Let's solve the given problem step-by-step: --- ## **Given:** Minimize \( F(A, B, C, D) = \sum m(2, 3, 4, 5, 6, 7) \) Variables: 4 variables (A, B, C, D) ### **Step 1: Write minterms in binary** | Decimal | Binary | |---------|---------| | 2 | 0010 | | 3 | 0011 | | 4 | 0100 | | 5 | 0101 | | 6 | 0110 | | 7 | 0111 | --- ## **a) Quine–McCluskey (Tabular) Method** ### **Group minterms by number of 1s:** - **One 1:** - 2 (0010) - 4 (0100) - **Two 1s:** - 3 (0011) - 5 (0101) - 6 (0110) - **Three 1s:** - 7 (0111) --- ### **Compare and combine minterms (1st pass):** #### **Group 1 and Group 2:** - 2 (0010) & 3 (0011): `001-` (covers 2,3) - 2 (0010) & 6 (0110): `0-10` (covers 2,6) - 4 (0100) & 5 (0101): `010-` (covers 4,5) - 4 (0100) & 6 (0110): `01-0` (covers 4,6) - 5 (0101) & 7 (0111): `01-1` (covers 5,7) - 3 (0011) & 7 (0111): `0-11` (covers 3,7) - 6 (0110) & 7 (0111): `011-` (covers 6,7) --- ### **List all prime implicants:** From your photo, the prime implicants are: | PI | Binary | Covered minterms | |----|---------|------------------| | P1 | 001- | 2, 3 | | P2 | 0-10 | 2, 6 | | P3 | 010- | 4, 5 | | P4 | 0-11 | 3, 7 | | P5 | 01-1 | 5, 7 | | P6 | 011- | 6, 7 | --- ## **b) Petrick's Method (Minimum Cover)** The prime implicant chart is already provided. ### **Prime Implicant Chart:** | | P1 | P2 | P3 | P4 | P5 | P6 | |------|----|----|----|----|----|----| | 2 | X | X | | | | | | 3 | X | | | X | | | | 4 | | | X | | | | | 5 | | | X | | X | | | 6 | | X | | | | X | | 7 | | | | X | X | X | Write the coverage for each minterm: - **2:** (P1 + P2) - **3:** (P1 + P4) - **4:** (P3) - **5:** (P3 + P5) - **6:** (P2 + P6) - **7:** (P4 + P5 + P6) The minimal cover is the product: \[ (P1 + P2) \cdot (P1 + P4) \cdot (P3) \cdot (P3 + P5) \cdot (P2 + P6) \cdot (P4 + P5 + P6) \] But notice that **P3** is essential (covers 4,5 uniquely), so **P3** must be included. After including \(P3\), minterms 4,5 are covered. So, update: - **2:** (P1 + P2) - **3:** (P1 + P4) - **6:** (P2 + P6) - **7:** (P4 + P5 + P6) Now, let's try to cover the rest with minimal prime implicants: - Choose **P3** (for 4,5) - To cover 2, 3, 6, 7, try combinations: Let's try **P2** (covers 2, 6): - **2:** covered by P2 - **3:** (P1 + P4) - **6:** covered by P2 - **7:** (P4 + P5 + P6) Now, cover 3 and 7: - Choose **P4** (covers 3, 7): So, our minimum cover is: **P2, P3, P4** --- ## **Final Answer:** ### **a) Prime Implicants:** - P1: 001- (2,3) - P2: 0-10 (2,6) - P3: 010- (4,5) - P4: 0-11 (3,7) - P5: 01-1 (5,7) - P6: 011- (6,7) ### **b) Minimum Cover:** - **P2 (0-10):** covers 2,6 - **P3 (010-):** covers 4,5 - **P4 (0-11):** covers 3,7 ### **Minimal Expression:** \[ F(A,B,C,D) = 0-10 + 010- + 0-11 \] Or, in variable form: - P2: 0-10 = \( A' C D' \) - P3: 010- = \( A' B C' \) - P4: 0-11 = \( A' C D \) So, \[ \boxed{F(A,B,C,D) = A' C D' + A' B C' + A' C D} \] Or, if you want to further factor: \[ = A' C (D' + D) + A' B C' = A' C + A' B C' \] \[ \boxed{F(A,B,C,D) = A' C + A' B C'} \] --- ### **Summary Table** | Step | Prime implicants | Minimum cover | Final Expression | |------|------------------|-------------------------|-------------------------| | a | P1, P2, P3, P4, P5, P6 | P2, P3, P4 | \(A'C D' + A'B C' + A' C D\) or \(A' C + A' B C'\) | --- Let me know if you need any clarifications!

