## ~luyu/comp5411-asgn3

comp5411-asgn3/main.tex -rw-r--r-- 25.8 KiB
3158dda4Luyu Cheng Add everything 8 months ago

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\documentclass[a4paper]{article}

\usepackage{amsmath,amsfonts}

\setlength{\oddsidemargin}{0in}
\setlength{\textwidth}{6.5in}
% \setlength{\topmargin}{0in}
% \setlength{\textheight}{8.7in}
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\usepackage[T1]{fontenc}
\usepackage{charter}
\usepackage{tikz}
\usepackage{caption}
\usepackage{subcaption}
\usepackage{multicol}
\usepackage{array}
\usepackage{amsmath}
\usepackage{cancel}
\usepackage{enumitem}
\usepackage{pgfplots}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{hyperref}
\usepackage{xcolor}
\usetikzlibrary{arrows.meta}

\def\comma{\text{,}}
\def\period{\text{.}}
\newcommand*{\transpose}{^{\mkern-1.5mu\mathsf{T}}}

\newtheorem*{lemma}{Lemma}
\newtheorem*{property}{Property}

\setcounter{section}{-1}

\title{Advanced Computer Graphics \\ \Large{Geometry Part -- Written Assignment}}
\author{Luyu Cheng, HKUST}

\begin{document}
\maketitle

\newcommand{\bc}{B\'{e}zier curve}
\newcommand{\bcs}{B\'{e}zier curves}
\newcommand{\ubs}[2]{\underset{#1}{\underbrace{#2}}}

\begin{enumerate}
\item
\begin{enumerate}
\item {\bfseries Endpoint interpolation property.} Compute the endpoints. \begin{alignat*}{1}
C(0) & = \frac16P_0 + \frac23P_1 + \frac16P_2 \neq P_0 \\
C(1) & = \frac16P_1 + \frac23P_2 + \frac16P_3 \neq P_3
\end{alignat*}
Therefore, the basis functions do not satisfy the endpoint interpolation property.

{\bfseries Partition of unity property.} All terms of $u$ are cancelled in the calculation.
\begin{align*}
\sum_{k=0}^3N_i(u) &= \frac16(1-u)^3 + \frac16(3u^3−6u^2+4) + \frac16(−3u^3+3u^2+3u+1)+\frac16u^3\\
&= \frac16\left(
\left(\cancel{-u^3}+\cancel{3u^2}-\cancel{3u}+1\right)
+\left(\cancel{3u^3}−\cancel{6u^2}+4\right)
+\left(−\cancel{3u^3}+\cancel{3u^2}+\cancel{3u}+1\right)
+\cancel{u^3}\right) \\
&= 1
\end{align*}
Therefore, the basis functions have the partition of unity property.

{\bfseries Convex hull property.} $C(u)$ is a weighted average of four control points and weights sum to 1 (i.e. is a convex combination of control points). Therefore, $C(u)$ belongs to the convex hull of control points.

% Derive the first derivative and the second derivative of C(u). Let C1(u) and C2(u) be two B-sline curves defined by P0P1P2P3 and P1P2P3P4, respectively. Find the first and second derivatives of the two B-spline curves at the joint and discuss the order of continuity there.
\item \label{it:1b} Here are the first and the second derivatives of $C(u)$. \begin{alignat*}{6}
C'(u) & = & \ubs{N'_0(u)}{-\frac{1}{2}(1-u)^2} P_0 & + & \ubs{N'_1(u)}{\left(\frac{3}{2}u^2 - 2u\right)} P_1 & + & \ubs{N'_2(u)}{\left(-\frac{3}{2}u^2+u+\frac{1}{2}\right)} P_2 & + \ubs{N'_3(u)}{\frac{1}{2}u^2} P_3 \\
C''(u) & = & \ubs{N''_0(u)}{(1-u)} P_0 & + & \ubs{N''_1(u)}{(3u-2)} P_1 & + & \ubs{N''_2(u)}{(-3u+1)} P_2 & + \ubs{N''_3(u)}{u} P_3
\end{alignat*}

