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The object of this puzzle is to construct a Minesweeper grid with a special property. The grid can be any shape, but it must have four labeled squares around the outer edge. Like this:

enter image description here

The four labeled squares must be labeled A, B, C, D, in that order going around the outer edge.

The objective is for a Minesweeper player to be able to conclude that A and C are the same, and B and D are the same, but not be able to conclude anything else. To be specific:

Place numbers in the grid to get a Minesweeper puzzle. This puzzle must have the following properties:

  1. There are no safe clicks. That is, for any unnumbered square, there is a solution in which that square is a mine.
  2. In any valid solution, A=C and B=D. (A=C means that A has a mine if and only if C does.)
  3. A valid solution must satisfy every combination of A=C and B=D simultaneously.

Again, the grid can be any shape. The grid can even have holes in it, but understand that "outer edge" does not include the edge neighboring a hole. A, B, C, D can be placed anywhere along the outer edge, but must be in that order. Aim for the smallest grid!

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  • $\begingroup$ When you say A=C .. do they count each other for determing number of mines next to ? So if there are 2 mines next to A. and 1 next to C .. does A show 2, or 3 ? does C show 1 or 3 ? $\endgroup$
    – Ditto
    Commented Mar 24, 2015 at 15:59
  • $\begingroup$ @Ditto What it means is that either A and C both have mines, or neither one does. What number they would show if they didn't have mines is not part of the puzzle. $\endgroup$
    – KSmarts
    Commented Mar 24, 2015 at 16:01
  • $\begingroup$ Ok, so they are different squares .. they just share a mine/no mine property .. got it :) $\endgroup$
    – Ditto
    Commented Mar 24, 2015 at 16:01
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    $\begingroup$ You're asking this to get some help completing your P=NP Minesweeper proof, aren't you? ;-) $\endgroup$
    – Kevin
    Commented Mar 24, 2015 at 20:10
  • $\begingroup$ Does the definition of the puzzle grid include the number of mines to look for (as in the minesweeper game), or can different solutions use different numbers of mines? $\endgroup$
    – Blckknght
    Commented Mar 25, 2015 at 1:45

3 Answers 3

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I think I got it now. Using a minesweeper developer program found on here, you can create, evaluate and try your own board. I have uploaded an image of my solution, or else this would've been a LOT of typing:

enter image description here

And an explanation about what you see in the image:

I've put the 4 labels (A, B, C and D) beside each questionmark where it should be. The red flags mark the spots of pre-discovered bombs.
Every pair (A-C and B-D) have their own 'wire' as it is called. Then an intersection to let these signals cross eachother, and arive at the other side.
For the first rule: there are no safe clicks, in all possible solutions this setup has got, any unnumbered square has been a mine at least once.
In èvery solution of this setup: it covers the A=C and B=D requirement. So both the second and third rule are covered too.

Discarded answers (Just for comment-reference)

1) .A   2)  A  3) AB
    1 .    D2B    DC
  D1.1B     C
  . 1
    C.

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    $\begingroup$ Maybe I made the rules unclear. Rule 2 says "In any valid solution, A=C and B=D." Your first option breaks this because there is a solution where A and D have mines and B and C don't. Your second option breaks this because anything is a solution. Is it clear how Rule 2 works now? $\endgroup$
    – Lopsy
    Commented Mar 24, 2015 at 15:04
  • $\begingroup$ @Lopsy Ok, I think it is clear now, let me (and others) work on a possible solution $\endgroup$
    – JBSregath
    Commented Mar 24, 2015 at 15:10
  • $\begingroup$ I don't think your new answer implies that A=C and B=D. For example, you could have A with a mine and the bottom-right dot has a mine, but C does not. $\endgroup$
    – KSmarts
    Commented Mar 24, 2015 at 15:58
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    $\begingroup$ What I'm saying is that there is a valid solution where A is a mine and the dot next to C is a mine, so C is not a mine. So A$\neq$C. $\endgroup$
    – KSmarts
    Commented Mar 24, 2015 at 16:10
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    $\begingroup$ Btw can we adapt this to make ABCD on the outer edge of the board to fulfill all the requirements? $\endgroup$
    – justhalf
    Commented Mar 30, 2015 at 21:26
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I think you might need to clarify what a "hole" is.

