Tricky Problem: Implementing F(A,B,C) = Sigma-m(1,2,5,7) Using a Multiplexer
Apply the MUX-implementation technique of assigning higher-order variables to select lines and deriving each data input as a function of the remaining variable, rather than assuming every data input must be a fixed 0 or 1.
Implement F(A,B,C) = Sigma-m(1,2,5,7) using a multiplexer, by selecting appropriate select lines and assigning appropriate data inputs.
Why This Problem Is Tricky
With three variables and a standard 4:1 MUX, only two variables (A and B) can be used as select lines, leaving C as a 'leftover' variable. The trap is assuming every data input line must be a constant 0 or 1, the way a solver might expect after only ever seeing MUX examples where the function happens to be constant within each select-line group. Here, at least one select-line group (AB=00) has a function value that actually depends on C, so its data input must be C itself (or C'), not a fixed constant — missing this is the single most common error in MUX-based implementation problems.
Step-by-Step Worked Solution
Choose A and B as the two select lines of a 4:1 MUX, leaving C as the variable whose value determines each data input. Build the full truth table for F(A,B,C) = Sigma-m(1,2,5,7): minterm 1 (A=0,B=0,C=1) -> F=1; minterm 2 (A=0,B=1,C=0) -> F=1; minterm 5 (A=1,B=0,C=1) -> F=1; minterm 7 (A=1,B=1,C=1) -> F=1; all other minterms (0,3,4,6) -> F=0.
Group the table by AB value and examine how F depends on C within each group. AB=00: C=0 gives minterm 0 (F=0); C=1 gives minterm 1 (F=1). F follows C exactly, so D0 = C. AB=01: C=0 gives minterm 2 (F=1); C=1 gives minterm 3 (F=0). F is the opposite of C, so D1 = C'. AB=10: C=0 gives minterm 4 (F=0); C=1 gives minterm 5 (F=1). F follows C, so D2 = C. AB=11: C=0 gives minterm 6 (F=0); C=1 gives minterm 7 (F=1). F follows C, so D3 = C.
The completed multiplexer implementation connects select lines S1=A, S0=B, and data inputs D0=C, D1=C', D2=C, D3=C.
Why the Obvious Wrong Answer Is Wrong
A common mistake is checking only ONE of the two rows in an AB group (for example, only the C=0 row of AB=00, which gives minterm 0 with F=0) and concluding the data input for that group must be the constant 0 — without checking the second row (C=1, minterm 1, F=1) where the function value is actually different. Since F changes between the two rows of that group, the correct data input is C itself, not a constant. Skipping the second row of any group is the specific trap this problem is built around, since three of the four groups here (AB=00, AB=10, AB=11) all happen to need the non-constant input C, making the pattern easy to miss if only spot-checked.
Code Examples
Every group is checked using BOTH of its two rows (C=0 and C=1), and the data input is set to whatever single-variable expression of C reproduces both F values in that group — a constant only when both rows happen to agree, and C or C' whenever they disagree. Expected result: F(A,B,C)=Sigma-m(1,2,5,7) fully implemented by a 4:1 MUX
D0=C, D1=C', D2=C, D3=C (select lines S1=A, S0=B)
A B C : minterm : F
0 0 0 : 0 : 0 -+
0 0 1 : 1 : 1 +-- AB=00: F follows C -> D0 = C
0 1 0 : 2 : 1 -+
0 1 1 : 3 : 0 +-- AB=01: F is opposite of C -> D1 = C'
1 0 0 : 4 : 0 -+
1 0 1 : 5 : 1 +-- AB=10: F follows C -> D2 = C
1 1 0 : 6 : 0 -+
1 1 1 : 7 : 1 +-- AB=11: F follows C -> D3 = C
Select lines: S1=A, S0=B
Data inputs: D0=C, D1=C', D2=C, D3=C4:1 MUX Implementation
D0=C --->+
D1=C' --->| 4:1
D2=C --->| MUX ---> F
D3=C --->+
^ ^
S1 S0
A BWhy This MUX Technique Matters
This select-line-plus-residual-variable technique is precisely how a Field-Programmable Gate Array's Look-Up Table (LUT) implements arbitrary logic functions internally — a LUT is, at its core, a large multiplexer selecting among stored truth-table bits. Getting the residual-variable assignment wrong when implementing logic this way produces a circuit that computes the wrong function for exactly the input combinations where the 'obvious' constant assumption failed.
D0 = 0 (checked only the AB=00, C=0 row and stopped) D0 = C (must check both the C=0 and C=1 rows of the AB=00 group)
The AB=00 group has F=0 when C=0 (minterm 0) but F=1 when C=1 (minterm 1) — the function is not constant within that group, so its correct data input is the variable C itself, not a fixed 0 or 1. Checking only one row of a group and assuming the other row matches is the exact trap this problem tests.
Implementing F(A,B,C)=Sigma-m(1,2,5,7) with A,B as MUX select lines, what should data input D0 (for AB=00) be? A. D0 = 0 B. D0 = 1 C. D0 = C D. D0 = C'
C. D0 = C — For AB=00, minterm 0 (C=0) gives F=0 and minterm 1 (C=1) gives F=1 — F exactly follows C within this group, so D0=C. Assuming D0=0 (option 1) comes from checking only the C=0 row and ignoring the C=1 row where F actually changes to 1.
When implementing an n-variable Boolean function with a multiplexer using n-1 select lines, what value(s) can each data input take? A. Only 0 or 1 B. Only the remaining (residual) variable or its complement C. 0, 1, the remaining variable, or its complement, depending on the function D. Only the remaining variable, never its complement
C. 0, 1, the remaining variable, or its complement, depending on the function — Each data input reproduces the function's behavior across the two rows of its select-line group. If both rows agree, the data input is the constant they agree on (0 or 1); if the rows disagree, the data input must be the residual variable itself or its complement, whichever matches the pattern of the two rows.
Implement F(A,B,C) = Sigma-m(0,3,4,7) using a 4:1 MUX with A and B as select lines.
AB=00: C=0 -> minterm 0, F=1; C=1 -> minterm 1, F=0 => D0=C'. AB=01: C=0 -> minterm 2, F=0; C=1 -> minterm 3, F=1 => D1=C. AB=10: C=0 -> minterm 4, F=1; C=1 -> minterm 5, F=0 => D2=C'. AB=11: C=0 -> minterm 6, F=0; C=1 -> minterm 7, F=1 => D3=C.
Build the same grouped truth table, checking both rows of every AB group.