linear feedback shift register cryptography

However, if the period is lower than this, we must be careful to ensure that the starting vector is indeed on the main cycle. Now, if the period is maximal, then every non-zero element must be part of the 'main' cycle, and so any initialization vector will lead to an LFSR of maximal period. It does matter that your initial condition is somewhere on this cycle, as for example initializing with the $0$ state leads to a cycle of only $1$ element. Since each time you 'step' the LFSR you are multiplying by $x$ within the appropriate field extension, to calculate the period it doesn't matter where on the cycle you start. Consider the tightness of the bounds described by K.G.: if the feedback polynomial is 'badly' chosen (for example $x^k-1$), the period may be much lower. In particular, this is the cycle generated if you initialize the LFSR with all the entries $0$ except for the last one set to $1$.įor a good LFSR, the period will be close to the maximum possible. If $g(x)$ is the feedback polynomial of an LFSR, then $e$ is the upper bound you seek. In general, the period of a polynomial $g(x)$ is the least integer $e$ such that $g(x)$ divides $x^e−1$. However, working out how tight this bound is can be much more difficult (as per Dilip's comment): Owing to the good statistical properties, large period and low implementation costs, LFSR have achieved wide acceptance in developing stream ciphers. When the LFSR reaches a state it has been in before, what happens to the output sequence? How many possible states are there? Now you have an upper bound Linear Feedback Shift Registers (LFSRs) have always received considerable attention in cryptography. That is, you can derive an upper bound using the method described by KG: and Dilip Sarwate have pretty much answered your question.

Linear feedback shift register cryptography