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A Gobo is caught for leaving the school for the lnternet bar to play computer games without permission. He is asked to solve a problem.


Put 4 mines in a 3 by 5 minesweeper grid to fill ALL the other cells with twos.

.....            22222
.....     =>     22222
.....            22222

There is only one solution.

Prove it.

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2 Answers 2

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Start by considering the leftmost three cells.

One of the leftmost three cells must be empty. (If not, then the row-2-col-2 cell is too big.) It must have two mines next to it; therefore there must be two mines in the left two columns. The same applies to the rightmost three cells.

So, two mines are in the left two columns and two are in the right two columns; the middle must therefore be empty.

If there are two mines in the second column, then either the top left or the bottom left corner is broken. Ditto for the fourth column. So there must be one mine in each column besides the middle.

To satisfy the middle column, the mines in columns 2 and 4 must be placed in the middle cells. To satisfy the rest of columns 2 and 4, the mines in columns 1 and 3 must also be placed in the middle row. And we're done! There's only one configuration, and it's "cells across the middle row except for the very center".

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  • $\begingroup$ Nice explanation - even made sense past midnight $\endgroup$
    – Avi
    Commented Nov 10, 2019 at 7:43
  • $\begingroup$ Added a new tag. Would you change the answer if it doesn't fit? $\endgroup$ Commented Nov 10, 2019 at 8:00
  • $\begingroup$ @Scratch---Cat I didn't use a computer for this, if that's what you're asking? $\endgroup$
    – Deusovi
    Commented Nov 10, 2019 at 8:01
  • $\begingroup$ Oh, sorry. The teacher in the story forced the Gobo to turn off the computer. $\endgroup$ Commented Nov 10, 2019 at 8:19
  • $\begingroup$ Answer this question if you like. $\endgroup$ Commented Nov 10, 2019 at 8:20
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I started by considering the corners.

A mine in each corner does not work, so at least 1 corner is empty.
There are two mines and an empty cell adjacent to that empty corner. That empty cell already has 2 adjacent mines, so its other neighbours are empty.

These two arrangements don't work because the nearest corner would be a 1.

 2 x . . .
 x 2 . . .
 1 ? . . .

 2 x . . .
 2 x . . .
 1 ? . . .

So to must be like this:

 2 2 2 . .
 x x 2 . .
 . . . . .

We only have two more mines. The following arrangement does not work:

 2 2 2 . x
 x x 2 . .
 . . . . x

So at least one of the right corners is empty. The same reasoning as above then produces:


 2 2 . 2 2
 x x . x x
 . . . . .

or

 2 2 . . .
 x x . x x
 . . . 2 2

Either way, the mines are placed in the same spots, and all the undetermined squares have two adjacent mines and will be twos when left empty.

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  • 1
    $\begingroup$ Sorry, you got Deusovi’d :( $\endgroup$
    – Avi
    Commented Nov 10, 2019 at 7:45
  • 2
    $\begingroup$ That happens all the time, but at least this time my reasoning is sufficiently different to make it still worth sharing. $\endgroup$ Commented Nov 10, 2019 at 7:48
  • $\begingroup$ Fair enough - it’s worth seeing $\endgroup$
    – Avi
    Commented Nov 10, 2019 at 7:49

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