Let's solve the problem step-by-step using the **Quine-McCluskey Method**. ### **1. The Function** Given: \[ f(a, b, c, d) = \sum m(1, 3, 4, 5, 6, 7, 10, 12, 13) + \sum d(2, 9, 15) \] Where: - \( m \) = minterms (where the function is 1) - \( d \) = don't cares ### **2. List all minterms and don't cares in binary** Let's write all minterms and don't cares as 4-bit binary numbers: | Decimal | Binary | Type | |---------|--------|-------------| | 1 | 0001 | Minterm | | 2 | 0010 | Don't care | | 3 | 0011 | Minterm | | 4 | 0100 | Minterm | | 5 | 0101 | Minterm | | 6 | 0110 | Minterm | | 7 | 0111 | Minterm | | 9 | 1001 | Don't care | | 10 | 1010 | Minterm | | 12 | 1100 | Minterm | | 13 | 1101 | Minterm | | 15 | 1111 | Don't care | ### **3. Initial Grouping (by number of 1s)** Let's group all minterms and don't cares by the number of 1's in binary. #### Group 1 (one 1): - 0001 (1) - 0010 (2, d) - 0100 (4) #### Group 2 (two 1's): - 0011 (3) - 0101 (5) - 0110 (6) - 1001 (9, d) - 1010 (10) - 1100 (12) #### Group 3 (three 1's): - 0111 (7) - 1101 (13) - 1111 (15, d) ### **4. Pairwise Combination** We combine terms that differ by only one bit, replacing the differing bit with a dash (-). #### Group 1 and Group 2 | Group 1 | Group 2 | Difference | Result | |----------|---------|------------|---------| | 0001(1) | 0011(3) | 2nd bit | 00-1 | | 0001(1) | 0101(5) | 2nd bit | 0-01 | | 0001(1) | 1001(9) | 1st bit | -001 | | 0010(2d) | 0011(3) | 4th bit | 001- | | 0010(2d) | 0110(6) | 2nd bit | 0-10 | | 0010(2d) | 1010(10)| 1st bit | -010 | | 0100(4) | 0101(5) | 4th bit | 010- | | 0100(4) | 0110(6) | 3rd bit | 01-0 | | 0100(4) | 1100(12)| 1st bit | -100 | #### Group 2 and Group 3 | Group 2 | Group 3 | Difference | Result | |-----------|-----------|------------|---------| | 0011(3) | 0111(7) | 2nd bit | 0-11 | | 0101(5) | 0111(7) | 3rd bit | 01-1 | | 0101(5) | 1101(13) | 1st bit | -101 | | 0110(6) | 0111(7) | 4th bit | 011- | | 1010(10) | 1111(15d) | 2nd, 4th | (no) | | 1010(10) | 1101(13) | bits 2, 3 | (no) | | 1010(10) | 1111(15d) | bits 2, 4 | (no) | | 1100(12) | 1101(13) | 4th bit | 110- | | 1100(12) | 1111(15d) | 3rd, 4th | (no) | ### **5. List All Prime Implicants** Now, collect all the unique combined terms from above, and note which minterms (including don't cares) each one covers. Let's list all the **combined terms** and the corresponding minterms (numbers in parentheses): 1. 00-1 : 0001 (1), 0011 (3) 2. 0-01 : 0001 (1), 0101 (5) 3. -001 : 0001 (1), 1001 (9d) 4. 001- : 0010 (2d), 0011 (3) 5. 0-10 : 0010 (2d), 0110 (6) 6. -010 : 0010 (2d), 1010 (10) 7. 010- : 0100 (4), 0101 (5) 8. 01-0 : 0100 (4), 0110 (6) 9. -100 : 0100 (4), 1100 (12) 10. 0-11 : 0011 (3), 0111 (7) 11. 01-1 : 0101 (5), 0111 (7) 12. -101 : 0101 (5), 1101 (13) 13. 011- : 0110 (6), 0111 (7) 14. 