# Distance of a Bézier curve and control points

Derive a formula (independent of $n$) for $H(P_n, B_n)$ if $B_n$ is the Bézier curve which is determined by the control points $p_0 := (−4, 0), p_1 = p_2 = . . . = p_{n−1} := (0, 4)$ and $p_n := (4, 0)$, and $P_n = \{p_0,\ldots,p_n\}$. Here $\quad H(A, B) \quad is \quad sup_{a∈A} (inf_{b∈B} d(a, b))$ for two non-empty subsets $A$, $B$ of $\mathbb{R}^2$ also $d(a, b)$ denotes the Euclidean distance of $a, b ∈ \mathbb{R}^2$.

• What is $P_n$? Is it supposed to be the same as $p_n$ ?? Probably not. What things do you know that we can use to solve the problem? – bubba Dec 10 '17 at 6:35
• @bubba : let denote the control polygon formed by $p_0, p_1, . . . , p_n by P_n$ . – GreenQuestioner Dec 10 '17 at 12:15
• maybe Frechet distance would be useful – GreenQuestioner Dec 10 '17 at 12:38
• Who is your teacher? He (or she) is doing a very good job of inventing new and interesting homework questions. Can you post some more of them, please, or post a link to an on-line source. – bubba Dec 10 '17 at 14:11
• @bubba : sure we will. – GreenQuestioner Dec 11 '17 at 11:10

It's obvious by symmetry that the maximum distance between the curve and its control points occurs at the mid-point $q = B_n(\tfrac12)$ of the curve. If you don't believe that it's obvious, you can reason as follows: the control points are symmetric about the $y$-axis, so the curve will also be symmetric. So, at the point where it crosses the $y$-axis, the curve tangent must be horizontal. This means that the line from $p_1, \ldots, p_{n-1}$ to $q$ is normal to the curve at $q$, so $q$ is the point on the curve that is closest to $p_1, \ldots, p_{n-1}$.
Let $\phi_0, \ldots, \phi_n$ be the Bernstein polynomials of degree $n$, then the point $q$ is given by $$q = \phi_0(\tfrac12)p_0 + \phi_1(\tfrac12)p_1 + \cdots + \phi_n(\tfrac12)p_n$$ So, using the coordinates of $p_0, \ldots, p_n$, the $y$-coordinate of the point $q$ is \begin{align} &(0)\phi_0(\tfrac12) + (4)\phi_1(\tfrac12) + (4)\phi_2(\tfrac12) + \cdots + (4)\phi_{n-1}(\tfrac12) + (0)\phi_n(\tfrac12) \\ = \;&4\big[\phi_0(\tfrac12) + \phi_1(\tfrac12) + \cdots + \phi_n(\tfrac12)\big] - 4\phi_0(\tfrac12) - 4\phi_n(\tfrac12) \\ = \;&4 - 4(1-\tfrac12)^n - 4(\tfrac12)^n \\ = \;&4 - (\tfrac12)^{n-3} \\ \end{align} So the desired distance is $(\tfrac12)^{n-3}$. Note that this is not independent of $n$; in fact, it decreases as $n$ increases. This should make sense -- putting more control points at the location $(0,4)$ will pull the curve more strongly towards this point.
• you have mentioned, "It's obvious by symmetry that the maximum distance between the curve and its control points occurs at the mid-point $q=B_n(\frac{1}{2})q=B_n(\frac{1}{2})$ of the curve." but you have not given any proof or reference to your claim, would you please let us know more about your statement. – GreenQuestioner Dec 11 '17 at 14:29
• How euclidean distance under folowing condition $sup_{a∈A} (inf_{b∈B} d(a, b))$ which is mentioned in question has been taken place in your answer? – GreenQuestioner Dec 11 '17 at 14:35