# What is the purpose of regime bits in posit encoding?

Why do we need regime bits in posit?

posit encoding: 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 $$2^{-52} \approx 2\cdot 10^{-16}$$
• I didn't follow through how the bits are used. In standard $64$ bit floating point there are$1$ sign bit, $11$ exponent bits, and $52$ mantissa bits. That gives a fractional accuracy of about $2^{-52}$ throughout the range and a range of exponent of $\pm 1023$. We could define the exponent to have $0$ followed by four bits for exponents close to zero. This would handle exponents $\pm 7$ but would give $58$ mantissa bits, so the fractional accuracy in this range would be 2^{-58}$. When you are outside this range, if you want to maintain the overall range Mar 30 '19 at 23:38 • you would have to put a$1$before the usual exponent, so the accuracy would go down to$2^{-51}$. Is this a good trade? If most of your numbers are within a factor$2^7$of$1\$ it is good. They are using some more complicated encoding, but the idea will be the same. Mar 30 '19 at 23:39