LinkedIn Tango #604 Answer & Analysis

Row 1 has Moons at R1C3 and R1C5 from the givens. Placing a Moon at R1C4 would create three Moons in a row (C3–C4–C5), forbidden by the rules. Therefore R1C4 must be a Sun. The bottom-edge between R1C4 and R2C4 is an Equal clue, so R2C4 must also be Sun.

Column 3 already contains three Moon givens (R1C3, R2C3, R5C3). To achieve the 3/3 balance in column 3, the remaining two cells in that column must be Suns.

After step 1, both row 1 and row 2 have Moons at C3 and C5 and a Sun at C4. Each of these rows therefore still needs one more Moon. There is an Equal clue linking R1C6 and R2C6, so the remaining Moon for both rows can be placed at column 6 — R1C6 = R2C6 = Moon.

The Cross between R5C2 and R6C2 means those two must be opposite symbols. Considering nearby row/column balances and avoiding forbidden triples leads to R6C2 = Moon and thus R5C2 = Sun. Also the Equal between R5C4 and R6C4 forces them to be identical; with surrounding constraints they become Moons.

With these key forced placements propagated, fill the remaining cells to satisfy 3 Suns / 3 Moons per row and column while avoiding three identical symbols in a row. The remaining placements follow from simple balance checks and local no-three constraints.