To minimize the Boolean function $Y(A, B, C, D)$ using the Quine-McCluskey method, we'll proceed with the following steps:

Problem Setup:

We are given the minterms for the function: $Y(A, B, C, D) = \sum m(0, 3, 4, 5, 6, 7, 11, 15)$ where $m$ stands for the minterms. This represents the Boolean function in terms of minterms.

Step 1: Binary Representation of Minterms

Each minterm will be converted to its 4-bit binary equivalent:

• $m(0)$ → 0000
• $m(3)$ → 0011
• $m(4)$ → 0100
• $m(5)$ → 0101
• $m(6)$ → 0110
• $m(7)$ → 0111
• $m(11)$ → 1011
• $m(15)$ → 1111

Step 2: Group Minterms by the Number of 1’s

Now, we group these minterms based on the number of 1’s in their binary representation:

• 0 ones: $0000$ (minterm 0)
• 2 ones: $0100, 0011$ (minterms 4, 3)
• 3 ones: $0101, 0110, 0111, 1011$ (minterms 5, 6, 7, 11)
• 4 ones: $1111$ (minterm 15)

Step 3: Prime Implicant Table (Pairwise Combination)

We now begin the process of combining the minterms where they differ by only one bit.

1. Combine 0000 (m0) with 0100 (m4):

   • Result: 0–00 (combined m0 and m4)

2. Combine 0011 (m3) with 0111 (m7):

   • Result: 0–11 (combined m3 and m7)

3. Combine 0101 (m5) with 0111 (m7):

   • Result: 01–1 (combined m5 and m7)

4. Combine 0110 (m6) with 0111 (m7):

   • Result: 011– (combined m6 and m7)

No further combinations are possible.

Step 4: Prime Implicant Chart

From the results of step 3, the prime implicants are:

• $0–00$ (covers $m0, m4$)
• $0–11$ (covers $m3, m7$)
• $01–1$ (covers $m5, m7$)
• $011–$ (covers $m6, m7$)
• $1011$ (covers $m11$)
• $1111$ (covers $m15$)

Step 5: Essential Prime Implicants

Using a prime implicant chart, we determine which minterms are covered by each implicant:

• $0–00$ → $m0, m4$
• $0–11$ → $m3, m7$
• $01–1$ → $m5, m7$
• $011–$ → $m6, m7$
• $1011$ → $m11$
• $1111$ → $m15$

Coverage:

• Minterms $m0$ and $m4$ are covered by $0–00$.
• Minterms $m3$ and $m7$ are covered by $0–11$ and $01–1$.
• Minterm $m11$ is covered only by $1011$.
• Minterm $m15$ is covered only by $1111$.

The essential prime implicants are:

1. $0–00$ (covers $m0, m4$)
2. $1011$ (covers $m11$)
3. $1111$ (covers $m15$)

Step 6: Final Simplified Expression

Now, we convert the prime implicants back to their Boolean forms:

• $0–00$ → $A'C'$ (since $A = 0$ and $C = 0$, while $B$ and $D$ are irrelevant)
• $0–11$ → $A'BD$ (since $A = 0$, $B = 1$, and $D = 1$)
• $01–1$ → $A'B'D'$ (since $A = 0$, $B = 1$, and $D = 0$)
• $1011$ → $ABCD'$
• $1111$ → $ABCD$

Thus, the simplified Boolean expression is:

$Y(A, B, C, D) = A'C' + A'BD + ABCD' + ABCD$

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Would you like any further clarification on this method or more details?

5 Related Questions:

1. How does the Quine-McCluskey method differ from Karnaugh Maps?
2. Can we use the Quine-McCluskey method for functions with more variables?
3. What are essential and non-essential prime implicants?
4. How can we optimize larger Boolean expressions efficiently?
5. How do don't-care conditions affect minimization in Quine-McCluskey?

Tip:

When minimizing Boolean functions, always ensure to cover all the minterms while avoiding redundant implicants.