base-2: why DECIMAL_DIG - DIG == 2 or 3?-CodePudding

base-2: why DECIMAL_DIG - DIG == 2 or 3?

Examples:

FLT_DECIMAL_DIG - FLT_DIG == 9 - 6 == 3
DBL_DECIMAL_DIG - DBL_DIG == 17 - 15 == 2
LDBL_DECIMAL_DIG - LDBL_DIG == 21 - 18 == 3
FLT16_DECIMAL_DIG - FLT16_DIG == 5 - 3 == 2

Extra: Is it guaranteed that on any given implementation using base-2 this difference will be 2 or 3?

CodePudding user response：

Yes, for base-2 radix, xxx_DECIMAL_DIG - xxx_DIG will be 2 or 3.

Informal proof

For a floating point type:

b is the base or radix or exponent representation (an integer > 1)
p is the precision (the number of base-b digits in the significand) (I assume this is > 1 and rational)

The xxx_DECIMAL_DIG values are defined as:

⎰ p log₁₀ b — if b is a power of 10
⎱ ⌈1 p log₁₀ b⌉ — otherwise

The xxx_DIG values are defined as:

⎰ p log₁₀ b — if b is a power of 10
⎱ ⌊(p - 1) log₁₀ b⌋ — otherwise

For b = 2, log₁₀ b ≈ 0.301 and is irrational (proof).
∴ p log₁₀ 2 is irrational (since p is rational and > 1).
∴ ⌈p log₁₀ 2⌉ - ⌊p log₁₀ 2⌋ = 1.
∴ ⌈1 p log₁₀ 2⌉ - ⌊p log₁₀ 2⌋ = 2. — ①
⌊p log₁₀ 2⌋ - ⌊(p - 1) log₁₀ 2⌋ ∈ {0, 1}. — ② (since 0 < log₁₀ 2 < 1)
∴ ⌈1 p log₁₀ 2⌉ - ⌊(p - 1) log₁₀ 2⌋ ∈ {2, 3}. — (from ① and ②) ∎

CodePudding user response：

Given definitions of xxx_DECIMAL_DIG and xxx_DIG we have
2 <= ceil(1 p*log10(2)) - floor((p-1)*log10(2)) <= 3
Simplifying:

2 <=     ceil(1 p*log10(2)) - floor((p-1)*log10(2)) <= 3
2 <= 1   ceil(  p*log10(2)) - floor((p-1)*log10(2)) <= 3
let q = p – 1
2 <= 1   ceil(  (1   q)   *log10(2)) - floor((p-1)*log10(2)) <= 3
2 <= 1   ceil(log10(2)   q*log10(2)) - floor(  q  *log10(2)) <= 3
let R = trunc(q*log10(2))
let r = trunc(q*log10(2)) – R, r is [0.0 …1.0)
log10(2) = 0.301…
2 <= (1   R   ceil(0.301…   r)) – (R   floor(r)) <= 3
2 <=  1       ceil(0.301…   r)       – floor(r)  <= 3
2 <=  1       ceil(0.301…   r)       – 0         <= 3
When r <= 1.0 - 0.301…
  2 <= 1   1 <= 3
Otherwise r > 1.0 - 0.301…
  2 <= 1   2 <= 3

Even though it is shown that the difference is 2 or 3, I suppose the next questions are why are
xxx_DECIMAL_DIG: ceil(1 p*log10(2))
and
xxx_DIG: floor((p-1)*log10(2))?