Rectangle Rooms strategy: check the candidates, then the uncovered cells.
Compare nine candidate rectangles, reproduce two locally legal dead ends, and use a blank corner cell to force the right room on a different checked board.
A rectangle can pass every placement check and still make the puzzle impossible to finish. The game checks the room you draw: one number, matching area and no overlap. It does not search ahead to certify your choice. This lesson turns that distinction into a practical method: list the possible rooms, eliminate conflicts, then inspect the blank cells that still need an owner.
We use a different clue arrangement from the beginner lesson. The exercise comes from the site's finite checked collection; it is a new worked example for this series, rather than a claim of a new game or a newly expanded puzzle bank. You can reproduce both failed choices and recover with Undo.
Start with candidates, not committed rooms
Choose Practice this example beside the article, or Load strategy board in the game. Confirm Edits 0 and these clues. A dot represents a blank cell:
· 4 · · · · · · · · 8 · 2 · 2 ·
R means row and C means column, counted from the top left. A candidate is an axis-aligned rectangle inside this grid that contains exactly one printed number and has the required area. Candidates on this list are alternatives, not rooms to draw together. The two twos are different clues, so record their locations as well as their values.
For each clue, factor its area into a height and width that fit the board, then slide each shape through every position containing that clue. Reject any position containing a second number. The eight can fit as 2 × 4 or 4 × 2; 1 × 8 and 8 × 1 do not fit. The four can fit as 1 × 4, 2 × 2 or 4 × 1. Each two has a horizontal or vertical domino shape.
| Clue | Candidate corners |
|---|---|
| 4 at R1C2 | R1C1–R1C4 R1C1–R2C2 R1C2–R2C3 R1C2–R4C2 |
| 8 at R3C3 | R2C1–R3C4 |
| 2 at R4C1 | R3C1–R4C1 R4C1–R4C2 |
| 2 at R4C3 | R4C2–R4C3 R4C3–R4C4 |
There are nine candidate rectangles in total: four for the four, one for the eight, and two for each two. This is the complete list on an empty board. Once a room is placed, discard candidates that overlap it; do not discard a candidate merely because its shape looks unusual.
Why the eight is forced
The only surviving 2 × 4 position for the eight spans rows 2–3 and all four columns. Rows 3–4 would include the bottom twos. Every 4 × 2 position containing the eight would also contain a bottom two, with some positions containing the top four as well. Therefore the eight has exactly one candidate: R2C1–R3C4.
A single candidate for a clue must occur in any complete solution, provided the candidate list is exhaustive and the current assumptions are valid. Placing this room covers eight cells. It eliminates three of the four's alternatives and the vertical room of the left two. The right two still has two locally compatible choices; the forced eight alone does not decide everything.
Experiment 1: an accepted square that blocks the forced room
Before the clean solution, Retry to begin at zero. Draw R1C1–R2C2, the square candidate for the four. It is accepted: one clue, area four, no overlap. Edits becomes 1 and four cells are covered. Now try R2C1–R3C4 for the eight. The game refuses it because it overlaps the square. Edits stays 1; a refusal does not remove the earlier room.
Reveal why the square is a dead end and how to undo it
The eight has no alternative room. Its only candidate overlaps the square in R2C1 and R2C2. Because every clue must receive a room, no complete partition can contain that square. The four's acceptance was a local check, not a correct-answer signal.
Undo last edit removes the accepted square, returning to Edits 0 and no rooms. The refused eight was never added, so there is no extra failed edit to undo first. Place the eight now and it succeeds at Edits 1, eight cells covered. Retry would instead clear the entire attempt while preserving these clues.
Inspect a blank cell when a clue still has two choices
With just the eight placed, look at R4C4. It is blank, but the rules still require a room to cover it. In the complete candidate list, the only rectangle containing this cell is R4C3–R4C4, belonging to the right two. That room is forced even though this clue still has two candidate shapes.
This is a different deduction from “one clue, one remaining candidate”: it is “one uncovered cell, one possible owner.” Check all clues that could reach the cell, rather than assuming it must belong to the nearest number. On this board the table makes the search complete.
Reveal a second wrong choice and the exact recovery checkpoint
From Edits 1 with only the eight, try the right two horizontally to the left: R4C2–R4C3. This is accepted at Edits 2, ten cells covered. R4C4 remains unassigned and has no nonoverlapping candidate left. The right two is already inside a room; making a second room containing it would overlap the first. A one-cell room at R4C4 would contain no number. This is a dead end even though ten cells are covered.
Undo once restores just the eight: Edits 1, eight cells covered. Then draw R4C3–R4C4, the right-facing room. You return to Edits 2 and ten covered cells, but now a compatible completion remains. The same edit and coverage counts can describe different positions; neither count is a correctness proof.
Run a clean four-room solution
Retry first so the following counts describe a fresh attempt. Select each pair of corner cells once. The letters name rooms in placement order.
- A: R2C1 → R3C4. Place the eight's only candidate, covering rows 2–3. Edits 1; eight cells covered.
- B: R4C3 → R4C4. Cover the bottom-right blank cell with its only possible owner, the right two. Edits 2; ten cells covered.
- C: R4C1 → R4C2. The left two's vertical candidate conflicts with A, so its bottom-row candidate is forced. Edits 3; twelve cells covered.
- D: R1C1 → R1C4. The top row is the only remaining room for the four. Edits 4; sixteen cells covered and the board is solved.
Reveal the final map, areas and a test before placing the eight
D D D D A A A A A A A A C C B B
The areas are A = 2 × 4 = 8, B = 1 × 2 = 2, C = 1 × 2 = 2 and D = 1 × 4 = 4. Their total is sixteen. Every room contains exactly one corresponding clue; they are disjoint and cover all sixteen cells. The printed puzzle has exactly one complete valid partition, checked independently.
After the win, Undo returns to Edits 3 and twelve covered cells: only the top row is missing. Undo three more times to reach zero. On a fresh board, can you force the right two without placing the eight first? Yes. The complete empty-board candidate list already shows that R4C4 has only one possible owner. Choosing a clue with few options is useful, but scanning blank boundary cells can be just as decisive.
Use the method within its limits
Keep a pencil ledger of candidate corners, remove only candidates excluded by a rule, and alternate clue checks with uncovered-cell checks. If an uncovered cell has no candidate, revisit a previous assumption; if it has one, that candidate is forced. If a clue has no candidate, the current assumptions are inconsistent. Several candidates can remain without any immediate single deduction, so this method is not a promise that every puzzle can be solved without branching.
An area sum of sixteen is only a consistency check. Nine candidates are not nine solutions. Local acceptance, covered-cell count and edit count do not establish that the remainder can be partitioned. A contradiction is persuasive only when the candidate list and the intermediate eliminations are complete.
Load strategy board restores this exact clue set; Retry this board clears the current attempt, pending corner and removal mode. New puzzle returns to the rotating seven-board collection and clears the practice address. A practice address reloads the selected clues after refresh, without saving rooms or history. Your choices stay in the browser tab. No account, timer, payment or advertisement action is needed.