In Security/ Cyphering: I know that it's bad to use a one time pad twice becuase it can be "easily" cracked. What I want to know is that if I have two cipher texts encoded with the same key (made up of groups of 6 digits each) and a list of the only workable six digit groups for the plaintext, how  can I determine the key? I know it is possible without brute force. Please only answer if you know what I'm talking about.
199873
This algorithm will run in O(n) time with respect to the number of workable six digit groups, assuming that you've got a good hash table implementation. Note that it's possible that this won't necessarily result in a single solution.\n\nPut all the six digit groups into a hash table\nfor each six digit group\n-----determine the pad necessary to encode it to the first ciphertext\n-----decrypt the second ciphertext with that pad\n-----if the decrypted second ciphertext is present in the hash table, then you've found a potential key\n\nI apologize for the dashes. Yahoo seems to eat whitespace at the beginning of lines
77829
I think the contributions thus far have hit upon the reason, but I'm not sure it is all tied together.\n\nChoosing any two-digit number and subtracting its separate digits will yield 81 for any two-digit number that starts with 9, 72 for a choice of 8x (8x = 89, 88, 87, 86, to 80), 63 for a choice of 7x, and so on.\n\n9*9 = 81\n8*9 = 72\n7*9 = 63\n6*9 = 54\netc\n\nIf you look at the corresponding numbers for ALL of these choices you will see the same symbol! So no matter where you start you will always get this answer.\n\nAlso, the symbols change every few refreshes of the page. So when you "Try Again?" you will get a fresh symbol. If you look at them closely you will see this happen as you go back and play again and again.\n\nThis is a very slick game, and even tho I know how it works, I still think it's way kewl.\n\nlol
615412
They are stored in a password file and they are encrypted, you won't be able to see what the file contents. It is not in a human readable form.
657079
There are different ways of doing this in either hardware or software.  Different vendors may approach this in different ways.  Based on your question, I am assuming you are talking about things like RSA SecurID tokens.  The server and the token are initiated with seed record and synched to each other prior to sending the token to the end user.  With these devices, you must also enter a PIN number when you access; this provides what is called two-factor authentication.  It is 2 factor because you must have the token and you must know your PIN number.  Knowledge of your PIN is useless without the token and the token is useless with out knowing the PIN.  When you initiate a session between your computer and the VPN 'server' and provide your credentials, your computer begins a key exchange process with the VPN host .  There are a number of key exchange protocols available such as Diffie-Helman; there are others.  These key exchange protocols get fairly in depth and usually deal with extremely large prime numbers (so large that only a computer can do the math).    Once the key exchange process is complete, each end of the connection can build an encrypted tunnel using those keys.\n\nIf you are interested in cryptography or key exchange processes, you can google things like Diffie Helman and Kerberos to get you started.  There are also many books on encryption technologies  out there, but be warned, some of them can get pretty deep into the mathmatics.
268688
sure. \n\nI am 27 and i do it as a hobby in my white hat work for \nMINTM purposes.\n\nbut I am not a think tank. so what kind of stuff are you looking for exactly.\n\nIn general first start is math. study the history of how simple encryption started. move on to enigma cause that's when stuff picks up, then a blend of steganography would surley excite you. \n\nas for actual decryption and or how to create a new algorithem, I recommend studying permutational algorithims with Mue and Limbda, that's always a good start.\n\nhope that's usefule. \n\ngoogle up the stuff you don't understand.\n\n8t3 i make m1st4k3s 0n purpos3 ;-)\n\nregards,\n\nCysmn
148340
Basically it works like this.  The RSA involves sophisticated function that uses prime numbers generated with a pair of numeric keys.  When you plug these two prime numbers into the equation, it encodes some text into some other text in a way that makes it very difficult to decipher without knowing both keys.  Easy to encode; nearly impossible to decode.\n\nNow, I keep one of these keys secret.  The other I give out to everyone else.  When they want to send me messages, they can encode their message using my public key and only I can decode it with my private key.  In the same way, when I want to send messages back I use the other person's public key and they decode with their own private key.\n\nNow, knowing the public key it is possible for me to guess at what values are possible for the private key.  And if we use small values for the keys, this wouldn't take any time at at to decrypt.  This is why the public and private keys (or rather the important factors of these keys) are extremely large numbers -- either less than 2^64 or better yet 2^128 or pehaps even larger.  \n\nIn the end, any code can be cracked but it'll take so long to crack by brute-force that we'll all be dead before we find out what the message was, even with very powerful, very fast computers.\n\nMore details at the link below.  Also more math, sorry.
