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In Olympus Has Fallen the terrorist leader Kang repeatedly threatens staff members to get their Cerberus access codes. After torturing them too much the president always gives in and instructs them to give him their code. He justifies this by saying that they won't get his code anyway, which is even more believable once Connor, his son, is out of the terrorists' grasp.

Yet at the end when there's only Kang and the president left he suddenly has all the neccessary codes. But how did he get the last required code? It's not very likely that the president just gave in, since he wasn't even tortured. Neither does it make much sense that he didn't need the codes from everybody, since in this case the president's apology of not giving away his own code wouldn't make much sense. But maybe I have overlooked something in the story.

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The movie shows them brute-forcing one letter at a time, which would only take seconds to test 36 letters until it worked, then move on to the next one. I hope this isn't the only thing keeping us from nuclear meltdown! –  user6003 Sep 3 '13 at 2:05
    
i dont think so. because if a 3rd code could be found while the two codes are known, then what is the necessity for third code.therefore there should be only two codes rather than three. –  user6378 Oct 9 '13 at 15:50

3 Answers 3

up vote 11 down vote accepted

It was mentioned in the movie that a brute-force attack to crack all 3 Cerberus codes would take an awful long time (not sure about the exact time-period). So Kang could not employ a brute-force attack to get the codes. However, once he had 2 codes, it was just a matter of time (forgive the pun) before he was able to break through, without essentially having the 3rd access code as the system's security was compromised.

It is safe to assume that Kang knew from the beginning that he would be able to access Cerberus once he had at least 2 access codes. His entire play was to make President Asher believe that he needed all 3 codes and it would all come down to the President at the end. This made the President give away the first 2 codes rather easily since he was confident that there was nothing Kang could do that would make him give up the 3rd code.

TL;DR Kang's Plan A was to get all 3 access codes, using Connor (the President's son) as the leverage for the President's code. His contingency plan was to acquire the other 2 codes and then attack the already compromised Cerberus.

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Although your answer seems quite reasonable and I upvoted it, still I dont get As only the President of USA knew the third code, how on earth could Kang possibly come to know the code beforehand? –  Mistu4u Aug 15 '13 at 12:53
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@Mistu4u He didn't. The last code was brute-forced. –  Tom Aug 15 '13 at 15:18
    
@Tom: Okay so your theory is At once 3 codes cant be broken but A single code can be broken in short time. Is that correct? –  Mistu4u Aug 15 '13 at 18:00
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@Mistu4u No, but if you watch the movie again you'll notice that they've been running a program trying to brute-force the codes from the start. They get lucky. Just like Basher said in his answer, the plan was to use the President's son against him for the 3rd code, but since they managed to brute-force it, they didn't have to. –  Tom Aug 15 '13 at 18:28
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Exactly my point Tom. –  KeyBrd Basher Aug 16 '13 at 6:35

Ok I looked in to this by watching the movie a few times. What I think is that they were in the room for pretty much the entire movie, so once they took control of the room and the computer, they started working on the codes. The code breaker had been split up into three different sets, splitting up the power. The President pretty much gave them what they needed. With 2 of the codes unlocked, the software was able to focus all its power to break the last code. Which in fact it had already been working on for pretty much the entire time.

If that guy is smart enough to infiltrate the South Korean security team and be the biggest terrorist on the planet, then he is smart enough to have backup plans. Like having the next-gen weaponry in place and covering the White House in C4. And also knowing about the interiors of the White House. Breaking one code, though it was random, should be pretty easy.

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I'm more inclined to believe that Cereberus required all three codes to authenticate and that Kang getting the last code was just a plot hole. If I recall correctly, one of the men in the conference room said it would take at least three days for all three codes to be cracked - obviously assuming a brute force attack. My memory is not too clear, but I think each code had 7 characters. This means 36^7 ~ 7.8x10^10 candidates, assuming all possible combinations of the (uppercase) Latin alphabet and the numbers 0-9 were possible. Assuming only permutations were possible, we'd still have 36!/29! ~ 4.2x10^10. Taking out codes consisting only of a string of a single character (e.g., AAAAAAA, BBBBBBB...), (36!/29!)-36 = negligible decrease = still a shitload of candidate codes to try. Unless the attack lucked out and generated the correct code within ~12 hours (give or take), making the above, assumption, it's unlikely that the system only required 2 codes - let alone 2 codes excluding the presidents. But hold on. We're talking about an extremely powerful government security system here that has the potential to leave the United States vulnerable to nuclear attacks. Why the hell would it be possible to brute force the system? Even moderately secure authentication systems limit the number of failed authentication attempts possible within a given time frame. Why would the American government make it possible to brute force their system? They wouldn't. Well, then maybe they were brute forcing the file containing the encrypted passwords? Maybe the encryption algorithm used to encrypt the passwords before storage was leaked to the Koreans? Maybe, m-maybe... maybe GOD gave them the passwords!

I just don't see the US government designing an security system with such major vulnerabilities. Just as a precaution, I'd design the encryption code so that all three codes are combined, encrypted, and verified together, thereby increasing the number of possible codes (possible codes = k^n, where k = number of characters to choose from, n = number of characters in the code).

It was a tense movie. That's all it was good for. The technical aspects of it were merely vehicles by which the writers carried the story. The only good things about this movie were the fact that it made me think about encryption, and seeing Gerard Butler's character follow through on his promise to stick his knife in Kang's face.

Adios.

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This doesn't really answer the question. And also, considering the fact that the system was only accessible via the bunker beneath the White House (which they, up until the attack, believed was impenetrable,) I'd say it's a fairly safe system for a Hollywood movie. –  Tom Aug 18 '13 at 18:48

protected by Community Oct 9 '13 at 23:32

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