big o - Shifting an LFSR loop in O(1) time? -


i'm wondering if there's way combine 2 concepts: lfsrs, , barrel shifters

i'm looking way to, in o(1) time, shift lfsr loop given number of shifts.

what i'm hoping find simple process have current state of lfsr , number of times want shift state parameters quick/easy process.

at first thought looking @ taps, moving taps on 1 , looking @ them again, finding shift in bit each time , appending onto end of course isn't o(1) , gets complicated if want shift many times tap "slide off" original lfsr state.

if not in o(1) time, there more efficient way multiple shifts doing each 1 individually?

in general, inclined answer no. reason if such algorithm existed, compute in o(1) time given bit in sequence generated lfsr. seems unlikely doable in general.

however, can precomputation speed things somewhat. note after fixed number of steps, state of each cell in lfsr linear combination of bits initial state. so, if precompute coefficients in linear combination each cell 1 step, 2 steps, 4 steps, 8 steps, etc, should able jump ahead many steps in relatively short time. of course, give useful speedup in "sliding off" cases mention before.


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