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  • $\begingroup$ Thanks! So my definition of RPTAS was equivalent to P. What if we change the definition as follows: an RPTAS does not work for arbitrary ranges $[v,(1+\epsilon)v]$, but only for ranges $[(1+\epsilon)^i,(1+\epsilon)^{i+1}]$ for $i\geq 0$. Then RPTAS is still contained in FPTAS. Is it now equivalent to P? $\endgroup$ Commented Dec 16, 2021 at 13:24
  • $\begingroup$ My guess is that by playing with $\epsilon$, you will be able to mimic the argument. $\endgroup$ Commented Dec 16, 2021 at 13:28
  • $\begingroup$ Indeed: assuming $v\geq 2$, we can take $\epsilon=v-1$. Then the RPTAS can check all the ranges $[v^i,v^{i+1}]$ for $0\leq i\leq \log_v V$. $\endgroup$ Commented Dec 16, 2021 at 13:49