Nonogram contradiction: test a possibility, prove one fact.
If filling an unknown square forces an impossible line, that square must be empty. Keep the trial separate, follow its consequences, name the exact conflict, then erase the whole trial before adding the fact you proved.
This original 5 × 5 teaching puzzle shows one such proof. It is separate from our playable puzzles. If clue groups or ordinary overlap are new to you, read the Nonogram introduction or the overlap examples and line helper first.
Start with nine proved marks
R1 means row 1; C1 means column 1. Read row clues from left to right and column clues from top to bottom. A clue of 1, 1 means two separate single filled squares, with at least one empty square between them.
Beginning with an unknown grid, eight ordinary line checks give the position shown here. There are nine proved marks and sixteen unknown squares. Question marks are still open to either filled or empty.
Now compare every legal placement in each of the five rows and five columns, respecting the marks already proved. All ten lines have no new shared mark. Each remaining unknown differs between at least two legal placements of its own line. Repeating that same line-by-line check gives nothing further here.
The clues and existing marks still agree. The next section asks what would happen under one additional assumption.
| Row numbers and clues | C11 | C23 | C33 | C421 | C511 |
|---|---|---|---|---|---|
| R13 | |||||
| R23 | |||||
| R33 | |||||
| R41 | |||||
| R52 |
How the starting marks follow from eight line checks
For the short patterns below, # means filled and . means empty. These are legal alternatives to compare, not extra marks to copy.
- R1, clue 3: the run can start at C1, C2 or C3. All three placements fill R1C3.
- R2, clue 3: the same overlap fills R2C3.
- R3, clue 3: the same overlap fills R3C3.
- C2, clue 3: a three-square run in five rows always covers the middle, so fill R3C2.
- R3, clue 3: with C2 and C3 filled, its only placements are
###..and.###.. Both leave R3C5 empty. - C3, clue 3: R1, R2 and R3 are already filled. The run is complete; mark R4C3 and R5C3 empty.
- C4, clues 2, 1: its legal top-to-bottom placements are
##.#.,##..#and.##.#. All fill R2C4. - R2, clue 3: with C3 and C4 filled, its only placements are
.###.and..###. Both leave R2C1 empty.
On a separate copy, assume R1C1 is filled
R1C1 is unknown in the starting position. Temporarily fill it on a copy or in separate notes. This is an assumption, not a proved move.
In every trial diagram, A identifies that assumption and T identifies a consequence that depends on it. Both use dashed borders. Unbadged filled and empty marks are the nine facts from the starting position.
Keep those two kinds of information separate until the trial has been resolved.
| Row numbers and clues | C11 | C23 | C33 | C421 | C511 |
|---|---|---|---|---|---|
| R13 | |||||
| R23 | |||||
| R33 | |||||
| R41 | |||||
| R52 |
R1 would have to be
###..R1 has clue 3. If C1 is filled, the three-square run must begin there: there is no room to extend left. It therefore fills C1–C3 and leaves C4–C5 empty.
Conditional additions: R1C2 filled; R1C4 and R1C5 empty. R1C3 was already proved.
Trial consequence 1: R1 completes Row numbers and clues C11 C23 C33 C421 C511 R13 R23 R33 R41 R52 ■ = filled · × = empty · ? = unknown. Trial only: A = provisional assumption; T = provisional consequence. Dashed cells are not proved. Unbadged marks were already proved. C2 would have to be
###..The trial now fills R1C2, while R3C2 was already filled. C2 has clue 3, so one uninterrupted group must cover rows 1–3. Rows 4 and 5 would be empty.
Conditional additions: R2C2 filled; R4C2 and R5C2 empty. These still depend on the R1C1 assumption.
Trial consequence 2: C2 completes Row numbers and clues C11 C23 C33 C421 C511 R13 R23 R33 R41 R52 ■ = filled · × = empty · ? = unknown. Trial only: A = provisional assumption; T = provisional consequence. Dashed cells are not proved. Unbadged marks were already proved. R2 would have to be
.###.In the trial, R2C2 joins the already proved fills at C3 and C4. That uses R2's entire group of 3, so R2C5 would be empty. R2C1 was already empty.
