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Thread: Random Number Analysis

  1. #16

    i want to...

    Hi Mike
    thank you for replying to my questions.
    Finally something that i know appear in this
    thread.I know RSA very well(to my knowledge),i have used
    it a lot of time.As i told you my first project is to prove to my
    boss that these random numbers are week.I think afer all your
    (Dimdrol,you and others) help i am very close to it.
    The secont project is to give him(my boos) a good random
    generator.I have thinked to use the RSA method described
    in chapter 5 of applied cryptography handbook.
    But before this i have to prove to my boss that the current
    algo used by his sister is really bad.
    The divisability rules found by dimdrol was very good
    because i could use it to show to my boss that I can find the
    next password with 10 hope in the worst case,but as you have seen I have found a contradiction to it.
    So i have to use another rule.
    If you know anything that could help me please help me.
    I have another question from you how did you find the LGC of

    thank you very much for your help

    I promise that I have read the FAQ and tried to use the Search to answer my question.

  2. #17
    Handbook of Applied Cryptography

    See Chapter 5 for BlumBlumShub

    And a C implementation is here:

    I promise that I have read the FAQ and tried to use the Search to answer my question.

  3. #18
    הבּרוּ נשׂאי כּלי יהוה mike's Avatar
    Join Date
    Mar 2001

    lcg passwords

    the differences between sequential seeds are
    not ALWAYS divisible by 10.
    So what? If you know one of them, then skip 9, generate four passwords, and test them. You expect to get it on the tenth, but sometimes it'll be a little different. Once you find which of the four work, skip 9 from that one and try again. Even in the worst case where you don't know the original seed, you only have to test 2 billion passwords (2^31). Depending on how long the test takes, it could be as little as an hour to try all possible passwords.

    Even if you didn't know how the passwords were generated, their length is too short: trying 10^14 different numbers= 2^46 is very doable.

    I looked up delphi random on google and poked around. 0x8088405+1 has a period of about 2^32 and is used all over the place, especially in Zip encryption.

    Again, you aren't going to get a secure system by accident. There are lots of ways to attack systems that most people don't even think of. What are you trying to protect with these passwords? Are you going to store them somewhere? How is that database protected?

  4. #19

    what i want

    thanks for your reply .
    about the project:
    the project is for the Telephone company.They will sel phone card
    with this passwords.Each person after picking up a public telephone will here a message that asks for the password
    then he have to dial the password ,if he dial a valid password
    he can dial any number he want.As you see if this password
    are regenerable my boss will dye because our company has
    investigated on this project.
    For the bruteforce that you sayed ,it is impossible because
    10 password is resonable to try on the phone but 40(4*10)
    is impossible.I have to give them a solution that is resonable.
    the seed is changed between 10 and 100 so if I add 9 and generate 4 password i have to do this 10 time in the worst case
    that gives me 4*10 password(not resonable).Correct me if I
    have not understand what you mean by adding 9 and generating 4 passwords.
    About the storage the passwords will be encrypted and stored
    in an oracle database.No one will have access to them,and the encryption is done by the encryption library of oracle 9i.

    thank you very much for your help,you teach me a lot of
    great things i really apreciate you.


    I promise that I have read the FAQ and tried to use the Search to answer my question.

  5. #20
    Hi akimp3

    Yes I knew what those differences between sequential seeds are not _always_ divisible by 10 ( it's seen from my example with seeds 66921289 and 66920498 ).
    That's why I said:
    "So it's possible to say what if one knows 1 password - he can find the next sequental password in at most 10 hops _with_a_very_good_probability."

    Let's make a statistical analysis of differences between sequental seeds.

    For the example ( these are not a real numbers, just for the illustration ):
    30, 70, 50, 100, 31, 50, 30, 70, 62, 40, 80, ....

    It's seen what there are much more neighbour differences which are divisible by 10 than others.

    To calculate exact probability of getting the next difference divisible by 10 we need to know the full set of differences, but even without knowing that, from the statistical analysis we can see what this probability is higher than 1/2 which is _very_ good ( in casino for example chances to win are much less ).

    And besides, you can find the next password not only in direct order, but in a reverce order also.

    I mean if you've found the seed say 66923456 - you can do 2 passes - in a direct order
    66923456 + 10, 66923456 + 20, 66923456 + 30, .. 66923456 + 100
    and in a reverse order
    66923456 - 10, 66923456 - 20, 66923456 - 30, ... 66923456 -100

    BTW: I think range 10 - 100 between seeds is too strict. I think it's possible to reduce it ( but you need to gather more statistics ).

    So you see - max. 20 hops (can be reduced I think) ( 10 in direct order and 10 in reverse order ) and you get a password _with_a_very_good_probability ( again )

    Regards, Dimedrol.
    I promise that I have read the FAQ and tried to use the Search to answer my question.

  6. #21

    Wink more statistics

    Dimedrol thank you very much for your help.
    here are more statisticd. I have find something in them
    but i am not sure if it usable.
    Download the attached text file.the file has all the passwords
    and their seeds in front of them.
    As you have told me the seeds are divisble by 10 in most case.But after a number of password the (difference-1) is
    divisible by 10.And after that the other passwords seeds are
    divisible by 10 since we arrive to another KEY password that
    the seed difference is not divisible by 10 and (difference-1)
    is ivisible by 10.I have tried to find the number of ordinary password between two KEY password.We can not count
    the first transition because we dont know maybe there is other
    passwords before it that have 5 as their last digit.
    but after that there is:
    15 number with 6 at their end
    17 number with 7 at their end
    15 number with 8 at their end
    19 number with 9 at their end
    18 number with 0 at their end
    Do you think that a rule exist between this numbers?

    thanks in advance

    Attached Files Attached Files
    I promise that I have read the FAQ and tried to use the Search to answer my question.

  7. #22
    הבּרוּ נשׂאי כּלי יהוה mike's Avatar
    Join Date
    Mar 2001

    use a war dialer

    How do you think telemarketers call people? By hand? If you know one seed that generates a number, you can set up a war dialer to try every seed after that until it finds one. According to the list below, you'd have to try at most 100.

  8. #23

    Thanks to all of you !

    thank to all of you that helped me on this project
    specially Mike and Dimedrol.
    I have shown the program to my boss and he has
    accepted that the algo is week.He asked me to write
    him a random number generator.Without your help
    i will never arrive to this point.
    This board,its moderator and all of its members are greats.
    if I had asked for help on other boards they would
    not have helped me and they would have treated me as person
    who ask for cr*ck. But here you have teach me a lot
    of thinks you have answered to all my stupids questions
    and you solved my problem.

    Thank to all of you
    I will never forget your help

    I promise that I have read the FAQ and tried to use the Search to answer my question.

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