This is a problem from Google Code Jam qualification round (which is over now). How to solve this problem?
Note: If you have a different method from the ones discussed in answers, please share it so we can expand our knowledge of the different ways to solve this problem.
Minesweeper is a computer game that became popular in the 1980s, and is still included in some versions of the Microsoft Windows operating system. This problem has a similar idea, but it does not assume you have played Minesweeper.
In this problem, you are playing a game on a grid of identical cells. The content of each cell is initially hidden. There are M mines hidden in M different cells of the grid. No other cells contain mines. You may click on any cell to reveal it. If the revealed cell contains a mine, then the game is over, and you lose. Otherwise, the revealed cell will contain a digit between 0 and 8, inclusive, which corresponds to the number of neighboring cells that contain mines. Two cells are neighbors if they share a corner or an edge. Additionally, if the revealed cell contains a 0, then all of the neighbors of the revealed cell are automatically revealed as well, recursively. When all the cells that don't contain mines have been revealed, the game ends, and you win.
For example, an initial configuration of the board may look like this ('*' denotes a mine, and 'c' is the first clicked cell):
*..*...**.
....*.....
..c..*....
........*.
..........
There are no mines adjacent to the clicked cell, so when it is revealed, it becomes a 0, and its 8 adjacent cells are revealed as well. This process continues, resulting in the following board:
*..*...**.
1112*.....
00012*....
00001111*.
00000001..
At this point, there are still un-revealed cells that do not contain mines (denoted by '.' characters), so the player has to click again in order to continue the game.
You want to win the game as quickly as possible. There is nothing quicker than winning in one click. Given the size of the board (R x C) and the number of hidden mines M, is it possible (however unlikely) to win in one click? You may choose where you click. If it is possible, then print any valid mine configuration and the coordinates of your click, following the specifications in the Output section. Otherwise, print "Impossible".
My Tried Solution:
So for the solution, you need to make sure that each non-mine node is in a 3x3 matrix with other non-mine nodes, or a 3x2 or 2x2 matrix if the node is on an edge of the grid; lets call this a 0Matrix. So any node in a 0Matrix have all non-mine neighbors.
Firstly, Check whether less mines are required, or less empty nodes
if(# mines required < 1/3 of total grid size)
// Initialize the grid to all clear nodes and populate the mines
foreach (Node A : the set of non-mine nodes)
foreach (Node AN : A.neighbors)
if AN forms a OMatrix with it's neighbors, continue
else break;
// If we got here means we can make A a mine since all of it's neighbors
// form 0Matricies with their other neighbors
// End this loop when we've added the number of mines required
else
// We initialize the grid to all mines and populate the clear nodes
// Here I handle grids > 3x3;
// For smaller grids, I hard coded the logic, eg: 1xn grids, you just populate in 1 dimension
// Now we know that the # clear nodes required will be 3n+2 or 3n+4
// eg: if a 4x4 grid need 8 clear nodes : 3(2) + 2
For (1 -> num 3's needed)
Add 3 nodes going horizontally
When horizontal axis is filled, add 3 nodes going vertically
When vertical axis is filled, go back to horizontal then vertical and so on.
for(1 -> num 2's needed)
Add 2 nodes going horizontally or continuing in the direction from above
When horizontal axis is filled, add 2 nodes going vertically
For example, say we have an 4x4 grid needing 8 clean nodes, here are the steps:
// Initial grid of all mines
* * * *
* * * *
* * * *
* * * *
// Populating 3's horizontally
. * * *
. * * *
. * * *
* * * *
. . * *
. . * *
. . * *
* * * *
// Populating 2's continuing in the same direction as 3's
. . . *
. . . *
. . * *
* * * *
Another Example: 4x4 grid with 11 clear nodes needed; output:
. . . .
. . . .
. . . *
* * * *
Another Example: 4x4 grid with 14 clear nodes needed; output:
// Insert the 4 3's horizontally, then switch to vertical to insert the 2's
. . . .
. . . .
. . . .
. . * *
Now here we have a grid that is fully populated and can be solved in one click if we click on (0, 0).
My solution works for most cases, but it didn't pass the submission (I did check an entire 225 cases output file), so I'm guessing it has some problems, and I'm pretty sure there are better solutions.
Let's first define N, the number of non-mine cells:
N = R * C - M
A simple solution is to fill an area of N non-mine cells line-by-line from top to bottom. Example for R=5, C=5, M=12:
c....
.....
...**
*****
*****
That is:
N / C rows with non-mines from top to bottom.N % C non-mines from left to right.There are only a few special cases you have to care about.
If N=1, any configuration is a correct solution.
If R=1, simply fill in the N non-mines from left-to-right. If C=1, fill N rows with a (single) non-mine.
If N is even, it must be >= 4.
If N is odd, it must be >= 9. Also, R and C must be >= 3.
Otherwise there's no solution.
If N is even and you can't fill at least two rows with non-mines, then fill the first two rows with N / 2 non-mines.
If N is odd and you can't fill at least two rows with non-mines and a third row with 3 non-mines, then fill the first two rows with (N - 3) / 2 non-mines and the third row with 3 non-mines.
If N % C = 1, move the final non-mine from the last full row to the next row.
Example for R=5, C=5, M=9:
c....
.....
....*
..***
*****
It is possible to write an algorithm that implements these rules and returns a description of the resulting mine field in O(1). Drawing the grid takes O(R*C), of course. I also wrote an implementation in Perl based on these ideas which was accepted by the Code Jam Judge.