This page contains all the theory used to develop this packaged. It can be resumed as a mix of set theory, analitic and parametric geometry on the plane \(\mathbb{R}^2\).

Hereafter we describe the main objects used in this study (Empty, Whole, Point, Curve and Shape), and how they interact with each other.

Point

Point is a 0-dimensional object defined by only a pair of real values \((x, \ y)\).

It can also be understood as a mathematical set of a single element, meaning \(\{(x, \ y)\}\)

Let \(p_0 = \left(x_0, \ y_0\right)\) and \(p_1 = \left(x_1, \ y_1\right)\) be two points. The operations between then are:

Inner product :

\[\langle p_0, \ p_1 \rangle = x_0 \cdot x_1 + y_0 \cdot y_1\]

Cross product :

\[p_0 \times p_1 = x_0 \cdot y_1 - x_1 \cdot y_0\]

Norm L2 of a point :

\[\|p_0\| = \sqrt{x_0^2 + y_0^2}\]

Curve

Curves are one-dimensional objects, that contains a infinite number of connected points.

A curve can be defined as implicit or parametrized by one variable.

Examples:

Implicit:
- Circle
\[C = \left\{\left(x, \ y\right) \in \mathbb{R}^2 : x^2 + y^2 = 1\right\}\]
- Straight line
\[C = \left\{\left(x, \ y\right) \in \mathbb{R}^2 : 2\cdot x + 5 \cdot y = 10\right\}\]
- Hyperbola
\[C = \left\{\left(x, \ \dfrac{1}{x}\right) \in \mathbb{R}^2 : x \in \mathbb{R}^{+}\right\}\]
Parametrized:
- Circle
\[C = \left\{\left(\cos t, \ \sin t\right) \in \mathbb{R}^2 : t \in \left[0, \ 2\pi\right] \subset \mathbb{R}\right\}\]
- Straight line
\[C = \left\{\left(2+t, \ 4+t\right) \in \mathbb{R}^2 : t \in \mathbb{R}\right\}\]
- Hyperbola
\[C = \left\{\left(\exp -t, \ \exp t\right) \in \mathbb{R}^2 : t \in \mathbb{R}\right\}\]

For this package * Segment is a \(C^{1}\) parametrized curve \(p(t)\) defined on a interval \(\left[0, \ 1\right]\)

PiecewiseCurve is a parametrized curve that is a sequence of segments.

The curve contains \(n\) segments, that are described by using knots \(\left[t_0, \ t_1, \ \cdots, \ t_n\right]\).

\[\begin{split}p(t) = \begin{cases}p_{0}(t) \ \ \ \ \ \text{if} \ t_{0} \le t \le t_{1} \\ p_{1}(t) \ \ \ \ \ \text{if} \ t_{1} \le t \le t_{2} \\ \vdots \\ p_{n-1}(t) \ \ \ \ \ \text{if} \ t_{n-1} \le t \le t_{n} \end{cases}\end{split}\]

JordanCurve is also a piecewise curve, but it’s continuous, closed and does not intersect itself.

Jordan curve

The jordan curve used in this package:

Is a Closed Curve that doesn’t intersect itself.
Is oriented, either counter-clockwise (positive) or clockwise (negative)
Divides the plane in two regions: Interior and exterior
Is either bounded, or can go to infinity only once.
Can be parametrized by piecewise analitic curves

Examples:

Counter-clockwise circle:

\[C = \left\{\left(\cos 2\pi t, \ \sin 2\pi t\right) \in \mathbb{R}^2 : t \in \left[0, \ 1\right]\right\}\]

Straight line:

\[C = \left\{\left(2+3t, \ 3-4t\right) \in \mathbb{R}^2 : t \in \mathbb{R}\right\}\]

Right hand of an Hyperbola

\[C = \left\{\left(\cosh t, \ \sinh t\right) \in \mathbb{R}^2 : t \in \mathbb{R}\right\}\]

Note

Although the jordan curve can go to the infinity once, the current implementation doesn’t allow it yet.