631360
Each word can be spelled using only one of the keys on a telephone keypad.\n\nThe other answer has already been posted.
234993
Hi Lauren!  Lauren here, lol!!!!  (yes thats my real name too).  the answer is........cdefgabc  - not to mention the sharps, flats, etc. which are the black keys.  Need more info?  e-mail me.  I too have a keyboard with 36 keys.  and I can go more into detail.   Good Luck Lauren,  sincerely, Lauren.....p.s. luv yer name!!!!!  LMAO !!
472661
Try tracing a very small example on paper to gain some insight into this.  Perhaps just two or three elements with at least two equal keys.  Then reverse the order of the two equal keys in the input set, and see if you still get the same result.  Tracing the algorithm like this will get you the answer in a manner that should give you some confidence.
367755
kbhit will return a integer to you, go to asciitable.com to find out what des each interger corresponds to what key.\n\ngood luck
535461
I understand what you are saying when you mention the permutations for 10 digits, but I think you're missing what it really means.  First, this isn't right to think of it this way to think about this.  Non-repeating doesn't mean that I can't find a string in the decimal expansion that's the same as an earlier string (for example, finding a piece 14159 later in pi isn't impossible).  What it means is that there is no point where the digits in pi *completely* repeat (i.e. it becomes something like 141591415914159.... on and on forever).  So you're right, it can include some kind of partial repeat, but non-repeating means it can't have a total repeat.\n\nYour comment about the permutations (not combinations) on the numbers {0,...,9} is right but it's used incorrectly also.  It's wrong because of what I mentioned above.  Basically it's assuming that once you pick a number you can't repeat it again in a 10 digit sequence, but that's clearly wrong (again, 14159 appears in pi, where 1 occurs twice).  So you have several right ideas here but you're combining them seomwhat incorrectly.\n\nHope this clears things up!!
527256
First answerer correct.  \n\nBut the second set is a function, because there are no repeated x values.  If you have a component of the first set corresponding to two different components of the second, then it isn't a function, e.g. in the second set, if the next pair were (15, 5) then it wouldn't be a function because of the double result for 15.  In the first set, if your misprinted pair were (1,2) then it wouldn't be a function, same reason.
405087
Ok this question has plauge computer users for quite some time we all know that passwords should be atleast 4 - 8 charaters long include caps , numbers , not be in the dictionary and be difficult to guess , the problem is how can we use all that and still remember the password , well it's easy  Take 2 items from your desk memory names maybe two items you can remember , now take an important date once agian something you can remember now you have three items seprate anyone could guess your password however if you mix them together you have a very long and difficult password for anyone or thing ( program) to guess and you can remember let me give and example\n\non my desk i have a stuffed cat and a dog and my birthday is in 1976 ( not really) i also like dogs more then cats\n\nif we say dogs are bigger then cats we can change dog to DOG and cats remains the same now also DOGS are better then cats so we start with that\n\nwe could make the pass\nDOGcat1976  now thats not all that strong so lets mix it up a bit\n\n19DcOaGt76  by intertwinging the letters and putting the first part of the date before everything else and the last part at the end we have a very secure password.
159682
Below you will find the specifications for the AES (Rijndael) and Three-DES algorithms.  \n\nAs for the writing of the programs themselves, I'm almost done with my CSE degree--sounds like your pain is just beginning :-D
455556
the only way you can get the key is to log on to the route/access point that is responsible for encrypting the data.\n\nwhen you are in there you can go to the wireless or security settings and read them from there.\n\nso basically ask the person who administers the network.. if you are meant to have the key they will give it to you... there is no point for you to get it any other way as any person with an ounce of sense will rotate the key to a new code on a regular basis.. not to mention Temporal Keys that do this automatically.
166058
See this webisite:\n\n"This is an essay I wrote for a mathematics essay prize. We had a number of possible topics to choose from and I choose cryptology, because I already had a passing interest in thanks to my attempts to code encryption algorithms for my computer programs. It won, by the way ;-). This is the original essay with a few corrected spelling mistakes (though there are probably some left)."\n\nLoads of info
450637
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)