Bleichenbacher's PKCS 1.5 Padding Oracle (Complete Case)
Cryptanalytic MVP award
This is an extraordinarily useful attack. PKCS#1v15 padding, despite being
totally insecure, is the default padding used by RSA implementations.
The OAEP standard that replaces it is not widely implemented. This attack
routinely breaks SSL/TLS.
This is a continuation of challenge #47; it implements the complete
Set yourself up the way you did in #47, but this time generate a 768
To make the attack work with a realistic RSA keypair, you need to
reproduce step 2b from the paper, and your implementation of Step 3
needs to handle multiple ranges.
The full Bleichenbacher attack works basically like this:
Starting from the smallest 's' that could possibly produce
a plaintext bigger than 2B, iteratively search for an 's' that
produces a conformant plaintext.
For our known 's1' and 'n', solve m1=m0s1-rn (again: just a
definition of modular multiplication) for 'r', the number of
times we've wrapped the modulus.
'm0' and 'm1' are unknowns, but we know both are conformant
PKCS#1v1.5 plaintexts, and so are between [2B,3B].
We substitute the known bounds for both, leaving only 'r'
free, and solve for a range of possible 'r' values. This
range should be small!
Solve m1=m0s1-rn again but this time for 'm0', plugging in
each value of 'r' we generated in the last step. This gives
us new intervals to work with. Rule out any interval that
is outside 2B,3B.
Repeat the process for successively higher values of 's'.
Eventually, this process will get us down to just one
interval, whereupon we're back to exercise #47.
What happens when we get down to one interval is, we stop blindly
incrementing 's'; instead, we start rapidly growing 'r' and backing it
out to 's' values by solving m1=m0s1-rn for 's' instead of 'r' or
'm0'. So much algebra! Make your teenage son do it for you! *Note:
does not work well in practice*