How to use K-map to prove a boolean equation?

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SUMMARY

The discussion focuses on using Karnaugh maps (K-maps) to simplify the boolean equation B + AB'C'D + AB'CD = B + AD. A K-map is defined as a table that organizes combinations of boolean variables A, B, C, and D, rather than a graph. Participants clarify that to fill in the K-map, one must identify where the variables equal 1 based on the given expression, ensuring that all relevant combinations are accurately represented in the table.

PREREQUISITES
  • Understanding of boolean algebra and expressions
  • Familiarity with Karnaugh maps (K-maps)
  • Knowledge of boolean variables and their combinations
  • Basic skills in logic simplification techniques
NEXT STEPS
  • Study the process of filling K-maps for various boolean expressions
  • Learn about the implications of boolean simplification in digital circuit design
  • Explore advanced K-map techniques for more than four variables
  • Investigate software tools for automating K-map simplifications
USEFUL FOR

Computer engineering students, digital circuit designers, and anyone involved in logic design and optimization will benefit from this discussion.

FallArk
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This is actually a computer engineering problem:
How to use k-map to show that B + AB’C’D + AB’CD = B + AD?
Isn't K-map a graph that shows combinations of input A, B, C and D?
What is B then?
 
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FallArk said:
This is actually a computer engineering problem:
How to use k-map to show that B + AB’C’D + AB’CD = B + AD?
Isn't K-map a graph that shows combinations of input A, B, C and D?
What is B then?

Hey FallArk! ;)

A k-map or Karnaugh map is not a graph - it's a table that looks like:
K-map_6%2C8%2C9%2C10%2C11%2C12%2C13%2C14_anti-race.svg


It's a method to simplify boolean expressions of a number of boolean variables (4 variables in the example).

B is just one of the 4 boolean variables.
The AB at the top represents all possible combinations of the 2 boolean variables A and B.
And the CD at the left represents all possible combinations of C and D.
If we make such tables for both the left hand side and the right hand side, we should find that the tables are identical.
 
I like Serena said:
Hey FallArk! ;)

A k-map or Karnaugh map is not a graph - it's a table that looks like:It's a method to simplify boolean expressions of a number of boolean variables (4 variables in the example).

B is just one of the 4 boolean variables.
The AB at the top represents all possible combinations of the 2 boolean variables A and B.
And the CD at the left represents all possible combinations of C and D.
If we make such tables for both the left hand side and the right hand side, we should find that the tables are identical.
Sorry for not wording it correctly, what I was trying to say is that how can I tell where to fill in the 1s and 0s in the map according to what is given to me?
 
FallArk said:
Sorry for not wording it correctly, what I was trying to say is that how can I tell where to fill in the 1s and 0s in the map according to what is given to me?

The rightmost 2 columns correspond to $A=1$, or just $A$.
The center columns correspond to $B$.

We have an expression that begins with $B + ...$.
That means that it's true if $B$ is.
So we can fill in all fields corresponding to $B=1$.

\begin{array}{c|c|c|c|}
{}_{\small{CD}}\backslash{}^{\small{AB}}&\tiny{00}&\tiny{01}&\tiny{11}&\tiny{10} \\
\hline
\tiny{00}&\ & 1&1&\ \\
\hline
\tiny{01}&\ & 1&1&\ \\
\hline
\tiny{11}&\ & 1&1&\ \\
\hline
\tiny{10}&\ & 1&1&\ \\
\hline
\end{array}

Now we need to add 1's for the other parts of the expression.

For instance $AD$ is represented by:
\begin{array}{c|c|c|c|}
{}_{\small{CD}}\backslash{}^{\small{AB}}&\tiny{00}&\tiny{01}&\tiny{11}&\tiny{10} \\
\hline
\tiny{00}& & & & \\
\hline
\tiny{01}& & & 1& 1\\
\hline
\tiny{11}& & & 1&1 \\
\hline
\tiny{10}& & & & \\
\hline
\end{array}

So $B+AD$ is:
\begin{array}{c|c|c|c|}
{}_{\small{CD}}\backslash{}^{\small{AB}}&\tiny{00}&\tiny{01}&\tiny{11}&\tiny{10} \\
\hline
\tiny{00}&0 &1 &1 &0 \\
\hline
\tiny{01}&0 &1 &1 &1\\
\hline
\tiny{11}&0 &1 &1 &1 \\
\hline
\tiny{10}&0 &1 &1 &0 \\
\hline
\end{array}
 
I like Serena said:
The rightmost 2 columns correspond to $A=1$, or just $A$.
The center columns correspond to $B$.

We have an expression that begins with $B + ...$.
That means that it's true if $B$ is.
So we can fill in all fields corresponding to $B=1$.

\begin{array}{c|c|c|c|}
{}_{\small{CD}}\backslash{}^{\small{AB}}&\tiny{00}&\tiny{01}&\tiny{11}&\tiny{10} \\
\hline
\tiny{00}&\ & 1&1&\ \\
\hline
\tiny{01}&\ & 1&1&\ \\
\hline
\tiny{11}&\ & 1&1&\ \\
\hline
\tiny{10}&\ & 1&1&\ \\
\hline
\end{array}

Now we need to add 1's for the other parts of the expression.

For instance $AD$ is represented by:
\begin{array}{c|c|c|c|}
{}_{\small{CD}}\backslash{}^{\small{AB}}&\tiny{00}&\tiny{01}&\tiny{11}&\tiny{10} \\
\hline
\tiny{00}& & & & \\
\hline
\tiny{01}& & & 1& 1\\
\hline
\tiny{11}& & & 1&1 \\
\hline
\tiny{10}& & & & \\
\hline
\end{array}

So $B+AD$ is:
\begin{array}{c|c|c|c|}
{}_{\small{CD}}\backslash{}^{\small{AB}}&\tiny{00}&\tiny{01}&\tiny{11}&\tiny{10} \\
\hline
\tiny{00}&0 &1 &1 &0 \\
\hline
\tiny{01}&0 &1 &1 &1\\
\hline
\tiny{11}&0 &1 &1 &1 \\
\hline
\tiny{10}&0 &1 &1 &0 \\
\hline
\end{array}

I get it now! So I just have to make sure all the 1's I fill in corresponds to what it given. Thanks!
 

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