To map the given Boolean expression on a Karnaugh map (K-map), we need to simplify the expression first and then plot it on the K-map. The given expression is:

$\overline{A} + A\overline{B} + A + A\overline{B}\overline{C}\overline{C}$

Step 1: Simplify the Expression

1. Combine like terms:

   $\overline{A} + A = 1 \quad (\text{Since A + }\overline{\text{A}} = 1)$

   Now the expression reduces to:

   $1 + A\overline{B} + A\overline{B}\overline{C}\overline{C}$

2. Since $1$ is in the expression, it means the entire expression will always be true regardless of the other terms. Therefore, the simplified expression is:

   $1$

Step 2: Plot on the K-Map

Since the simplified expression is $1$, it means that all cells in the K-map will be filled with 1. Here is the 3-variable K-map representation:

\hline 00 & 01 & 11 & 10 \\ \hline 0 & 1 & 1 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 \\ \hline \end{array}$$ ### Explanation Since the simplified expression is always true, every cell in the K-map is filled with 1. --- Do you want further details or have any questions? Here are some related questions to expand your understanding: 1. What are the basics of Karnaugh maps and their use? 2. How to minimize Boolean expressions using K-map? 3. How do you plot an expression directly onto a K-map without simplification? 4. How to solve a Boolean expression using different methods, like Quine-McCluskey? 5. What is the significance of using Karnaugh maps in digital logic design? **Tip:** Always try to simplify the Boolean expression before mapping it on a K-map for easier representation.