Simplify the following expression using 𝐾 −map;
i. 𝑓(𝑎, 𝑏, 𝑐, 𝑑) = ∑𝑚(0, 1, 2 ,3, 4, 6, 8, 9, 10, 11, 12, 14).
ii. F(A, B, C, D) = Σm(0, 1, 2, 5, 7, 8, 9, 10, 13, 15)

To simplify the given Boolean expressions using Karnaugh maps (K-maps), we'll proceed step by step for both parts:

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### i. \( f(a, b, c, d) = \sum m(0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 14) \)

We will construct a 4-variable K-map for this function. The variables are \(a\), \(b\), \(c\), and \(d\), so we will create a 4x4 grid to represent all the combinations of these variables. The minterms provided represent where the output is 1.

#### K-map for \( f(a, b, c, d) \):

| ab\cd | 00  | 01  | 11  | 10  |
|------|-----|-----|-----|-----|
| **00** | 1 (m0)  | 1 (m1)  | 1 (m3)  | 1 (m2)  |
| **01** | 1 (m4)  | 0  | 1 (m6)  | 1 (m5)  |
| **11** | 1 (m12) | 1 (m13) | 1 (m15) | 0  |
| **10** | 1 (m8)  | 1 (m9)  | 1 (m11) | 1 (m10) |

#### Step-by-Step Grouping:

Now, we'll group the adjacent ones (1's) in pairs, quads, or octets. Let's look for groups of 1's in the K-map.

- **Group 1:** A quad of 1's in the top-left corner: \( m0, m1, m2, m3 \) → Simplifies to \( \overline{a} \)
- **Group 2:** A quad of 1's in the second row: \( m4, m6, m8, m10 \) → Simplifies to \( \overline{c} \)
- **Group 3:** A pair of 1's in the third column: \( m8, m9 \) → Simplifies to \( \overline{b} \cdot d \)
- **Group 4:** A pair of 1's in the fourth row: \( m12, m14 \) → Simplifies to \( b \cdot \overline{c} \)

#### Final Simplified Expression:
Combining all the groups, the simplified Boolean expression for \( f(a, b, c, d) \) is:
\[
f(a, b, c, d) = \overline{a} + \overline{c} + \overline{b} \cdot d + b \cdot \overline{c}
\]

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### ii. \( F(A, B, C, D) = \sum m(0, 1, 2, 5, 7, 8, 9, 10, 13, 15) \)

Now, let's construct a 4-variable K-map for the second function \( F(A, B, C, D) \), based on the minterms provided.

#### K-map for \( F(A, B, C, D) \):

| AB\CD | 00  | 01  | 11  | 10  |
|------|-----|-----|-----|-----|
| **00** | 1 (m0)  | 1 (m1)  | 1 (m3)  | 1 (m2)  |
| **01** | 0  | 1 (m5)  | 1 (m7)  | 0  |
| **11** | 0  | 1 (m9)  | 0  | 1 (m8)  |
| **10** | 1 (m13) | 0  | 1 (m15) | 0  |

#### Step-by-Step Grouping:

Now, let's group the 1's.

- **Group 1:** A quad in the top-left corner: \( m0, m1, m2, m3 \) → Simplifies to \( \overline{A} \)
- **Group 2:** A pair of 1's in the second row: \( m5, m7 \) → Simplifies to \( \overline{B} \cdot D \)
- **Group 3:** A pair of 1's in the third column: \( m8, m9 \) → Simplifies to \( C \cdot \overline{D} \)
- **Group 4:** A pair of 1's in the last row: \( m13, m15 \) → Simplifies to \( A \cdot \overline{C} \)

#### Final Simplified Expression:
Thus, the simplified Boolean expression for \( F(A, B, C, D) \) is:
\[
F(A, B, C, D) = \overline{A} + \overline{B} \cdot D + C \cdot \overline{D} + A \cdot \overline{C}
\]

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Would you like more details on the K-map grouping or need further clarification? Here are some related questions that may help deepen your understanding:

1. What are the key differences between using a 3-variable and a 4-variable K-map?
2. How do you identify and combine adjacent minterms in a K-map?
3. Can you explain how to simplify Boolean expressions using Boolean algebra directly as an alternative to K-maps?
4. What are the advantages of using K-maps for Boolean simplification?
5. How do you handle “don’t care” conditions in K-maps, and how can they help in simplification?

**Tip:** When working with K-maps, always start by marking the cells for minterms that correspond to 1's, then systematically group them into the largest possible groups (pairs, quads, or octets) to minimize the expression efficiently.

Simplify the following expression using K-map: i. f(a, b, c, d) = Σm(0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 14). ii. F(A, B, C, D) = Σm(0, 1, 2, 5, 7, 8, 9, 10, 13, 15)

Simplify the Boolean functions f(a, b, c, d) = Σm(0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 14) and F(A, B, C, D) = Σm(0, 1, 2, 5, 7, 8, 9, 10, 13, 15) using K-maps. Learn step-by-step logic simplification techniques for minimizing expressions.

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