Now check C5, the bordered column in this diagram. Its first three squares would all be empty.
Trial consequence 3: R2 completes; check C5 Row numbers and clues C11 C23 C33 C421 C511 R13 R23 R33 R41 R52 ■ = filled · × = empty · ? = unknown. Trial only: A = provisional assumption; T = provisional consequence. Dashed cells are not proved. Unbadged marks were already proved. Column C5 has an outer border for the conflict check.
C5 cannot fit its clues 1, 1
Read C5 from top to bottom: ...??. R1C5 and R2C5 are empty only under the trial; R3C5 was already proved empty. That leaves just R4C5 and R5C5 available.
Two separate single fills need at least three positions: filled, empty, filled. Two remaining adjacent squares cannot hold them. Filling both would make a group of 2, not the clues 1, 1. No placement of C5 satisfies its clues under this assumption.
Assume R1C1 filled → R1 completes → C2 completes → R2 completes → C5 cannot fit 1, 1. Therefore R1C1 cannot be filled.
The conflict uses the clues and the trial deductions; it does not require looking at the completed picture.
Discard the entire trial. Prove R1C1 empty.
Return to the original nine-mark starting position. Erase the assumption and every T mark. A consequence of a failed assumption does not become a fact just because it looked useful.
An unknown monochrome Nonogram square must be filled or empty. Filling R1C1 led to an impossibility, so R1C1 is empty. Add only that mark to the original board.
The reset board contains ten proved marks and fifteen unknowns. It differs from the starting board in exactly one square, R1C1. For example, R2C2 is unknown again.
| Row numbers and clues | C11 | C23 | C33 | C421 | C511 |
|---|---|---|---|---|---|
| R13 | |||||
| R23 | |||||
| R33 | |||||
| R41 | |||||
| R52 |
Ordinary line checks can now continue. R1 has two possible placements, .###. and ..###; both fill R1C4. With R1C4 and R2C4 filled, C4's clues 2, 1 force R3C4 empty. Continue crossing between lines from those proved facts.
Reveal the completed teaching puzzle and continuation
After the clean reset, these ordinary line checks finish this particular puzzle. Each new mark is shared by every legal placement of the named line.
- R1: 2 legal placements. New proved mark: R1C4 filled.
- C4: 2 legal placements. New proved mark: R3C4 empty.
- R3: 1 legal placement. New proved mark: R3C1 filled.
- C1: 1 legal placement. New proved marks: R4C1 empty; R5C1 empty.
- R5: 1 legal placement. New proved marks: R5C2 empty; R5C4 filled; R5C5 filled.
- C2: 2 legal placements. New proved mark: R2C2 filled.
- R2: 1 legal placement. New proved mark: R2C5 empty.
- C4: 1 legal placement. New proved mark: R4C4 empty.
- C5: 1 legal placement. New proved marks: R1C5 filled; R4C5 empty.
- R1: 1 legal placement. New proved mark: R1C2 empty.
- R4: 1 legal placement. New proved mark: R4C2 filled.
| Row numbers and clues | C11 | C23 | C33 | C421 | C511 |
|---|---|---|---|---|---|
| R13 | |||||
| R23 | |||||
| R33 | |||||
| R41 | |||||
| R52 |
What the trial does, and does not, establish
A contradiction rules out the assumption only when the starting marks and every conditional deduction are sound. Name an impossible clue, run or gap; difficulty finding another move is not itself a contradiction.
If a trial produces no contradiction, that alone proves nothing. Do not promote its provisional marks to facts. This one example also does not show that every stalled Nonogram has a short contradiction proof or that this method will finish every puzzle.
The example was checked by enumerating the possible five-square lines and independently checking the complete grid. Its clues have one solution. The lesson's refutation rests on the C5 conflict, rather than an assumption that the answer is unique.