how secure is your password?

Posted on September 30, 2013 by KG

passGive it a go.

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11 Responses to how secure is your password?

  1. KG says:
    September 30, 2013 at 19:54

    Mind you, I got that result using one of my multiple word, randomly generated passwords…..
    http://world.std.com/~reinhold/diceware.html
    But that will be useless if the service provider hands over your information to the NSA, won’t it?

  2. Ronbo says:
    October 1, 2013 at 03:02

    As I’ve said before, when I worked for NSA over 30 years ago at the main campus Ft. Meade, Md., the super secret agency had massed scores of the world’s largest, fastest and most advanced computers in the basement of the main building who were supplying the 35,000 intelligence workers with all the news worth knowing from the four corners of the earth – with exceptions for American Homeland and The First Cousins – UK, Australia, New Zealand and Canada.

    Today everyone in the world could be an NSA target under the unofficial motto of, “In God We Trust, all others we monitor.”

    I hope everyone understands the creation of the Internet was not altruistic on the part of America, as the cables that pass thru our Homeland shoots all the raw data from all over the world to Ft.Meade.

  3. Seneca III says:
    October 1, 2013 at 05:24

    This may be too big :cry:
    PGP AND THE RSA SYSTEM.
    In the early 1970’s Ralph Merkle, Whitfield Diffie and Martin Hellman proposed a new kind of code called public key cryptosystem. Basically they postulated that an encoding system that is very hard to undue (Ian Stewart in the 1996 edition of his book From Here to Eternity described it as like trying to ‘unboil’ an egg), then there is no harm in releasing its details.
    Essentially any code takes a message M and produces a coded form f(M). To decode the message a recipient has to find the inverse function f-1 such that:
    f-1 (f (M)) = M.
    If it is very easy to compute f we call it a trapdoor function but only where also f-1 is very hard to calculate. However a trapdoor function in this sense isn’t very practical because the legitimate user finds it just as hard to decode as an enemy does. The final twist is to define f in such a way that a single extra piece of (secret) information makes the calculation of f-1 easy and this is the only bit of information you can’t tell the enemy.
    The RSA system designed by Ted Rivest, Adi Shamir and Leonard Adleman in 1977 is based on the fact that it is easy to multiply two prime numbers but extremely difficult to divide the result again into two prime numbers particularly if they are large primes.
    [For example the number n = 267 – 1 which resolves into:
    193707721 x 761838257287, uniquely.]
    So, begin with two large primes, p and q that are kept secret but release to the public their product n = pq and a further number e the encoding key, then it is enough to encode block by block ‘B’ into the number C = Be (mod(≡) n) and C is the coded message. To decode you need to know the decoding key d so that de is congruent to 1 ≡ (p-1)(q-1) then Cd ≡ B (mod n), so you can recover B from C by a similar process.
    The point is that whilst everybody knows n and e they don’t know p and q and so can’t work out (p-1)(q-1) and therefore can’t find d. The designer of the code and the intended recipient whom has been informed both know p and q and can find d.
    If we were doing it just between the two of us, and by hand, God forbid, this is how we could go about it:
    First, I choose two large prime numbers, referred to as p and q. I then take a number e that is not divisible by either p or q. The number e, together with the product of p and q is my public key.
    Next I calculate a number d that is the mathematical inverse of e when calculating modulo (p-1)(q-1). This means that if you multiply e by d, and you divide the result by (p-1) times (q-1), you end up with 1. The number d, together with the product of p and q is my private key.
    The security of the RSA system lies in the fact that it is very difficult to compute p and q given only n. If you were able to find p and q, you could calculate d just like I did and then you could read any message I sent you.
    For (ease of) example I choose two small prime numbers p = 5 and q = 11. You then choose n = 7, because 7 is not divisible by 5 or 11. I now must find a number d that, multiplied by 7, results in 1 modulo 40. After all, (5-1) x (11-1) = 4 x10 = 40.
    Using trial and error I discover that 23 times 7 equals 161, which is 4 times 40 plus 1. Hence I will use d = 23. I would probably use the extended Euclidian algorithm to efficiently compute the right value for d.
    Now you receive a copy of my public key, which is the number n = 7 and the number 55 (p times q). You would now like to send me the number “2”. You now compute 2 to the power of 7 modulo 55 as follows:
    27 mod 55 = 128 mod 55 = 2 x 55 + 18 = 18 mod 55
    When I receive the message “18” from you, I raise the message to the power 23, also modulo 55. This gives me:
    1823 mod 55 = 181 x 182 x184 x 1816 mod 55 = 18 x 49 x 36 x 26 mod 55 = 825552 = 15010 x 55 + 2 = 2 mod 55.
    And I would know what that meant. Simples! (I think )
    Any numbers can be sent this way, perhaps a set ‘A’ designating every element of the ASCII code for example, and thus complex messages, e-mails for instance, can be sent in a virtually unbreakable code.
    I say ‘virtually’ because all of the above calculations and functions can be and are done by computer in the real world and provided that computer remains secure the messages are secure, but bear in mind that governments around the world spend a lot of time, money and computing power trying to intercept and crack such codes. As do crooks and hackers.
    By the way, guys, it’s a long time since I’ve done this sort of thing and as I have banged it off rather quickly my examples are probably littered with mathematical errors and omissions. However it’s the principles involved that I have so inadequately tried to explain and I hope that there, at least, I have made sense.

    Best Regards, S III.

  4. KG says:
    October 1, 2013 at 05:41

    That’s very kind of you, Seneca, but I was lost after “this may be too big”. :mrgreen:
    I’ve used PGP but the “engine room” is (and by the looks of your explanation always will be) a mystery.

    • Seneca III says:
      October 3, 2013 at 05:35

      I’m with you there, Wabbit. I wrote that several years ago when I still had a couple of clues as to what I was talking about. Unfortunately these days all of my mental effort is devoted to trying to figure out “why the ****”, when I am halfway upstairs, “I am going up here in the first place?” :cry:

      Rgds, S III.

      P.S. But the thing that really depresses me is that I now have think twice before I fart. Old age is not for wimps. :mrgreen:

  5. RossN says:
    October 1, 2013 at 07:47

    My password would take 10 days to crack.
    Whew! Good, my wife will be able to get access if I get suddenly struck dead.

  6. Darin says:
    October 1, 2013 at 10:21

    Well,they don’t want your password,at least the governments don’t.They don’t need them,they just tag your ISP number and track it.They can see which sites you visted and how long you were there and that’s all they really need to put you on a list and not the Christmas one.

  7. JustBob says:
    October 3, 2013 at 02:52

    It would take a desktop PC about 12 trillion years to crack your password

    All my “passwords” are Passphrases.

  8. KG says:
    October 3, 2013 at 04:46

    The two I care about are seven word phrases, Bob. As for the rest I suspect any protection they give from gummint snoops is illusory.