Takuzu / Binairo › Guide
Stuck on Takuzu? Use the Distinct-Lines Rule
When pairs, sandwiches and line counts stop in classic Takuzu, compare a nearly finished line with a completed parallel line. If its two missing symbols could copy that line, reject the duplicate ending and use the other balanced ending.
This rule applies to classic Takuzu / Binairo, not Solvedle's Tango. Compare rows with rows and columns with columns. A row matching a column is allowed. A unique solution and distinct lines are separate properties.
When does this two-gap deduction apply?
- The reference line is complete.
- The parallel line has exactly two gaps and matches the reference in every known position.
- The reference has opposite symbols at the two gap positions.
- Balance requires one of each symbol in those gaps. The matching order would duplicate the reference, so swap them.
The examples use suns as 0 and moons as 1, like Solvedle's online game. All three source puzzles are 6×6 medium boards. Their setup reaches a point with no pair, sandwich or completed-count deduction available before distinct-line reasoning is used.
1. Compare row 5 with row 2
| Line | Before | Reason |
|---|---|---|
| row 2 | 001101 | Completed reference |
| row 5 | 001··1 | Known positions match; two gaps remain |
The missing cells R5C4, R5C5 need one 0 and one 1. Copying the reference would make identical rows. Therefore R5C4 is sun (0); R5C5 is moon (1). The second value also follows from the line's balance once the first is fixed.
Show the board after filling both gaps
Reconstruct the position from its starting clues
- The suns at R1C2, R1C4 surround a gap. Another sun in that gap would make three consecutive suns. Place a moon (1) at R1C3.
- The moons at R4C1, R6C1 surround a gap. Another moon in that gap would make three consecutive moons. Place a sun (0) at R5C1.
- The adjacent suns at R5C1, R5C2 cannot be extended by a third sun. Place a moon (1) at R5C3.
- The line R1C1, R2C1, R3C1, R4C1, R5C1, R6C1 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R2C1, R3C1.
- The adjacent suns at R2C1, R2C2 cannot be extended by a third sun. Place a moon (1) at R2C3.
- The adjacent moons at R2C3, R2C4 cannot be extended by a third moon. Place a sun (0) at R2C5.
- The line R2C1, R2C2, R2C3, R2C4, R2C5, R2C6 already contains 3 suns. Every unfilled cell in it must be a moon to balance the line. Place a moon (1) at R2C6.
- The adjacent suns at R1C2, R2C2 cannot be extended by a third sun. Place a moon (1) at R3C2.
- The moons at R3C2, R3C4 surround a gap. Another moon in that gap would make three consecutive moons. Place a sun (0) at R3C3.
- The line R1C2, R2C2, R3C2, R4C2, R5C2, R6C2 already contains 3 suns. Every unfilled cell in it must be a moon to balance the line. Place a moon (1) at R4C2, R6C2.
- The adjacent moons at R4C1, R4C2 cannot be extended by a third moon. Place a sun (0) at R4C3.
- The adjacent moons at R6C1, R6C2 cannot be extended by a third moon. Place a sun (0) at R6C3.
- The adjacent moons at R2C4, R3C4 cannot be extended by a third moon. Place a sun (0) at R4C4.
- The adjacent suns at R4C3, R4C4 cannot be extended by a third sun. Place a moon (1) at R4C5.
- The line R4C1, R4C2, R4C3, R4C4, R4C5, R4C6 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R4C6.
- The suns at R4C6, R6C6 surround a gap. Another sun in that gap would make three consecutive suns. Place a moon (1) at R5C6.
2. Compare row 6 with row 3
| Line | Before | Reason |
|---|---|---|
| row 3 | 101100 | Completed reference |
| row 6 | 10·1·0 | Known positions match; two gaps remain |
The missing cells R6C3, R6C5 need one 0 and one 1. Copying the reference would make identical rows. Therefore R6C3 is sun (0); R6C5 is moon (1). The second value also follows from the line's balance once the first is fixed.
Show the board after filling both gaps
Reconstruct the position from its starting clues
- The adjacent moons at R1C2, R1C3 cannot be extended by a third moon. Place a sun (0) at R1C4.
- The line R1C1, R1C2, R1C3, R1C4, R1C5, R1C6 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R1C5.