Certainly! Based on the provided image and the minterms \(2, 3, 4, 5, 6, 7\), I'll provide a detailed step-by-step solution, including all calculations, explanations, and the final minimized Boolean expression. --- ## Step 1: List the Minterms in Binary Form | Minterm | Decimal | Binary (A B C D) | Number of 1s | |----------|---------|------------------|--------------| | 2 | 2 | 0010 | 1 | | 3 | 3 | 0011 | 2 | | 4 | 4 | 0100 | 1 | | 5 | 5 | 0101 | 2 | | 6 | 6 | 0110 | 2 | | 7 | 7 | 0111 | 3 | --- ## Step 2: Group Minterms Based on the Number of 1s - **Group 1 (1 one):** 2 (0010), 4 (0100) - **Group 2 (2 ones):** 3 (0011), 5 (0101), 6 (0110) - **Group 3 (3 ones):** 7 (0111) --- ## Step 3: Find Prime Implicants by Combining Minterms ### Combining minterms from adjacent groups: **Group 1 & 2:** - **2 (0010)** & **3 (0011):** Difference in last bit (D): 0 vs 1 → Combine: **001-** (Covers minterms 2 & 3) - **2 (0010)** & **6 (0110):** Difference in B: 0 vs 1 → Combine: **0-10** (Covers minterms 2 & 6) - **4 (0100)** & **5 (0101):** Difference in D: 0 vs 1 → Combine: **010-** (Covers minterms 4 & 5) - **4 (0100)** & **6 (0110):** Difference in C: 0 vs 1 → Combine: **01-0** (Covers minterms 4 & 6) **Group 2 & 3:** - **3 (0011)** & **7 (0111):** Difference in B: 0 vs 1 → Combine: **0-11** (Covers minterms 3 & 7) - **5 (0101)** & **7 (0111):** Difference in C: 0 vs 1 → Combine: **01-1** (Covers minterms 5 & 7) - **6 (0110)** & **7 (0111):** Difference in D: 0 vs 1 → Combine: **011-** (Covers minterms 6 & 7) --- ## Step 4: List Prime Implicants | Prime Implicant | Binary Pattern | Minterms Covered | |-------------------|------------------|------------------| | **PI1:** 001- | 001- | 2, 3 | | **PI2:** 0-10 | 0-10 | 2, 6 | | **PI3:** 010- | 010- | 4, 5 | | **PI4:** 0-11 | 0-11 | 3, 7 | | **PI5:** 01-1 | 01-1 | 5, 7 | | **PI6:** 011- | 011- | 6, 7 | --- ## Step 5: Construct Prime Implicant Chart | Minterm | P1 | P2 | P3 | P4 | P5 | P6 | |----------|----|----|----|----|----|----| | **2** | X | X | | | | | | **3** | X | | | X | | | | **4** | | | X | | | | | **5** | | | X | | X | | | **6** | | X | | | | X | | **7** | | | | X | X | X | --- ## Step 6: Find Minimum Cover (Using Petrick's Method) Identify essential prime implicants: - **Minterm 4** is only covered by **PI3** → **PI3** is essential. - **Minterm 2** is covered by **PI1** and **PI2** → choose one, but both are options. - **Minterm 6** is covered by **PI2** and **PI6**. - **Minterm 3** is covered by **PI1** and **PI4**. - **Minterm 5** is covered by **PI3** and **PI5** (but **PI3** already selected). - **Minterm 7** is covered by **PI4**, **PI5**, **PI6**. ### Step 6.1: Select essential prime implicants: - **PI3** (covers 4, 5) - **PI1** (covers 2, 3) - **PI6** (covers 6, 7) or **PI4/PI5** (but choosing **PI6** simplifies options) ### Step 6.2: Final minimal set: - **PI1:** covers minterms 2, 3 - **PI3:** covers minterms 4, 5 - **PI6:** covers minterms 6, 7 This covers all minterms with minimal prime implicants. --- ## **Step 7: Write the Boolean Expression for Each Prime Implicant** | Prime Implicant | Binary Pattern | Corresponding Expression | Explanation | |-------------------|------------------|-------------------------------------|----------------------------------------------------------------| | **PI1:** 001- | 001- | \(\overline{A} \overline{B} C\) | A=0, B=0, C=1, D=don't care | | **PI3:** 010- | 010- | \(\overline{A} B \overline{C}\) | A=0, B=1, C=0, D=don't care | | **PI6:** 011- | 011- | \(\overline{A} B C\) | A=0, B=1, C=1, D=don't care | --- ## **Step 8: Write the Final Minimized Expression** \[ F = \text{PI1} + \text{PI3} + \text{PI6} \] \[ F = \overline{A} \overline{B} C + \overline{A} B \overline{C} + \overline{A} B C \] ### **Factor further if desired:** Notice \(\overline{A} B \overline{C} + \overline{A} B C = \overline{A} B (\overline{C} + C) = \overline{A} B \times 1 = \overline{A} B\). Thus, the simplified Boolean expression is: \[ \boxed{ F = \overline{A} \overline{B} C + \overline{A} B } \] --- ## **Final Result:** \[ \boxed{ \boxed{ F(A, B, C, D) = \overline{A} \overline{B} C + \overline{A} B } } \] This is the minimized Boolean expression representing the logic for the minterms 2, 3, 4, 5, 6, and 7. --- ## **Summary:** - **Prime Implicants Selected:** \(\overline{A} \overline{B} C\), \(\overline{A} B \overline{C}\), \(\overline{A} B C\) - **Final Expression:** \(\overline{A} \overline{B} C + \overline{A} B\) Feel free to ask if you'd like further clarification!