Prove the $C_0$ continuity at the joint of $C_1$ and $C_2$. \begin{alignat*}{3}
C_1(1) & =
\frac16P_1 + \frac23P_2 + \frac16P_3 && =
C_2(0) && =
\frac16P_1 + \frac23P_2 + \frac16P_3
\end{alignat*}

Prove the $C_1$ continuity at the joint of $C_1$ and $C_2$. \begin{alignat*}{3}
C'_1(1) & = &
\cancel{-\frac{1}{2}(1-1)^2 P_0} + \left(\frac{3}{2} - 2\right) P_1 + \cancel{\left(-\frac{3}{2}+1+\frac{1}{2}\right)} P_2 + \frac{1}{2} P_3 & =
-\frac12P_1 + \frac12P_3 \\
C'_2(0) & = &
-\frac{1}{2}(1-0)^2 P_1 + \cancel{\left(\frac{3}{2}\cdot 0 - 0\right)} P_2 + \left(-\frac{3}{2}+0+\frac{1}{2}\right) P_3 + \cancel{\frac{1}{2}\cdot 0 P_4} & =
-\frac12P_1 + \frac12P_3
\end{alignat*}

Prove the $C_2$ continuity at the joint of $C_1$ and $C_2$. \begin{alignat*}{3}
C''_1(1) & = &
\cancel{(1-1) P_0} + (3-2) P_1 + (-3+1) P_2 + 1 P_3 & =
P_1 -2 P_2 + P_3 \\
C''_2(0) & = &
(1 - 0)P_1 + (0 - 2)P_2 + (0 + 1)P_3 + \cancel{0 P_3} & =
P_1 - 2P_2 + P_3
\end{alignat*}

Therefore, the joint of B-spline curves is $C^2$ continuity.

% What properties of the basis functions lead to the order of continuity being independent of the location of the control points?
\item If we concatenate the basis functions into a piecewise function $N(x)$. $N(x)$ is $C^2$ continuous at every joints. The $C^2$ continuity makes the order of continuity is independent of the location of control points. \begin{equation}
\label{eq:piecewise-basis-functions}
N(x)=\begin{cases}
N_0(r) & x=4n+0+r\\
N_1(r) & x=4n+1+r\\
N_2(r) & x=4n+2+r\\
N_3(r) & x=4n+3+r\\
\end{cases}, n\in \mathbb Z, 0\le r<1.
\end{equation} We can discover the property by observing the calculation in solution \ref{it:1b}. We can find that \begin{displaymath}
\begin{array}{ccc}
N_1(1) = N_0(0), & N'_1(1) = N'_0(0), & N''_1(1) = N''_0(0); \\
N_2(1) = N_1(0), & N'_2(1) = N'_1(0), & N''_2(1) = N''_1(0); \\
N_3(1) = N_2(0), & N'_3(1) = N'_2(0), & N''_3(1) = N''_2(0).
\end{array}
\end{displaymath}
We can also find these relations on the plot (Figure \ref{fig:1c}) of the basis functions and their derivatives. Note that values at endpoints of functions are the same.

