Independent dMAT preparation
Latin Squares: solve the target with constraints
Learn row-and-column elimination with an original 5×5 grid, a verified target, common mistakes and a short exercise.
On this page
Look for the smallest useful deduction. A Latin square is a constraint problem: every placed letter removes that letter from other cells in its row and column. You often need one intermediate cell, not a complete grid.
The official format uses a 5 × 5 grid with one occurrence of each available letter per row and column.
A candidate-intersection method
- Identify the letters missing from the target’s row.
- Identify the letters missing from its column.
- Keep only letters missing from both. One candidate forces the cell; two candidates mean more information is needed.
- Scan a nearby blank with fewer candidates. Use only a forced placement, then return to the target.
“Not ruled out yet” does not mean “proved correct.” If B and D are both possible, resist choosing the one that makes a visually pleasing pattern.
Worked original grid
These are original Aptitrail teaching examples. They are independent of official exercises and are not an official difficulty benchmark.
| Row | Column 1 | Column 2 | Column 3 | Column 4 | Column 5 |
|---|---|---|---|---|---|
| 1 | A | · | C | ? | E |
| 2 | B | C | · | E | · |
| 3 | C | D | E | · | B |
| 4 | D | · | A | · | C |
| 5 | E | A | B | C | D |
Reveal the target and the two deductions
Row 1 is missing B and D. Column 2 already contains C, D and A, so its missing letters are B and E. Row 1 already has E; therefore row 1, column 2 must be B. Now D is the only letter missing from row 1, so the target at row 1, column 4 is D.
You need only these two deductions. Do not assume a diagonal rule or an alphabetical pattern. Automated enumeration confirms that every valid completion of this original grid puts D at the target.
Why the neighbouring cell helps
Initially the target row lacks B and D, while its column lacks A, B and D. Their intersection is B or D, so the target is not immediately forced. Column 2 is more informative: B or E are missing, and row 1 excludes E. Placing B there removes B from the entire first row. That single placement resolves the target.
The reasoning depends only on the displayed clues. Although the grid was constructed from a valid square, you are not asked to infer its construction pattern.
Common mistakes
- Applying Sudoku rules: there are no smaller boxes to check and no diagonal uniqueness rule in this task.
- Guessing a cyclic pattern: an alphabetical sequence in one row does not establish the rest of the square.
- Forgetting a deduction: after mentally placing B, treat it as present when checking the next cell.
- Completing everything: once the target is forced, unnecessary placements add memory load.
A small exercise
In the same grid, determine row 4, column 2. Can you explain it using the placement already proved in row 1?
Reveal the answer and justification
E. The given letters in column 2 are C, D and A. Its blanks therefore contain B and E. The worked deduction proves B at row 1, column 2, leaving E at row 4, column 2. This answer is the same in every valid completion.
In your next drill, practise saying a concise reason: “column missing B/E; row excludes E; therefore B.” Once the method is familiar, hold only the candidates needed for your current step.
By the Aptitrail editorial team. AI-assisted original teaching examples checked with automated tests. These checks do not constitute an independent expert review. Report a correction.