Approximating Bézier curves with least square polygons

Abstract

Approximating Bézier curve by polygon is a traditional technique in computer-aided geometric design, and there are several methods for constructing such polygons. In this paper, we propose to approximate Bézier curve by the least square polygon (LSP). LSP was derived by least square fitting which minimizes the integral of approximating errors. We provide a closed formula for the vertices of LSP. Because of its optimization method, LSP has lower approximation errors than existing approximating polygons. This observation is verified by many numerical examples we tested and several typical examples are included in this paper.

Appendices

Appendix A. The computation of $\textbf{c}_j$

$$\begin{aligned} \textbf{c}_{ j }= & {} \int _{0}^{\frac{1}{n}} \left[ nt \textbf{P}\left( t+t_{ j -1}\right) + ( 1- nt) \textbf{P}\left( t+t_j \right) \right] {\textrm{d}}t \\= & {} \int _0 ^{\frac{1}{n}}\sum \limits _{g = 0 }^{k} B^k_g \left( t+t_{j -1}\right) nt{{\varvec{p}}}_g{\textrm{d}}t \\{} & {} + \int _0 ^{\frac{1}{n}}\sum \limits _{g = 0 }^{k} B^k_g \left( t+t_j \right) (1-nt){{\varvec{p}}}_g{\textrm{d}}t. \\= & {} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \left[ \int _0 ^{\frac{1}{n}} B^k_g\left( t+t_{j -1}\right) nt{\textrm{d}}t \right. \\{} & {} \left. + \int _0 ^{\frac{1}{n}} B^k_g \left( t+t_j \right) (1-nt){\textrm{d}}t\right] . \end{aligned}$$

Let $t+t_{j-1} = u$ and $t+t_j = u$ in the two integrals respectively, we have

$$\begin{aligned} \textbf{c}_{j}= & {} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \left[ \int _{t_{j-1}} ^{t_j} B^k_g(u) (nu-j+1){\textrm{d}}u \right. \\{} & {} \left. + \int _{t_j}^{t_{j+1}} B^k_g(u) (1-nu+j){\textrm{d}}u \right] \\= & {} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \left[ n \int _{t_{j-1}} ^{t_j} B^k_g(u) u {\textrm{d}}u +(1-j)\int _{t_{j-1}} ^{t_j} B^k_g(u) {\textrm{d}}u \right] \\{} & {} + \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \left[ - n\int _{t_j}^{t_{i+1}} B^k_g(u) u {\textrm{d}}u + (1+j)\int _{t_j}^{t_{i+1}} B^k_g(u) {\textrm{d}}u\right] . \\= & {} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \frac{n(g+1)}{k+1} \left( \int _{t_{j-1}}^{t_j} B^{k+1}_{g+1}(u) {\textrm{d}}u - \int _{t_j}^{t_{j+1}} B^{k+1}_{g+1}(u) {\textrm{d}}u\right) \\{} & {} + \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \left[ \int _{t_{j-1}}^{t_{j+1}} B^k_g(u) {\textrm{d}}u - j\int _{t_{j-1}}^{t_j} B^k_g(u) {\textrm{d}}u+ j\int _{t_j}^{t_{j+1}} B^k_g(u) {\textrm{d}}u \right] . \end{aligned}$$

By the equation $\int _{0}^t B^k_g(u) {\textrm{d}}u = \frac{1}{k+1} \sum \nolimits _{i = g+1 }^{k+1} B^{k+1}_{i} (t)$ [13], $\textbf{c}_{j }$ can be computed as

$$\begin{aligned} \textbf{c}_{j}= & {} \frac{n }{(k+1)(k+2)} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g (g+1) \\{} & {} \sum \limits _{i=g+2}^{k+2} \left[ 2B^{k+2}_{i}\left( t_{j}\right) - B^{k+2}_{i}\left( t_{j-1}\right) -B^{k+2}_{i}\left( t_{j+1}\right) \right] \\{} & {} + \frac{1}{k+1 } \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+1}^{k+1} \left[ B^{k+1}_{i}\left( t_{j+1}\right) - B^{k+1}_{i}\left( t_{j-1}\right) \right] \\{} & {} + \frac{j}{k+1 }\sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+1}^{k+1} \left[ B^{k+1}_{i}\left( t_{j+1}\right) - 2B^{k+1}_{i}\left( t_{j}\right) + B^{k+1}_{i}(t_{j-1}) \right] \\= & {} \frac{n }{(k+1)(k+2)} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g (g+1) \\{} & {} \sum \limits _{i=g+2}^{k+2} \left[ 2B^{k+2}_{i}\left( t_{j}\right) - B^{k+2}_{i}\left( t_{j-1}\right) -B^{k+2}_{i}\left( t_{j+1}\right) \right] \\{} & {} + \frac{1}{k+1 } \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+1}^{k+1} \left[ (j+1)B^{k+1}_{i}\left( t_{j+1} \right) \right. \\{} & {} \left. - 2j B^{k+1}_{i}\left( t_{j}\right) +(j-1) B^{k+1}_{i}\left( t_{j-1}\right) \right] \end{aligned}$$

By $B^{k+1}_{i}(t) = \frac{i+1}{t(k+2)}B^{k+2}_{i}(t)$, It’s easy to compute

$$\begin{aligned}{} & {} \frac{1}{k+1 } \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+1}^{k+1} \left[ (j+1)B^{k+1}_{i}\left( t_{j+1}\right) \right. \\{} & {} \qquad \left. - 2j B^{k+1}_{i}\left( t_{j}\right) +(j-1) B^{k+1}_{i}\left( t_{j-1}\right) \right] \\{} & {} \quad = \frac{n}{(k+1)(k+2)} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+1}^{k+1} (i+1) \left[ B^{k+2}_{i+1}\left( t_{j+1} \right) \right. \\{} & {} \qquad \left. - 2 B^{k+2}_{i+1}\left( t_{j} \right) + B^{k+2}_{i+1}\left( t_{j-1} \right) \right] \\{} & {} \quad = \frac{n}{(k+1)(k+2)} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+2}^{k+2} i \left[ B^{k+2}_{i}\left( t_{j+1} \right) \right. \\{} & {} \qquad \left. - 2 B^{k+2}_{i} \left( t_{j} \right) + B^{k+2}_{i}(t_{j-1}) \right] \end{aligned}$$

Denote $\Delta _{j,i,n}^{k+2} = B^{k+2}_{i}\left( t_{j+1} \right) - 2 B^{k+2}_{i} \left( t_{j} \right) + B^{k+2}_{i}(t_{j-1})$, we have

$$\begin{aligned} \textbf{c}_{j}= & {} \frac{n }{(k+1)(k+2)} \sum \limits _{g = 0 }^{k} {{\varvec{p}}}_g \sum \limits _{i=g+2}^{k+2} (i-g-1) \Delta _{j,i,n}^{k+2}. \end{aligned}$$

Appendix B. Examples

Example 1

$\{ (3.60, 1.34); (-0.44. 7.28); ( 11.24,9.10); ( 7.41, 1.83 )\}$;

Example 2

$\{ (3, 1); (1, 2); (2, 7); (5, 9); (9, 8); (10, 2); (8, 0); (5, 0) \}$;

Example 3

$\{ (0, 0); (0, 4); (1, 8); (5, 10); (10, 10); (10, 6); (10, 2); (8, 0); (3, 0); (0, 0)\}$;

Example 4

$\{(1,1); (2, 7); (6, 10); (9, 6); (4, 6); (3,3); (6, 5); (7, 5); (9, 4); (8, 1); (5, 1); (6, 4); (4, 2)\}$.