Normalizing from [0.5 - 1] to [0 - 1]

Question

I'm kind of stuck here, I guess it's a bit of a brain teaser. If I have numbers in the range between 0.5 to 1 how can I normalize it to be between 0 to 1?

Thanks for any help, maybe I'm just a bit slow since I've been working for the past 24 hours straight O_O

Answer 1

Others have provided you the formula, but not the work. Here's how you approach a problem like this. You might find this far more valuable than just knowning the answer.

To map [0.5, 1] to [0, 1] we will seek a linear map of the form x -> ax + b. We will require that endpoints are mapped to endpoints and that order is preserved.

Method one: The requirement that endpoints are mapped to endpoints and that order is preserved implies that 0.5 is mapped to 0 and 1 is mapped to 1

a * (0.5) + b = 0 (1)
a * 1 + b = 1     (2)

This is a simultaneous system of linear equations and can be solved by multiplying equation (1) by -2 and adding equation (1) to equation (2). Upon doing this we obtain b = -1 and substituting this back into equation (2) we obtain that a = 2. Thus the map x -> 2x - 1 will do the trick.

Method two: The slope of a line passing through two points (x1, y1) and (x2, y2) is

(y2 - y1) / (x2 - x1).

Here we will use the points (0.5, 0) and (1, 1) to meet the requirement that endpoints are mapped to endpoints and that the map is order-preserving. Therefore the slope is

m = (1 - 0) / (1 - 0.5) = 1 / 0.5 = 2.

We have that (1, 1) is a point on the line and therefore by the point-slope form of an equation of a line we have that

y - 1 = 2 * (x - 1) = 2x - 2

so that

y = 2x - 1.

Once again we see that x -> 2x - 1 is a map that will do the trick.

Answer 2

Subtract 0.5 (giving you a new range of 0 - 0.5) then multiply by 2.

double normalize( double x )
{
    // I'll leave range validation up to you
    return (x - 0.5) * 2;
}

Answer 3

To add another generic answer.

If you want to map the linear range [A..B] to [C..D], you can apply the following steps:

Shift the range so the lower bound is 0. (subract A from both bounds:

[A..B] -> [0..B-A]

Scale the range so it is [0..1]. (divide by the upper bound):

[0..B-A] -> [0..1]

Scale the range so it has the length of the new range which is D-C. (multiply with D-C):

[0..1] ->  [0..D-C]

Shift the range so the lower bound is C. (add C to the bounds):

[0..D-C] -> [C..D]

Combining this to a single formula, we get:

       (D-C)*(X-A)
X' =   -----------  + C
          (B-A)

In your case, A=0.5, B=1, C=0, D=1 you get:

       (X-0.5)
X' =   ------- = 2X-1
        (0.5)

Note, if you have to convert a lot of X to X', you can change the formula to:

       (D-C)         C*B - A*D
X' =   ----- * X  +  ---------  
       (B-A)           (B-A)

It is also interesting to take a look at non linear ranges. You can take the same steps, but you need an extra step to transform the linear range to a nonlinear range.

Answer 4

× 2 − 1

should do the trick

Answer 5

Lazyweb answer: To convert a value x from [minimum..maximum] to [floor..ceil]:

General case:

normalized_x = ((ceil - floor) * (x - minimum))/(maximum - minimum) + floor

To normalize to [0..255]:

normalized_x = (255 * (x - minimum))/(maximum - minimum)

To normalize to [0..1]:

normalized_x = (x - minimum)/(maximum - minimum)

Answer 6

You could always use clamp or saturate within your math to make sure your final value is between 0-1. Some saturate at the end, but I've seen it done during a computation, too.