You can solve it by the applying the following logic. Consider the places A and B:

  ?  ?  ?  C  A  B
  2  2  2  1  2  2
  2  X  2  ·  1  X
  2  X  2  1  2  2
  1  1  1  1  X  1
  ·  ·  ·  1  1  1

The number 2 below B forces a mine to be either in A or B. That means, that the 2 blow A is aloready satisfied (it already touches one mine, and now another one from either A or B).
Thus, C can be safely opened:

  ?  ?  ?  1  ?  ?
  2  2  2  1  2  2
  2  X  2  ·  1  X
  2  X  2  1  2  2
  1  1  1  1  X  1
  ·  ·  ·  1  1  1

Now we can apply the same logic to the left side:

  A  B  C  1  ?  ?
  2  2  2  1  2  2
  2  X  2  ·  1  X
  2  X  2  1  2  2
  1  1  1  1  X  1
  ·  ·  ·  1  1  1

By the same logic, C can be opened (because either A or B contains one mine):

  ?  ?  1  1  ?  ?
  2  2  2  1  2  2
  2  X  2  ·  1  X
  2  X  2  1  2  2
  1  1  1  1  X  1
  ·  ·  ·  1  1  1

Now from this state, the board can be solved easily again.

at that point you have also an information how much mines are there, maybe you could use it

stuck at the same place
i don't know how this should be solvable without wild guessing

My algorithm is stuck with this board:
? ? ? ? ? ?
2 2 2 1 2 2
2 x 2 0 1 x
2 x 2 1 2 2
1 1 1 1 x 1
0 0 0 1 1 1

In fact, I can't find how can it continue resolving from there safely. This is the first test so I guess I have something wrong?