LinkedIn Tango #584 Answer & Analysis

R1 already has Moons at C3 and C4 (cells 2 and 3). Placing a Moon at R1C5 would create three consecutive Moons across C3–C5, which is illegal. Therefore R1C5 must be a Sun; this is a forced placement that starts reducing possibilities in row 1 and nearby columns.

There is an Equal relation linking R5C2 (cell25) and R6C2 (cell31). Equal means these two must be identical. Considering column 2's givens (a Moon at R2C2 and a Sun at R3C2) and the need for 3 of each in that column, the only consistent choice that balances the column with other deductions is to set both R5C2 and R6C2 to Moon. This pair choice helps lock other cells in those rows.

Cells R6C3 and R6C4 are linked by a right-edge Equal clue (they must match). Column 4 already has two given Suns (R2C4 and R4C4). To reach the required three Suns in column 4 without breaking the no-three rule elsewhere, the matching pair R6C3–R6C4 resolve as two Suns (R6C3 = Sun and R6C4 = Sun). This uses an equal clue plus column balance to force the bottom-row pair.

Column 5 already has Moons at R2C5 and R3C5. Since R1C5 was placed Sun in step 1, the column still needs one more Moon. A Cross clue links R5C5 and R6C5 (they must be opposite). Because the remaining Moon in column 5 must be placed at R6C5 to satisfy counts, that forces R5C5 to be the opposite symbol (Sun). Use Cross + column balance to get both R6C5 = Moon and R5C5 = Sun.

After the forced placements above, fill the remaining cells by checking each row and column needs (three Suns and three Moons) and avoiding any three identical symbols in a row. The remaining cells fall into place through simple balance and no-three-in-a-row eliminations. Cross-check every Equal/Cross clue as you fill to avoid contradictions.