Given the following cipher text:

LE-eheadty.AC2lta oiwyGC24

What possible ciphers could it be?

closed as off-topic by Steffen Ullrich, schroeder Apr 28 '17 at 6:30

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  • Recovered German U-Boat transmission using the Enigma Machine? – Gypsy Spellweaver Apr 28 '17 at 4:30
  • Sorry @GypsySpellweaver we don't have a key to crack it. The text is a cipher text. no key – Timothy Wong Apr 28 '17 at 4:31
  • 2
    Unless it's worth a million or life-and-death, it's probably not worth cracking. Too short, and with no context, to even attempt an analysis. The first comment was meant to show the futility without the key. – Gypsy Spellweaver Apr 28 '17 at 4:37

Literally anything. Ciphertext is arbitrary binary data, usually grouped into bytes. Text, including the characters you mention, are also bytes. There are no ciphers seriously used in the last thirty years that can't produce all of those characters in their ciphertext, and plenty of older ones can too.

It's atypical for an arbitrary ciphertext to not have a single character that isn't a printable ASCII value even in a relatively short string (the odds of this happening are less than 1 in 2^N, where N is the length), but if somebody is aiming for that outcome it'd be pretty easy to achieve. Even if the message itself is fixed, or the key is, there's no reason you couldn't adjust the other of those two plus things like the cipher primitive, the mode of operation (where relevant), the IV (where relevant), and the padding (where relevant) to get a ciphertext string matching that criterion.

The length may be more informative; 26 bytes (if indeed those are the entire ciphertext) is wrong for pretty much all symmetric block ciphers (which mostly have a block size of 8 or 16 bytes) unless of course they're in a counter-based mode of operation (which makes them like stream ciphers). It's possible it's something asymmetric, though in that case it's weirdly short. The most likely option is a stream cipher, of which there are only a relative handful in common use. Of course, this doesn't help you decrypt it unless you have the key or at least some other source of useful information.

To sum up, there's no way to know.

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