Same here, dealing with two dimensional arrays in C# is unpleasant.
I think also the actual and expected values are to be switched :D

okay, I'll see with Donald to add at least one fixed test (I'm actually surprised it's not already there)

(double post...)

note: seems you didn't do RS4 yet. Don't forget to purge your RS5 solution before posting for RS4 when you'll be done with it (to not spoil the hardest one) ;)

np.
Be careful with /*...//...\n..*/ too ;) (just in case)

according to the rules, here is what your purged program should look like:

R1(RFR7F(F5))R5FR4((RR6F8P4)2((()2F6 )6RLLL2)2(F5 )P9)8LF(F6P7F)1F
P5P9F(L2)7R
L5F2FP6Fp0F
FR7qp4p0(R1F()7)FR(F1)6Rp5p374qL0 L0FFF0qq(LFR)L5RF0
R7P0FqLFL3(F4)F0L2L4F3L2Fp5L
FF R(F)p2(()R9L())8L3LL p9LFL1R0RF7L2F R LFL8R1L8qP5RqP9qF L3R5
RL5(L8)LL8P5P4RF(RR4)RLp6p3(P7LF1F)8L8FFFL3
(P9P7R8)7P0LqP4RR3L3(
R)8 L LqFP7FR P0F(( R)3(P7))F(R0L7FRL)R2L9L(RF(L1L(FRL3)FP9) )F6R6(((R7 )F0 ) )8L4p7

RLL3p7p708()5LF5F4()RFR1RR5RLqqqp9 R9(FRR4F9L6)0R3(R4)1F(
(
R)7LF9)0RFF4L0P6FqP0LL2LP9L FRF1L4F7

your code isn't managing the comments correctly.

EDIT: when you post about a problem, don't forget to provide the exact behavior on the output side: what's the assertion/error message, for instance?

Found it: there was a tiny mistake in one of my regexp.

Now compliant with your answers. Sorry for the troubles. ;)

okay, there is a problem, effectively. I'll take a look at this (but I don't have time for now)

Hi,

Your algo isn't smart enough yet:

Input       ||  Expected  || Variant 1 || Variant 2 (incomplete but resolvable)
            ||            ||           || the border has sufficient mines
------------||------------||-----------||------------------------------------            
0 0 0 1 x   || 0 0 0 1 x  || 0 0 0 1 x || 0 0 0 1 x
0 1 1 2 2   || 0 1 1 2 2  || 0 1 1 2 2 || 0 1 1 2 2
1 2 x 1 0   || 1 2 x 1 0  || 1 2 x 1 0 || 1 2 x 1 0
? ? 2 1 0   || ? M 2 1 0  || M ? 2 1 0 || ? M 2 1 0
? ? 2 1 2   || ? ? 2 1 2  || ? M 2 1 2 || ? ? 2 1 2
? ? 2 x 2   || ? A 2 x 2  || ? A 2 x 2 || ? A 2 x 2
? ? ? 2 2   || ? M ? 2 2  || ? ? ? 2 2 || ? ? M 2 2
? ? ? 1 0   || ? ? M 1 0  || ? ? M 1 0 || ? ? ? 1 0

if you look at the different versions upper, you'll see that you can actually safely open the position A, and that will lead you to the expected board.

No problem with that test, except that it uses assertFalse, not assertTrue, and does it on purpose... ;)
You should check again the "en passant" rule, it's a tricky one.