How to solve Shikaku

1. Divide the whole grid into rectangles (squares count). Every cell belongs to exactly one room.
2. Each room contains exactly one number.
3. That number is the room's area: its width times its height, counted in cells.

A number can only become rectangles whose sides multiply to it. A 6 can be 1×6, 6×1, 2×3 or 3×2. A 4 can be 1×4, 4×1 or 2×2. A prime number such as 5 or 7 can only be a straight line, and a 1 is a single cell. The fewer shapes a number has, the easier it is to place.

Look at numbers near an edge or a corner, numbers hemmed in by other numbers, and large numbers in tight spaces. A room may never contain a second number, and it can't leave the grid, so many shapes are ruled out at once. Often a corner number has only one shape that fits.

Every cell must end up in some room. Pick an empty cell and ask which numbers could possibly stretch a room over it. If only one can, that cell belongs to that number, and any shape that misses the cell can be crossed out.

Every room you draw takes cells away from its neighbours, so their options shrink. After each placement, look again at the numbers next to it: one of them usually has a single shape left. Good solves are chains of small, certain steps, not guesses.

Here is a small 5×5 floor plan with six numbers. Rows are counted from the top and columns from the left. Each picture shows the rooms drawn so far, with the newest one outlined.

1. 6

   3

   6

   2

   3

   5

   The empty floor plan. The numbers add up to 25, which is exactly the number of cells, as they always do.

2. 6

   3

   6

   2

   3

   5

   1. Start with the 5 in the bottom-right corner. 5 is prime, so its room is a straight line of five cells. Lying flat along the bottom row, it would also contain the 3, so it must stand upright: the whole right-hand column.

3. 6

   3

   6

   2

   3

   5

   2. Now look at the empty cell in row 1, column 3 (dashed). Any room for the 3 stays in that 3's own row or column, so it can't reach this cell, and the 6 in row 3 would have to take in a second number to get there. Only the 6 in row 1 can cover it, and of its shapes only the 2×3 block across the top-left does.

4. 6

   3

   6

   2

   3

   5

   3. The 3 in row 2 now has just one shape left that avoids the rooms already drawn and the 2 below it: a vertical line from row 1 to row 3 in column 4.

5. 6

   3

   6

   2

   3

   5

   4. The 6 in row 3 is boxed in: the first 6 sits above it, the 3's column is to its right, and a tall 3×2 shape would take in the 3 in the bottom row. The only fit is the 2×3 block covering rows 3 and 4.

6. 6

   3

   6

   2

   3

   5

   5. The 2 in row 4 has one free neighbour left, the cell below it, so its room is that vertical pair.

7. 6

   3

   6

   2

   3

   5

   6. The 3 in the bottom row fills the last three cells. Every cell is covered, each room holds one number, and each number matches its area: solved.

• Add up all the numbers: they always total the number of cells. If your rooms leave a gap, a room somewhere is the wrong shape.
• Stuck? Hunt for a cell only one number can reach. It breaks most deadlocks.
• On a keyboard, move with the arrow keys, press Enter on one corner and Enter again on the opposite corner to draw a room.
• Practise on past puzzles in the archive. They don't affect your streak or the standings.