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\begin{document}
\title{Spline Axioms}
\author{Fassio Blatter}
\maketitle

\chapter{Spline Approximations}
A set of vectors $\begin{pmatrix}x \\ y\end{pmatrix} \in X$ within any parametrisation bounds $t\in[a,b]$ can be translated, rotated and scaled by $M^{3x3}$, so that $\iota_{X}(a)=\begin{pmatrix}0 \\ 0\end{pmatrix}$ and $\iota_{X}(b)=\begin{pmatrix}1 \\ 0\end{pmatrix}$. An approximation (arc) $A$ of arbitrary polynomial degree is required.\\\\
For example, approximating a straight line has polynomial degree one.
\begin{equation}
line(t):
\begin{pmatrix}
1 \\
t
\end{pmatrix}
\underbrace{\begin{pmatrix}
1 & -1\\
0 & 1
\end{pmatrix}}_{Jordanian}
(\underbrace{\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}}_{M^{2x2}}
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix})
=
\begin{pmatrix}
x \\
y
\end{pmatrix}
\end{equation}
Note: Because $S$ and $E$ are already at $\begin{pmatrix}0\\0\end{pmatrix}$ and $\begin{pmatrix}1\\0\end{pmatrix}$, the matrix $M^{2x2}$ of affine transformations is the identity matrix.\\\\
The more complicated case of a quadratic bezier spline
\begin{equation}
quad(t):
\begin{pmatrix}
1 \\
t \\
t^2
\end{pmatrix}
\underbrace{\begin{pmatrix}
1 & -1 & 0\\
0 & 1 & -1\\
0 & 0 &1
\end{pmatrix}}_{Jordanian}
(M^{3x3}*\underbrace{\begin{pmatrix}
x(S) & x(C_{1}) & x(E) \\
y(S) & y(C_{1}) & y(E) \\
0 & 0 & 0
\end{pmatrix}}_{\text{Bezier Control Points}})
=
\begin{pmatrix}
x \\
y \\
0
\end{pmatrix}
\end{equation}
Note: $M^{NxN}$ has Rank 2\\\\
The Jordanian is composed of the sum of identity matrices in upper right sub-spaces.
The characteristic polynomials of the diagonal and bidiagonal are composed of binomial coefficients.
\begin{align}
\chi_{\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}} = \begin{pmatrix}3 \\ 0\end{pmatrix}*t^0 - \begin{pmatrix}3 \\ 1\end{pmatrix}*t^1 + \begin{pmatrix}3 \\ 2\end{pmatrix}*t^2 - \begin{pmatrix}3 \\ 3\end{pmatrix}*t^3\\
\chi_{\begin{pmatrix}-1 & 0\\0 & -1\end{pmatrix}} = \begin{pmatrix}2 \\ 0\end{pmatrix}*t^0 + \begin{pmatrix}2 \\ 1\end{pmatrix}*t^1 + \begin{pmatrix}2 \\ 2\end{pmatrix}*t^2
\end{align}
Note: It is a sparse approximation.\\\\
If start and end points are limited to be in the transversality $S,E \in Y$, the maximal polynomial degree can be reduced to four.
\begin{equation}
quart(t):
\begin{pmatrix}
1 \\ t \\ t^2 \\ t^3 \\ t^4
\end{pmatrix}
\underbrace{\begin{pmatrix}
1 & -1 & 0 & 0 & 0\\
0 & 1 & -1 & 0 & 0\\
0 & 0 &1 & -1 & 0\\
0 & 0 & 0 & 1 & -1\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}}_{Jordanian}
(M^{5x5}*H^{5x5})
=
\begin{pmatrix}
x \\ y \\ 0 \\ 0 \\ 0
\end{pmatrix}
\end{equation}
Note: We are only interested in planar coordinates, the tri- and quadridiagonals of the Jordanian can be ignored. It is a polynomial of degree 4 in a 2 dimensional subspace (Sparse).\\\\
Because of associativity, an inverse of the control points can be left multiplied if it would exist. Because of associativity, the inverse of the Jordanian can be left multiplied.
\begin{equation}
quart(t):
\begin{pmatrix}
1 \\ t \\ t^2 \\ t^3 \\ t^4
\end{pmatrix}
=
\underbrace{\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
0 & 0 & 1 & 1 & 1\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}}_{Jordanian^{-1}}
(M^{5x5}H^{5x5})^{-1}
\begin{pmatrix}
x \\ y \\ 0 \\ 0 \\ 0
\end{pmatrix}
\end{equation}

