Tricky Problem: Minimizing F(A,B,C,D) = Σm(0,1,2,3,8,9,10,11) on a 4-Variable K-map

Recognize that a K-map's top and bottom rows (and left and right columns) are logically adjacent by wraparound, so two 4-cell groups that look separate can actually merge into a single 8-cell group with a simpler result.

Minimize the four-variable function F(A,B,C,D) = Σm(0,1,2,3,8,9,10,11) using a Karnaugh map. The minterm list is deliberately chosen so that the correct minimal grouping is not visually obvious on a standard K-map layout unless the solver remembers that the map's edges wrap around.

Why This Problem Is Tricky

On a standard 4-variable K-map, the rows are ordered by AB in Gray-code sequence: 00, 01, 11, 10 (top to bottom). Minterms 0-3 (A=0,B=0, all C,D) occupy the top row, and minterms 8-11 (A=1,B=0, all C,D) occupy the bottom row. Between those two rows sit the AB=01 and AB=11 rows, which are entirely empty (all 0s) in this function. Visually, the two filled rows look separated by two empty rows in between — which makes it tempting to treat them as two independent 4-cell groups rather than noticing they are directly adjacent to each other by wraparound (the K-map's top edge and bottom edge are logically the same boundary, exactly like the Gray-code sequence wrapping from 10 back to 00).

Missing that wraparound adjacency is the single most common way to under-minimize this function.

Step-by-Step Worked Solution

Place all eight minterms on a 4-variable K-map with AB as rows (order 00,01,11,10) and CD as columns (order 00,01,11,10). Minterms 0,1,2,3 fill the entire AB=00 row (all four CD combinations). Minterms 8,9,10,11 fill the entire AB=10 row (all four CD combinations). The AB=01 and AB=11 rows are entirely 0.

Because the K-map wraps around, row AB=00 (top) and row AB=10 (bottom) are adjacent — moving off the top edge lands you on the bottom edge, just as moving off the right edge of any row lands you on its left edge. That means the 4-cell group in row AB=00 and the 4-cell group in row AB=10 can be combined into a single 8-cell group covering all of AB=00 and all of AB=10.

Within that combined 8-cell group, examine which variables stay constant across every one of the eight minterms: A varies (0 in the top row, 1 in the bottom row) — not constant, so A drops out. B is 0 in both rows (AB=00 and AB=10 both have B=0) — B stays constant at 0, so B survives as B'. C varies across all four values in each row (0 and 1 both occur) — not constant, drops out. D likewise varies — not constant, drops out.

Only B is constant (always 0) across the full 8-cell group, so the group reduces to the single literal B'. Therefore F = B'.

Why the Obvious (Under-Minimized) Answer Is Wrong

A solver who does not use the wraparound adjacency will instead group the two rows as two separate 4-cell quads: minterms 0,1,2,3 (A=0,B=0, C,D free) simplify to A'B', and minterms 8,9,10,11 (A=1,B=0, C,D free) simplify to AB'. Combining them with an OR gives F = A'B' + AB', which is logically correct — it evaluates to the same truth table as B' — but it is not the minimal SOP form, because it still uses two 2-literal product terms and an OR gate instead of the single literal B' that the properly merged 8-cell group yields.

The deeper lesson is that A'B' + AB' is itself further reducible by the same complement law used in Problem 1 (A'B' + AB' = B'(A' + A) = B'), so failing to spot the K-map wraparound simply means redoing, by algebra, work the map was supposed to do visually — and on a real K-map with many more minterms, that missed adjacency can hide far larger and less obvious opportunities to merge groups.

