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.