01
0
X
1
0
1
0
11
0
0
1
0
1
0
10
0
0
1
0
Note that both f1 and f2 are valid implementations of f, but are not algebraically
equal to each other since the don’t cares are used differently when finding both f1
and f2.
Page 25
Multiple Output/Function
Minimization
 Sometimes, we might have multiple functions
that require minimization.
 We can minimize each function individually, but
sometimes the overall circuit implementation
will be better if we consider both functions
simultaneously.
 Basically, we will try to find common product
terms useful for both functions and then
share them.
Page 26
Example of Multiple Output/Function
Minimization
Consider the following 2, four input functions and the minimal implementation of
each function separately.

x1x2
x3x4
00
01
11
10
x1x2
x3x4
00
0
1
1
0
01
0
1
1
1
11
0
1
1
0
00
01
11
10
00
0
0
0
0
01
1
0
1
1
11
1
0
1
1
10
0
1
0
0
10
0
0
0
0

The functions have no product terms in common. The cost of f_1 is 1 OR + 3 AND +
10 gate inputs = 14. The cost of f_2 is 1 OR + 3 AND + 11 gate inputs = 15.

Therefore, to implement the entire circuit the cost is 29.
Page 27
Example of Multiple Output/Function
Minimization
Consider trying to include some product terms that appear in both K-Maps.

x1x2
x3x4



x1x2
x3x4
00
01
11
10
00
0
0
0
0
01
1
1
1
0
11
1
1
1
0
10
0
1
0
0
00
01
11
10
00
0
0
0
0
01
1
0
1
1
11
1
0
1
1
10
0
1
0
0
The individual expressions for f1 and f2 are not the best, but…
If we consider that 2 product terms are shared, the entire circuit requires 2 OR +
4 AND + 17 gate inputs.
The total cost of the entire circuit is 23.
Page 28
Multiple Outputs Illustrated
Page 29
Multiple-Output Minimization
▶ Multiple output minimization
- Two output functions
F = ∑ ( 3,6,7 ) = XY + YZ
G = ∑ ( 0,1,3 ) = X´Y´ + X´Z
ⅰ) Two independent single output design
30
Multiple-Output Minimization
Shared term
f = X Y + X´Y Z
g = X´Y´ + X´Y Z
Same input combination = shared term
31
Some More Terminology

Useful to be aware of some terminology…

When we have a function and we are told the output is 0 or 1 for every possible
input, the function is called completely specified.

When we have a function and are told the output is 0 or 1 for only some of the
inputs, we can assume the other inputs have don’t cares as outputs. It the situation
that we have don’t care outputs, the function is incompletely specified.

Consider the minterms for any function. We can separate the minterms in to three
sets:
 The on-set is those minterms which require the output to be 1.
 The off-set is those minterms which require the output to be 0.
 The dc-set is those minterms which have an output of X (don’t care).
Page 32
Implicants, Prime Implicants and Covers

A product term is called an Implicant if the logic function outputs a 1 for all
minterms in the product term.


An Implicant is called a Prime Implicant if the removal of any literal from the
implicant results in a new product term that is not an implicant.



E.g., all rectangles in a K-Map that contain only 1s are implicants.
E.g., Removing any literal from a product term in a K-Map corresponds to
increasing the size (doubling the area) of the rectangle.
If we cannot increase the size of the rectangle without including a 0, then the
implicant is prime.
A cover is a collection of implicants that account for all cases in which the function
is 1.

We can always form a cover using only prime implicants.
Page 33
Example of Implicants
x1x2
x3
00
01
11
10
0
1
1
0
0
1
1
1
1
0

In the K-Map, product terms x1’ and x2 x3 are prime implicants.

Product term x1’ x2’ is an implicant, but it is not a prime implicant.
Page 34
Essential Prime Implicants

A prime implicant is called an Essential
Prime Implicant if it includes a minterm that
in not found in any other prime implicant.

In realizing any function using only prime
implicants, all essential prime implicants
must be included!
 If
not, we would “miss” terms for which the
function outputs a 1.
Page 35
Essential prime implicants
YZ
WX
1
1
0
0
1
1
1
0
0
0
1
1
0
0
X
0
1
Z



If any group contains a minterm that is not also covered by another
overlapping group, then that is an essential prime implicant.
Essential prime implicants must appear in the MSP, since they contain
minterms that no other terms include.
Our example has just two essential prime implicants:
 The red group (w’y’) is essential, because of m0, m1 and m4.
 The green group (wx’y) is essential, because of m10.
36
Covering the other minterms
Y
W
1
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
X
Z



Finally pick as few other prime implicants as necessary to ensure that
all the minterms are covered.
After choosing the red and green rectangles in our example, there are
just two minterms left to be covered, m13 and m15.
 These are both included in the blue prime implicant, wxz.
 The resulting MSP is w’y’ + wxz + wx’y.
The black and yellow groups are not needed, since all the minterms are
covered by the other three groups.
37
Methodical Example…

The following K-Map has 6 prime implicants.
4
x1x2
x3x4
of them are essential, 2 are not essential.
00
01
11
10
00
1
0
1
1
01
0
1
0
0
11
1
1
1
0
10
0
0
1
1
x1x2
x3x4
00
01
11
10
00
1
0
1
1
01
0
1
0
0
11
1
1
1
0
10
0
0
1
1
Page 38
Karnaugh Maps

A logic function with no essential prime implicants
Several different minimal sums
F = ∑ (1,5,7,9,11,15 ) :
z
2 minimal sums
39
Why all this Terminology?


Understanding implicants, primes and essentials help us
form a more methodical procedure for minimization of logic
functions.
Covers using prime implicants typically have low cost. So to
minimize, we should…
 Generate prime implicants.
 Identify essential prime implicants.
 Include the essential prime implicants in the cover.
 If the function is covered, stop.
 Include as few non-essential prime implicants to finish
covering the function.
 And then stop…
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Textbook Sections

K-Maps, SOP and POS minimization, Don’t
cares and comments on implicants can be
found in Chapter 3, Sections 3.1 through 3.5
of the text book.
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