Karnaugh Map (K-Map) Simplification: A Fully Solved Example

A K-map is just a truth table redrawn so that boxes which differ by exactly one variable sit next to each other. That single redraw is what lets you spot simplifications by eye instead of grinding through Boolean algebra identities.

The function we'll simplify

Take a 3-variable function defined by its minterms:

F(A, B, C) = Σm(0, 2, 4, 5, 6)

That means F = 1 for minterms 0, 2, 4, 5, and 6, and 0 everywhere else. Written out as a truth table:

Drawing the K-map

A 3-variable K-map has one variable (A) on the rows and two (B, C) across the columns — and the columns must be ordered 00, 01, 11, 10, not 00, 01, 10, 11. That ordering (Gray code) is the entire trick: it guarantees any two side-by-side cells differ in exactly one bit.

              BC=00   BC=01   BC=11   BC=10
      A=0  |    1       0       0       1     |
      A=1  |    1       1       0       1     |

Placing each minterm: A=0,BC=00 is minterm 0 (→1), A=0,BC=01 is minterm 1 (→0), A=0,BC=11 is minterm 3 (→0), A=0,BC=10 is minterm 2 (→1), and similarly for row A=1 using minterms 4, 5, 7, 6.

Grouping the 1s — and the wraparound everyone forgets

The rule: group adjacent 1s into rectangles of size 1, 2, 4, 8... (powers of two), making each group as large as possible. The map wraps around — the leftmost column (BC=00) is adjacent to the rightmost column (BC=10), because in Gray code order 10 sits right next to 00 again once you go around. Most students only look for groups within the visible grid and miss this.

Look at all four corners here: A=0,BC=00, A=0,BC=10, A=1,BC=00, A=1,BC=10 — that's minterms 0, 2, 4, 6, and every one of them is 1. Because of the wraparound, these four cells are mutually adjacent and form a valid group of 4. What do they have in common? Check each: 000, 010, 100, 110 — the last bit (C) is 0 in all four, while A and B each take every combination freely. A group of 4 that agrees only on one variable being 0 collapses to that variable's complement: this group is simply C′.

One 1 is left uncovered: minterm 5 (A=1,B=0,C=1). It's adjacent to minterm 4 (A=1,B=0,C=0, already covered) — group them into a pair. What do minterms 4 and 5 share? 100 and 101 — A=1 and B=0 in both, while C varies. A group of 2 that agrees on two variables gives a two-literal term: AB′.

Every 1 in the map is now covered by one of these two groups, so the minimized expression is their sum:

F = C′ + AB′

Verifying it's actually correct

Don't trust a K-map answer without checking it against the original minterms. C′ is 1 exactly when C=0, which covers minterms 0, 2, 4, 6. AB′ is 1 exactly when A=1, B=0, which covers minterms 4 and 5. Their union is {0, 2, 4, 5, 6} — exactly the five minterms F was defined on, with minterm 4 correctly covered by both (overlapping groups are allowed; missing or extra minterms are the actual bug to watch for).

Compare the result to the unsimplified sum-of-products you'd get by reading the truth table directly — five three-literal terms (A′B′C′ + A′BC′ + AB′C′ + AB′C + ABC′) — versus the simplified C′ + AB′: two terms, at most two literals each. That reduction is the entire point: fewer gates, fewer inputs, less propagation delay on a real circuit.

For your viva

1. Column order for a K-map must follow Gray code (00, 01, 11, 10) so adjacent cells differ by exactly one bit — this is not optional formatting, it's what makes grouping valid.
2. The map wraps around top-to-bottom and left-to-right; corner and edge cells can be adjacent to each other.
3. A group of size 2ⁿ eliminates n variables from that group's term — the variables that stay constant across the whole group are what survive in the simplified expression.
4. Always make groups as large as possible, and always verify the final expression covers exactly the original minterms — no more, no less.

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