Not quite sure I fully understand your question, but let me share some thoughts:
- Provably safe encryption requires a key of the same length
k
(in bits) as the message (n
), with 2^n
possible keys and decryptions of the same ciphertext.
- If you take a key of length
k=n-1
instead, you'll have half as many keys and possible decryptions, and have 1
extra bit of the message to check for plausibility. Best case, you can discard half of the possible decryptions as implausible.
- If you take a key of length
k=n-2
instead, ... and so on.
- If you take a key of length
k=1
, you've got two possible decryptions, one of which is correct while the other is more or less recognizably incorrect.
The expectancy value for getting more than one plausible messages for a given key length then roughly depends on two factors:
a. n-k
, and
b. the total number of plausible messages of length n
.
Notice that, to detect if a decrypted message is plausible or not, there has to be redundancy in the plain text in the first place. If there were no redundancy in the plain text, it would be indistinguishable from pure random content.
This redundancy may exist inside the message, like it is given in any natural language text for example, or it may exist outside of the message itself which is the case in the more general known plaintext attacks.
However, measuring the number of plausible messages may prove to be difficult, and definitely depends on the kind of message. The (typical) plaintext's entropy might give a rough estimate. This is a factor that's independent of the encryption algorithm or key.