Karnaugh Maps
References:
Chapters 4 and 5 in Digital Principles (Tokheim)
Chapter 3 in Introduction to Digital Systems (Palmer and Perlman)
PHY 201 (Blum) 1

Review: Expressing truth tables



Every truth table can be expressed in terms of the basic Boolean operators AND, OR and
NOT operators.

E.g. using sum of products or product of sums.
The circuits corresponding to those truth tables can be build using AND, OR and NOT
gates, which can be made out of transistors.
In the sum of products approach, the input in each line of a truth table can be expressed in terms of AND’s and NOT’s ( though we will only need the rows that have 1 as an output ).
PHY 201 (Blum) 2

Algebra  Gate

A’ means NOT A high  low  high input low output
PHY 201 (Blum)
Red probe indicator
3

Algebra  Gates

AB means A AND B
PHY 201 (Blum) 4

Algebra  Gates

A+B means A OR B
PHY 201 (Blum) 5

Review: Line by Line
Inputs
Expression
A B
0
0
1
1
0
1
0
1
(Not A) AND
(NOT B)
(Not A) AND B
A AND (NOT
B)
A AND B
A
A
´
´
B
AB ´
AB
´
B
A ´ B ´ is true for the first line and false for the rest
A ´ B is true for the second line and false for the rest
AB ´ is true for the third line and false for the rest
A ´ B ´ is true for the fourth line and false for the rest
This is not yet a truth table. It has no outputs.
PHY 201 (Blum) 6

Writing the expression



To express a truth table as a sum of products
( minterm expression ), take the input lines that correspond to true (high, 1) outputs.
Write the expressions for each of those input lines (as shown on the previous slide).

This step will involve NOTs and ANDs
Then feed all of those expressions into an OR gate.
PHY 201 (Blum) 7

Example 1
1
1
0
1
1
0
0
A
0
0
1
1
0
1
0
1
B
0
PHY 201 (Blum)
1
0
1
0
1
1
0
C
0
1
0
0
0
1
Out
1
0
1
8

Example 1 (Cont.)



A’B’C’ + A’BC’ + AB’C + ABC
The expression one arrives at in this way is known as the sum of products.


You take the product (the AND operation) first to represent a given line.
Then you sum (the OR operation) together those expressions.
It’s also called the minterm expression.
PHY 201 (Blum) 9

Simplifying Boolean algebra expressions



Recall that (A’B’C + A’BC’ + A’BC + AB’C’ + AB’C +
ABC’ + ABC) and (A+B+C) correspond to the same truth table.
Before building a circuit that realizes a Boolean expression, we would like to simplify that expression as much as possible.
Fewer gates means

Fewer transistors



Less space required
Less power required (less heat generated)
More money made
PHY 201 (Blum) 10

A few fundamental theorems




A + 1 = 1
A + 0 = A
A·1 = A
A·0= 0




A + A = A
A·A = A
A + A’ = 1
A·A’ = 0
PHY 201 (Blum) 11

1
1
0
1
1
0
0
A
0
A Trivial Simplification Example
Expressions
0
1
1
0
1
0
1
B
0
1
0
1
0
1
1
0
C
0
0
1
1
0
1
Out
0
0
1 A’ B C’
A’ B C
A B C’
A B C
PHY 201 (Blum) 12

Simplifying a trivial example





A ´ BC ´ + A ´ BC + ABC ´ + ABC
A ´ B (C ´ + C) + AB (C ´ + C)
A ´ B + AB
(A ´ + A) B
B
C+C’ means C OR (NOT C)
In other words, we don’t care about C
PHY 201 (Blum) 13

How simplification occurs


Note that simplification occurs when two terms differ by only one factor.

For example, the terms A ´ BC ´ and A ´ BC have A’B in common and differ only in the C factor.


A’BC’ + A’BC  A’B(C’+C)  A’B
If the two terms differ by more than one factor, there is no simplification

For example, the terms A’BC’ and A’B’C have A’ in common and differ in the B and C factors
A’BC’ + A’B’C  A’(BC’ + B’C)  no simplification
PHY 201 (Blum) 14

PHY 201 (Blum)
1
1
0
1
1
0
0
A
0
Majority Rules Example
0
1
1
0
1
0
1
B
0
1
0
1
0
1
1
0
C
0
Majority
0
0
0
1
1
1
0
1
15

1
1
0
1
1
0
0
A
0
PHY 201 (Blum)
Row Expressions
0
1
1
0
1
0
1
B
0
1
0
1
0
1
1
0
C
0
Row expressions
A’B’C’
A’B’C
A’BC’
A’BC
AB’C’
AB’C
ABC’
ABC
16

Majority rules (sum of products) without simplification

A ´ BC + AB ´ C + ABC ´ + ABC
NOTs
OR
ANDs
PHY 201 (Blum) 17

Majority Rules: Boolean Algebra
Simplification





A ´ BC + AB ´ C + ABC ´ + ABC

The term A’BC can be combined with ABC since they differ by one and only one term