\begin{figure}
\centering
\begin{subfigure}{0.31\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\centering
\begin{axis}[
width=0.8\textwidth,
height=1.2\textwidth,
scale only axis,
domain=0:1,
xmin=0, xmax=1,
ymin=-0.5, ymax=1.0,
grid,
minor x tick num=1,
minor y tick num=1,
]
\end{axis}
\end{tikzpicture}
\caption{The plot of \textcolor{red}{$N_0(u)$}, \textcolor{teal}{$N_1(u)$}, \textcolor{blue}{$N_2(u)$}, and \textcolor{purple}{$N_3(u)$}.}
\end{subfigure}
\hfill
\begin{subfigure}{0.31\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\centering
\begin{axis}[
width=0.8\textwidth,
height=1.2\textwidth,
scale only axis,
domain=0:1,
xmin=0, xmax=1,
ymin=-0.75, ymax=0.75,
grid,
minor x tick num=1,
minor y tick num=1,
]
\end{axis}
\end{tikzpicture}
\caption{The plot of \textcolor{red}{$N'_0(u)$}, \textcolor{teal}{$N'_1(u)$}, \textcolor{blue}{$N'_2(u)$}, and \textcolor{purple}{$N'_3(u)$}.}
\end{subfigure}
\hfill
\begin{subfigure}{0.32\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\centering
\begin{axis}[
width=0.8\textwidth,
height=1.2\textwidth,
scale only axis,
domain=0:1,
xmin=0, xmax=1,
ymin=-0.5, ymax=1,
grid,
minor x tick num=1,
minor y tick num=1,
]
\end{axis}
\end{tikzpicture}
\caption{The plot of \textcolor{red}{$N''_0(u)$}, \textcolor{teal}{$N''_1(u)$}, \textcolor{blue}{$N''_2(u)$}, and \textcolor{purple}{$N''_3(u)$}.}
\end{subfigure}
\caption{The plot of the basis functions and their derivatives. Note that values at endpoints of these functions are the same.}
\label{fig:1c}
\end{figure}
\end{enumerate}
\item
\begin{enumerate}
\item \label{it:2a} Suppose the points of $C(u)$ are $P_0=(0,0)$, $P_1=(1,4)$, and $P_2=(5,3)$. We have
\begin{alignat*}{6}
P_0^{(1)} &= \left(1-\frac{1}{4}\right)P_0       &&+ \frac{1}{4}P_1       &&= (\frac{1}{4},1) \\
P_1^{(1)} &= \left(1-\frac{1}{4}\right)P_1       &&+ \frac{1}{4}P_2       &&= (2,\frac{15}{4}) \\
P_0^{(2)} &= \left(1-\frac{1}{4}\right)P_0^{(1)} &&+ \frac{1}{4}P_1^{(1)} &&= (\frac{11}{16},\frac{27}{16})
\end{alignat*}
Figure \ref{fig:2-a} illustrates $C(u)$ and $Q(0.25)$. \footnote{Figures in this report are plotted with Ti$k$Z.}

\begin{figure}
\centering
\begin{tikzpicture}
\draw[loosely dotted] (0,0) grid (5.5,4.5);
% \path[use as bounding box] (-2,-1) rectangle (5,5);
\draw[->] (-0.2,0) -- (5.5,0) node[right] {$x$};
\draw[->] (0,-0.25) -- (0,4.5) node[above] {$y$};
\foreach \x/\xtext in {1/1, 2/2, 3/3, 4/4, 5/5}
\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {1/1, 2/2, 3/3, 4/4}
\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
(0,0) parabola (0.1*1,0.01*1) |- (0,0);
\draw[color=gray] (0,0) -- (1,4) -- (5,3);
% P0, (1/3 P0+2/3 P1), (2/3 P1+1/3 P2), P2
\draw[thick] (0,0) .. controls (2/3,2.664) and (2.331,3.663) .. (5,3);
\foreach \p/\ptext/\o in {
(0,0)/$P_0$/{above left},
(1,4)/$P_1$/above,
(5,3)/$P_2$/right,
(0.6875,1.6875)/$Q(0.25)$/{right}} {
\draw[fill=black] \p circle (2pt) node[\o] {\ptext};
}
\end{tikzpicture}
\caption{The quadratic Bezier curve in problem \ref{it:2a}}
\label{fig:2-a}
\end{figure}

