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I started getting interested when I noticed that some situations, like this one, are unsolvable.
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Notes:
  • in each example, assume undrawn neighboring squares are numbers (not affecting the count)
  • a - in an example means 'don't care' - in other words, a numbered square whose value doesn't affect the squares being shown

Heuristics:

  1. (Easy) if a numbered square's neighboring exposed mine squares number the same as the number, expose all unexposed neighbors. In this example we can expose a, b, c and d.

X 2 a
X 3 b
c d X

The next sections require you to generate sets of 'must be true' (MBT) and 'must be false' (MBF statements for each numbered square, e.g....

1 a
2 b
2' c
3 d
X e

@ MBT MBF
1 {A or B} {A and B}
2 {A and B} or {A and C} or {B and C} {A and B and C}
2' {B and C} or {B and D} or {C and D} {B and C and D}
3 {C and D} or {C and E} or {D and E} { C and D and E}

Note, at 2' we really have {B and C and `D}, {B and D and `C}, {C and D and `B} where ` means 'not'

2. Walk through the MBT statements for the entire puzzle. If two 'or' MBT's differ by a letter, then you can eliminate that letter. This is like finding a least common denominator. For example here we can eliminate 'c' (but we can't know whether it's a or b)...

1 a
2 b
X c

@ MBT
1 A or B
2 A or B or C

3. Compare all the MBT statements with all the MBF statements. If any match, you can eliminate them. For example, we can eliminate {A and B}...

1 a
2 b
2' c
3 d
X e

@ MBT MBF
1 {A or B} {A and B}
2 {A and B} or {A and C} or {B and C} {A and B and C}
2' {B and C} or {B and D} or {C and D} {B and C and D}
3 {C and D} or {C and E} or {D and E} { C and D and E}


Logic examples

1.

1 X X X 2
3 5 a 4 2
X X b 3 X
X   c X 2

@ proves
5 a or b
4 a or b
3 a or b or c

Therefore, 'c' has no mine.

2.

1 X X X 2
3 5 a 4 2
X X b 3 X
X   3' X 2

@ proves
5 a or b
4 a or b
3' b or c

Therefore, inconclusive - nothing to do.

3.

X 2 1 1
X 4 2 X
a b c 2

@ proves
4 {a and b} or {a and c} or {b and c}
2 b or c

Therefore,

  • we can eliminate {b and c} - it contradicts b or c
  • we know we must have 'a'; {a and b} or {a and c} == {a} + {b or c}
  • stuck! Indeterminate!

4. TODO: indeterminate example where you compute the odds and take the best case

-- MattWalsh - 22 Feb 2002

Topic revision: r5 - 15 Sep 2002 - MattWalsh
 
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