I can instead create a FAQ about my code.

Q: Why do you use random() instead of first 8 bytes of SHA256 of the transaction hash as described in the algorithm?
A: Because SHA256 is designed to generate values that are uniformly distributed.

Q: What happens if a==b? Why don't you process this situation?
A: The original algorithm described in wiki doesn't give the answer either. But the possibility of this event is proportional to (2^-64)² (because 8 first bytes of sha256 have 64 bits) which is about 10^-38 so it is safe in this code to simple skip such cases for clarity. If you are feeling pedantic you can add the line
in the end of the code and it will always (well, ok, with an estimated probability of 1-NBLOCKS*(10^-38) = 1-10^-32) give you zero.