Mark the temporary assumption
A short Takuzu contradiction starts from a position whose existing entries you trust. Choose one empty cell and temporarily assign 0 or 1. Everything already established stays fixed. Every later move needs a reason that follows from those entries and the assumption.
Use a paper copy and circle the assumed value. Number its consequences so you can remove them together. Our game has Undo and Redo, but it has no editable pencil-note mode. Paper makes the distinction between an established entry and a temporary branch easier to keep visible.
The example below is an original teaching position with five given digits. Dots in the board mark empty cells. R1C3 means row 1, column 3, counting from the top left. This position deliberately has multiple completions. We exhaustively checked all 56, and every one has R1C1 = 1. You can therefore examine the proof without relying on a promised unique solution.
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | · | 0 | · | · | · | 0 |
| 2 | · | · | · | 0 | · | · |
| 3 | · | · | 1 | · | 1 | · |
| 4 | · | · | · | · | · | · |
| 5 | · | · | · | · | · | · |
| 6 | · | · | · | · | · | · |
Follow only forced consequences
Assume R1C1 = 0. Row 1 now starts 00, so R1C3 must be 1 to avoid three zeros. Look down column 3. R1C3 and the given R3C3 are both 1, which forces the cell between them, R2C3, to be 0.
Row 2 now contains zeros at C3 and C4. R2C5 must be 1. In column 5, that new 1 sits directly above the given R3C5 = 1. R1C5 must consequently be 0. Each move crosses into a line affected by the previous placement.
Follow the table in order on a fresh copy. There is only one temporary choice. The remaining values are compulsory under that choice. If a later placement merely looks convenient, leave it empty. A second unproved choice would require its own separate analysis before you could blame the original assumption.
| Step and cell | Value | Reason |
|---|---|---|
| Assume R1C1 | 0 | Temporary starting choice |
| 1. R1C3 | 1 | R1C1 and given R1C2 are 00 |
| 2. R2C3 | 0 | R1C3 and given R3C3 are 1_1 |
| 3. R2C5 | 1 | R2C3 and given R2C4 are 00 |
| 4. R1C5 | 0 | R2C5 and given R3C5 are 11 |
Identify the exact contradiction
Row 1 has become 001_00. Its zeros occupy C1, C2, C5 and C6. A six-cell row can contain only three zeros, so these four zeros are a concrete contradiction. No value in the remaining blank can repair the excess.
This establishes that R1C1 cannot be 0. Since the cell permits only two values, R1C1 must be 1. The proof has a precise endpoint even though most of the board remains empty. Completing the puzzle would add work without strengthening this deduction.
Other valid endpoints include a horizontal or vertical triple, two identical completed rows, or two identical completed columns. A line with no possible legal completion also qualifies, provided you checked every candidate compatible with the established entries. Diagonal triples do not count under classic rules. Two unfinished lines that currently look similar do not automatically form a duplicate.
The argument assumes the starting entries are correct and at least one legal completion exists. If earlier player entries were already inconsistent, a failed branch could reflect that earlier mistake. Our completion check supplies that missing safeguard for this teaching position.
Recognize a branch that has not failed
Reset to the five given digits. This time assume R2C5 = 0. Row 2 has 00 at C4 and C5, forcing R2C3 = 1. Together with given R3C3 = 1, this forces R1C3 = 0. The adjacent zeros at R1C2 and R1C3 then force R1C1 = 1.
Row 1 now has three zeros at C2, C3 and C6. Its remaining cells C4 and C5 must both be 1. The completed row is 100110, which has equal counts and no triple. This short trace has produced no contradiction.
You cannot reject R2C5 = 0 because you stopped here. You also cannot certify that starting value merely because these consequences stayed legal. For this particular teaching position, the complete witness below independently shows that the assumption can survive all the way to a legal board. Check its rows and columns, including uniqueness.
A surviving completion establishes possibility. Proving a value compulsory would require ruling out its alternative. Keep that distinction when a binary puzzle assumption and conflict exercise ends without a visible conflict.
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| 2 | 0 | 1 | 1 | 0 | 0 | 1 |
| 3 | 0 | 0 | 1 | 0 | 1 | 1 |
| 4 | 1 | 1 | 0 | 1 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 0 | 1 |
| 6 | 1 | 0 | 1 | 0 | 1 | 0 |
Undo the branch and keep the proved value
Return to the original position after the failed R1C1 = 0 branch. Erase all four temporary consequences and the assumption. Enter only R1C1 = 1 as the newly established result. Any other placement needs a fresh justification from the restored position.
The game’s Check button flags visible rule conflicts. A clean check does not certify that every entry belongs to a solution. If your branch ends because you cannot find another move, record it as unresolved and restore the board. You can choose a shorter branch later or look for a direct deduction.
For the worked proof, a sufficient paper record is R1C1 = 0, then R1C3 = 1, R2C3 = 0, R2C5 = 1, R1C5 = 0. Beside it, write that row 1 contains four zeros. That final count lets another player verify exactly why you kept R1C1 = 1.
Rules and example verification
BinaryPuzzle.com’s classic rules provide the balance, adjacency and uniqueness constraints used here. Both branch traces and the teaching position are our own examples. Exhaustive row enumeration found 56 legal completions, none with R1C1 = 0. The displayed witness was checked against every given digit and all three constraints.