# Curve between 0 and 1 for contrast filtering

I was editing an image by increasing the contrast. For this I used the formula $$(\tanh(\mathtt{CONTRAST\_FACTOR}*(x-0.5))+1)/2$$ This formula has limits at 0 and 1, with y=0.5 at x=0.5.

The problem for my situation is that y is only limited between 0 and 1, but I want a formula that has the following properties:

• the curve is continuous (at least between 0 and 1)
• the line goes through $$(0,0)$$, $$(0.5,0.5)$$ and $$(1,1)$$
• I want a constant (like contrast factor above) that can influence the slope around $$(0.5,0.5)$$, which basically determine the curviness.

Does anyone know a formula like that?

• Extending the approach you already have: Use a scaled, shifted $\tanh^{-1}$ to map $(0,1)$ to $(-\infty,\infty)$, multiply by the contrast factor, then apply $\tanh$ as before. See e.g. math.stackexchange.com/a/898792/856
– user856
Dec 21, 2018 at 5:26

Maybe a cubic Bézier curve: $$y=3hx(1-x)^2 +3(1-h)x^2(1-x) + x^3$$

This is just the Bézier curve with coefficients $$0,h,1-h,1$$. These are then combined with the cubic Bernstein polynomials $$(1-x)^3$$, $$3x(1-x)^2$$, $$3x^2(1-x)$$, $$x^3$$ to produce the formula above.

You can vary $$h$$ between $$0$$ and $$1$$ to adjust the shape of the curve. I suspect that you might want the curve to be monotone increasing, which will happen if $$h$$ lies between $$0$$ and $$\tfrac13$$.

• Could you explain your formula a bit like the other answer did (more than just how to use it I mean) Dec 21, 2018 at 15:53
• Standard Bézier curve stuff. The formula gives the real-valued Bézier curve with coefficients $0, h,1-h,1$. These coefficients are combined with the cubic Bernstein polynomials to give the formula. Dec 23, 2018 at 12:23

Let $$s$$ be the slope at $$(0.5,0.5)$$. You desire $$f'(0.5)=s$$, $$f(k)=k, k\in\{0,0.5,1\}$$.

A cubic seems to fit:

$$f(x)=ax^3+bx^2+cx+d$$ $$f'(x)=3ax^2 + 2bx + c$$ Plugging in each of the points, we have a system of equations: $$d=0$$ $$\dfrac{a}{8}+\dfrac{b}{4}+\dfrac{c}{2}=\dfrac{1}{2}$$ $$a+b+c=1$$ $$\dfrac{3a}{4}+b+c=s$$

which has solution:

$$a=4-4s, \ b=-3(2-2s), \ c=3-2s, \ d=0$$

So the desired cubic is:

$$f(x)=(4-4s)x^3-3(2-2s)x^2+(3-2s)x$$

See HERE for an animation over different values of $$s$$.

• I think that I am looking for the inverse of this formula, because I need s to be greater than 1 while still having a function that strictly increases between x of 0 and 1 Dec 21, 2018 at 15:58
• Check $1\le s\le 1.5$ It switches concavity and keeps monotonicity on your interval Dec 21, 2018 at 19:08
• lets see... can you extend the formula so that the slope at (0,0) and (1,1) is always 0? Dec 21, 2018 at 20:50
• That would require a higher-degree polynomial for manipulation purposes at least (this is only true for $s=1.5$ with this model), try degree five. Introduce that $f'(0)=0,f'(1)=0$ and solve the larger system Dec 21, 2018 at 21:27