The BC should look like this
$$BC(m,k,n) = (c,n')$$
$$BC^{-1}(c,k,n') = (m,n)$$
with the additional variable $n$ not symmetric, so in almost all cases:
$$n \not= n'$$
$m$... plaintext
$c$... encrypted value of $m$
$k$... key
$n$ would have the same block size as $m,c$
Given now a fixed $m,k$ we assume to calculate the BC for every possible $n$. Doing this we should also get that many different $c$ as results.
E.g. if $m,n$ has a block size of 128 bit and we calculate it for all $2^{128}$ values of $n$ (using the same $m,k$) we will get $2^{128}$ different $c$ as result.
Q: Is there any Block cipher (mode) like this? (BC with just a not symmetric key also works)
Further information:
If a BC is used in ECB mode (so no $n$) with a fixed key $k$ and we calculate it for every possible input $m$ (in block-size length) we also get that many different encrypted values $c$.
But if we use a fixed input and calculate AES with every possible key (with block-size length) we will get in mean only $(1-1/e)\approx 63\%$ of the values (afaik).
I'm looking for a way to include all values but make the inverse infeasible to calculate.
Also if:
$$BC(m_1,k,n_1) = (c_1,n_1')$$
$$BC(m_2,k,n_2) = (c_2,n_2')$$
is known it should still be infeasible to calculate ($n_*'$ and $n_°$ in)
$$BC^{-1}(c_1,k,n_*') = (m_2,n_*)$$
$$BC(m_1,k,n_°) = (c_2,n_°')$$
In each instance only a plaintext of a single block size will be used.
If the forward direction was computed the backward direction should also be known.
