Question:
The figure below shows the Karnaugh map (K-map) for a truth table with 4 variables \(A\), \(B\), \(C\) and \(D\). Which one of the following expressions represents the minimum sum of products? Note: The empty cells in the K-map are 0's.
AB / CD 00 01 11 10 00 01 1 1 1 11 1 1 1 10 1
The figure below shows the Karnaugh map (K-map) for a truth table with 4 variables \(A\), \(B\), \(C\) and \(D\). Which one of the following expressions represents the minimum sum of products? Note: The empty cells in the K-map are 0's.
| AB / CD | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | ||||
| 01 | 1 | 1 | 1 | |
| 11 | 1 | 1 | 1 | |
| 10 | 1 |
Show Hint
Find the largest group first (BC), then cover the leftover cells m5, m10, m12 with the smallest possible pairs.
Updated On: Jul 16, 2026
- \(BC + \bar{A}BD + AC\bar{D} + AB\bar{D}\)
- \(AB\bar{C}\bar{D} + \bar{A}B\bar{C}D + A\bar{B}C\bar{D} + BC\)
- \(BC + AC\bar{D} + \bar{A}BD + AB\bar{C}\bar{D}\)
- \(AB\bar{D} + \bar{A}BD + AC\bar{D} + BCD + BC\bar{D}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Read the 1's off the K-map.
Using row order \(AB = 00, 01, 11, 10\) and column order \(CD = 00, 01, 11, 10\), the cells marked \(1\) are at \((A,B,C,D) = (0,1,0,1), (0,1,1,1), (0,1,1,0), (1,1,0,0), (1,1,1,1), (1,1,1,0), (1,0,1,0)\). In minterm numbers (with \(A\) as the most significant bit) these are \(m_5, m_6, m_7, m_{10}, m_{12}, m_{14}, m_{15}\).
Step 2: Find the largest possible group first.
Look at the four cells where \(B=1\) and \(C=1\), letting \(A\) and \(D\) vary freely: \(AB=01,CD=11\) (\(m_7\)), \(AB=01,CD=10\) (\(m_6\)), \(AB=11,CD=11\) (\(m_{15}\)), \(AB=11,CD=10\) (\(m_{14}\)). All four are marked \(1\), so this is a valid quad, and it collapses to the 2-literal term \(BC\).
Step 3: Cover the leftover 1's with pairs.
After removing \(m_6, m_7, m_{14}, m_{15}\), the minterms \(m_5, m_{10}, m_{12}\) still need covering. \(m_5\) (\(AB=01,CD=01\)) sits next to \(m_7\) (\(AB=01,CD=11\)); both have \(A=0, B=1, D=1\) with \(C\) changing, giving the pair \(\bar{A}BD\). \(m_{10}\) (\(AB=10,CD=10\)) sits next to \(m_{14}\) (\(AB=11,CD=10\)); both have \(A=1, C=1, D=0\) with \(B\) changing, giving the pair \(AC\bar{D}\). \(m_{12}\) (\(AB=11,CD=00\)) sits next to \(m_{14}\) (\(AB=11,CD=10\)); both have \(A=1, B=1, D=0\) with \(C\) changing, giving the pair \(AB\bar{D}\).
Step 4: Combine the groups.
Every marked \(1\) (\(m_5,m_6,m_7,m_{10},m_{12},m_{14},m_{15}\)) is now covered by one of the four groups \(BC\), \(\bar{A}BD\), \(AC\bar{D}\), \(AB\bar{D}\), and none of these groups can be merged into anything bigger, so this is the minimum sum of products:
\[ F = BC + \bar{A}BD + AC\bar{D} + AB\bar{D} \]
Step 5: Why the other options are wrong.
Option (B) uses 3 and 4-literal terms like \(AB\bar{C}\bar{D}\) for cells that group into much larger, cheaper terms, so it is not minimal.
Option (C) swaps in \(AB\bar{C}\bar{D}\) in place of \(AB\bar{D}\), covering only \(m_{12}\) alone instead of merging \(m_{12}\) with \(m_{14}\), so it needlessly uses a 4-literal term where a 3-literal one works.
Option (D) adds a redundant term \(BCD\) that is already fully covered by \(BC\), which is not allowed in a minimum expression.
Final Answer:
The minimum sum of products is \(BC + \bar{A}BD + AC\bar{D} + AB\bar{D}\).
\[ \boxed{BC + \bar{A}BD + AC\bar{D} + AB\bar{D}} \]
Using row order \(AB = 00, 01, 11, 10\) and column order \(CD = 00, 01, 11, 10\), the cells marked \(1\) are at \((A,B,C,D) = (0,1,0,1), (0,1,1,1), (0,1,1,0), (1,1,0,0), (1,1,1,1), (1,1,1,0), (1,0,1,0)\). In minterm numbers (with \(A\) as the most significant bit) these are \(m_5, m_6, m_7, m_{10}, m_{12}, m_{14}, m_{15}\).
Step 2: Find the largest possible group first.
Look at the four cells where \(B=1\) and \(C=1\), letting \(A\) and \(D\) vary freely: \(AB=01,CD=11\) (\(m_7\)), \(AB=01,CD=10\) (\(m_6\)), \(AB=11,CD=11\) (\(m_{15}\)), \(AB=11,CD=10\) (\(m_{14}\)). All four are marked \(1\), so this is a valid quad, and it collapses to the 2-literal term \(BC\).
Step 3: Cover the leftover 1's with pairs.
After removing \(m_6, m_7, m_{14}, m_{15}\), the minterms \(m_5, m_{10}, m_{12}\) still need covering. \(m_5\) (\(AB=01,CD=01\)) sits next to \(m_7\) (\(AB=01,CD=11\)); both have \(A=0, B=1, D=1\) with \(C\) changing, giving the pair \(\bar{A}BD\). \(m_{10}\) (\(AB=10,CD=10\)) sits next to \(m_{14}\) (\(AB=11,CD=10\)); both have \(A=1, C=1, D=0\) with \(B\) changing, giving the pair \(AC\bar{D}\). \(m_{12}\) (\(AB=11,CD=00\)) sits next to \(m_{14}\) (\(AB=11,CD=10\)); both have \(A=1, B=1, D=0\) with \(C\) changing, giving the pair \(AB\bar{D}\).
Step 4: Combine the groups.
Every marked \(1\) (\(m_5,m_6,m_7,m_{10},m_{12},m_{14},m_{15}\)) is now covered by one of the four groups \(BC\), \(\bar{A}BD\), \(AC\bar{D}\), \(AB\bar{D}\), and none of these groups can be merged into anything bigger, so this is the minimum sum of products:
\[ F = BC + \bar{A}BD + AC\bar{D} + AB\bar{D} \]
Step 5: Why the other options are wrong.
Option (B) uses 3 and 4-literal terms like \(AB\bar{C}\bar{D}\) for cells that group into much larger, cheaper terms, so it is not minimal.
Option (C) swaps in \(AB\bar{C}\bar{D}\) in place of \(AB\bar{D}\), covering only \(m_{12}\) alone instead of merging \(m_{12}\) with \(m_{14}\), so it needlessly uses a 4-literal term where a 3-literal one works.
Option (D) adds a redundant term \(BCD\) that is already fully covered by \(BC\), which is not allowed in a minimum expression.
Final Answer:
The minimum sum of products is \(BC + \bar{A}BD + AC\bar{D} + AB\bar{D}\).
\[ \boxed{BC + \bar{A}BD + AC\bar{D} + AB\bar{D}} \]
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