% Problem 2 (b)
\item Let $P_0$, $P_1$, and $P_2$ be the points of a \bc. \begin{displaymath}
\begin{alignedat}{1}
P_0^{(1)} &= (1-t)P_0 + tP_1 \\
P_1^{(1)} &= (1-t)P_1 + tP_2 \\
P_0^{(2)} &= (1-t)P_0^{(1)} + tP_1^{(1)} \\
&= (1-t)\left((1-t)P_0 + tP_1\right) + t\left((1-t)P_1 + tP_2\right) \\
&= \ubs{\beta_0}{(1-t)^2} P_0 + \ubs{\beta_1}{2t(1-t)}P_1 + \ubs{\beta_2}{t^2} P_2
\end{alignedat}
\end{displaymath}
Therefore, the basis functions are \begin{displaymath}
\begin{alignedat}{1}
\beta_0 &= t^2 \\
\beta_1 &= 2t(1-t) \\
\beta_2 &= (1-t)^2.
\end{alignedat}
\end{displaymath}

% Problem 2. (c) Express the quadratic Bezier curves and their first derivative in matrix form.
\item We can derive the matrix form of quadratic \bcs\space as follows. \begin{displaymath}
\begin{alignedat}{1}
C(t) & = (1-t)^2 P_1 + 2(1-t)t P_2 + t^2 P_3 \\
& = (1 - 2t + t^2)P_1 + (0 + 2t - 2t^2)P_2 + (0 + 0t + t^2)P_3 \\
& = \left[\begin{matrix}
1 - 2t + t^2 \\
0 + 2t - 2t^2 \\
0 + 0t + t^2
\end{matrix}\right]^{\transpose} \left[\begin{matrix}
P_1 \\ P_2 \\ P_3
\end{matrix}\right] \\
& = \left[\begin{matrix}
1 \\ t \\ t^2
\end{matrix}\right]^{\transpose} \left[\begin{matrix}
1  & 0  & 0 \\
-2 & 2  & 0 \\
1  & -2 & 1
\end{matrix}\right] \left[\begin{matrix}
P_1 \\ P_2 \\ P_3
\end{matrix}\right]
\end{alignedat}
\end{displaymath}
We can derive the matrix form of the first derivative of quadratic \bcs\space as follows. \begin{displaymath}
\begin{alignedat}{1}
C'(t) & = 2t P_1 + 2(1-2t)P_2 + 2(t-1)P_3 \\
& = (0+2t) P_1 + (2-4t) P_2 + (-1+2t) P_3 \\
& = \left[\begin{matrix}
1 \\ t
\end{matrix}\right]^{\transpose} \left[\begin{matrix}
0 & 2  & -1 \\
2 & -4 & 2
\end{matrix}\right] \left[\begin{matrix}
P_1 \\ P_2 \\ P_3
\end{matrix}\right]
\end{alignedat}
\end{displaymath}

% Problem 2. (d) What are the conditions for two quadratic Bezier curves to join at C1 continuity in terms of the placement of the certain control points?
\item Suppose we have two quadratic \bcs, $C_1(t)$ and $C_2(t)$. The points of $C_1(t)$ are $P_1$, $P_2$, and $P_3$. The points of $C_2(t)$ are $P_3$, $P_4$, and $P_5$.  Expand and reduce $C_1(u)$ and $C_2(t)$, we will have
\begin{alignat*}{4}
C_1(t) & = (P_1-2P_2+P_3)t^2+2(-P_1+P_2)t+P_1\\
C_2(t) & = (P_4-2P_5+P_6)t^2+2(-P_4+P_5)t+P_4 .
\end{alignat*}
Their derivatives are
\begin{alignat*}{4}
C'_1(t) & = 2t(P_1-2P_2+P_3)+2(-P_1+P_2)\\
C'_2(t) & = 2t(P_4-2P_5+P_6)+2(-P_4+P_5) .
\end{alignat*}

Let $C_1(1)=C_2(0)$, we will have
\begin{alignat*}{3}
(\cancel{P_1}-\cancel{2P_2}+P_3)+\cancel{2(-P_1+P_2)}+\cancel{P_1} & = P_4 \\
P_3 & = P_4
\end{alignat*}

