Why offset is calculated as `2^{n-1} - 1` instead of `2^{n-1}` for floating point exponent representation

Question

I'm trying to understand why offset K in binary offset notation is calculated as 2^{n-1}-1 instead of 2^{n-1} for floating point exponent representation. Here is my reasoning for 2^{n-1}.

Four bits can represent values in the range [-8;7], so 0000 represents -8. An offset from zero here is 8 and can be calculated as 2^{n-1}. Using this offset we can define representation of any number, for example, the number 3.

What number do we need to add to -8 to get 3? It's 11, so 3 in offset binary is represented as 1011. And the formula seems to be number to represent + offset.

However, the real formula is number to represent + offset - 1, and so the correct representation is 1010. Can someone please explain why we also subtract additional one?

Answer 1

I am posting this as an answer to better explain my thougths, but even though I'll quote the standard a few times, I haven't found an explicitly stated reason.

In the following, I'll refer to the IEEE 754 standard (and succesive revisions) for floating point representation, even if OP doesn't mention it (if I'm wrong, please, let me know).

The question is about the particular representation of the exponent in a floating point number.

In subclause 3.3 Sets of floating-point data is said (emphasis mine):

The set of finite floating-point numbers representable within a particular format is determined by the following integer parameters:

― b = the radix, 2 or 10
― p = the number of digits in the significand (precision)
― emax = the maximum exponent e
― emin = the minimum exponent e

emin shall be 1 − emax for all formats.

Later it specifies:

The smallest positive normal floating-point number is b^emin and the largest is b^emax×(b − b^{1 − p}). The non-zero floating-point numbers for a format with magnitude less than b^emin are called subnormal because their magnitudes lie between zero and the smallest normal magnitude.

In 3.4 Binary interchange format encodings:

Representations of floating-point data in the binary interchange formats are encoded in k bits in the following three fields (...):
a) 1-bit sign S
b) w-bit biased exponent E = e + bias
c) (t=p−1)-bit trailing significand field digit string T=d₁ d₂ ... d_{p − 1} ; the leading bit of the significand, d₀, is implicitly encoded in the biased exponent E
(...)
The range of the encoding’s biased exponent E shall include:
― every integer between 1 and 2^w − 2, inclusive, to encode normal numbers
― the reserved value 0 to encode ±0 and subnormal numbers
― the reserved value 2^w − 1 to encode ± ∞ and NaNs.

For example a 32-bit floating point number has those parameters:

k, storage width in bits                      32
p, precision in bits                          24
emax, maximum exponent e                     127
emin, minimum exponent e                    -126
bias, E − e                                  127
w, exponent field width in bits                8
t, trailing significand field width in bits   23

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In this Q&A is pointed out that: "The purpose of the bias is so that the exponent is stored in unsigned form, making it easier to do comparisons."

Considering the above mentioned 32-bit floating point representation a normal (not subnormal) number has an encoded biased exponent E in the range between 1 and 254.

The reason behind the particular choice of the range -126, 127 for the exponent could be, in my opinion, to extend the range of representable numbers: very low numbers are represented by subnormals so a bigger (even if only by one) maximum exponent can take care of the big ones.