Minesweeper Puzzle: A New Challenge

[mattmns]

Not as interesting as the other, but I had never seen it before, until I lost to it a while ago

Mainly looking at the left center.


http://img280.echo.cx/img280/46/minesweepertest5ax.gif

 

[neurocomp2003]

2 solutions would be revealed bu in teh midleft four 3s the bombs are on teh middle 2.

 

[mattmns]

Yes that is right!

 

[Hurkyl]

I see that one all the time! It all boils down to you have one number which must have 2 more bombs next to it, and they both can't be next to the other number which can only have one more bomb next to it. (Thus marking the third square down as a bomb, there must be a bomb in one of the next two squares, and thus the next square is clear)

Using this with a |122 or a |112 on the edge was one of the first things I learned.

 

[mattmns]

You are right it probably is more common. I guess it was the first time I actually noticed it. And Hurkyl what are your times? Maybe you can take whozum's spot as the resident minesweeper king

 

[StatusX]

It seems like there are at least 9 mines left, not 8 like the counter says, and its impossible to tell where two of them go without more information. Am i missing something?

 

[mattmns]

Do not bother looking at the counter, it actually says there are 38 left. I was mostly interested in the middle left side

 

[Hurkyl]

I don't play it enough to do it quickly. =( I've gotten the coveted 1 or 2 seconds on beginner, though! (Don't remember which I got.. it was just two clicks though, I think)

 

[mattmns]

That is quite coveted. One day I spent about 30 mines trying to 1 click it, but it is way too boring. My best beg is 4 seconds

 

[ExecNight]

The biggest puzzle for me is ;

30 non-explored squares, and 38 mines to find...Unless we know how many mines there are to be found really, can't do much bout it...

 

[mattmns]

Yes ignore that. There are 40 mines on the board. I marked 2, the program auto flagged the others. But when the program (minesweeper clone) auto flags, the number at the top left does not go down.

 

[Alkatran]

Interesting problem: Assuming perfect play, what is the chance of beating minesweeper on advanced (or expert, whatever the hardest is called)?

I ask because there's nothing more annoying than playing well and ending up with a 50/50 chance at the end.

 

[mattmns]

That is a difficult question. First what do you define as a game, meaning after what point do you consider a game of minesweeper a game? When it is solvable? Because the solver has to make a few guesses at the start of a game to get it going, usually.

Saying that the game has a big spot or two open and seems solvable, I would say if there were no 50/50 guesses then it would be solvable 95+% of the time.

With 50/50 guesses though, I guess we need to find out what the odds of getting a game with a 50/50 guess, and what about games with worse odds, or multiple guessing. Seems like there are numerous things to consider.

 

[Greg825]

maybe a slightly simpler question would be what is the average sum of all numbers on a given board size x by y with mine count z.

 

[Jimmy Snyder]

maybe a slightly simpler question would be what is the average sum of all numbers on a given board size x by y with mine count z.

Suppose z = 1 (a single bomb on the board) and x and y both greater than or equal to 3. Then there are x * y possible layouts.

For those layouts where the bomb is in a corner, the sum of all numbers is 3. There are 4 such layouts, for a subtotal count of 3 * 4 = 12.

For those layouts where the bomb is along an edge, but not in a corner, the sum of all numbers is 5. There are 2 * (x + y) - 8 such layouts for a subtotal count of 10 * (x + y) - 40.

For those layouts where the bomb is not along an edge and not in a corner, the sum of all numbers is 8. There are (x - 2) * (y - 2) such layouts for a subtotal count of 8 * (x - 2) * (y - 2).

Total count is 8xy - 6x - 6y + 4
The average count is total count divided by number of layouts:

average count = 8 - 6/y - 6/x - 4/xy

As x and y increase to infinity, the average number goes to 8. This just says that as the board gets larger, the bomb is less likely to be on an edge or in a corner.

The calculations for two or more bombs are much more complicated because bombs can be next to each other. However, just on logical grounds, I think that for a fixed z, as x and y increase to infinity, the average number goes to 8z. The reason is that as x and y increase to infinity, on average the bombs get sparser and sparser so that on average they rarely are next to each other. And at the same time, they are less and less likely to be on the edge or in a corner.