\chapter{Representation}

A quartic spline is fully defined by $S,E$ and three on spline control points $c_{1},c_{3} \in A;c_{2} \in C_{4}$. The corresponding Bezier control points $q_{1},q_{2} \in Q_{3}$ are obtained by substituting $t=1/4$,$t=2/4$ and $t=3/4$ with $c_{1}$,$c_{2}$ and $c_{3}$ in the quartic linear equation.
\begin{align}
H=\begin{bmatrix}
x(S) & x(q_{1}) & x(C_{4}) & x(q_{2}) & x(E) \\
y(S) & y(q_{1}) & y(C_{4}) & y(q_{2}) & x(E) \\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\\
representation: S \times C_{4} \times E \rightarrow A \times V \times A\\
S,C_{4},E \in A; S,V,E \in \iota_{T}\\
(S,C,E)\mapsto(c_{1},v_{1},c_{3})
\end{align}\\
The four quadratic splines $\gamma_{H_{1}} ... \gamma_{H_{4}}$ are defined by $S,E$ and on spline control points $c_{1},c_{2},c_{3},c_{5},c_{6},c_{7} \in A;c_{4} \in C_{3}$. The corresponding Bezier control points $q_{1},q_{2},q_{3},q_{4} \in Q_{1}$ for the pieces $H_{1},H_{2},H_{3},H_{4}$ are obtained by substituting $t=0,t=1/2$ and $t=1$ with $c_{1},c_{3},c_{5},c_{7}$ in each of the four quadratic linear equations.
\begin{align}
H_{1}=
\begin{bmatrix}
x(S) & x(q_{1}) & x(c_{2})\\
y(S) & y(q_{1}) & y(c_{2})\\
0 & 0 & 0
\end{bmatrix};
H_{2}=
\begin{bmatrix}
x(c_{2}) & x(q_{2}) & x(c_{4})\\
y(c_{2}) & y(q_{2}) & y(c_{4})\\
0 & 0 & 0
\end{bmatrix}\\
H_{3}=
\begin{bmatrix}
x(c_{4}) & x(q_{3}) & x(c_{6})\\
y(c_{4}) & y(q_{3}) & y(c_{6})\\
0 & 0 & 0
\end{bmatrix};
H_{4}=
\begin{bmatrix}
x(c_{6}) & x(q_{4}) & x(E)\\
y(c_{6}) & y(q_{4}) & y(E)\\
0 & 0 & 0
\end{bmatrix}
\end{align}
\section{Geometric Meaning}
The axis connecting $C_{4}$ to $V$ is the focal axis in a perfect half circle. In an oval which can not be ruled out, $C_{4}$ is the only point intersecting the focal axis with a tangent on-spline. If the $S_{2}\cap E_{2}$ are bent outwards from the perfect half circle position, there are a total of three focal axes in a quartic spline, which is the default for a quartic spline. Because the Stopeight\_Analyzer algorithm is catching the bends starting from the limit of $straightness$, this case can be ruled out. Therefore, the number of $card(C_{4})$ per chart $U_{m}$ is reduced from $3$ to $1$.

\chapter{Inverse Representation}
In Representation\_Parallelogram.nb and Representation\_Circles.nb there are currently two representations evaluated. The parallelogram construction favors half-circles, while the more computationally extensive circles positions the two omitted $C_{4}$ closer togther for ovals, etc.\\\\
The inverse of the representation reconstructs $S,C_{4},E$ from the left and right on-spline control points $q \in A_{4}$ and the intersection of the angle-bisection in the triangle $SCE$ with the line $SE$.
\begin{align}
representation^{-1}: A \times V \times A \rightarrow S \times C_{4} \times E\\
(c_{1},v_{1},c_{3})\mapsto(S,C_{4},E)
\end{align}
\section{Geometric Meaning}
If the representation itself is re-interpreted to define a quartic spline of its own $S=c_{1},C_{3}=v_{1},E=c_{3}$, the resulting geometric figure is a sort of reflection surface, with the area narrowing down when repeating this procedure. The focal axis of each subsequent quartic spline rotates depending on how far towards $S$ or $E$, the single $C_{4}$'s are stretched. The inverse expands this area.\\\\
Axiom: The re-interpretation is a derivative of the representation (polynomial).

\section{Representation\_Circles}
The point $C_{4}$ can be relatively easy to reconstruct under the assumption that the line $InterL,InterR$ is the angle bisection of the triangles $WHT,InterL,C_{4}$ and $WHT,InterR,C_{4}$. However, this is not accurate. It can be used as initial value to solve the geometric problem of 3 circles of equal radius. $CircleL$ intersects $CircleWHT$ at $E$. $CircleR$ intersects $CircleWHT$ at $S$. Because the tangent through $C_{4}$ is inaccurate as a result of the bisection not being accurate, the centers $TL$ of $CircleL$ and $TR$ of $CircleR$ have to be adjusted marginally. The third circle $CircleWHT$ is positioned randomly, but has to satisfy the condition that $WHT$ lies on the diagonal that connects the two intersections $S$ and $E$. Extending from $WHT$, $InterL$ and $interR$ lie on the diagonals of $CircleL$ and $CircleR$.

\chapter{Signals}
A quartic representation is sufficient for pen-stroke analysis. Two possible implementations are provided. Going from two quartic to a quintic requires $n-3$ on-spline control points, $n$ stands for the polynomial degree. In a Spiral, it would make sense to position these at the $C_{4}$ of the two Cliffs with $t_{5}=1/3$ and $t_{5}=2/3$. For two circles, the control points would have to be at $t_{6}=1/4,t_{6}=2/4$ and $t_{6}=3/4$. The number of quartic splines required for a polynomial of higher degree doubles with every step $4...n$. The control points for a Swell would have to be at the $C_{4}$'s.\\\\
Axiom: Factorisation requires maximal polynomial degree, so Frame-bounds have to be $S,E$ of Compact Covers.
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