  A
D   B
  C

This satisfies the puzzle's requirements.

There are no safe spaces; all of the spaces are mines. Because none of the spaces are adjacent to each other, none of them can be a "numbered square"; the value of this number would be 0, making it an empty (or safe) space.

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  • $\begingroup$ This could be a solution indeed, but there might be a possibility this is not allowed because none of the sides are touching, thus not a minesweeper puzzle. I'll leave it up to Lopsy $\endgroup$
    – JBSregath
    Commented Mar 24, 2015 at 15:27
  • $\begingroup$ It follows the rules laid out in the challenge and it follows the rules of Minesweeper as they are modified to include holes. It's really just a question of what makes a "smaller grid". 4 non-hole squares is smaller than 8 non-hole squares, but they take up more space if you include the holes in the calculation. $\endgroup$
    – Ian MacDonald
    Commented Mar 24, 2015 at 15:31
  • $\begingroup$ How does this not have the "anything is a solution" problem that @JBSregath's second answer does? $\endgroup$
    – KSmarts
    Commented Mar 24, 2015 at 15:33
  • $\begingroup$ @KSmarts: See the second spoiler block. $\endgroup$
    – Ian MacDonald
    Commented Mar 24, 2015 at 15:34
  • $\begingroup$ The intention of rule 3 is that there should be a solution where (for example) A and C are both mines and B and D are both safe. That is, all 4 mine-settings of A, B, C, D satisfying A=C and B=D extend to a solution. I see now that I phrased this rule badly, sorry to the solvers so far. $\endgroup$
    – Lopsy
    Commented Mar 24, 2015 at 15:42
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I'm probably twisting the rules a bit, but here goes. "The grid can be any shape". Well, my grid is a torus! (And it kinda looks like a swastika, but let's ignore that.)

C+A D
  1 +
D1+1B
+ 1
B C+A

The labels are shown twice to indicate where the grid "wraps around". The grid itself uses 11 squares.

I realize that this probably isn't what is intended, but I thought of this and decided to post it while working on a "real" answer.

Pre-edit answer:
How about this:

1A1
D B
1C1

The space in the middle is a "hole". If A has a mine, then B and D do not, which means that C does. Similarly, if A does not have a mine, then B and D do, so C does not. Thus, this satisfies all three conditions.

It uses a 3x3 grid, with a total of 8 squares.

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  • $\begingroup$ @Ian Macdonald and KSmarts: I think Lopsy means that for évery combination a solution has to exist. Both AC and BD have mines, AC have mines and BD are clear, AC are clear and BD have mines, both AC and BD are clear $\endgroup$
    – JBSregath
    Commented Mar 24, 2015 at 15:23
  • $\begingroup$ @JBSregath: no, not quite. The reason why he claimed your answer did not follow rule 2 was that there existed a valid combination of ABCD that did not satisfy rule 2. No such rule-breaking combination exists in this answer (or mine). $\endgroup$
    – Ian MacDonald
    Commented Mar 24, 2015 at 15:28
  • $\begingroup$ @Ian Macdonald: And rule #3: For any mine-setting of A, B, C, and D that satisfies A=C and B=D, this setting extends to a solution of the whole grid. This means a valid solution has to exist for each of these settings: both AC and BD mines, AC mines and BD clear, AC clear and BD mines, both AC and BD clear $\endgroup$
    – JBSregath
    Commented Mar 24, 2015 at 15:32
  • $\begingroup$ @JBSregath I can see how it might mean that. I'll wait to hear from Lopsy, though, because that makes it a lot harder. $\endgroup$
    – KSmarts
    Commented Mar 24, 2015 at 15:34
  • $\begingroup$ Yes, JBSregath is correct. Feel free to edit the puzzle (I am on a phone which makes editing very annoying.) $\endgroup$
    – Lopsy
    Commented Mar 24, 2015 at 15:37

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