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

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The discussion focuses on using Karnaugh maps (k-maps) to simplify the Boolean expression B + AB’C’D + AB’CD and demonstrate its equivalence to B + AD. A k-map is clarified as a table, not a graph, that organizes combinations of four Boolean variables: A, B, C, and D. The participants explain how to fill in the k-map by identifying where to place 1s and 0s based on the given expression. Specifically, they highlight that since the expression starts with B, all fields corresponding to B=1 should be filled. Additional 1s are added for the terms involving AD. The final k-map representation confirms the equivalence of the two expressions, emphasizing the importance of accurately placing the values according to the Boolean logic involved.
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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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