How to decode and what is the use?(Math!)

@giulatestacoglione Don't write your nonsense in the forum.

@tromeus heil hitler! Si signore! Ave cesare....pf

Interesting code...
I would have though it was
MUJL SOFTWARE ABTORED.
Or something like that.

Or else
MKJL SOFTWARE ABTORED JOIN HTML CLICK TIME FIVE OF HQ.
I was a bit off 😂😂😂😂

😏😎

@tromeus

Of course, I'm not amazed that it is googlable, many textbook problems are, but where is the fun then? I definitely prefer to solve the riddle by myself.

And the other method you explain... well, when we use many alphabets to encrypt the message, it is intuitively named as the polyalphabetic substitution cipher. In this case probably even the polyalphabetic Caesarean cipher, because you (according to your description) never shuffle the alphabet, only shift it. Obviously, such a cipher is breakable by frequency analysis means only when the message is significantly longer than the key (like a few hundred times). A message doesn't tend to be, for example, 500 times longer than a book, so everything seems safe.

Yet, is it - as you write - "by all means unbreakable"? Well, not really. The clear weakness is that both the coder and the decoder need a particular book to work with the message. If you catch him or her with that book, you don't even need torture to divulge the key, everything falls apart by itself. I have heard about at least one such spectacular spy failure.

Moreover, even without such help, you should be able to break it only by mathematical analysis. I'll show you a sketch of the method. We will try to decode the key and the message. Let's look at the first letter of the encrypted message. Say it's J, the tenth letter of the alphabet. The whole alphabet has 26 letters. Thus, we know that the sum of alphabet positions of letters M1 (first in the message) and K1 (first in the key) is 10 or 36. Maybe M1='A' and K1='I', maybe M1='U' and K1='Q', there are 26 possibilities. Now we look at the second letter of the encrypted message. Of course, there are 26 possibilities again, which gives on summary 676 options after two letters. If we expect both the key and the message to be proper English texts, however, we already notice that most of these 676 options make no sense (for example, in about 50 of them there will be no U after Q). The tree of possibilities begins to narrow itself. After a few letters, it should narrow to just one. At times it can widen (when some words of the key and the message end at the same position), but only temporarily and controlably. And when you decode something like a hundred letters, you already have a fair chance to find the key in some database of literature.

All in all, it is true that the method you proposed is sometimes reported to be used by various agents, but for me it seems pretty risky. With any access to the computer, I would definitely go for RSA instead. What do you think?

RSA can be broken by NSA: the key may not exceed the length the supercomputer of NSA can crack.
Any code can be broken provided sufficient code sample and enough time.
The famous enigma code of the Second World War is a famous example.

@Otienimous Hello!

My example in #9 was to exemplify that no frequency analysis is going to work when there is not a 1-1 correspondence of encoded-decoded letters, ever.

Of course, if somebody steals the shared key code utilized by both parties it's over as you mention.

But, here is another example to elucidate my point: Suppose that you intercept a message that reads: "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" Is there a way to decrypt it with frequency analysis without the key?

I challenge everyone to try and decrypt a single word: "AAAAAAA" with frequency analysis.
Here a more sophisticated method was used for the encryption.
The text was:

"No
Lover's
Message
Passes
Via
Black
Holes",

where instead of using all letters of the text, we used only the first letter of each individual word: "NLMPVBH" ( The encoding key )

In order to decode we proceed as follows with our key at hand:

N is the 14th letter of the alphabet, so the first "A" is shifted 14 places to the left ( because encoding shifts to the right ) to become an "M".
L is the 12th letter of the alphabet, so the second "A" is shifted 12 places left to become an "O".
And so on...

The above example demonstrates clearly how any attempt of frequency analysis is hampered by the absence of 1-1 correspondence.

@tpr

RSA method, unlike all the other previous ones, has an asymmetrical complexity - coding is incomparably easier than breaking, even if you know the method - just as multiplying 13x89 is incomparably easier than finding prime divisors of 1157. With a bit of abilities, you can use your laptop to create a RSA key unbreakable for the NSA supercomputers.

@tromeus

I'm not sure whether you actually read my text carefully. I wrote that you can use frequency analysis only when the message is much longer than the key. You can do nothing with intercepted message like AAAAAAA, but if the message has a few thousand letters and you expect the key to have only a few letters (like NLMPVBH), you can break the code by applying frequency analysis to shorter messages comprised of every second letter, every third letter and so on, until you guess the right length.

I also proposed a method of decrypting the message which you know to be coded with a book as a key (unknown, long key and long message, both in known language). Have you noticed this fragment?

@Otienimous

I have read your message. The key must have at least as many letters as the message ( ideally, many more than the message ) . There is also randomness involved, for example if AAAAAAA translates To MONKEYS, in the first instance, do not expect every mention of the word MONKEYS in the enciphered text translate to AAAAAAA again, because other letters will be used for the encoding next time ( and not the very first same word NLMPVBH. )

But instead of a text and a book you could use any sequence of the digits of Pi ( in blocks of two digits ) after the 44654894121 position say.