Consider the affine enciphering transformation:
f(x) = 9x + 4 mod 49
on single letter message blocks of the ordinary English alphabet A - Z
and the letter b for “space”.
How many more letters can we add to our alphabet without any
further changes to our encoding procedure?
Would this just be 26 x 2 = 52
52 - 49 = 3, hence 3 more letters could be added?
Thanks in advance!
Again I believe this is a number theoretic question.I have never had a formal course in cryptography, but I know a little bit of number theory.
Originally Posted by hunkydory19
I think this question is essentially asking, find the largest number of x's such that is a bijection. Since .
This means there are 49 values of x which will map to a unique (9x mod 49).
Since we have already added 27 letters, we can add 22(=49-27) more letters to the alphabet without any further changes to our encoding procedure.
Again, feel free to point my mistakes. Since I am not entirely sure whether the utility of the affine enciphering stems from its bijective nature. It looks like common sense, but I am not sure :(