# Known plaintext attack on 3 round SIMON block cipher

I am attempting to write a program to perform a linear approximation attack on a reduced round version of the SIMON block cipher, but I am stuck on how to actually apply the linear approximation to get the subkeys for each round. I found this paper, in which the following high-bias linear approximations are listed:

P[ F(X)_i = X_(i-2) ]                        = 3/4
P[ F(X)_i = X_(i-2) ^ X_(i-1) ]              = 3/4
P[ F(X)_i = X_(i-2) ^ X_(i-8) ]              = 3/4
P[ F(X)_i = X_(i-2) ^ X_(i-1) ^ X_(i-8) ]    = 1/4


Where F(X) is the nonlinear function in the SIMON algorithm and F(X)_i is the ith bit of the output of said function, and X_n is the nth bit of the input to the function. This all makes sense to me. I am stuck, however, on how to leverage these linear approximations to derive the subkeys for each round. How do I go from the linear approximation to a program that finds the subkeys? If anyone could help me close the gap in my understanding, or at least point me in the right direction, I would be very grateful. Thanks so much!

Edit: originally posted on Stack Overflow

The paper also discusses linear characteristics for multiple rounds:

We present several approaches to produce linear characteristics for SIMON32/64 and present the best known linear characteristic for 11-round SIMON 32/64 with the bias of $2^{-16}$. We then extend this characteristic to 13 rounds of the cipher.

We present several approaches to produce an LC for SIMON32/64, as a case study, and present the best known 11-round LC for this cipher with the bias of $2^{-16}$ (expendable to 13 rounds of the cipher).

The linear approximations you list seem to be from equation 3. SIMON is a Feistel cipher. By combining the equation that describes one round of Feistel, the authors of the paper get the following linear expression for an entire round:

$$(P_R)_2 \oplus (K^1)_2 \oplus (X^1_L)_2 = (P_L)_0,$$

which holds with probablity $3/4$. Here $X^1 = X^1_L || X^1_R$ is the output of the first round, and $P = P_L || P_R$ is the plaintext. The notation $(X)_i$ is used for the $i$th bit of $X$. Note that the expression is written here for the first round, but it equally applicable to any round $i$, when written in the following form:

$$(X^{i-1}_R)_2 \oplus (K^i)_2 \oplus (X^i_L)_2 = (X^{i - 1}_L)_0,$$

where we replaced $P$ by $X^{i - 1}$ and $X^1$ by $X^i$. This can of course also be written as

$$(X^{i-1}_R)_2 \oplus (K^i)_2 \oplus (X^{i - 1}_L)_0 = (X^i_L)_2.$$

This is still true with probability $3/4$.

If we "substitute" (pile up) this last equation into a 3 round Feistel network (shown by Figure 3 in the paper), we get:

$$(X^{i - 1}_R)_2 \oplus (K^i)_2 \oplus (X_L^{i -1})_0 = (X^{i + 2}_R)_0 \oplus (K^{i + 2})_2 \oplus (X_L^{i + 1})_2,$$

or:

$$\Sigma_K \oplus (X^{i - 1}_R)_2 \oplus (X_L^{i -1})_0 \oplus (X^{i + 2}_R)_0 \oplus (X_L^{i + 1})_2 = 0,$$

with $\Sigma_K = (K^i)_2 \oplus (K^{i + 2})_2$.

The paper goes on to use this expression for more rounds, but we can stop here. The above expression is a linear characteristic for 3 rounds of the cipher. Since we know with which probability the above expression holds (depending on the parity of $\Sigma_K$), we can start obtaining key bits. This document explains that part well:

The process followed involves partially decrypting the last round of the cipher. Specifically, for all possible values of the target partial subkey, the corresponding ciphertext bits are exclusive-ORed with the bits of the target partial subkey and the result is run backwards through the corresponding S-boxes. This is done for all known plaintext/ciphertext samples and a count is kept for each value of the target partial subkey. The count for a particular target partial subkey value is incremented when the linear expression holds true for the bits into the last round’s S-boxes (determined by the partial decryption) and the known plaintext bits. The target partial subkey value which has the count which differs the greatest from half the number of plaintext/ciphertext samples is assumed to represent the correct values of the target partial subkey bits. This works because it is assumed that the correct partial subkey value will result in the linear approximation holding with a probability significantly different from 1/2. (Whether it is above or below 1/2 depends on whether a linear or affine expression is the best approximation and this depends on the unknown values of the subkey bits implicitly involved in the linear expression.) An incorrect subkey is assumed to result in a relatively random guess at the bits entering the S-boxes of the last round and as a result, the linear expression will hold with a probability close to 1/2.

The above applies to SPNs, but for a Feistel network the prinicple is the same.

• So if I am reading this correctly, the last equation you gave is relating the 2nd bit and 0th bits of different parts of the input/output/subkey, right? So if I want to find an approximation for the nth bit of the subkey can I just substitute n and n-2 in the proper places in that equation and will the approximation hold? That was one of the things in that paper that confused me, it seems that they just pulled the number 2 out of thin air for the subkey bit. – dylanrb123 Apr 21 '15 at 15:12
• @dylanrb123 No, the approximation works specifically and only with these bits. If you want to attack other bits of the subkeys, you need to use another linear approximation (the paper refers to several others). In practice, you usually need a bunch of expressions for a successful attack. – Aleph Apr 21 '15 at 16:15