Let $C'_1(1)=C'_2(0)$, we will have
\begin{alignat*}{3}
2(P_1-2P_2+P_3)+2(-P_1+P_2) & = 2(-P_4+P_5) \\
-P_2 + P_3 & = -P_4 + P_5 \\
\Longrightarrow P_3 = P_4 & = \frac{P_2 + P_5}2
\end{alignat*}
Therefore, the conditions are $P_3 = P_4 = \frac{P_2 + P_5}2$, which means that $P_3$ and $P_4$ are the midpoint of $P_2$ and $P_5$.

% Problem 2. (e) Given a sequence of four points B0, B1, B2, B3. Explain how you might construct a C1 continuous quadratic Bezier spline comprising two Bezier curves that approximates the given points (i.e., none of the given points need to be interpolated). Draw a sketch of your construction.
\item \label{it:2e} Let \begin{displaymath}
\begin{array}{ccccc}
P_0 = \frac{2}{3} B_0+\frac{1}{3} B_1\text{,} &
P_1 = B_1\text{,} &
P_2 = \frac{B_1+B_2}{2}\text{,} &
P_3 = B_2\text{,} &
P_4 = \frac{1}{3} B_2+\frac{2}{3} B_3\text{.}
\end{array}
\end{displaymath}
Let $P_0, P_1, P_2$ be control points of the first \bc\space and $P_1, P_2, P_3$ be control points of the second \bc. Figure \ref{fig:2e} shows an example of the construction.

\begin{figure}
\centering
\begin{tikzpicture}
\draw[loosely dotted] (0,0) grid (8,4);
% \path[use as bounding box] (-2,-1) rectangle (5,5);
\draw[->] (-0.2,0) -- (8.5,0) node[right] {$x$};
\draw[->] (0,-0.25) -- (0,4.5) node[above] {$y$};
\foreach \x/\xtext in {1/1, 2/2, 3/3, 4/4, 5/5, 6/6, 7/7, 8/8}
\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {1/1, 2/2, 3/3, 4/4}
\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
(0,0) parabola (0.1*1,0.01*1) |- (0,0);
\draw[color=gray] (3,0) -- (1,3) -- (4.5,3.5) -- (8,4) -- (6,0);
% P0, (1/3 P0+2/3 P1), (2/3 P1+1/3 P2), P2
\draw[thick] (2+1/3,1) .. controls (5/3,2) and (13/6,19/6) .. (4.5,3.5);
\draw[thick] (4.5,3.5) .. controls (1.5+16/3,3.5/3+8/3) and (16/3+2,8/3) .. (4+8/3,4/3);
\foreach \p/\ptext/\o in {
(3,0)/{$\textcolor{red}{B_0}$}/{above right},
{(2+1/3,1)}/{$P_0$}/{above right},
(1,3)/{$\textcolor{red}{B_1}$,$P_1$}/{above left},
(4.5,3.5)/$P_2$/above,
(8,4)/{$\textcolor{red}{B_2}$,$P_3$}/above,
{(4+8/3,4/3)}/{$P_4$}/{above left},
(6,0)/{$\textcolor{red}{B_3}$}/{above left}} {
\draw[fill=black] \p circle (2pt) node[\o] {\ptext};
}
\end{tikzpicture}
\caption{The quadratic B\'{e}zier spline in problem \ref{it:2e}. The given points are in red.}
\label{fig:2e}
\end{figure}

% Problem 2. (f) Repeat part (e) to construct a C2 continuous cubic Bezier spline comprising two Bezier curves for a minimum number of given points.
\item \label{it:2f} Before the construction, we will introduce the conditions of a $C^2$ continuous cubic B\'ezier spline.