110- : 1100 (12), 1101 (13) Now, check if any of these can be combined again. - For example, 00-1 and 0-01 differ in the 2nd and 3rd bit, so can't be combined. - 010- and 01-0 differ in the 3rd and 4th bit. - 0-10 and -010 differ in 1st and 2nd bit. - 01-1 and 01-0 differ only in 4th bit: So can be combined as 01--, covering 0101 (5), 0110 (6), 0111 (7), 0100 (4). Let's check which ones can be combined further. #### Second Level Combination - 010- (4,5) & 01-1 (5,7): 010- (4,5) = 0100,0101; 01-1 = 0101,0111; overlap at 0101 only. Can't combine. - 01-1 (5,7) & 01-0 (4,6): 01-1 = 0101,0111; 01-0 = 0100,0110; overlap at 01-- (covers 0100,0101,0110,0111). So, **01--** (0100,0101,0110,0111) is a prime implicant. Check for others: - 110- (12,13), 1100 and 1101: only differ in 4th bit, so already in prime form. - 0-10 (2d,6) & -010 (2d,10): 0-10 = 0010, 0110; -010 = 0010, 1010; overlap at 0010 only. Can't combine. So, **no further combinations**. Now, let's list all prime implicants, including the combined one: - 01-- (0100, 0101, 0110, 0111): covers minterms 4,5,6,7 - 110- (1100,1101): covers minterms 12,13 - -010 (0010, 1010): covers minterm 10 (and don't care 2) - 00-1 (0001,0011): covers minterms 1,3 - 0-01 (0001,0101): covers minterms 1,5 - -001 (0001,1001): covers minterm 1 (and don't care 9) - 001- (0010,0011): covers minterm 3 (and don't care 2) - 0-10 (0010,0110): covers minterm 6 (and don't care 2) - 010- (0100,0101): covers minterms 4,5 - 01-0 (0100,0110): covers minterms 4,6 - -100 (0100,1100): covers minterms 4,12 - 0-11 (0011,0111): covers minterms 3,7 - 01-1 (0101,0111): covers minterms 5,7 - -101 (0101,1101): covers minterms 5,13 - 011- (0110,0111): covers minterms 6,7 But many of these are **subsumed** by the larger group **01--** (4,5,6,7). ### **6. Prime Implicant Chart** Let's make a chart with minterms **only** (not don't cares): Minterms to be covered: - 1 (0001) - 3 (0011) - 4 (0100) - 5 (0101) - 6 (0110) - 7 (0111) - 10 (1010) - 12 (1100) - 13 (1101) Now, let's tabulate which prime implicants cover which minterms. | Prime Implicant | 1 | 3 | 4 | 5 | 6 | 7 | 10 | 12 | 13 | |-----------------|---|---|---|---|---|---|----|----|----| | 01-- | | | X | X | X | X | | | | | 110- | | | | | | | | X | X | | -010 | | | | | | | X | | | | 00-1 | X | X | | | | | | | | Now, check which columns (minterms) have only one X (i.e., essential prime implicants): - 1: covered only by 00-1 - 3: covered only by 00-1 - 4: covered only by 01-- - 5: covered only by 01-- - 6: covered only by 01-- - 7: covered only by 01-- - 10: covered only by -010 - 12: covered only by 110- - 13: covered only by 110- **ALL minterms are covered by only one prime implicant each!** ### **7. Final Minimal Sum-of-Products Solution** The essential prime implicants are: - **00-1** : covers minterms 1,3 - **01--** : covers minterms 4,5,6,7 - **-010** : covers minterm 10 - **110-** : covers minterms 12,13 Let's write these as Boolean expressions: - 00-1: **A' B' C D** - 01--: **A' B C'** (since 01-- = A' B C x x) - -010: **B' C D'** (since -010 = x 0 1 0) - 110-: **A B C'** (since 110- = 1 1 0 x) ### **8. Final Answer** **Step-by-step summary:** - List minterms and don't cares in binary - Group by number of 1's - Combine to find prime implicants - Make prime implicant chart - Identify essential prime implicants - Write final minimal SOP **Final minimal sum-of-products:** \[ f(a, b, c, d) = A'B'C D + A' B C' + B' C D' + A B C' \] or in plain English: \[ f(a, b, c, d) = \boxed{A'B'C D + A' B C' + B' C D' + A B C'} \] --- **If you need Karnaugh map steps or want the expression in other forms, let me know!**