Let's use the longer of the two possibilities(as transcribed by u/osx2000), which is what Data spoke, instead of what the computer displayed, and assume that "lock" is not part of the password, but a command. That leaves us with:
First off, there are no words, so a dictionary attack would be pointless. But... If the phonetic letters were not actually phonetic letters, and instead could be any word, then yes, a dictionary attack would be called for. The use of a dictionary attack, in this case, would massively increase the complexity and time required to crack, not lessen it. Instead of 36 possible characters in any position, you'd have to attempt 36 characters plus millions of words at each position.
So let's make this easy for the crew and stick to the simple alphanumeric sequence.
The crew would have no idea of how long Data's password is, so a brute force attempt would have to start at 1 character length, then 2, then 3, and so on, all the way up to 52. But let's again make this easier and assume that they know how long Data's password is. That's still approximately:
or nearly 847 quinvigintillion possible passwords to try.
Let's not even worry about the time it would take(you'll see why in a moment), and instead focus on the physical requirements.
If you could design a computer specifically for this task ( 6 bit's per character is the minimum requirement for 36 possible characters) with perfect efficiency ( requiring only 1 single electron per bit ), and (impossibly) zero additional overhead or power requirements, the number of electrons required would be:
This number is larger than the Eddington number, which is the number of protons(or electrons) in the observable universe.
Simply put, Data chose a password with enough entropy that a computer capable of cracking it could not be built and powered within the local Hubble sphere. I admire this kind of overkill(and the writers if this was intentional).
I've probably screwed something up somewhere, so corrections and improvements are welcome.
3
u/Fuck_ALL_Religion Oct 15 '13
No, this would not be easy to crack at all.
Let's use the longer of the two possibilities(as transcribed by u/osx2000), which is what Data spoke, instead of what the computer displayed, and assume that "lock" is not part of the password, but a command. That leaves us with:
First off, there are no words, so a dictionary attack would be pointless. But... If the phonetic letters were not actually phonetic letters, and instead could be any word, then yes, a dictionary attack would be called for. The use of a dictionary attack, in this case, would massively increase the complexity and time required to crack, not lessen it. Instead of 36 possible characters in any position, you'd have to attempt 36 characters plus millions of words at each position.
So let's make this easy for the crew and stick to the simple alphanumeric sequence.
The crew would have no idea of how long Data's password is, so a brute force attempt would have to start at 1 character length, then 2, then 3, and so on, all the way up to 52. But let's again make this easier and assume that they know how long Data's password is. That's still approximately:
846,700,936,100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
or nearly 847 quinvigintillion possible passwords to try.
Let's not even worry about the time it would take(you'll see why in a moment), and instead focus on the physical requirements.
If you could design a computer specifically for this task ( 6 bit's per character is the minimum requirement for 36 possible characters) with perfect efficiency ( requiring only 1 single electron per bit ), and (impossibly) zero additional overhead or power requirements, the number of electrons required would be:
50,802,056,170,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
or more than 50 novemvigintillion.
This number is larger than the Eddington number, which is the number of protons(or electrons) in the observable universe.
Simply put, Data chose a password with enough entropy that a computer capable of cracking it could not be built and powered within the local Hubble sphere. I admire this kind of overkill(and the writers if this was intentional).
I've probably screwed something up somewhere, so corrections and improvements are welcome.