Given the function:
def md5(r, n):
i = 0
while i != n:
r = hashlib.md5(r[:9]).hexdigest()
i+=1
return r
Results:
md5("00000000000000000000000000000000", 0) #00000000000000000000000000000000
md5("00000000000000000000000000000000", 1) #4c93008615c2d041e33ebac605d14b5b
md5("00000000000000000000000000000000", 2) #c6246d2a39695fb74971b09ced30874f
md5("4c93008615c2d041e33ebac605d14b5b", 1) #c6246d2a39695fb74971b09ced30874f
md5("00000000000000000000000000000000", 2017) #57ab27ac3626d4565913485ff42b037b
Is there a way to calculate:
md5("00000000000000000000000000000000", 2017201720172017)
Without dying first, of course.
md5("00000000000000000000000000000000", 20172017)
It takes me 84 seconds, so 2017201720172017 is 100000001 times that. Approximately 2 million hours.
I don't use xrange
or range
since the value 2017201720172017 is too big.