5 ms·

> The number of floats between 0 and 1 is the same as the number of floats between 1 and 3 No, the number of floats between 0 and 1 is (approximately) the same

by a1369209993 2y ago

> The number of floats between 0 and 1 is the same as the number of floats between 1 and 3 No, the number of floats between 0 and 1 is (approximately) the same as the number of floats between 1 and positive infinity. And this is the correct way for it work: 1/x has roughly the same range and precision as x, so you don't need (as many) stupid obfuscatory algebraic transforms in your formulas to keep your intermediate values from over- or under-flowing.

llm_trw 2y ago
Floating points numbers have a fix precision mantissa, and a fixed precision exponent. So you have xxxxx E xxx as an example of a 5 bit mantissa and 3 bit exponent. You have 2^5 floating point numbers for each possible exponent. So no, you're wrong. For exponent 0 you have 2^5, and for exponent 1, 10 and 11 you then have the same. The exponent 0b (0d) then contain the same number of possible floating mantissas as does 1b (1d), 10b (2d) and 11b(3d). Which means that there are as many mantissas between [0,1) as there are between [1,3)
- nh23423fefe 2y ago
  why do you think the range [0,1) is represented by one exponent?
  - llm_trw 2y ago
    Because it is half the range expressed in 1 bit of exponent, the same way that [1,3) is half the range expressed in 2 bits of exponent. I'd used [0,2) and [2,4) but that would confuse people used to thinking in base 10, which includes the OP author apparently.
- greg7mdp 2y ago
  > for exponent 1, 10 and 11 you then have the same Right. But the exponent is signed as well, so you have the same number of exponent values mapping to values between -1 and 1 (roughly exponent values from -127 to -1) as to all other values (between -inf and -1 and between 1 and inf), exponent values from 1 to 126.
anchpop 2y ago
I think you're right. I produced a small rust program to test it by iterating over all 32-bit floats: https://gist.github.com/anchpop/30f119efd8d4c29a6a99202d57a509b5 https://gist.github.com/anchpop/30f119efd8d4c29a6a99202d57a5... Output: Count between 0 and 1: 1065353215 Count between 1 and +inf: 1073741824 Ratio: 1.0 But a more theoretical approach will probably be needed to see if the same ratio exists for 64 bit floats.

llm_trw 2y ago

Floating points numbers have a fix precision mantissa, and a fixed precision exponent. So you have xxxxx E xxx as an example of a 5 bit mantissa and 3 bit exponent. You have 2^5 floating point numbers for each possible exponent. So no, you're wrong. For exponent 0 you have 2^5, and for exponent 1, 10 and 11 you then have the same. The exponent 0b (0d) then contain the same number of possible floating mantissas as does 1b (1d), 10b (2d) and 11b(3d). Which means that there are as many mantissas between [0,1) as there are between [1,3)

nh23423fefe 2y ago
why do you think the range [0,1) is represented by one exponent?
- llm_trw 2y ago
  Because it is half the range expressed in 1 bit of exponent, the same way that [1,3) is half the range expressed in 2 bits of exponent. I'd used [0,2) and [2,4) but that would confuse people used to thinking in base 10, which includes the OP author apparently.
greg7mdp 2y ago
> for exponent 1, 10 and 11 you then have the same Right. But the exponent is signed as well, so you have the same number of exponent values mapping to values between -1 and 1 (roughly exponent values from -127 to -1) as to all other values (between -inf and -1 and between 1 and inf), exponent values from 1 to 126.

anchpop 2y ago

I think you're right. I produced a small rust program to test it by iterating over all 32-bit floats: https://gist.github.com/anchpop/30f119efd8d4c29a6a99202d57a509b5 https://gist.github.com/anchpop/30f119efd8d4c29a6a99202d57a5... Output: Count between 0 and 1: 1065353215 Count between 1 and +inf: 1073741824 Ratio: 1.0 But a more theoretical approach will probably be needed to see if the same ratio exists for 64 bit floats.