THE SETUP
- Latin square: every digit 1–N appears exactly once per row and column. On a 5×5 grid, each row and column contains 1, 2, 3, 4, 5.
- Inequalities: wherever a < or > sits between two adjacent cells, the cell on the > side must hold a larger digit.
The inequalities are the primary tool for placing digits early. The Latin-square constraint closes the puzzle once the inequalities have narrowed most cells.
TECHNIQUE 1 — COUNT ARROWS PER CELL
For every cell, count two things:
- S (smaller): how many other cells must be smaller than this cell, following arrows outward. (This cell is on the > side of S arrows.)
- B (bigger): how many other cells must be bigger than this cell, following arrows outward. (This cell is on the < side of B arrows.)
Then the cell's digit must be at least S+1 (to have room for S smaller values below it) and at most N−B (to have room for B larger values above it).
5×5 grid, a cell with S=2 and B=1:
Min = 2+1 = 3 Max = 5−1 = 4 Range: 3 or 4
TECHNIQUE 2 — LOOK FOR FORCED CELLS FIRST
When min = max (the range collapses to one value), the digit is forced — place it immediately.
The best cells to check first are those at the ends of long chains:
- A cell that must be bigger than 4 others on a 5×5 grid: min = 4+1 = 5. It must be 5.
- A cell that must be smaller than 4 others: max = 5−4 = 1. It must be 1.
On a 5×5, any cell at the bottom of a chain of length 4 is automatically 1. Any cell at the top of a chain of length 4 is automatically 5. These are your first placements every solve.
TECHNIQUE 3 — TRACE THE FULL CHAIN, NOT JUST ONE ARROW
Arrows often form multi-step chains: A < B < C < D. To count S and B correctly, follow the chain all the way to each end — not just the immediate neighbour.
For cell B in A < B < C < D:
- S = 1 (only A must be smaller)
- B = 2 (both C and D must be bigger)
- Range: min = 2, max = 5−2 = 3 → B is 2 or 3
If you only read the single arrow B < C, you'd compute B = 1 and a max of 4 — a much wider range that forces nothing. Full chain tracing is the difference between a useful bound and an unhelpful one.
TECHNIQUE 4 — LATIN-SQUARE ELIMINATION
Once a digit is placed in any cell, remove it from the candidates of all other cells in the same row and column. This often narrows a 2-candidate range to a single value.
After placing a digit, also check whether the inequality constraint from that cell now forces its neighbour: if you just placed 4 and the arrow says the cell to the right must be bigger, that cell can only be 5.
THE SOLVING ORDER
- For every cell, trace all chains through it and compute S and B. Calculate min = S+1 and max = N−B.
- Place all cells where min = max (forced by chain length alone).
- Apply Latin-square elimination in rows/columns with newly placed digits.
- For each newly placed digit, check if adjacent inequality constraints now force a neighbour.
- Go back to step 1 — new placements tighten chain bounds.
Steps 1–4 alone solve most beginner and intermediate Inequality grids. Harder puzzles require narrowing to a 2-candidate range and using Latin-square exclusion to resolve — never guessing.
THE BEGINNER TRAP
Reading arrows one at a time and not chaining. A single > between two cells tells you only a direction — it doesn't force a value unless the rest of the chain bounds the range tightly. Beginners who stop at single arrows hit a wall immediately and conclude the puzzle needs guessing. It doesn't.
Always trace from the end of the chain inward. Start by finding the cell that must be largest (no cell is forced bigger than it) and the cell that must be smallest (no cell is forced smaller than it) in each row. Those extremes place first and the chain unravels from both ends.