My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum

LinkBack Thread Tools Display Modes
July 8th, 2015, 06:07 AM   #1
Joined: Jul 2015
From: London

Posts: 6
Thanks: 0

Truth table AND THE Karnaugh Map

I have answered regarding I'm not sure if it is correct. I have done the truth table and done the Boolean expression. I need someone to check my answers and correct me if I made a mistake. Questions and scenario:

The food dispenser will become active when the blue and green lights are on
(irrespective of whether the red light is on or off); when the blue light is off and the green light is on (irrespective of whether the red light is on or off); and when only the blue light is on. All other combinations will not provide food.

(a) Draw a truth table representing each light as a Boolean variable, showing all combinations of the variables and the output A which indicates whether the dispenser is active or not.
(b) Using the truth table, write down a Boolean expression that will activate the dispenser when appropriate.
(c) Use a Karnaugh Map to find the simplest representation of the expression.

My answers:

a) for the truth table click on the attachment

b) The Boolean expression that I have written down were:
Output = r’gb + r’gb’ + r’g’b + rgb + rgb’

c) I'm struggling with the Karnaugh map. I need someone to explain it to me and show me how to do it.
Attached Files
File Type: pdf Truth table.pdf (3.9 KB, 1 views)
assu94 is offline  
July 9th, 2015, 05:41 PM   #2
Senior Member
Joined: Dec 2007

Posts: 687
Thanks: 47

I took a look at it, and it seems that you simply arrange a matrix $\displaystyle m$x$\displaystyle n$ with pairs of literals if m and n are both even, or with a single literal and its negation, say, in the column if n is odd. Example:

you have a formula with A, B and C being literals, then

\begin{array}{r | r | c | c |}
&&C&\sim\!C \\
\hline \\
&&0&1 \\
\hline \\
AB&11&& \\
\hline \\
\sim\!AB&01&& \\
\hline \\
A\sim\!B&10&& \\
\hline \\

where $\displaystyle \sim\!AB$ means $\displaystyle \neg A\wedge B$, and the binary are the respective values. If you'd have got A, B, C and D, then your matrix would be:

\begin{array}{r | r | c | c | c | c |}
&&CD&\sim\!CD&C\sim\!D&\sim\!C\sim\!D \\
\hline \\
&&11&01&10&00 \\
\hline \\
AB&11&&&& \\
\hline \\
\sim\!AB&01&&&& \\
\hline \\
A\sim\!B&10&&&& \\
\hline \\

It is really just a scheme to ease visualization.
al-mahed is offline  

  My Math Forum > College Math Forum > Abstract Algebra

karnaugh, map, table, truth

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
karnaugh map goose78 Applied Math 1 October 25th, 2014 05:57 AM
Making a truth table (logics) Janbaas Applied Math 1 March 18th, 2014 06:40 PM
Given truth table with 3 variables, find boolean expression unwisetome3 Applied Math 2 February 3rd, 2014 04:22 PM
Truth Table,validity... john616 Applied Math 2 February 19th, 2012 01:33 PM
Truth Table problem Akhanoth Applied Math 5 October 29th, 2011 08:38 AM

Copyright © 2019 My Math Forum. All rights reserved.