Shape

Shape is bi-dimensional object that can be obtained from union and intersection of the internal regions of some Jordan Curves.

For this package, the shapes are classified in three types:

SimpleShape: Defined by only one JordanCurve. It’s the interior area if the jordan is counter-clockwise, otherwise it’s the exterior area
ConnectedShape: It’s the intersection of some SimpleShape
DisjointShape: It’s the union of SimpleShape and ConnectedShape

Simple Shape

If the oriented jordan curve is counter-clockwise, the shape is positive. If it’s clockwise, then we say the shape is negative.

An integral is made to compute the absolute value of the area, and the sign is accordingly with the jordan curve orientation.

Connected Shape

Is similar to a simple shape but has holes.

It can be described as the intersection of some Simple shapes

\[C = \bigcup_{i} S_i\]

From construction, max of only one of \(S_i\) is positive. The order used is:

Simple shape with positive area comes first
Then order the rest in increasing order

Disjoint Shape

It’s the union of some disjoint Simple and Connected shapes.

It can be described as the union

\[D = \bigcup_{i} C_i\]

From construction, max of only one of \(C_i\) is negative The order used is:

Connected/Simple shape with negative area comes first
Then order the rest in decreasing order

Integral

One of the main uses of shapepy is to compute integrals. It can assume two forms:

Line integrals : When the integration occurs over a curve, one-dimensional integral

\[I = \int_{C} f(x, \ y) \ ds\]

\[I = \int_{C} \langle g(x, \ y) , \ ds\rangle\]

\[I = \int_{C} g(x, \ y) \times \ ds\]

Shape integrals : Bidimensional

\[I = \int_{S} f(x, \ y) \ dx \ dy\]

The computation of the integral can change depending on the function and on the curve/shape. Here we show how we compute

Curve integrals

Three types of line integrals were given.

\[I = \int_{C} f(x, \ y) \ ds = \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} f(x(t), \ y(t)) \ \|p'(t)\| \ dt\]

\[I = \int_{C} \langle g(x, \ y) , \ ds\rangle = \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} \langle g(x(t), \ y(t)), \ p'(t) \rangle \ dt\]

\[I = \int_{C} g(x, \ y) \times \ ds = \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} g(x(t), \ y(t)) \times p'(t) \ dt\]

Shape integrals

The shapes are classified in Simple, Connected and Disjoint.

If \(S\) is disjoint, then it’s the union of subshapes \(S_i\), it’s transformed

\[I = \int_{\cup_i S_{i}} f(x, \ y) \ dS = \sum_{i} \int_{S_i} f(x, \ y) \ dS\]

If \(S\) is connected, then it’s the intersection of simple shapes \(S_i\), it’s transformed

\[I = \int_{\cup_i S_{i}} f(x, \ y) \ dS = \sum_{i} \int_{S_i} f(x, \ y) \ dS\]

Meaning, the integral over Connected or Disjoint are transformed into integrals over Simple shapes.

The strategy to integrate over a simple shape is transform the integral over the area, into a integral over the jordan curve (its boundary) by using Green’s theorem

\[\int_{S} \left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) dx \ dy = \int_{C} P \ dx + Q \ dy\]

Without loss of generality, take \(\alpha \in \mathbb{R}\) a constant, and set

\[P(x, \ y) = \left(\alpha - 1 \right) \int f(x, \ y) \ dy\]

\[Q(x, \ y) = \alpha \cdot \int f(x, \ y) \ dx\]

If \(f(x, \ y)\) is polynomial then

\[f(x, \ y) = x^{a} \cdot y^{b}\]

Hence

\[P = \dfrac{\alpha-1}{b+1} \cdot x^{a} \cdot y^{b+1}\]

\[Q = \dfrac{\alpha}{a+1} \cdot x^{a+1} \cdot y^{b}\]

In special, take \(\alpha = (a+1)/(a+b+2)\) and the integral is transformed

\[P \ dx + Q \ dy = \dfrac{x^{a} \cdot y^{b}}{a+b+2} \cdot \left(x \ dy - y \ dx\right)\]

\[I = \int_{S} x^{a} y_{b} \ dx \ dy = \dfrac{1}{a+b+2} \int_{C} x^{a} y^{b} \cdot \left(x \ dy - y \ dx\right)\]

Since every parametric curve is divided in \(n\) intervals, it’s written

\[I = \dfrac{1}{a+b+2} \sum_{k=0}^{n-1} \int_{t_k}^{t_{k+1}} x(t)^a \cdot y(t)^2 \cdot p(t) \times p'(t) \ dt\]

This integral is easier computed by using One-dimensional integration.

