Let's solve the given Boolean function using **Petrick’s method** step by step: ### **Given:** \( f(a, b, c, d) = \sum m(0, 1, 3, 4, 6, 7, 8, 9, 14, 15) \) --- ## **Step 1: List Minterms in Binary** | Decimal | Binary | a | b | c | d | |---------|---------|---|---|---|---| | 0 | 0000 | 0 | 0 | 0 | 0 | | 1 | 0001 | 0 | 0 | 0 | 1 | | 3 | 0011 | 0 | 0 | 1 | 1 | | 4 | 0100 | 0 | 1 | 0 | 0 | | 6 | 0110 | 0 | 1 | 1 | 0 | | 7 | 0111 | 0 | 1 | 1 | 1 | | 8 | 1000 | 1 | 0 | 0 | 0 | | 9 | 1001 | 1 | 0 | 0 | 1 | | 14 | 1110 | 1 | 1 | 1 | 0 | | 15 | 1111 | 1 | 1 | 1 | 1 | --- ## **Step 2: Find All Prime Implicants** Let's use a Karnaugh map (K-map) for simplification: ``` cd 00 01 11 10 ab 00 0 1 3 - 01 4 - 7 6 11 - - 15 14 10 8 9 - - ``` ### **Groupings (Prime Implicants):** - **Group 1:** (0, 1, 4, 5) — Not possible, as 5 is not given. - **Group 2:** (6, 7) — Both present. - **Group 3:** (14, 15) — Both present. - **Group 4:** (8, 9) — Both present. - **Group 5:** (0, 4, 8) — (All present) - **Group 6:** (1, 3) — Both present. - **Group 7:** (3, 7) — Both present. - **Group 8:** (0, 1) — Both present. Let's write all possible prime implicants (covering all minterms): 1. (0, 1, 8, 9): \( b'c'd' \) (minterms: 0, 1, 8, 9) 2. (0, 4, 8): \( a'c'd' \) (minterms: 0, 4, 8) 3. (6, 7, 14, 15): \( bcd \) (minterms: 7, 15); \( bc d' \) (minterms: 6, 14) 4. (3, 7, 15): \( a'b'd \) (minterms: 3, 7, 15) 5. (1, 3): \( a'b'c'd \) (minterms: 1, 3) 6. (4, 6): \( a'bc'd' \) (minterms: 4, 6) 7. (8, 9): \( ab'c'd' \) (minterms: 8, 9) 8. (14, 15): \( abcd' \) (minterms: 14, 15) But to use Petrick's method, let's move to a systematic chart. --- ## **Step 3: Prime Implicant Chart** Let’s denote each prime implicant as: - \( P_1 \): \( b'c'd' \) — 0, 1, 8, 9 - \( P_2 \): \( a'c'd' \) — 0, 4, 8 - \( P_3 \): \( bc d' \) — 6, 14 - \( P_4 \): \( bcd \) — 7, 15 - \( P_5 \): \( a'b'd \) — 3, 7, 15 - \( P_6 \): \( abcd' \) — 14, 15 Let’s construct the prime implicant chart: | Minterm | \( P_1 \) | \( P_2 \) | \( P_3 \) | \( P_4 \) | \( P_5 \) | \( P_6 \) | |---------|-----------|-----------|-----------|-----------|-----------|-----------| | 0 | X | X | | | | | | 1 | X | | | | | | | 3 | | | | | X | | | 4 | | X | | | | | | 6 | | | X | | | | | 7 | | | | X | X | | | 8 | X | X | | | | | | 9 | X | | | | | | | 14 | | | X | | | X | | 15 | | | | X | X | X | --- ## **Step 4: Essential Prime Implicants** - Minterm 1 is covered only by \( P_1 \) → **\( P_1 \) is essential** - Minterm 3 is covered only by \( P_5 \) → **\( P_5 \) is essential** - Minterm 4 is covered only by \( P_2 \) → **\( P_2 \) is essential** - Minterm 6 is covered only by \( P_3 \) → **\( P_3 \) is essential** - Minterm 9 is covered only by \( P_1 \) → already covered - Minterm 14 is covered by \( P_3 \), \( P_6 \) - Minterm 15 is covered by \( P_4 \), \( P_5 \), \( P_6 \) - Minterm 7 is covered by \( P_4 \), \( P_5 \) So after selecting all essential prime implicants (\( P_1, P_2, P_3, P_5 \)), let's see what's left: - 7: \( P_4 \), \( P_5 \) (but \( P_5 \) is selected, so covered) - 14: \( P_3, P_6 \) (but \( P_3 \) is selected, so covered) - 15: \( P_4, P_5, P_6 \) (but \( P_5 \) is selected, so covered) So all minterms are covered by the essential prime implicants. --- ## **Step 5: Final Minimal Solution** So the minimal solution is: \[ f(a, b, c, d) = P_1 + P_2 + P_3 + P_5 \] Substitute back: - \( P_1: b'c'd' \) - \( P_2: a'c'd' \) - \( P_3: bc d' \) - \( P_5: a'b'd \) \[ \boxed{ f(a, b, c, d) = b'c'd' + a'c'd' + bc d' + a'b'd } \] --- ## **Step 6: Are There Other Minimal Solutions?** Try removing any implicant and see if all minterms are still covered: - If you remove any one, some minterms are left uncovered. **Thus, this is the unique minimal solution.** --- ## **Final Answer:** \[ \boxed{ f(a, b, c, d) = b'c'd' + a'c'd' + bc d' + a'b'd } \] This is the minimum solution using Petrick’s method.