\begin{lemma}
Suppose a B\'{e}zier spline has \bcs\space $C_1(t)$ and $C_2(t)$. The control points of $C_1(t)$ are $P_1$, $P_2$, $P_3$, and $P_4$. The control points of $C_2(t)$ are $Q_1$, $Q_2$, $Q_3$, and $Q_4$. If their joint is $C^2$ continuous, we have $P_4=Q_1$, $P_2-P_1=Q_4-Q_3$, and $(P_3-P_2)+(P_3-P_4)=(Q_2-Q_1)+(Q_2-Q_3)$.
\end{lemma}

We can construct a $C^2$ continuous cubic B\'{e}zier spline (containing two \bcs) with at least \emph{three} given points. Suppose given points are $B_0$, $B_1$, and $B_2$. From the lemma, we know that we need to build 7 control points. The process is as follow.

\begin{enumerate}
\item Let $P_0 = B_0$ and $P_6 = B_2$. If $B_0$ and $B_2$ should not be interpolated, we can also choose $P_0$ and $P_6$ to be nearby points on line segments $\overline{B_0B_1}$ and $\overline{B_2B_1}$ respectively.
\item Select $P_1$ and $P_5$ respectively on line segments $\overline{B_0B_1}$ and $\overline{B_1B_2}$ such that $\left|\overline{P_1B_1}\right|=\left|\overline{B_1P_5}\right|$.
\item Let $P_2=\frac{P_1+B_1}2$, $P_4=\frac{B_1+P_5}2$, and $P_3=\frac{P_2+P_4}2$.
\end{enumerate}

The first cubic \bc\space has control points $P_0$, $P_1$, $P_2$, and $P_3$. The second cubic \bc\space has control points $P_3$, $P_4$, $P_5$, and $P_6$. From Figure \ref{fig:2f} depicts three example of the construction.