Same for AB’C and ABC
Same for ABC’ and ABC
In logic, ABC = ABC + ABC + ABC
A ´ BC+AB ´ C+ABC ´ +ABC +ABC+ABC
A ´ BC+ABC + AB ´ C+ABC + ABC ´ +ABC
(A ´ +A)BC + A(B ´ +B)C + AB(C ´ +C)
BC + AC + AB
PHY 201 (Blum) 18

Majority rules after simplification
PHY 201 (Blum) 19

Majority Rules Comparison
Gates: 3 NOTs, 4 3-input
ANDs, 1 4-input OR
PHY 201 (Blum)
Gates: 0 NOTs, 3 2-input
ANDs, 1 3-input OR
20

Simplifying made easy



Simplifying Boolean expressions is not always easy.
So we introduce next a method (a
Karnaugh or K map) that is supposed to make simplification more visual.
The first step is to rearrange the inputs into what is called “Gray code” order.

Here, Gray is a guy not a color.
PHY 201 (Blum) 21

Frank Gray in Wikipedia
PHY 201 (Blum) 22

PHY 201 (Blum) 23

Gray code



In addition to binary numbers, there is another way of representing numbers using
1’s and 0’s.

Put another way, there is another useful ordering of the combinations of 1’s and 0’s.
It is not useful for doing arithmetic, but has other purposes.
In Gray code the numbers are ordered such that consecutive numbers differ by one bit only.
PHY 201 (Blum) 24

1
1
0
0
0
0
1
1
PHY 201 (Blum)
Gray code (Cont.)
1
1
1
1
0
0
0
0
0
1
1
0
0
1
1
0
Each row different by one bit only
25

Constructing Gray code ( a.k.a. reflected binary code )
0
1
PHY 201 (Blum) 26

Reflect lower bits and add 0’s in front of the original rows and 1’s in front of the new rows
Add 0’s
0
Lower bits
0
0 1
Add 1’s
1
1
1
0
Reflect lower bits through red line
PHY 201 (Blum) 27

Add
0’s
Add
1’s
Reflect lower bits and 0’s then
1’s in front (again)
1
1
0
0
0
0
1
1
1
1
1
1
0
0
0
0
0
1
1
0
0
1
1
0
Reflect lower bits through red line
PHY 201 (Blum) 28

An important property


In gray-code order, two consecutive rows of a truth table differ by one bit only.
Thus if a truth table is put in gray code order and if two consecutive rows contain a 1, then a simplification of the
Boolean expression is possible.

A term like X + X’ can be factored out.
PHY 201 (Blum) 29

Trivial Example in Gray code
Note:
Gray code ordered inputs
1
1
0
1
1
0
0
A
0
1
0
1
1
0
0
1
B
0
1
1
0
0
0
1
1
C
0
1
0
1
1
0
Out
0
0
1
PHY 201 (Blum) 30

Improving



Some combinations that differ only by a single bit are not in consecutive rows.
Thus there may be a simplification associated with such a combination and we might miss it.
So we put some of the inputs in as columns.
PHY 201 (Blum) 31

1
1
0
1
1
0
0
A
0
PHY 201 (Blum)
Two rows that differ by one bit but are not consecutive
1
0
1
1
0
0
1
B
0
1
1
0
0
0
1
1
C
0
1
0
1
1
0
Out
0
0
1
32

A row-column version
Place the C inputs across the top.
This output corresponds to the input A=0, B=1 and C=0
All inputs are filled in with light blue.
0
1
1
A
0
B\C
0
1
1
0
1
1
0
0
0
1
1
0
1
0
In this version, more inputs differing by one bit only are in adjacent positions.
PHY 201 (Blum) 33

Karnaugh-map

This way of arranging truth tables combined with the rules for simplifying
Boolean expressions goes by the name
Karnaugh map or K map.

Named for Maurice Karnaugh.
PHY 201 (Blum) 34

Maurice Karnaugh
PHY 201 (Blum) 35

The rules


Put the truth table into a form with inputs in Gray code order.
Then one identifies output “blocks” (as large as possible).

A block must be a rectangle containing 1’s and
only 1’s.


