What is the purpose of regime bits in posit encoding?

Question

Why do we need regime bits in posit?

posit encoding:

Answer 1

The key is the sentence about tapered accuracy. Standard floating point allocates a fixed number of bits to the exponent and the mantissa. Every float in range is represented with the same fractional accuracy. Back when floats were $32$ bits, one standard was one bit for the sign, eight bits for the exponent, and $23$ bits for the mantissa, so the fractional accuracy was about $2^{-23}\approx 10^{-7}$. For IEEE $64$ bit floats there are $52$ bits in the mantissa, so the fractional accuracy is about $2^{-52} \approx 2\cdot 10^{-16}$

Tapered accuracy is essentially data compression on the exponent. Exponents near zero are more common than those near the end of the range. You use short bit strings to represent small exponents at the price of longer strings to represent large exponents. That leaves more bits for the mantissa when the exponent is small and fewer when the exponent is large. The posit bits are one implementation of this compression, which the author is claiming can be (almost) as fast as standard floating point. If small exponents can be represented with only six bits instead of eleven, you have five more bits of accuracy in the mantissa when the exponent is in that range.