|Résumé||We present the first linear lower bound for the number of bits required to be accessed in the worst case to increment an integer in an arbitrary space-optimal binary representation. The best previously known lower bound was logarithmic. It is known that a logarithmic number of read bits in the worst case is enough to increment some of the integer representations that use one bit of redundancy, therefore we show an exponential gap between space-optimal and redundant counters.
Our proof is based on considering the increment procedure for a space optimal counter as a permutation and calculating its parity. For every space optimal counter, the permutation must be odd, and implementing an odd permutation requires reading at least half the bits in the worst case. The combination of these two observations explains why the worst-case space-optimal problem is substantially different from both average-case approach with constant expected number of reads and almost space optimal representations with logarithmic number of reads in the worst case.
https://arxiv.org/abs/1607.00242 ; ICALP 2017 |