A tiny generalization in my humble view would be choosing elements from
a sequence with a probability p by a statistically good PRNG. Suppose
one chooses p=0.1, pseudo-randomly pick a starting point and "decimate"
thus the digit sequence of Pi, is there any conceivable yet practical
way that an anylist could succeed to do prediction in such cases?
Thanks,
M. K. Shen
Absolutely. Although it does increase the difficulty, it does not change an
insecure PRNG to a cryptographically secure PRNG.
Joe
It may be noted however that there is an "indirectness" involved, i.e.
the insecure PRNG employed is not directly used to encrypt (xor with
the plaintext), so that the analyst can't get its bits in order to
break it.
M. K. Shen
If you decimate the output of a LFSR (which is "an insecure PRNG") you get a
cryptographically secure PRNG (self-shrinking LFSR).
Cristiano
No you don't. There are attacks against the SSG.
Greg.
--
Greg Rose
232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C
There are attacks against many ciphers, but it doesn't mean that they are
not cryptographically secure.
Here:
http://en.wikipedia.org/wiki/Self-shrinking_generator#Cryptanalysis
I read that there is an attack against the SSG which requires 2^(0.7*L)
steps. If you take, say, L=256 or longer, the time needed to break that SSG
will be very big. I would call that SSG cryptographically secure PRNG.
Cristiano