Rectangle Rooms practice: find shared cells before choosing a room.
Try a new five-room challenge with no initial single candidate. Find the cells shared by two possible sixes, then check a complete solution and a reversible wrong assumption.
What can you deduce when no clue has only one candidate room? This challenge asks you to use cells shared by alternative rectangles before deciding a complete room. It has a new clue arrangement, different from the beginner and strategy examples, with five rooms instead of four. “Advanced” describes this exercise's extra reasoning step, not a measured difficulty rating.
Try the new board before revealing the deductions
Choose Practice this example beside the article, or Load advanced board in the game. Confirm Edits 0 and the following four rows. A dot is a blank cell; rows and columns start at the top left.
· · · 3 6 · · · · · 2 · 2 · · 3
The areas total sixteen: 3 + 6 + 2 + 2 + 3 = 16. A finished answer must cover every cell with disjoint rectangles, each containing exactly one matching number. The two threes and two twos refer to different clue locations; do not treat equal numbers as interchangeable rooms.
- List every room for the six at R2C1. Which cells belong to its room in every alternative, even though you do not yet know its final shape?
- The three at R1C4 could extend left or down. Use your answer about the six to exclude one direction. Continue until you have five rooms.
- After solving, undo the last room. Which cells should become unassigned? Predict the coverage count before looking at it.
You may write candidate corners on paper, but the game commits whole rooms only. It does not mark guaranteed membership cells separately or check that an accepted room permits a complete solution. Selecting opposite corners proposes a room; arrows move keyboard focus, and Enter or Space selects. Retry clears your attempt while keeping these clues.
Reveal the complete candidates and the shared-cell deduction
For each clue, enumerate the factor shapes that fit the four-by-four grid, keep those containing that clue, and exclude any containing a second number. On the empty board there are twelve candidate rectangles:
| Clue location | Candidate corners |
|---|---|
| 3 at R1C4 | R1C2–R1C4 R1C4–R3C4 |
| 6 at R2C1 | R1C1–R2C3 R1C1–R3C2 |
| 2 at R3C3 | R2C3–R3C3 R3C2–R3C3 R3C3–R3C4 R3C3–R4C3 |
| 2 at R4C1 | R3C1–R4C1 R4C1–R4C2 |
| 3 at R4C4 | R2C4–R4C4 R4C2–R4C4 |
Every clue has at least two candidates. Every cell occurs in at least two candidate rectangles too, so an immediate one-candidate scan does not start this puzzle. Count rectangles here, not just distinct owning clues.
The six has two candidates, both anchored at R1C1: a two-row, three-column rectangle or a three-row, two-column rectangle. The other six-cell shape positions are excluded by the board edges or another printed clue. Their shared cells are R1C1, R1C2, R2C1 and R2C2. Because the six must choose one of these two candidates, all four shared cells must belong to its eventual room.
In particular, R1C2 cannot belong to the top-right three. That three's horizontal candidate, R1C2–R1C4, uses R1C2 and is therefore impossible. Its vertical candidate R1C4–R3C4 is forced. This uses overlap between alternatives to prove a cell's owner without committing the six yet.
Do not draw just the four shared cells as a room. R1C1–R2C2 contains the six but has area four, so the game refuses it at Edits 0. Shared membership is a deduction about part of a future room, not a new area clue and not a separate completed room.
Finish the partition and explain each elimination
Try to finish from the forced top-right room before opening the next answer. Keep each deduction tied to a printed clue or a previously justified room. If you explored other accepted placements, Retry before comparing the clean route's edit counts.
Reveal the five-room route, coverage ledger and final map
- A: R1C4 → R3C4. The shared-cell argument forces the top three downward. Area 3 × 1 = 3. Edits 1; three cells covered.
- B: R4C2 → R4C4. The bottom-right three cannot extend upward, since R2C4–R4C4 overlaps A. Its bottom-row candidate is forced. Area 1 × 3 = 3. Edits 2; six cells covered.
- C: R3C1 → R4C1. The bottom-left two cannot extend right into B. Its vertical candidate is forced. Area 2 × 1 = 2. Edits 3; eight cells covered.
- D: R1C1 → R2C3. The six's three-row candidate R1C1–R3C2 would overlap C at R3C1. Its two-row candidate is now forced. Area 2 × 3 = 6. Edits 4; fourteen cells covered.
- E: R3C2 → R3C3. The middle two's upward option overlaps D, rightward option overlaps A and downward option overlaps B. Only its leftward option remains. Area 1 × 2 = 2. Edits 5; sixteen cells covered, five rooms, solved.
D D D A D D D A C E E A C B B B
Check 3 + 3 + 2 + 6 + 2 = 16, one clue in each room and no overlap. The map accounts for all sixteen cells. This printed clue set has exactly one complete valid partition, independently enumerated; the route explains why each of its rooms is forced.
At the win, cell controls lock and focus moves to Undo last edit. Undo removes E, returning to Edits 4 and fourteen covered cells. The only missing cells are R3C2 and R3C3. Replacing E solves again. Ten corner selections produced five edits because only the second selection commits each room.
Test the wrong assumption, then recover
Retry this board to reach zero. Deliberately place the top three horizontally, R1C2–R1C4. It is locally accepted at Edits 1 and covers three cells. Now try both candidates for the six: R1C1–R2C3 and R1C1–R3C2. Predict whether either can be added, and how many Undo actions are needed to recover.
Reveal both conflicts and the exact Undo result
The two-row six would overlap the wrong three in R1C2 and R1C3. The three-row six would overlap it in R1C2. Both attempts are refused, leaving Edits 1 and three covered cells. Those are all of the six's candidates, so no complete solution can retain the horizontal three.
Undo once removes the accepted three and returns to zero; the two refusals added no edits. Draw A vertically to restart the justified chain. The same three covered cells at Edits 1 can instead be a valid first step if you choose A, but that is a different set of cells. Counts and local acceptance do not certify an answer.
Check that the deduction really applies
Shared-cell reasoning is valid only after enumerating every remaining candidate of that clue. Omitting a possible rectangle can create a false intersection. Candidates for a single clue are alternatives, not rooms to place together. If a clue has no candidate, the current assumptions are inconsistent; do not conclude that every cell is shared.
This example needs no guessing once the complete candidate list is known. That is a property of these clues and this explanation, not a promise for all rectangle puzzles. The game does not display the paper candidate ledger or search ahead for you.
Load advanced board restores this exact challenge. Retry clears rooms, history, pending corner and removal mode on the current board. New puzzle moves through the original seven-board collection and removes the practice parameter; this extra teaching challenge is loaded separately. Refreshing its practice address restores the clues with no saved attempt. No login, timer, payment or advertisement action is required.