TECHNIQUE 3 OF 4
Compare overlapping clues by subtracting shared squares
Two clues may be inconclusive on their own but decisive together. Name their unresolved neighbours and compare the shared set with the extra squares.
The rule
After subtracting proved mines, suppose one clue requires a mines in set S and another requires b mines in a larger set T that contains every square of S. The extra squares T minus S contain exactly b minus a mines. If that difference is zero they are all safe; if it equals the number of extra squares they are all mines. Partial overlap without containment is not enough for this subtraction.
A common mistake
Subtracting two clue numbers without comparing their actual neighbouring sets ignores unshared cells. The method depends on one unresolved set being a subset of the other and on using remaining counts rather than original numbers.
TRY IT 1 · mines-subset-1
| C1 | C2 | C3 | C4 | C5 | C6 | |
|---|---|---|---|---|---|---|
| R1 | ||||||
| R2 | ||||||
| R3 | ||||||
| R4 | ||||||
| R5 | ||||||
| R6 |
What can you justify?
Use the compare overlapping clues by subtracting shared squares technique. What does it prove about R5C6? Treat the displayed starting marks as established facts, except flags, which the explanation must justify from revealed clues.
Written explanation
Step 1: clue 1 at R1C4 still requires 1 mine among R2C5 after the established safe cells and proved mines are accounted for. Player flags are not used as evidence. Therefore R2C5 is a mine.
Step 2: clue 1 at R1C5 still requires 0 mines among R1C6, R2C6 after the established safe cells and proved mines are accounted for. Player flags are not used as evidence. Therefore R2C6 is safe.
Step 3: clue 2 at R3C5 requires 1 mines in {R3C6, R4C6}; clue 1 at R4C5 requires 1 in its superset {R4C6, R3C6, R5C6}. Subtracting leaves 0 mines in the 1 extra square. Therefore R5C6 is safe.
TRY IT 2 · mines-subset-2
| C1 | C2 | C3 | C4 | C5 | C6 | |
|---|---|---|---|---|---|---|
| R1 | ||||||
| R2 | ||||||
| R3 | ||||||
| R4 | ||||||
| R5 | ||||||
| R6 |
What can you justify?
Use the compare overlapping clues by subtracting shared squares technique. What does it prove about R4C1? Treat the displayed starting marks as established facts, except flags, which the explanation must justify from revealed clues.
Written explanation
clue 1 at R6C2 requires 1 mines in {R6C1, R5C1}; clue 2 at R5C2 requires 2 in its superset {R5C1, R4C1, R6C1}. Subtracting leaves 1 mines in the 1 extra square. Therefore R4C1 is a mine.
Partial overlap is not set containment
After all established mines are subtracted, one clue says unresolved squares A and B contain one mine. Another says B and C contain one mine. These are the only constraints in this local exercise.
The tempting move
Subtract the equal counts and conclude that C is safe.
Why it fails
Neither set contains the other. Both B mined with A and C safe, and B safe with A and C mined, satisfy the two clues. C changes between these arrangements.
What is actually justified
The equations imply that A and C have the same mine status, but they do not prove either safe or mined. The subset rule requires full containment of one unresolved set in the other.
Transfer the idea
After accounting for proved mines, one clue says A and B contain exactly one mine. Another says A, B and C contain exactly one mine. What do you know about C, and can you choose which of A or B is the mine?
Check your reasoning
C is safe: A and B already supply the one mine required by the larger set. These two statements still do not distinguish A from B. Additional information is needed before selecting either one as the mine.