Minimize the three variable logic function using the following K-Map.

Concept

In Karnaugh map minimization, any two cells representing minterms that differ in exactly one variable are adjacent and can be grouped together; larger groups of 2, 4, 8, ... cells are also valid as long as every cell in the group is marked 1. Every variable that changes value across a group is eliminated, while variables that stay constant across the group appear in the group's product term (uncomplemented if constant at 1, complemented if constant at 0). The minimized SOP expression is the smallest set of such groups (prime implicants) that together cover every 1-cell at least once.

Applying to This K-Map

For f(A, B, C) = Σm(0, 2, 5, 7), plot the four 1-cells on the 3-variable map (columns AB, rows C):

1. Group 1: minterms 0 and 2 are adjacent, differing only in B. A stays at 0 and C stays at 0 across the pair, so B is eliminated, giving A'C'.

2. Group 2: minterms 5 and 7 are adjacent, differing only in B. A stays at 1 and C stays at 1 across the pair, so B is eliminated, giving AC.

3. No 4-cell group is possible here: minterms 1, 3, 4, and 6 are all 0, so the pair {0, 2} cannot merge with {5, 7} or with any other cell.

Combining the two groups gives the minimized expression A'C' + AC.

Cross-Check

Checking every input combination against A'C' + AC confirms it equals 1 exactly at (A, B, C) = (0, 0, 0), (0, 1, 0), (1, 0, 1), and (1, 1, 1) — minterms 0, 2, 5, and 7 — and 0 at every other combination, matching the function exactly.

Answer

The minimized expression is A'C' + AC.