Question: is the encrypted file size determinic by original file? What’s the formula?

[fyears]

Hi all,

Is there any way to compute the encrypted file size quickly by using math formula based on the original file size? Assuming no recipients are set during the process. Even more, any reverse formula from the encrypted to the original?

I am asking this because I am building a file backup tool and want to detect some file changes quickly by comparing file sizes. If the computed size from unencrypted version 2 is different from the size of previously encrypted version 1, I could say that the two versions must be different. Some sort of shortcut logic.

Thanks!

[fyears]

According to the man page: https://github.com/FiloSottile/age/blob/main/doc/age.1.ronn :

age encrypted files are binary and not malleable, with around 200 bytes of overhead per recipient, plus 16 bytes every 64KiB of plaintext.

So I am assuming the formula should be: encrypted =ceil(original *1.0/ (64*1024))*16+ original. Is it correct? Any other thing I am missing?

Update: After reading https://github.com/C2SP/C2SP/blob/main/age.md , there is obviously at least 3 lines of header I am missing