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Karnaugh Maps

References:Chapters 4 and 5 in Digital Principles (Tokheim)Chapter 3 in Introduction to Digital Systems (Palmer and Perlman)

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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 LineInputs Expression A B0 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.

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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.

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Example 1A B C Out0 0 0 10 0 1 00 1 0 10 1 1 01 0 0 01 0 1 11 1 0 01 1 1 1

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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.

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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 Expressions0 0 0 00 0 1 00 1 0 1 A’ B C’0 1 1 1 A’ B C1 0 0 01 0 1 01 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

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Majority Rules Example

A B C Majority0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1

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Row ExpressionsA B C Row expressions0 0 0 A’B’C’0 0 1 A’B’C0 1 0 A’BC’0 1 1 A’BC1 0 0 AB’C’1 0 1 AB’C1 1 0 ABC’1 1 1 ABC

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Majority rules (sum of products) without simplification A´BC + AB´C + ABC´ + ABC

NOTs

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

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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.

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Gray code (Cont.) 0 0 00 0 10 1 10 1 01 1 01 1 11 0 11 0 0

Each row different by one bit only

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Constructing Gray code (a.k.a. reflected binary code)

01

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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 00 11 11 0

Lower bits

Reflect lower bits through red line

Add 0’s Add 1’s

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Reflect lower bits and 0’s then 1’s in front (again)

0 0 00 0 10 1 10 1 01 1 01 1 11 0 11 0 0

Reflect lower bits through red line

Add 0’s

Add 1’s

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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 Out0 0 0 00 0 1 00 1 1 10 1 0 11 1 0 11 1 1 11 0 1 01 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 Out0 0 0 00 0 1 00 1 1 10 1 0 11 1 0 11 1 1 11 0 1 01 0 0 0

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A row-column version

A B\C 0 10 0 0 00 1 1 11 1 1 11 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

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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.

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Maurice Karnaugh

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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.

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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.

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W X Y Z Output0 0 0 0 10 0 0 1 00 0 1 0 00 0 1 1 00 1 0 0 10 1 0 1 10 1 1 0 10 1 1 1 01 0 0 0 11 0 0 1 01 0 1 0 01 0 1 1 01 1 0 0 11 1 0 1 11 1 1 0 11 1 1 1 0

Karnaugh Example

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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’

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Example in Karnaugh (identifying block in gray code truth table)

Z 0 1 1 0W 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

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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 0W 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

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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 0W 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

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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 0W 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

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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).

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From Binary order to Gray code order

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From Binary order to Gray code order

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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