Function from Integers to the [0,1] float interval_问答_开发者

Given an integer I would like to produce a unique floating point number in the interval [0,1]. (this number will be used as id). The problem I found with all functions I have thought about, is that they encounter duplicates before running out of integer values. For example if f(a:int):float = 0.a then f(16000) = 0.16 and f(16001) = 0.16001. But since it is floating point, 0.16 and 0.16001 may be represented the same. In other words, I need a function that produ开发者_JS百科ce not only unique numbers, but also numbers that are represented uniquely (at least forthe C++ integer domain).

I know the answer is dependent on the size of integer and floating point in a specific environment, but if you can give an example for specific sizes it will still be helpful.

As someone else pointed out you can simply cast an int to a float of the same size to get a unique float (with some post-filtering for NaN and -0 and Inf). However, that will not meet your requirement of being in [0,1]. In fact, you can use that relation to show that there are not enough floats in [0,1] to represent the set of integers. If you use double then the mantissa is easily large enough for a 32-bit int and an expression like I / (double)INT_MAX should be sufficient (obviously allow for unsignedness if you need to).

If 23 bits are enough, you could build a float from int n like this:

float f;
*((__int32*)&f) = (0x7E<<23) | (n & 0x7FFFFF); // exponent=0x7E-0x7F=-1, fraction=n

It will map integers [0, 2^23-1] into floats [0.5, 1.0) according to the IEEE floating point standard.

If 23 bits are not enough, you have a solution at the link that you cited yourself. There are some other important notes, by the way, be sure to understand all the limitations.

use 1/(float)n it's a good function for distribution, and if n is zero set it 0.

This is just like the problem of hashing functions and strings. Given that the size of the floating point precision is smaller than that supported by an int type, you're inevitably going to have collisions.

The mathematically correct way is to go for:

f(x) = 0 where x = 0; f(x) = 1/x where x > 0 and x < limit (limit could be 10000000000, depending on your platform's float sizes)

Otherwise you're trying to fit in more values than you can, which is impossible.

Alternatively, (since you're dealing with graphics), if you want to cater for the whole range of integers, and you could live with rounding 2 close integers to the same value, you could divide the value by 2 (or 3 or by how many you want) to scale down to fit in the whole int range required. Dividing by 2 would double the range supported.

So 16000 and 16001 would both end up with 1/8000.

I guess it depends on your situation, but you can't fit 1million people in a stadium of 500,000 seats, and expect everyone to have a unique seat number.. you'll have collisions.

Just cast the integer to a float, assuming they are the same size this will produce an unique float for every integer.

int id = 16;    
float f_id = (float) id;

EDIT: I just read that you'd like the floats to be in [0, 1] range - that would considerably reduce number of possible ids, probably it's best just to use the integer as an id, not the float that is dependent on in-memory representation as you pointed out.

When an integer value is represented by 32 bits and a float value is represented by 32 bits, then it is obviously impossible to have every possible integer map to a unique float. The simple solution is to use a double for the ID or a short for the value.

The key is to use a floating point type for which the significand has more bits than the integer type.