Ciphertext help ?

[Deleted member 61728]

What is the hardest to break cipher that can be used to encrypt plain text using just a pen and paper no computer involved and no mathematician required ?

 

[Energize]

One time pad, impossible to break.

 

[EndlessSky]

One time pad is the way, into a spot of light terrorism?

 

[Deleted member 61728]

Thought it was,but how long would the worlds fastest supercomputer take to decrypt a properly encrypted plain text message of say 10 lowercase letters done using the one time pad cipher without knowing the key?

 

[Reaper 392]

Thought it was, but how long would the worlds fastest supercomputer take to decrypt a properly encrypted plain text message of say 10 lowercase letters done using the one time pad cipher without knowing the key?

it couldnt be done without the key. i've just had a little read of the one time pad method and wiki explains it very well. it also explains why no computer could crack it without the key:

Example

Suppose Alice wishes to send the message "HELLO" to Bob. Assume two pads of paper containing identical random sequences of letters were somehow previously produced and securely issued to both. Alice chooses the appropriate unused page from the pad. The way to do this is normally arranged for in advance, as for instance 'use the 12th sheet on 1 May', or 'use the next available sheet for the next message'. The material on the selected sheet is the key for this message. Each letter from the pad will be combined in a predetermined way with one letter of the message. It is common, but not required, to assign each letter a numerical value: e.g. "A" is 0, "B" is 1, and so on. In this example, the technique is to combine the key and the message using modular addition. The numerical values of corresponding message and key letters are added together, modulo 26. If key material begins with "XMCKL" and the message is "HELLO", then the coding would be done as follows:

      H       E       L       L       O  message
   7 (H)   4 (E)  11 (L)  11 (L)  14 (O) message
+ 23 (X)  12 (M)   2 (C)  10 (K)  11 (L) key
= 30      16      13      21      25     message + key
=  4 (E)  16 (Q)  13 (N)  21 (V)  25 (Z) message + key (mod 26)
      E       Q       N       V       Z  → ciphertext

If a number is larger than 25, then the remainder after subtraction of 26 is taken in modular arithmetic fashion. This simply means that if your computations "go past" Z, you start again at A.
The ciphertext to be sent to Bob is thus "EQNVZ". Bob uses the matching key page and the same process, but in reverse, to obtain the plaintext. Here the key is subtracted from the ciphertext, again using modular arithmetic:

       E       Q       N       V       Z  ciphertext
    4 (E)  16 (Q)  13 (N)  21 (V)  25 (Z) ciphertext
-  23 (X)  12 (M)   2 (C)  10 (K)  11 (L) key
= -19       4      11      11      14     ciphertext — key
=   7 (H)   4 (E)  11 (L)  11 (L)  14 (O) ciphertext — key (mod 26)
       H       E       L       L       O  → message

Similar to the above, if a number is negative then 26 is added to make the number positive.
Thus Bob recovers Alice's plaintext, the message "HELLO". Both Alice and Bob destroy the key sheet immediately after use, thus preventing reuse and an attack against the cipher.

Attempt at cryptanalysis
Suppose Eve intercepts Alice's ciphertext: "EQNVZ". If Eve had infinite computing power, she would quickly find that the key "XMCKL" would produce the plaintext "HELLO", but she would also find that the key "TQURI" would produce the plaintext "LATER", an equally plausible message:

    4 (E)  16 (Q)  13 (N)  21 (V)  25 (Z) ciphertext
−  19 (T)  16 (Q)  20 (U)  17 (R)   8 (I) possible key
= −15       0      −7       4      17     ciphertext-key
=  11 (L)   0 (A)  19 (T)   4 (E)  17 (R) ciphertext-key (mod 26)

In fact, it is possible to "decrypt" out of the ciphertext any message whatsoever with the same number of characters, simply by using a different key, and there is no information in the ciphertext which will allow Eve to choose among the various possible readings of the ciphertext

 

[Energize]

Thought it was,but how long would the worlds fastest supercomputer take to decrypt a properly encrypted plain text message of say 10 lowercase letters done using the one time pad cipher without knowing the key?

impossible to break.

 

[Deleted member 61728]

Cia,mi6,nsa etc they must break it often surely ? I wonder how i guess a combination of really powerful computers and a really big dictionary?

 

[J.B]

Trouble with the one time pad is key distribution and synchronisation. Hence the conspiracy theories around number stations.

 

[Energize]

Cia,mi6,nsa etc they must break it often surely ?

No, it's mathematically impossible to break. You're creating random data and subtracting it from the plaintext, so you're left with a ciphertext of purely random data. You could try every possible key and you would never know what the plaintext is.

Despite what tv programs would have you believe the government cannot decrypt strong encryption like AES and Serpent, which is why these algorithms are used by the goverment in the first place!

 

[Ev0]

With the one time pad anyone trying to decrypt it has no idea when they've hit on the correct plaintext.

in a big long document then it can be obvious when you've decrypted it as it'll start making sense

But for 10 lower case chars, well that could be any number of words or phrases, so how would you know which was the right plaintext?

 

[Energize]

in a big long document then it can be obvious when you've decrypted it as it'll start making sense

It isn't because there will be a quadrillion plaintexts for that ciphertext that make sense.

 

[gobbo]

As long as you obey these four rules:

* The key must be as long or longer than the message you are encrypting
* The key must be truly random (don't use a computer to generate it!)
* The key and plaintext must be in digits, letters or binary
* The key is only used once and then disposed of by both sender and receiver

It is impossible for any computer given infinite time and infinite processing resources to be able to break OTP, however, if you don't obey them, it's not OTP and in theory could be broken

 

[Ev0]

It isn't because there will be a quadrillion plaintexts for that ciphertext that make sense.

well true lol, I guess if you had some idea of the subject of the ciphertext, but even then it could have many plaintexts that make sense

 

[Energize]

There may be extremely limited circumstances where you know it's a certain type of file and there are only a few possible conforming plaintexts for that filetype and size, but knowing all that about the plaintext you wouldn't need the ciphertext anyway making any kind of encryption pointless.