Code Examples

The systematic check walks every one of the four variables across the merged 8-cell group and keeps only the ones that never change value across the whole group. Only B is constant (always 0), so it is the only literal that survives, appearing complemented because its constant value is 0. Expected result: F = B' (a single-literal result covering all 8 given minterms)

Minterms: 0,1,2,3,8,9,10,11

Row AB=00: m0,m1,m2,m3  -> all 1s
Row AB=01: (none)       -> all 0s
Row AB=11: (none)       -> all 0s
Row AB=10: m8,m9,m10,m11 -> all 1s

Row AB=00 and row AB=10 are adjacent by wraparound (top/bottom edges touch).
Merge into one 8-cell group covering both rows, all CD columns.

Check each variable across the 8-cell group:
  A: 0 in one row, 1 in the other -> varies -> eliminated
  B: 0 in both rows               -> constant (B') -> kept
  C: 0 and 1 both occur           -> varies -> eliminated
  D: 0 and 1 both occur           -> varies -> eliminated

Result: F = B'

4-Variable K-map with Wraparound Grouping

Treating rows 00 and 10 as separate quads misses the wraparound merge into a single octet.

CD
             00  01  11  10
        00 [ 1 ] [ 1 ] [ 1 ] [ 1 ]   <- AB=00 row
   AB   01 [ 0 ] [ 0 ] [ 0 ] [ 0 ]
        11 [ 0 ] [ 0 ] [ 0 ] [ 0 ]
        10 [ 1 ] [ 1 ] [ 1 ] [ 1 ]   <- AB=10 row

  Top row (AB=00) and bottom row (AB=10) touch by wraparound,
  forming one 8-cell group where only B stays constant (B=0) -> F = B'

Why Wraparound Adjacency Matters in Practice

Hand-minimizing K-maps builds the intuition needed to sanity-check what an automated logic synthesis tool produces; a designer who consistently misses wraparound adjacencies will systematically overestimate the gate count a piece of combinational logic actually needs, and will struggle to verify or debug synthesis output when the tool legitimately produces a smaller circuit than the designer expected by hand.

F = A'B' + AB' A'B' + AB' is a logically correct but non-minimal answer that comes from treating the two filled rows as separate 4-cell groups. Recognizing the top-bottom wraparound adjacency merges them into one 8-cell group, and the resulting single-literal answer, B', is the true minimal SOP form.

What is the minimal SOP form of F(A,B,C,D) = Σm(0,1,2,3,8,9,10,11)? A. F = A'B' + AB' B. F = B' C. F = B' + C' D. F = A' + B'

B. F = B' — Minterms 0-3 and 8-11 together fill the entire AB=00 and AB=10 rows for all values of C and D. Those two rows are adjacent by K-map wraparound, so they merge into a single 8-cell group. Across that group only B is constant (B=0), so the fully minimized result is F = B'. A'B' + AB' is the same function but uses two terms instead of recognizing the merge into one.

In a standard 4-variable K-map with AB rows ordered 00, 01, 11, 10 (top to bottom), which two rows are adjacent to each other by wraparound? A. Rows AB=00 and AB=11 B. Rows AB=01 and AB=11 C. Rows AB=00 and AB=10 D. Rows AB=01 and AB=10

C. Rows AB=00 and AB=10 — The Gray-code row order 00, 01, 11, 10 is arranged so that any two physically adjacent rows differ in exactly one bit. The top row (00) and bottom row (10) also differ in exactly one bit (the B bit), and the K-map is defined so that the top and bottom edges wrap around and touch — making rows 00 and 10 adjacent even though they are drawn furthest apart on the page.

Minimize F(A,B,C,D) = Σm(0,2,4,6,8,10,12,14) on a 4-variable K-map.

Every listed minterm has D=0 (0,2,4,6,8,10,12,14 are all even numbers, meaning bit D is 0), while A, B, and C each take on both 0 and 1 across the list. This is the column-wraparound analog of the worked problem: the CD=00 and CD=10 columns (which are adjacent by left-right wraparound) merge across all four AB rows into one 8-cell group where only D is constant. F = D'.

Look at the least-significant variable, D, in every one of the listed minterms — is it always the same value?