2
$\begingroup$

I am reading up on whitebox cryptography and have trouble understanding how are ciphers implemented as one lookup table?

Assuming my plaintext is just 4 bits so the size of my lookup table should be $2^4$, what does such a lookup table look like?

Specifically, this is the slide I am looking at, I found it on Google.

enter image description here

$\endgroup$
4
$\begingroup$

For a fixed key, a (block) cipher is a reversible transformation of a plaintext set to ciphertext set. Usually (and in the slide) these sets are identical, and consist of all the exactly $n$-bit strings. This set has $2^n$ elements, and is often noted $\{0,1\}^n$.
Note: in this notation $\{0,1\}$ designates the set with the two elements $0$ and $1$, and as usual raising a set to the $n$ means we are considering the set of all $n$-tuples with each element in the base set.

One possible implementation of a that transformation is a table $T$ of the cipher's output for each of its input. This table has one entry for each input, thus $2^n$ entries. Each entry is an $n$-bit string, thus uses $n$ bits. The table thus uses $2^n\cdot n$ bits.

To use the cipher defined by way of the table, one converts the $n$-bit input to an integer $j$ with $0\le j<2^n$, fetches the table at index $j$ that is $T[j]$, obtains an $n$-bit value, and that's the cipher's output.

Addition per comment: if the adversary possesses a box (of whatever color) implementing that transformation, we must assume the $n$-bit inputs and outputs, and their correspondence, are observed when the box is used. That might not be a total disaster if the adversary can't economically reproduce the box, and/or enumerating all input/outputs is impossible (e.g. requires too much time/energy, or exhausts a usage counter).

$\endgroup$
  • $\begingroup$ thank you for replying, could you explain to me why is it noted {0,1}^n ? I understand why a set has 2^n elements, if the block is a 32bit block the number of possible permutation is 2^32. $\endgroup$ – laycat Mar 22 '15 at 8:56
  • $\begingroup$ I think this is also the part I do not understand, if you were to convert a string and look it up from an internal table, in a white-box model, wouldn't the adversary be able to get the $n-bit value that correlates with the input and output? $\endgroup$ – laycat Mar 22 '15 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.