Polygonal

In special, if \(S\) is a polygon, the integrals can be simplified even more. The curve can be divided into \(n\) segments that connects two vertices \(V_k\) and \(V_{k+1}\).

\[p(t) = (1-t) \cdot V_{k} + t \cdot V_{k+1}\]

\[x \ dy - y \ dx = p \times p' = V_{k} \times V_{k+1}\]

\[I = \int_{S} x^a y^b \ dx \ dy = \sum_{k=0}^{n-1} \dfrac{V_{k} \times V_{k+1}}{a+b+2} \underbrace{\int_{0}^{1} x^{a} y^{b} dt}_{I_{a,b,k}}\]

The right integral can be expanded and then use the integral of beta function

\[\int_{0}^{1} \left(1-t\right)^{a} \cdot t^b \ dt = \dfrac{1}{a+b+1} \cdot \dfrac{1}{\binom{a+b}{a}}\]

\[(a+b+1) \binom{a+b}{a} \cdot I_{a,b,k} = \sum_{i=0}^{a}\sum_{j=0}^{b}\binom{i+j}{j}\binom{a+b-i-j}{b-j}x_{k}^{a-i}x_{k+1}^{i}y_{k}^{b-j}y_{k+1}^{j}\]

\[\dfrac{(a+b+2)!}{a! \cdot b!} \cdot I = \sum_{k=0}^{n-1} \left(x_{k}y_{k+1}-x_{k+1}y_{k}\right)\sum_{i=0}^{a}\sum_{j=0}^{b} X_{k,i} \cdot M_{ij} \cdot Y_{k,j}\]

With

\[M_{ij} = \binom{i+j}{j}\binom{a+b-i-j}{b-j}; \ \ \ \ X_{k,i} = x_{k}^{a-i} \cdot x_{k+1}^{i}; \ \ \ \ Y_{k, j} = y_{k}^{b-j}y_{k+1}^{j}\]

def integrate(vertices: np.ndarray, a: int, b: int) -> float:
    vertices = np.array(vertices)
    if vertices.ndim != 2 or vertices.shape[1] != 2:
        raise ValueError(f"Invalid vertices! {vertices.shape}")
    matrix = np.zeros((a + 1, b + 1), dtype="int64")
    for i in range(a + 1):
        for j in range(b + 1):
            matrix[i, j] = sp.binomial(i + j, i) * sp.binomial(a + b - i - j, b - j)
    shiverts = np.roll(vertices, shift=-1, axis=0)
    cross = vertices[:, 0] * shiverts[:, 1] - vertices[:, 1] * shiverts[:, 0]
    xvand0 = np.vander(vertices[:, 0], a + 1)
    xvand1 = np.vander(shiverts[:, 0], a + 1, True)
    yvand0 = np.vander(vertices[:, 1], b + 1)
    yvand1 = np.vander(shiverts[:, 1], b + 1, True)
    soma = np.einsum("k,ki,ki,ij,kj,kj", cross, xvand0, xvand1, matrix, yvand0, yvand1)
    denom = (a + b + 2) * (a + b + 1) * sp.binomial(a + b, a)
    return soma / denom

Polynomial

If \(S\) is a polygonomial by parts, then for an interval \(\left[t_k, \ t_{k+1}\right]\)

\[x_{k}(t) = x_0 + x_1 \cdot t + \cdots + x_q \cdot t^q\]

\[y_{k}(t) = y_0 + y_1 \cdot t + \cdots + y_q \cdot t^q\]

Contains

Deciding if a set \(A\) is (or not) a subset of \(B\) is not a trivial. This section describes the algorithms to decide it.