- The suns at R2C1, R2C3 surround a gap. Another sun in that gap would make three consecutive suns. Place a moon (1) at R2C2.
- The line R2C1, R2C2, R2C3, R2C4, R2C5, R2C6 already contains 3 suns. Every unfilled cell in it must be a moon to balance the line. Place a moon (1) at R2C6.
- The adjacent suns at R1C1, R2C1 cannot be extended by a third sun. Place a moon (1) at R3C1.
- The moons at R3C1, R3C3 surround a gap. Another moon in that gap would make three consecutive moons. Place a sun (0) at R3C2.
- The line R1C1, R2C1, R3C1, R4C1, R5C1, R6C1 already contains 3 suns. Every unfilled cell in it must be a moon to balance the line. Place a moon (1) at R6C1.
- The line R1C2, R2C2, R3C2, R4C2, R5C2, R6C2 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R4C2, R6C2.
- The adjacent suns at R1C4, R2C4 cannot be extended by a third sun. Place a moon (1) at R3C4.
- The adjacent moons at R3C3, R3C4 cannot be extended by a third moon. Place a sun (0) at R3C5.
- The line R3C1, R3C2, R3C3, R3C4, R3C5, R3C6 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R3C6.
3. Compare column 2 with column 6
| Line | Before | Reason |
|---|---|---|
| column 6 | 010110 | Completed reference |
| column 2 | ·10·10 | Known positions match; two gaps remain |
The missing cells R1C2, R4C2 need one 0 and one 1. Copying the reference would make identical columns. Therefore R1C2 is moon (1); R4C2 is sun (0). The second value also follows from the line's balance once the first is fixed.
Show the board after filling both gaps
Reconstruct the position from its starting clues
- The moons at R1C3, R1C5 surround a gap. Another moon in that gap would make three consecutive moons. Place a sun (0) at R1C4.
- The suns at R2C1, R2C3 surround a gap. Another sun in that gap would make three consecutive suns. Place a moon (1) at R2C2.
- The suns at R2C3, R2C5 surround a gap. Another sun in that gap would make three consecutive suns. Place a moon (1) at R2C4.
- The line R2C1, R2C2, R2C3, R2C4, R2C5, R2C6 already contains 3 suns. Every unfilled cell in it must be a moon to balance the line. Place a moon (1) at R2C6.
- The adjacent moons at R3C4, R3C5 cannot be extended by a third moon. Place a sun (0) at R3C3.
- The line R3C1, R3C2, R3C3, R3C4, R3C5, R3C6 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R3C2.
- The suns at R5C1, R5C3 surround a gap. Another sun in that gap would make three consecutive suns. Place a moon (1) at R5C2.
- The moons at R6C1, R6C3 surround a gap. Another moon in that gap would make three consecutive moons. Place a sun (0) at R6C2.
- The adjacent moons at R6C3, R6C4 cannot be extended by a third moon. Place a sun (0) at R6C5.
- The line R6C1, R6C2, R6C3, R6C4, R6C5, R6C6 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R6C6.
- The adjacent suns at R2C3, R3C3 cannot be extended by a third sun. Place a moon (1) at R4C3.
- The adjacent moons at R2C4, R3C4 cannot be extended by a third moon. Place a sun (0) at R4C4.
- The line R1C4, R2C4, R3C4, R4C4, R5C4, R6C4 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R5C4.
- The adjacent suns at R5C3, R5C4 cannot be extended by a third sun. Place a moon (1) at R5C5.
- The moons at R3C5, R5C5 surround a gap. Another moon in that gap would make three consecutive moons. Place a sun (0) at R4C5.
- The adjacent suns at R4C4, R4C5 cannot be extended by a third sun. Place a moon (1) at R4C6.
- The line R1C6, R2C6, R3C6, R4C6, R5C6, R6C6 already contains 3 moons. Every unfilled cell in it must be a sun to balance the line. Place a sun (0) at R1C6.
When should I avoid this shortcut?
Do not treat two unfinished lines as identical just because their currently filled cells agree. Do not compare a row to a column. If there are more than two gaps, additional reasoning is needed; the two-gap proof above does not establish an entire longer ending. Never import this rule into Tango.
Follow the full Takuzu walkthrough, download classic binary printables, or compare Takuzu, Binairo and Binoxxo. The publisher's binary-puzzle rules explain the underlying constraints.