Certainly! Let's approach this problem step-by-step, providing detailed calculations and reasoning for each part. --- ## **Given:** Boolean function: \[ f(a, b, c, d) = \sum m(0, 1, 3, 4, 6, 7, 8, 9, 14, 15) \] which means the minterms where the function outputs 1 are the decimal numbers listed. --- ## **Step 1: Convert minterms to binary** | Decimal | Binary | a | b | c | d | |---------|---------|---|---|---|---| | 0 | 0000 | 0 | 0 | 0 | 0 | | 1 | 0001 | 0 | 0 | 0 | 1 | | 3 | 0011 | 0 | 0 | 1 | 1 | | 4 | 0100 | 0 | 1 | 0 | 0 | | 6 | 0110 | 0 | 1 | 1 | 0 | | 7 | 0111 | 0 | 1 | 1 | 1 | | 8 | 1000 | 1 | 0 | 0 | 0 | | 9 | 1001 | 1 | 0 | 0 | 1 | | 14 | 1110 | 1 | 1 | 1 | 0 | | 15 | 1111 | 1 | 1 | 1 | 1 | --- ## **Step 2: Find Prime Implicants** ### **Method:** Use Karnaugh Map (K-map) to group minterms and find prime implicants. ### **Construct the 4-variable K-map:** Label the rows by \(a, b\) and columns by \(c, d\): | | 00 | 01 | 11 | 10 | |-----|-----|-----|-----|-----| | 00 | 0 | 1 | 3 | - | | 01 | 4 | - | 7 | 6 | | 11 | - | - | 15 | 14 | | 10 | 8 | 9 | - | - | ### **Filling minterms:** - 0 at (00,00) - 1 at (00,01) - 3 at (00,11) - 4 at (01,00) - 6 at (01,10) - 7 at (01,11) - 8 at (10,00) - 9 at (10,01) - 14 at (11,10) - 15 at (11,11) --- ## **Step 3: Find prime implicants via grouping** ### **Identify groups:** - **Group 1:** Minterms 0, 1, 8, 9 — form a 4-group - **Group 2:** Minterms 4, 6, 7, 14, 15 — form larger groups - **Others:** Minterms 3, 6, 14, 15, etc. ### **Prime implicants:** Let's list all prime implicants: 1. **Group covering minterms 0, 1, 8, 9** - These share the common literals: - \(a' b' c' d'\) for 0 - \(a' b' c' d\) for 1 - \(a' b' c d'\) for 8 - \(a' b' c d\) for 9 The common part is \(a' b'\), with \(c'\) and \(d\) varying. **Prime implicant:** \[ P_1: a' b' \] covers minterms 0, 1, 8, 9. 2. **Group covering minterms 4, 6, 7** - 4: 0100 - 6: 0110 - 7: 0111 Common in \(a=0\), \(b=1\), \(c=1\), \(d\) varies. **Prime implicant:** \[ P_2: a' b c \] covers 4, 6, 7. 3. **Group covering minterms 14, 15** - 14: 1110 - 15: 1111 Common: \(a=1, b=1, c=1\), \(d\) varies. **Prime implicant:** \[ P_3: a b c \] covers 14, 15. 4. **Group covering minterms 3, 7, 15** - 3: 0011 - 7: 0111 - 15: 1111 Common in \(d=1\), \(b=1\), \(a\) varies, \(c\) varies. **Prime implicant:** \[ P_4: c d \] covers 3, 7, 15. 5. **Group covering minterms 6, 14** - 6: 0110 - 14: 1110 Common: \(c=1\), \(d=0\), \(a\) varies, \(b=1\) **Prime implicant:** \[ P_5: b c d' \] covers 6,14. --- ## **Step 4: Prime implicant chart** | Minterm | \(P_1\) | \(P_2\) | \(P_3\) | \(P_4\) | \(P_5\) | |---------|---------|---------|---------|---------|---------| | 0 | X | | | | | | 1 | X | | | | | | 3 | | | | X | | | 4 | | X | | | | | 6 | | | | | X | | 7 | | X | | X | | | 8 | X | | | | | | 9 | X | | | | | | 14 | | | X | | X | | 15 | | | X | X | X | --- ## **Step 5: Essential prime implicants** - **Minterm 0**: Only covered by \(P_1\) → **\(P_1\) is essential** - **Minterm 1**: only \(P_1\) → covered - **Minterm 3**: only \(P_4\) → essential - **Minterm 4**: only \(P_2\) → essential - **Minterm 6**: only \(P_5\) → essential - **Minterm 7**: covered by \(P_2, P_4\) → choose either - **Minterm 14**: covered by \(P_3, P_5\) → choose either - **Minterm 15**: covered by \(P_3, P_4, P_5\) → choose one ### **Select essential prime implicants:** \[ \boxed{ \text{Essential prime implicants: } P_1, P_2, P_4, P_5 } \] --- ## **Step 6: Final minimal cover** Since all minterms are covered by these prime implicants, the minimal cover consists of: \[ f(a, b, c, d) = P_1 + P_2 + P_4 + P_5 \] ### **Expressing each implicant in variables:** - \(P_1: a' b'\) - \(P_2: a' b c\) - \(P_4: c d\) - \(P_5: b c d'\) --- ## **Final Boolean Expression:** \[ \boxed{ f(a, b, c, d) = a' b' + a' b c + c d + b c d' } \] --- ## **Summary:** - **Prime Implicants:** \(a' b'\), \(a' b c\), \(c d\), \(b c d'\) - **Minimal Sum of Products (SOP):** \[ \boxed{ f(a, b, c, d) = a' b' + a' b c + c d + b c d' } \] - **Number of terms:** 4 (minimum possible using Petrick's method). --- ## **Note:** This detailed approach shows how to systematically identify prime implicants, essential primes, and derive the minimal sum expression using Petrick's method and Karnaugh Map grouping. --- **Let me know if you'd like further explanations or simplifications!**