PHY 201 (Blum)1 Karnaugh Maps References: Chapters 4 and 5 in Digital Principles (Tokheim) Chapter 3...
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Transcript of PHY 201 (Blum)1 Karnaugh Maps References: Chapters 4 and 5 in Digital Principles (Tokheim) Chapter 3...
PHY 201 (Blum) 1
Karnaugh Maps
References:Chapters 4 and 5 in Digital Principles (Tokheim)Chapter 3 in Introduction to Digital Systems (Palmer and Perlman)
PHY 201 (Blum) 2
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).
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Algebra Gate A’ means NOT A
high
low
high input
low output
Red probe indicator
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Algebra Gates AB means A AND B
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Algebra Gates A+B means A OR B
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Review: Line by Line
InputsExpression
A B
0 0 (Not A) AND (NOT B) A´B´ A´B´ is true for the first
line and false for the rest
0 1 (Not A) AND B A´B A´B is true for the second
line and false for the rest
1 0 A AND (NOT B) AB´ AB´ is true for the third
line and false for the rest
1 1 A AND B AB 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) 7
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) 8
Example 1
A B C Out
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
PHY 201 (Blum) 9
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) 10
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
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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
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A Trivial Simplification Example A B C Out Expressions
0 0 0 0
0 0 1 0
0 1 0 1 A’ B C’
0 1 1 1 A’ B C
1 0 0 0
1 0 1 0
1 1 0 1 A B C’
1 1 1 1 A B C
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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
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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) 15
Majority Rules Example
A B C Majority
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
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Row Expressions
A B C Row expressions
0 0 0 A’B’C’
0 0 1 A’B’C
0 1 0 A’BC’
0 1 1 A’BC
1 0 0 AB’C’
1 0 1 AB’C
1 1 0 ABC’
1 1 1 ABC
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Majority rules (sum of products) without simplification
A´BC + AB´C + ABC´ + ABCNOTs
ANDs
OR
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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
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Majority rules after simplification
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Majority Rules Comparison
Gates: 3 NOTs, 4 3-input ANDs, 1 4-input OR
Gates: 0 NOTs, 3 2-input ANDs, 1 3-input OR
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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.
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Frank Gray in Wikipedia
PHY 201 (Blum) 23
PHY 201 (Blum) 24
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) 25
Gray code (Cont.)
0 0 0
0 0 1
0 1 1
0 1 0
1 1 0
1 1 1
1 0 1
1 0 0
Each row different by one bit only
PHY 201 (Blum) 26
Constructing Gray code (a.k.a. reflected binary code)
0
1
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Reflect lower bits and add 0’s in front of the original rows and 1’s in front of the new
rows
0 0
0 1
1 1
1 0
Lower bits
Reflect lower bits through red line
Add 0’s
Add 1’s
PHY 201 (Blum) 28
Reflect lower bits and 0’s then 1’s in front (again)
0 0 0
0 0 1
0 1 1
0 1 0
1 1 0
1 1 1
1 0 1
1 0 0
Reflect lower bits through red line
Add 0’s
Add 1’s
PHY 201 (Blum) 29
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.
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Trivial Example in Gray code
A B C Out
0 0 0 0
0 0 1 0
0 1 1 1
0 1 0 1
1 1 0 1
1 1 1 1
1 0 1 0
1 0 0 0
Note: Gray code ordered inputs
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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.
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Two rows that differ by one bit but are not consecutive
A B C Out
0 0 0 0
0 0 1 0
0 1 1 1
0 1 0 1
1 1 0 1
1 1 1 1
1 0 1 0
1 0 0 0
PHY 201 (Blum) 33
A row-column version
A B\C 0 1
0 0 0 0
0 1 1 1
1 1 1 1
1 0 0 0
Place the C inputs across the top.
All inputs are filled in with light blue.
In this version, more inputs differing by one bit only are in adjacent positions.
This output corresponds to the input A=0, B=1 and C=0
PHY 201 (Blum) 34
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) 35
Maurice Karnaugh
PHY 201 (Blum) 36
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) 37
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) 38
W X Y Z Output
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 1
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
Karnaugh Example
PHY 201 (Blum) 39
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) 40
Example in Karnaugh (identifying block in gray code truth table)
Z 0 1 1 0
W X\Y 0 0 1 1
0 0 1W’X’Y’Z’
0 0 0
0 1 1W’XY’Z’
1W’XY’Z
0 1W’XYZ’
1 1 1WXY’Z’
1WXY’Z
0 1WXYZ’
1 0 1WX’Y’Z’
0 0 0
PHY 201 (Blum) 41
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 1 0
W X\Y 0 0 1 1
0 0 1W’X’Y’Z’
0 0 0
0 1 1W’XY’Z’
1W’XY’Z
0 1W’XYZ’
1 1 1WXY’Z’
1WXY’Z
0 1WXYZ’
1 0 1WX’Y’Z’
0 0 0
PHY 201 (Blum) 42
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’ Z 0 1 1 0
W X\Y 0 0 1 1
0 0 1W’X’Y’Z’
0 0 0
0 1 1W’XY’Z’
1W’XY’Z
0 1W’XYZ’
1 1 1WXY’Z’
1WXY’Z
0 1WXYZ’
1 0 1WX’Y’Z’
0 0 0
PHY 201 (Blum) 43
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 1 0
W X\Y 0 0 1 1
0 0 1W’X’Y’Z’
0 0 0
0 1 1W’XY’Z’
1W’XY’Z
0 1W’XYZ’
1 1 1WXY’Z’
1WXY’Z
0 1WXYZ’
1 0 1WX’Y’Z’
0 0 0
PHY 201 (Blum) 44
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) 45
From Binary order to Gray code order
PHY 201 (Blum) 46
From Binary order to Gray code order
PHY 201 (Blum) 47
Online References http://www.facstaff.bucknell.edu/
mastascu/eLessonsHTML/Logic/Logic3.html
http://www.cs.usm.maine.edu/~welty/karnaugh.htm
http://en.wikipedia.org/wiki/Frank_Gray_(researcher)
http://en.wikipedia.org/wiki/Maurice_Karnaugh