# Are ciphertext-only attacks on LFSRs possible?

Reading about LFSR, I know that breaking an LFSR by knowing it's design and having enough (plaintext, ciphertext) pairs is an relatively easy task but let's assume we know the design of LFSR and a just a long ciphertext, which is the result of xoring plaintext and LFSR output. Is there any way to decrypt the ciphertext?

For example suppose we have an LFSR with primitive characteristic polynomial $x^{51}+x^{9}+1$ and a 5600 bit ciphertext, is there any way to obtain plaintext which we already know is an English plaintext?

One brute-force approach is to try all $2^{51}$ possible seed values, xoring ciphertext with LFSR output and check if resulting plaintext has normal english letter entropy, but this approach gets harder and harder having a larger LFSR degree.

Things get more difficult as the plaintext becomes less redundant. With 7-bit ASCII, noticing that space, digits, common punctuation and all lowercase letters are in range $[\mathtt{0x20}\dots\mathtt{0x3F}]\cup[\mathtt{0x60}\dots\mathtt{0x7F}]$, we know that every such 7-bit symbol has bit with weight 5 set, and there is fair chance that 51 such characters occur in a row somewhere in the 800-character plaintext. A guess of where allows to decode the rest, and it is easy to detect if the guess was right or not, thanks to characteristics of English text.