Brain Warm-up Similar to puzzles from Mensa Math & Logic Puzzles
The following sets of characters represent games of minesweeper. Each number represents the number of mines that are adjacent (horizontally, vertically, or diagonally) to it. Your job is to replace ten of the question-marks in each set with one mine each to prove the numbers accurate. There is only one possible solution for each set.
1?1????1 ????????
?11111?1 0?1?0??1
1?1????1 ??1??1??
?1??1?1? ??????1?
?1??1??1 ?3?2?1?0
??1???1? ?????2??
1??????1 ??13????
?1?1?1?? ?0????0?
  1. tom von kercey (solved: 1-12-2005 12:09pm)
  2. Steve Klein (solved: 1-12-2005 3:23pm)
  3. sunil (solved: 1-12-2005 4:24pm)
  4. Erin Harris (solved: 1-12-2005 4:37pm)
  5. Pete Lamonica (solved: 1-12-2005 11:21pm)
1?1*??*1 ??*?????
*11111?1 0?1?0??1
1?1????1 ??1??1*?
?1?*1?1* *??*??1?
*1??1??1 *3?2*1?0
??1???1? *????2??
1??*?*?1 ??13*???
*1?1?1?* ?0?**?0?
On Board #1, a good strategy is to find places that that mines cannot possibly go first. For example, there must be a mine in (1,2) or (2,1) to satisfy the (1,1) square, Thus there cannot be a mine in (3,1) because both (1,2) and (2,1) are adjacent to (2,2). There cannot be a mine in (3,7) because then (1,8) could not be satisfied without violating (2,8). I could go on and on, but eventually you will X-out enough squares that a few mines are forced. If you keep track of the “must be here or here” rules then a chain reaction usually results. I was not able to chain reaction the entire problem one, but I did reach a point where many 1’s had to be covered by just a few mines and so their locations were the obviously central ones.

Board #2 has some extra help in the form of 0’s. 0’s can have all their adjacent squares marked mine-less immediately. Another helpful hint to use on this board is the “there is only one solution rule.” This rule means that any time you could have a mine in more then one place without making a difference to the rest of the puzzle then those places must either be completely filled with mines of devoid of them because there is “only one solution”. The comes into play with the 3 in (5,2) A mine could go in (4,1), (5,1), or (6,1) and not be adjacent to any other numbered squares. This means that these three squares must be empty or full of mines. Thus, if you reach a point (as I did) in which there are no longer three available spaces to the right, top or bottom of the 3 you can now be sure that all three of the above mentioned squares should be filled. This pattern occurs in other places as well.
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