Prove a safe move before taking it

For a conventional Minesweeper clue, subtract already proven adjacent mines from its number. If zero mines remain, its other unknown neighbors are safe. If the remaining mine count equals the remaining unknown count, every one of those neighbors is a mine. Otherwise the clue alone may not decide.

Use a precise clue model

The DLC announces a Minesweeper challenge. The following are original exercises using conventional Minesweeper rules: a clue counts mines among its adjacent cells, including diagonals. They do not establish the DLC’s actual adjacency, action or damage rules.

Each example is an independent local neighborhood. A and B are the named neighbors being discussed; all other adjacent squares are already known safe. In the first example, B is a proven mine, not a flag placed as a guess.

Use mines still needed = clue − proven adjacent mines. Then count the neighbors whose status is still unknown. Do not subtract the same mine twice or include a cell outside that clue’s neighborhood.

Example 1: clue 1, B proven mined

A?1B

1 − 1 proven = 0 mines still needed — A is safe: the one adjacent mine is already accounted for by B.

Example 2: clue 2, no proven mines

A?2B?

2 − 0 proven = 2 mines still needed across 2 unknown cells — A and B are both mines: every unknown neighbor must hold one.

Example 3: clue 1, no proven mines

A?1B?

1 − 0 proven = 1 mine still needed across 2 unknown cells — Ambiguous: either A or B is the mine. This clue alone proves neither cell.

Reveal each conclusion when you have your own answer. The next section writes the same three cases out in full.

Three local examples, three different conclusions

A clue of 1 with one proven mine

The clue touches A and B. B is already proven to contain the one adjacent mine; A is the only remaining unknown. The remaining count is 1 − 1 = 0, so A is safe.

A clue of 2 with two unknown neighbors

A and B are its only unknown neighbors, with no proven adjacent mine. It still needs two mines and has exactly two unknown cells. Both A and B must be mines. Opening either would contradict the clue’s count.

A clue of 1 with two unknown neighbors

A and B are its only unknown neighbors, with no proven adjacent mine. One mine is needed across two cells. Either A is the mine or B is the mine; this clue alone proves neither cell safe.

The difference between the first and third examples is evidence about B. Drawing a flag on B does not create that evidence.

A safe deduction procedure

  1. Select one revealed clue and enumerate exactly its adjacent cells.
  2. Separate known safe, proven mine and still unknown cells.
  3. Subtract only the proven mines from the clue.
  4. If the remainder is zero, mark the unknowns as safe in your reasoning.
  5. If the remainder equals the unknown count, mark those unknowns as mines.
  6. If neither rule applies, keep them unknown and inspect another clue.

A negative remainder or a remainder larger than the number of unknowns indicates an inconsistent record under this model. Recheck a previous assumption or neighborhood before acting; it is not an instruction to force a move.

When no move is forced

“Not proven to be a mine” does not mean safe. In the third example, both assignments satisfy the single clue, so choosing A is a guess unless another observation constrains it.

Additional clues can help, but only when their neighborhoods and shared cells are tracked exactly. Keep the local record small and inspect one constraint at a time rather than treating the whole board as a pattern to memorize.

Before transferring these deductions to the DLC, confirm what its numbers mean, which cells count as adjacent and which actions reveal, mark or damage a cell. This page teaches clue reasoning; it does not supply a verified underground board solution or an automatic solver.

Sources and further reading