The simplification rules require that the number
of 1’s in a block should be a power of 2 (1, 2,
4, 8, …).
However, a given output 1 can belong to more
than one block.
PHY 201 (Blum) 36

Wrapping



There are still cases in which inputs differing by only one bit are not adjacent (e.g. the first and last row).
Imagine that the rows wrap around, so for instance, a block can include the top and bottom rows (without intermediate rows).
Similarly for columns.
PHY 201 (Blum) 37

1
1
1
1
1
1
1
0
0
1
0
0
0
0
0
W
0
PHY 201 (Blum)
0
1
1
1
1
0
0
1
1
0
0
1
0
0
1
X
0
Karnaugh Example
1
0
0
1
1
0
0
1
1
1
1
0
0
1
0
Y
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
Z
0
0
1
1
1
0
1
0
1
0
0
Output
1
0
1
0
0
1
38

Karnaugh Example (Unsimplified
Boolean algebra expression)

WXY’Z + W’XY’Z +
WX’Y’Z’ + W’X’Y’Z’ +
WXYZ’ + WXY’Z’ +
W’XY’Z’ + W’XYZ’
PHY 201 (Blum) 39

Example in Karnaugh (identifying block in gray code truth table)
Z 0 1
0
0
1
1
W X\Y 0
0
PHY 201 (Blum)
1
1
0
0
1
W’X’Y’Z’
1
W’XY’Z’
1
WXY’Z’
1
WX’Y’Z’
0
1
W’XY’Z
1
WXY’Z
0
0
0
0
1
1
0
0
1
0
1
W’XYZ’
1
WXYZ’
0
40

For Yellow Group: W and X inputs change; Y and Z inputs don’t change from zeros. Group represented by Y’Z’
Z 0 1
0
0
1
1
W X\Y 0
0
PHY 201 (Blum)
1
1
0
0
1
W’X’Y’Z’
1
W’XY’Z’
1
WXY’Z’
1
WX’Y’Z’
0
1
W’XY’Z
1
WXY’Z
0
0
0
0
1
1
0
0
1
0
1
W’XYZ’
1
WXYZ’
0
41

For Red Group: W and Z inputs change; X input does not change from 1; Y input does not change from 0. Group represented by XY’
0
0
1
1
Z 0
W X\Y 0
0
PHY 201 (Blum)
1
1
0
1
0
1
W’X’Y’Z’
1
W’XY’Z’
1
WXY’Z’
1
WX’Y’Z’
0
1
W’XY’Z
1
WXY’Z
0
1
1
0
0
0
0
0
1
0
1
W’XYZ’
1
WXYZ’
0
42

For Green group: W and Y inputs change; X input does not change from 1; Z input does not change from 0. Group represented by XZ’
Z 0 1
0
0
1
1
W X\Y 0
0
PHY 201 (Blum)
1
1
0
0
1
W’X’Y’Z’
1
W’XY’Z’
1
WXY’Z’
1
WX’Y’Z’
0
1
W’XY’Z
1
WXY’Z
0
1
1
0
0
0
0
0
1
0
1
W’XYZ’
1
WXYZ’
0
43

Result




Y’Z’ + XY’ + X Z’
A block of size two eliminates one Boolean variable; a block of four eliminates two Boolean variables; and so on.
To find the expression for a block, identify the inputs for that block that don’t change, AND them together, that’s your expression for the block.
Obtain an expression for each block and OR them together. Every 1 must belong to at least one block
(even if it is a block onto itself).
PHY 201 (Blum) 44

From Binary order to Gray code order
PHY 201 (Blum) 45

From Binary order to Gray code order
PHY 201 (Blum) 46

Online References



 http://www.facstaff.bucknell.edu/masta scu/eLessonsHTML/Logic/Logic3.html http://www.cs.usm.maine.edu/~welty/k arnaugh.htm http://en.wikipedia.org/wiki/Frank_Gray
_(researcher) http://en.wikipedia.org/wiki/Maurice_Ka rnaugh
PHY 201 (Blum) 47

Combinations of Resistors
Series, Parallel
PHY 202 (Blum) 48

Analyzing a combination of resistors circuit



Look for resistors which are in series (the same current passing through one must pass through the other) and replace them with the equivalent resistance (R eq
= R
1
+ R
2
).
Look for resistors which are in parallel (both the tops and bottoms are connected by wire and only wire) and replace them with the equivalent resistance (1/R eq
= 1/R
1
+ 1/R
2
).
Repeat as much as possible.
PHY 202 (Blum) 49

Look for series combinations
R eq
=3k

R eq
=3.6 k

50 PHY 202 (Blum)

Look for parallel combinations
R eq
= 1.8947 k

R eq
= 1.1244 k

51 PHY 202 (Blum)

Look for series combinations
PHY 202 (Blum)
R eq
= 6.0191 k

52

Look for parallel combinations
R eq
= 2.1314 k

53 PHY 202 (Blum)

Look for series combinations
R eq
= 5.1314 k

PHY 202 (Blum) 54

Equivalent Resistance
PHY 202 (Blum)
I = V/R = (5 V)/(5.1314 k

) = 0.9744 mA
55

Backwards 1
V= (3)(.9744) = 2.9232
PHY 202 (Blum)
V= (2.1314)(.9744) = 2.0768
56

Backwards 2
PHY 202 (Blum)
V = 2.0768=I (3.3)
I=0.629mA
V = 2.0768=I (6.0191)
I=0.345mA
57

Backwards 3
V=(.345)(3)=1.035
V=(.345)(1.8947)
=0.654
V=(.345)(1.1244)=0.388
58 PHY 202 (Blum)