LinkedIn Mini Sudoku #271 Answer & Analysis

Box (rows 3–4, cols 1–3) already has 3,4,5 in row 3 and 6 in row 4 from the givens. That box is therefore missing {1,2}. Those two empty cells lie in row 4, and row 4 itself already contains two of the digits (3,4,5,6). So row 4 must take {1,2} in its two empty spots. Check column 1: R2C1 is a given 1, so R4C1 cannot be 1 and must be 2. The remaining cell in that box (R4C3) must be 1.

Row 3 has givens 3,4,5,1 so it's missing {2,6} and these empty cells are in cols 4 and 6 (the central-right box). Column 6 already contains a given 6 in row 5 (R5C6), so R3C6 cannot be 6 and must be 2. That forces R3C4 to be 6.

Column 4 now contains 2 (R1), 4 (R2), 6 (R3), 3 (R4) and 5 (R5) — after placing R3C4=6 and using the givens — so the only missing digit in column 4 is 1. That leaves R6C4 = 1.

With R6C4=1 and the givens R5C4=5, R5C6=6, the bottom-right box needs {2,3,4}. Use row 6 and column checks: row 6 currently has 1 (now placed) and cannot have duplicates. Eliminations give R6C5=3 and R6C6=4, which then sets R5C5=2 (only remaining in row 5).

With many central and bottom cells determined, the top two rows reduce to single candidates in several places. Use column elimination and the box constraints to place the remaining digits in rows 1 and 2. For example, once columns 1–3 and box placements are set, R1C1–R1C6 and R2C2,R2C4–R2C6 fall into place by straightforward elimination.