\begin{figure}
\centering
\begin{subfigure}{0.55\textwidth}
\begin{tikzpicture}[scale=0.9]
\draw[loosely dotted] (0,0) grid (8.5,4);
% \path[use as bounding box] (-2,-1) rectangle (5,5);
\draw[->] (-0.2,0) -- (8.5,0) node[right] {$x$};
\draw[->] (0,-0.25) -- (0,4.5) node[above] {$y$};
\foreach \x/\xtext in {1/1, 2/2, 3/3, 4/4, 5/5, 6/6, 7/7, 8/8}
\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {1/1, 2/2, 3/3, 4/4}
\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
(0,0) parabola (0.1*1,0.01*1) |- (0,0);
\draw[color=gray] (1,1) -- (2,2) -- (3,3) -- (4,4) -- (5,3) -- (6,2) -- (8,0);
\draw[color=gray] (3,3) -- (5,3);
\draw[thick] (1,1) .. controls (2,2) and (3,3) .. (4,3);
\draw[thick] (4,3) .. controls (5,3) and (6,2) .. (8,0);
\foreach \p/\ptext/\o in {
(1,1)/{$\textcolor{red}{B_0}$,$P_0$}/{below},
(2,2)/{$P_1$}/{above left},
(3,3)/$P_2$/{above left},
(4,4)/{$\textcolor{red}{B_1}$}/above,
(4,3)/{$P_3$}/above,
(5,3)/{$P_4$}/{above right},
(6,2)/{$P_5$}/{above right},
(8,0)/{$\textcolor{red}{B_2}$,$P_6$}/{above right}} {
\draw[fill=black] \p circle (2pt) node[\o] {\ptext};
}
\end{tikzpicture}
\caption{An example.}
\end{subfigure}
\begin{subfigure}{0.44\textwidth}
\centering
\begin{tikzpicture}[scale=0.9]
\draw[loosely dotted] (0,0) grid (6.5,4);
% \path[use as bounding box] (-2,-1) rectangle (5,5);
\draw[->] (-0.2,0) -- (6.5,0) node[right] {$x$};
\draw[->] (0,-0.25) -- (0,4.5) node[above] {$y$};
\foreach \x/\xtext in {1/1, 2/2, 3/3, 4/4, 5/5, 6/6}
\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {1/1, 2/2, 3/3, 4/4}
\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
(0,0) parabola (0.1*1,0.01*1) |- (0,0);
\draw[color=gray] (2.5,0) -- (1.8315935,1.336813) -- (1.1657968,2.6684065) -- (0.5,4) -- (1.959856,3.7080288) -- (3.419712,3.4160576) -- (5.5,3);
\draw[color=gray] (1.1657968,2.6684065) -- (1.959856,3.7080288);
\draw[thick] (2.5,0) .. controls (1.8315935,1.336813) and (1.1657968,2.6684065) .. (1.5628264,3.1882176);
\draw[thick] (1.5628264,3.1882176) .. controls (1.959856,3.7080288) and (3.419712,3.4160576) .. (5.5,3);
\foreach \p/\ptext/\o in {
(2.5,0)/{$\textcolor{red}{B_0}$,$P_0$}/{above right},
(1.8315935,1.336813)/{$P_1$}/{left},
(1.1657968,2.6684065)/$P_2$/{left},
(0.5,4)/{$\textcolor{red}{B_1}$}/above,
(1.5628264,3.1882176)/{$P_3$}/{above left},
(1.959856,3.7080288)/{$P_4$}/{above right},
(3.419712,3.4160576)/{$P_5$}/{above right},
(5.5,3)/{$\textcolor{red}{B_2}$,$P_6$}/{below}} {
\draw[fill=black] \p circle (2pt) node[\o] {\ptext};
}
\end{tikzpicture}
\caption{Another example.}
\label{fig:2fb}
\end{subfigure}
\begin{subfigure}{\textwidth}
\centering
\begin{tikzpicture}[scale=0.9]
\draw[loosely dotted] (0,0) grid (12,5);
% \path[use as bounding box] (-2,-1) rectangle (5,5);
\draw[->] (-0.2,0) -- (12,0) node[right] {$x$};
\draw[->] (0,-0.25) -- (0,4.5) node[above] {$y$};
\foreach \x/\xtext in {1/1, 2/2, 3/3, 4/4, 5/5, 6/6, 7/7, 8/8, 9/9, 10/10, 11/11}
\draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {1/1, 2/2, 3/3, 4/4, 5/5}
\draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
(0,0) parabola (0.1*1,0.01*1) |- (0,0);
\draw[color=gray] (0.5,2) -- (5.9285774,3.986065) -- (7.3142886,4.4930325) -- (8.7,5) -- (9.435513,3.7208471) -- (10.171026,2.441694) -- (11,1);
\draw[color=gray] (7.3142886,4.4930325) -- (9.435513,3.7208471);
\draw[thick] (0.5,2) .. controls (5.9285774,3.986065) and (7.3142886,4.4930325) .. (8.374901,4.10694);
\draw[thick] (8.374901,4.10694) .. controls (9.435513,3.7208471) and (10.171026,2.441694) .. (11,1);
\foreach \p/\ptext/\o in {
(0.5,2)/{$\textcolor{red}{B_0}$,$P_0$}/{below},
(5.9285774,3.986065)/{$P_1$}/{above left},
(7.3142886,4.4930325)/$P_2$/{above left},
(8.7,5)/{$\textcolor{red}{B_1}$}/above,
(8.374901,4.10694)/{$P_3$}/above,
(9.435513,3.7208471)/{$P_4$}/{above right},
(10.171026,2.441694)/{$P_5$}/{above right},
(11,1)/{$\textcolor{red}{B_2}$,$P_6$}/{below}} {
\draw[fill=black] \p circle (2pt) node[\o] {\ptext};
}
\end{tikzpicture}
\caption{Yet another example.}
\label{fig:2fc}
\end{subfigure}
\caption{Three examples of the quadratic B\'{e}zier spline construction in problem \ref{it:2f}. The given points are in red.}
\label{fig:2f}
\end{figure}
\end{enumerate}
\end{enumerate}

\end{document}