How it works
Board representation. This board can be represented like:
abcd e13f ghij klmn
#Field Rules
So we have a 1 and a 3 and we know there are 6 mines in total on the board. This can be represented as:
a+b+c+e+g+h+i = 1
b+c+h+i+d+f+j = 3
a+b+c+d+e+f+g+h+i+j+k+l+m+n = 6
This is what I call rules (the FieldRule class in my code)
#Field Groups
By grouping fields into which rules they are in, it can be refactored into:
(a+e+g) + (b+c+h+i) = 1
(b+c+h+i) + (d+f+j) = 3
(b+c+h+i) + (d+f+j) + (k+l+m+n) = 6
These groups I call Field Groups (The FieldGroup class in the code)
#AnalyzeFactory
The AnalyzeFactory class stores a collection of rules, and when it is getting solved it begins by splitting the fields into groups, then creates a GameAnalyze object to do the rest of the work.
GameAnalyze. It starts by trying to simplify things (we'll come to that later), when it can't do so any more it picks a group and assign values to it. Here I pick the (a+e+g) group. I find that it's best to start with a small group.
(a+e+g) = 0 is chosen and a new instance of GameAnalyze is created, which adds (a+e+g) = 0 to its knownValues.
Simplify (FieldRule.simplify method). Now we remove groups with a known value and try to deduce new known values for groups.
(a+e+g) + (b+c+h+i) = 1
(a+e+g) is known, so (b+c+h+i) = 1 remains which makes the rule solved. (b+c+h+i) = 1 is added to knownValues. Next rule:
(b+c+h+i) + (d+f+j) = 3
(b+c+h+i) = 1 is known so we have left (d+f+j) = 2, making also this rule solved and another FieldGroup known. Last rule:
(b+c+h+i) + (d+f+j) + (k+l+m+n) = 6
The only unknown remaining here is (k+l+m+n) which after removing the other groups has to have the value 3, bBecause (b+c+h+i) + (d+f+j) = 1 + 2.
#Solution
So what we know is:
(a+e+g) = 0
(b+c+h+i) = 1
(d+f+j) = 2
(k+l+m+n) = 3
As all rules have been solved and all groups have a value, this is known as a solution (Solution class).
Doing the same for (a+e+g) = 1 leads, after simplification to another solution:
(a+e+g) = 1
(b+c+h+i) = 0
(d+f+j) = 3
(k+l+m+n) = 6 - 3 - 1 = 2
#Calculating combinations for solutions
Now we have two solutions where all the groups have values. When a solution is created, it calculates the combinations possible for that rule. This is done by using nCr (Binomial coefficient).
For the first solution we have:
(a+e+g) = 0 --> 3 nCr 0 = 1 combination
(b+c+h+i) = 1 --> 4 nCr 1 = 4 combinations
(d+f+j) = 2 --> 3 nCr 2 = 3 combinations
(k+l+m+n) = 3 --> 4 nCr 3 = 4 combinations
Multiplying these combinations we get 143*4 = 48 combinations for this solution.
As for the other solution:
(a+e+g) = 1 --> 3 nCr 1 = 3
(b+c+h+i) = 0 --> 4 nCr 0 = 1
(d+f+j) = 3 --> 3 nCr 3 = 1
(k+l+m+n) = 2 --> 4 nCr 2 = 6
3 * 1 * 1 * 6 = 18 combinations.
So a total of 48 + 18 = 66 combinations.
#Probabilities
The total combinations where a field in the (k+l+m+n) group is a mine is:
In first solution: 3 mines, 4 fields, 48 combinations for the solution.
In second solution: 2 mines, 4 fields, 18 combinations in solution.
To calculate the probability we take this value divided by the total combinations of the entire board and we get: