Quantcast
Viewing all articles
Browse latest Browse all 4

Answer by Daniel S for Majority-based feedback shift register

I would hesitate to claim a better lower bound than 1.

We note that for a majority vote feedback shift register (hereafter MVFSR) if there are $2k+1$ taps and if at any point the fill of the register has at most $k$ zeros (respectively at most $k$ ones) then the register will subsequently fill up with ones (respectively with zeros) and hit a cycle of 1.

Likewise, one feels that there is a lot of substantialism in that MVFSR with sparse fills with lots of zeros are likely to produce zero feedbacks and dense fills with lots of ones are likely to produce one feedbacks. Heuristically this would drive the fill density away from 1/2 and towards the above degeneracy.

It's possible to produce MVFSRs with taps set in arithmetic progression that degenerate into interleavings of smaller registers and so which hit stable cycles of length equal to the number of smaller registers, but this does not feel like a good idea.

One can also note that the map from fill-to-fill for an MVFSR has many two-to-one maps which places a poor upper bound on the final cycle length. In general fills where $k+1$ of the first $2k$ taps are zero or one (i.e. where there are not exactly $k$ zeros and $k$ ones in the first $2k$ tap positions) are fills where the oldest tap bit is irrelevant to the feedback and so we create a two-to-one.


Viewing all articles
Browse latest Browse all 4

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>