# A minesweeper to annoy 4 logicians

This puzzle is an original, I'm sure it shares a similar idea with some other logic problems. However, I worked out the details of this one. I haven't put the [no-computer] tag because computer aid search can make the puzzle more accessible. A solution MUST NOT be based on purely computational methods, but rather on logical deduction you can explain and work out by hand.

Some good friends of mine recently became obsessed with logic and I can't take anymore of their games. It's always about Alice, Bob and Charlie somehow. I decided to craft my own little game that will keep them quiet for the rest of night. The base is that of a minesweeper but with a small twist. Take a look at the following grid:

What I plan on doing is handing this out to my four friends and tell them:

• "There are 4 bombs in this grid"
• "I've put flags on some cells. I can assure you that on those cells, there are no bombs and they all have at least one bomb in the 8 surrounding cells"
• "Behind the flags, there is at least four 1s, two 2s and one 3. Of course there can be more"

Then I will hand out to each one of them a special kind of glasses that reveal flagged cells that have the same value as the type of the glasses. So for example, I will give to player 1 the glasses of type 1, that can see cells that are flagged AND with a value of 1.

In this minesweeper, the goal is not to avoid bombs but rather to find the bombs. I will ask them to hand me a map of where they think the bombs are, and I will compare with the correct one. That gives me the advantage that if they are not sure of the emplacements of the bombs, i.e. multiple possibilities remain, I can swap with another valid map of the bombs and make my favorite logicians eat their hats. I should note that the flags are always in the same position as in the picture.

I figured it will be more fun to make them play by teams of 2, as it maximizes my chance to see them fail, but let them pick the teams knowing who has which glasses. At this point, I thought of 2 variants:

• In the first variant, within a team they can only communicate by saying whether or not they know all the bomb spots;
• In the second variant, the two members of one team can share their glasses, and use all the information of the pair to spot the bombs;

The question I have, is which one should I give them, to ensure at least one team can fail for at least one grid. Can you help me?

Hint 1:

I will not be able to play more than 22 rounds of this game. If this is too much let me know I can add some constraint to make it down to 10.

• I thought I had a solution. I'm reworking it. Commented May 1 at 4:55
• "There is at least 4 flagged cells surrounded by exactly 1 bomb, 2 by exactly 2 and 1 by exactly 3" Does this mean that there are exactly two flags surrounded by two bombs, or does it mean that there are at least two flags surrounded by two bombs? Commented May 1 at 12:22
• @LeppyR64 at least, just thought the repetition was too heavy, ill put it in Commented May 1 at 13:09
• @Fluorine, I probably agree that the repetition is too heavy. :) Thanks for the clarity though. Commented May 1 at 13:56
• @Truej I'm thinking of resizing the problem to make it a bit easier, are you still working on a solution? Commented May 2 at 16:10

Awesome puzzle. This is only a partial solution. I didn't go through all the cases, just some, but I used no computers, so I will just summarize the results for these cases.

I will refer to points on the grid by coordinates, row first, then column. Let's start with some deductions of where the bombs can be.

The flag at (5,4) tells us that there needs to be a bomb at either (4,3) or (4,4). The flag at (1,4) tells us that there needs to be a bomb at either (2,3) or (2,4).

Next I will separate two cases. Either there is a bomb at (3,1) or there isn't. I only looked at the case where there is no bomb there. It felt like too many cases for pen-and-paper otherwise.

If there is no bomb at (3,1) then the flag at (2,1) tells us that there has to be a bomb at either (1,1) or (1,2). The flag at (4,1) needs a bomb nearby as well, so we found rough locations for all 4 bombs.

There needs to be a flag with at least 3 bombs next to it.

This means the bomb near the flag at (4,1) needs to be on (4,2), this will guarantee a 3 for flag (3,3) no matter where exactly the bombs from the first spoiler are and possibly a second 3 for the flag (3,2).

We need 4 flags that have only one bomb next to it.

The flags at (2,1), (1,4), (4,1) and (5,4) will all get a 1 no matter how the remaining possibilities are resolved. So that condition is automatically satisfied.

For each of the 3 bombs not found yet we have two options giving us 8 cases to consider in total. There is always at least one flag with a 2 at (5,3), one of the 8 cases leads to no further 2s so has to be excluded.

The other 7 cases are all possible. The logician with the 2-glasses can distinguish between the 7 cases so always knows where all the bombs are on their own. There are no 4s , so the 4-glasses give no information. The 1-glass and the 3-glass can find all the bombs if they work together but not on their own.

To get a full solution one would also have to consider the case where the is a bomb at (3,1) and do a similar analysis but I haven't done that.