One-Time Pad Encryption With No Pad?
thepooleboy writes: "The Globe and Mail has an article about a Toronto area company that has perfected 'Unbreakable Encryption' using the Vernam Cipher." The idea is to use as a one-time pad a large number generated by equations sent with an initial (proprietary) exchange which takes place when users connect to an equipped server. Since real one-time pads' numbers are by definition random and known in advance to both sender and receiver, though, the company seems to be playing fast-and-loose with their terms.
Actually, a correctly used one-time pad is unbreakable. The true randomness of the pad cannot be calculated, and if it's never reused, you have no clues as to how to calculate the encryption.
However, this scheme isn't a one-time pad. It's a function, with parameters encrypted with a standard encryption algorithm. If you break the algorithm used to exchange the parameters, you've broken the whole code. It's certainly no better than anything else out there.
Attempts to get around the fundamental limits of data encryption (and data compression, and a lot of other software fundamentals) remind me of all the pointless efforts to build a Perpetual Motion Machine. "Yeah, the smart guys say energy is "conserved", but anybody with any common sense can see if you just tweak this gearbox this way..."
They have a program which generates new keys for each subsequent transaction, and they claim that this counts as a "one-time pad".
Nonsense -- a one-time pad is only secure because there is provably no way to figure out the keys without a copy of the codebook (assuming they were generated through appropriate random means).
As long as a program is producing the keys, they will exist in a particular sequence. All you need to do is figure out at which point in the random sequence you are, and then you can generate the rest of the sequence easily, allowing you to eavedrop on the conversation.
Admittedly, the article was fluff, but key-hopping doesn't significantly increase the difficulty of breaking encryption. Unless there is something else behind this that I'm missing, this is another "Compress random data by 99%! For real this time!"
ZFS: because love is never having to say fsck
finding their website was non trivial on google
its here
http://www.prescient.net/
no, a vernam cipher is the only form of unbreakable encryption. It happens like this: you have a stream of extremely random bits. And you have to make sure they are really really random, no pseudo random number generators. Say it's coming from a satelite up in space that measures radioactive particles(this was proposed in a paper not too long ago). Now the satellite streams these bits down to earth, so anybody can access them. Alice and Bob want to communicate securely over an insecure channel. So the agree on a series of bits to encrypt with. This can be anything from "every other bit" to a large polynomial function that says which bits to use. So every bit the function designates as an encrypted bit is used to XOR any message Alice and Bob use to communitacte. So, Alice computes bit random bit number x to encrypt bit y. She does XOR(x,y)->c and sends it to Bob. Bob also has this formula and performs the calculation to find which bit number x to use, then performs XOR(c,x)->y. The key is keeping the bit number function secret. Now, why is this secure? because anybody listening on the channel doesn't know the function(hopefully) and if your bits are truely random there is *no* way to distinguish whether any given bit can be 0 or 1. Try all the combinations for 0 or 1 in the message you want, but every permutation you want will look like the correct decryption.
- "Never let a computer tell me shit." - DelTron Zero
..."unsinkable" is to "ship"
The client generates a series of random numbers to use as an encryption key. This is number is exchanged with the server through a secure process known only to Prescient, the server uses it to encrypt any information
Ha! The fools! Just send your message through this secure process. No need for the one-time-pad nonsense! QED.
Given infinite time, a monkey will eventually bang out the contents of the OTP.
Sure. The question is: How will you know it when you see it?
The monkey will bang out every possible n-bit sequence. Applying them all to the n-bit encrypted message will give you every possible n-bit message. So you'll get all of the following:
So, how, exactly, will you know when you've found *the* message?
That's why an OTP is provably unbreakable. Because every pad is equiprobable. And that's why no algorithmically-generated pseudo-random sequence can be used for a one-time pad.
Note to ACs: I usually delete AC replies without reading them. If you want to talk to me, log in.
Ummmm... comparing asymmetric encryption to symmetric encryption (of which a one-time pad is a subset) with key-lengths is like comparing apples to oranges.
This much is right.
In asymmetric encryption, your security is in your keyspace... every bit doubles the time to search the keyspace.
This much is nowhere near right. According to our best estimates at the present time, it'll take on the order of 2**80 operations to factor out RSA-1024. It'll take on the order of 2**128 operations to factor out RSA-3072.
Adding two thousand bits doesn't increase the difficulty by 2**2048... only 2**48. Asymmetric crypto does not double in difficulty with each added bit.
In symmetric encryption, security is all about the keys; symmetric encryption is so easy to do that you can try millions of keys a second, as opposed to thousands or hundreds, so you HAVE to have a big keyspace.
This is not correct. In fact, it's downright astonishingly wrong. The problem is you're assuming symmetric and conventional, non-ECC asymmetric keyspaces are both flat (they're not). But if they were flat, then asymmetric crypto would have a keyspace multiple orders of magnitude larger. Which is the opposite of what you're asserting here.
Conventional, non-ECC asymmetric keys are so huge because most of the keys are weak. Let's compare DES to RSA. Is 0xFA810DD0 a legitimate 64-bit DES key? Yes. (Note: DES only uses 56 of those bits for key material; the other 8 are used for parity.) Is 0xFA810DD0 a legitimate 64-bit RSA key? No. Why? Because 0xFA810DD0 is an even number, which makes it much, much easier to factor.
Conventional, non-ECC asymmetric keyspaces are so huge partially (not exclusively) because most of the keys in that keyspace are unusable. Symmetric keyspaces are so small partially (not exclusively) because most of the keys in that keyspace are usable.
A keyspace in which all (or the overwhelming majority of) keys possess equal strength is called a "flat" keyspace. A keyspace in which some keys are stronger or weaker is... well, non-flat.
But, most symmetric encryption algorithms allow you to get it partly right; if the key is partly right, you get a partly decoded message, so the search algorithm is linear instead of exponential.
This is so wrong that it staggers the imagination. Claude Shannon established some principles back in the 1940s which still guide cipher development today. One of these is called the avalanche effect. The idea behind the avalanche effect is that a single one-bit error, anywhere in the enciphering/deciphering process, will affect the output of half the bits in the entire e/d process.
Go ahead. Use Blowfish with a 40-bit key. (There are lots of Blowfish implementations out there; if you want one, email me and I'll send you one.) Encrypt it with one 40-bit key, and then decrypt it with a key that's only one bit different. You'll get absolute, total, gibberish. You'll get gibberish because Blowfish is a well-designed cipher and avalanches properly.
But wait--it gets even worse. Only a chump runs a cipher in electronic codebook mode. Usually, ciphers are run in a block-chaining mode, where every subsequent block gets XORed with the prior block. So if you have a one-bit error in your process, that will affect half the bits of the block... which then create errors in half the bits of the next block... which avalanche... which propagate their error forwards, on and on and on... etcetera.
You get the idea.
(All of the above information can be found in either Bruce Schneier's Applied Cryptography, 2nd Ed or Menezes, Oorschot and Vanstone's Handbook of Applied Cryptography.)
Bill Stewart
New Fast-Compression-only CPR http://preview.tinyurl.com/dy575ks