Solution techniques for Akari puzzles (Light-up)

A cell is free if it is:

white,
not illuminated,
without a dot,
without a bulb.

A white cell is visible from another white cell if:

they are on the same row or column,
there are no black cells along the path.

The neighborhood of a cell is made of immediately adjacent cells, vertically and horizontally.

Not illuminated cells

Case 1. If a cell X is free and there are no other free visible cells, then the cell X must contain a bulb.

Case 2. If a cell X is not illuminated, it contains a dot, there's only one free visible cell Y, then the free cell Y must contain a bulb.

Isolated numeric constraints

Given a black cell ( ) C with a numeric constraint (n_C), you need to count in its neighborhood ( ) the number of bulbs (b_C) and the number of free cells (f_C). The missing bulbs (m_C) are the difference between the numeric constraint and the bulbs counted in the neighborhood (m_C = n_C - b_C).

Case 3. m_C == 0. There are no missing bulbs in the neighborhood. Then all the free cells in the neighborhood must be marked with a dot.

Case 4. m_C == f_C. There are as many missing bulbs as free cells in the neighborhood. Then all the free cells in the neighborhood must contain a bulb.

Case 5. f_C == m_C + 1. There's one more free cell than missing bulbs in the neighborhood. Then only a single dot has to be marked. The cells in diagonal direction ( ), between two free cells in the neighborhood, will be certainly illuminated. Then each cell (in diagonal direction) between two free cells in the neighborhood must be marked with a dot.



	C


		•
	2
		•

Cross numeric constraints

Given two black cells C and D, both with a numeric constraint, positioned close to each other in a diagonal direction, then you need to count the number of missing bulbs (m_C, m_D) and the number of free cells (f_C, f_D) in the neighborhoods of C and D, respectively. As in the case 5, it must be f_C == m_C + 1 (there's one more free cell than missing bulbs). Moreover the shared cells in the neighborhoods (adjacent to both C and D) ( ) must be free.


		D
	C

Case 6. m_D == 1. There is only one missing bulb in the neighborhood of D. The two shared cells contain a bulb and a dot. Then you need to apply to C the same counting as in cases 3 and 4, but skipping the shared cells in both the neighborhoods of C and D and subtracting a bulb from the final result. This means you need to put m_C - 1 bulbs (and then dots) in the remaining f_C - 2 free cells in the neighborhood of C.


		1
	3


		1
¤	3
	¤

Case 7. m_C == 1. The shared cells are the only two free cells in the neighborhood of C. The two free cells have to contain a bulb and a dot. Thus you need to apply to D the same counting as in cases 3 and 4, but skipping the shared cells in both the neighborhoods of C and D and subtracting a bulb from the final result. This means you need to put m_D - 1 bulbs (and then dots) in the remaining f_D - 2 free cells in the neighborhood of D.


		1
¤	3
	¤

		•
		1	•
¤	3
	¤

Cul-de-sac

Case 8. Given a free cell X, you try to put a bulb in it and then apply all obvious moves (cases 1-7). If the resulting configuration contains an error (violated constraints or bulbs illuminating each other), then the cell X must be marked with a dot.

Case 9. Given a free cell X, you try to mark it with a dot and then apply all obvious moves (cases 1-7). If the resulting configuration contains an error (violated constraints or bulbs illuminating each other), then the cell X must contain a bulb.

Invariant

Case 10. Given a free cell X, you try to mark it with a dot and then apply all obvious moves (cases 1-7). In the same free cell X, you try to put a bulb and then apply all obvious moves (cases 1-7). If the two alternative resulting configurations are equals in certain cells, then those cells have an invariant content, which can be definitely applied to the game board.

If a cell Y, in both possible configurations of X, is marked with a dot or is illuminated (i.e. in both cases it cannot contain a bulb), then on the game board the cell Y must be definitely marked with a dot.

If a cell Y, in both possible configurations of X, contains a bulb, then on the game board the cell Y must definitely contain a bulb.

Michele Tomaiuolo @ UniPR.it