I am not sure what this is even called in math terms, but here I go...
i have a range from 0 to 1
and I want to take the value of 1 and make that 0, and 0 and make that 1. and if I have .1 make it .9 and .9 make it .1
So I am not sure how to do this. And/or what it's called.
Thanks!
I am not sure what this is even called in math terms, but here I go...
i have a range from 0 to 1
and I want to take the value of 1 and make that 0, and 0 and make that 1. and if I have .1 make it .9 and .9 make it .1
So I am not sure how to do this. And/or what it's called.
Thanks!
it would be y= 1.0 - x
It's called linear interpolation (Lerp).
Really? It sounds like just a complement like fir said. Linear interpolation requires 2 points (may be farther than 1 unit from the origin) and a delta between 0 and 1, and the original post doesn't name any of that.
The 2 points are p0 = 1 and p1 = 0 (values at time t = 0 and t = 1 respectively).
He mentioned specific values for t in his post (t = 0.1 and 0.9).
Lerp is the general case for this
p = (1 - t) * p0 + t * p1 = p0 - t * (p0 - p1)
Plugging in p0 = 1 and p1 = 0 we get
p = 1 - t(1 - 0) = 1 - t
which is what fir said too (with variables named differently, y == p and x == t)
From reading his post, he wants 0 --> 1, 1 --> 0 and 0.1 --> 0.9 and 0.9 --> 0.1.
Indeed, y = 1.0 - x
You can use it to determine diagonals on slopes if you have square tiles, for example.
I'd read his post again then ;)
He wants
0.0 -> 1.0
0.1 -> 0.9
0.9 -> 0.1
1.0 -> 0.0
I agree with Diego that it's.. kind of a stretch to call this linear interpolation. It's just a simple complement, and just happens to be a special case of linear interpolation with a = 1, b = 0 (and probably a special case of plenty of other transforms). But anyway it's good to have a reference to it since most likely MARS_999 will need it soon 
Why stop at linear interpolation? We can complicate things further and think of it as polynomial interpolation: There is a unique polynomial of degree up to 3 such that f(0)=1, f(0.1)=0.9, f(0.9)=0.1 and f(1)=0.
