Let's solve the K-map step by step: ### **Step 1: Write the Variables** Let: - \( A, B \) are the row variables (across the top) - \( C, D \) are the column variables (down the side) ### **Step 2: Identify the 1's and Don't-care (x)** From the K-map: - 1's are at: (01,00), (01,01), (11,10) - x (don't care) at: (11,01) ### **Step 3: Fill the K-map with Variable Combinations** | C D \ A B | 00 | 01 | 11 | 10 | |-----------|----|----|----|----| | 00 | 0 | 1 | 0 | 0 | | 01 | 1 | 1 | 0 | 0 | | 11 | 0 | x | 1 | 0 | | 10 | 0 | 0 | 1 | 0 | ### **Step 4: Group the 1's (and possibly x's, if it helps)** #### **Group 1 (Largest possible group):** - Group the two 1's at (01,01) and (01,00). - These are adjacent horizontally. For (01,01) and (01,00): - Row: 01 → \( C \overline{D} \) - Columns: 01 and 00 → \( B \overline{A} \) and \( \overline{A}\overline{B} \), but since only B changes, \( \overline{A} \) is common. So, the group is: \( \overline{A} C \overline{D} \) #### **Group 2 (Vertical group):** - Group the 1's at (01,01) and (11,10) using the don't care at (11,01) to form a group of two vertically. For (01,01) and (11,01): - Column: 01 → \( \overline{A} B \) - Rows: 01 and 11 → \( C \overline{D} \) and \( C D \) → C is common So, the group is: \( B C \) #### **Group 3 (Single 1 left):** - (11,10): Row 11, Column 10 (11,10): - Row: 11 → \( C D \) - Column: 10 → \( A \overline{B} \) So, term: \( A \overline{B} C D \) #### **Group 4 (Single 1 left):** - (10,11): Row 10, Column 11 (10,11): - Row: 10 → \( C \overline{D} \) - Column: 11 → \( A B \) So, term: \( A B C \overline{D} \) But looking again, (10,11) and (11,11) can be grouped together (wrap around). But (11,11) is 1, so yes. But in the map, (10,11) is 1, (11,11) is 1. Group: (10,11) and (11,11): - Columns: 11 → \( A B \) - Rows: 10 and 11 → \( C \overline{D} \), \( C D \) → C common. So, term: \( A B C \) ### **Step 5: Write the Final SOP Expression** Combine all groups found: \[ F(A,B,C,D) = \overline{A} C \overline{D} + B C + A B C \] This is the simplest SOP form. --- ## **Final Answer** \[ \boxed{F(A,B,C,D) = \overline{A} C \overline{D} + B C + A B C} \] --- Let me know if you want to see the grouping highlighted on the K-map!