Basically either \(A\) and \(B\) can assume the forms of Empty, Point, Curve, Shape, Whole

That means, \(5 \times 5 = 25\) possibilities. The table here after reduces the quantity of verifications to 6, which are represented by the empty spaces

	Empty	Point	Curve	Shape	Whole
Empty	T	T	T	T	T
Point	F				T
Curve	F	F			T
Shape	F	F	F		T
Whole	F	F	F	F	T

The next sections verifies these 6 missing verifications.

Point in Point

Point is a 0-dimensional object, and contains only one element : the point itself. Therefore, a point contains another point if, and only if the points are equal.

\[A \subset B \Leftrightarrow A = B\]

Point in Curve

To verify if a curve contains a point \(q\), we compute the projection of this point in the curve.

The projection is computed by finding \(t^{\star}\) such minimizes the distance square:

\[D^2(t) = \|p(t) - q\|^2\]

It’s a positive convex function and therefore there is at least one minimum of this function. Its minimum can occurs at the nodes \(t_k\) or when the derivative is zero:

\[\dfrac{d}{dt} D^2 = 0 \Leftrightarrow \langle p(t)-q, \ p'(t)\rangle = 0\]

This equation has at least one solution, but it may have infinite (take a circle as example).

Once the solutions are found, one can compute the distance between the point and the projected point. If this projected point is equal to the point itself, then the point is on the curve.

Point in Shape

The Lebesgue Density function is used to determine if the shape contains the point.

Basically this function tells if a point is inside the shape, or outside, or at the boundary:

If \((x, \ y)\) is at the interior, then \(w(x, \ y) = 1\)
If \((x, \ y)\) is at the exterior, then \(w(x, \ y) = 0\)
If \((x, \ y)\) is at the boundary,:math:0 < w(x, y) < 1

Hence, the verification happens as:

If shape is simple:
- If \(w = 0\), then shape doesn’t contain the point
- If \(w > 0\) and shape is closed, then
- If \(w = 1\), then shape contains the point
- If \(0 < w < 1\), then

Note

The possibility of using the ray-casting algorithm was considered, but it’s Arguments against it are : depending on the direction of the ray, the result can vary. If it touches a vertex, etc.

Also, having a mesure of how acute an angle is very useful.

Curve in Curve

Check if a curve is inside another curve can be done by parts

Check if the vertices of A’s curve are inside the B’s curve.
That uses Point in Curve
Check if the basis functions are the same.

Curve in Shape

Checking if a curve is inside a shape is made by parts.

If shape is disjoint: Check if the curve is inside any subshape
If shape is connected: Check if the curve is inside all subshapes
If shape is simple:
1. Checks if all vertices are inside the shape.
  The vertices are the curve evaluated at knots. This uses the Point in Shape.
2. Find the intersection between the curve and the shape’s curve.
  If they don’t intersect, the curve is inside the shape. If they do intersect, continue
3. Find the parameters where the two curves intersect.
  Compute the midpoints. For each midpoint, check if the midpoint is inside the shape.

Shape in Shape

There are three shape classifications: Simple, Connected and Disjoint. Checking A in B have 9 possibilities, which is reduced:

If \(A\) is disjoint, then
\[\bigcup_i A_i \subset B \Leftrightarrow \text{all}_{i}\left(A_i \subset B\right)\]
If \(A\) is simple or connected, and \(B\) is disjoint, then
\[A \subset \bigcup_{i} B_i \Leftrightarrow \text{any}_{i}\left(A \subset B_i\right)\]
If \(A\) is simple or connected, and \(B\) is connected, then
\[A \subset \bigcap_{i} B_i \Leftrightarrow \text{all}_{i}\left(A \subset B_i\right)\]
If \(A\) is connected, and \(B\) is simple, then
\[\bigcap_i A_i \subset B \Leftrightarrow \text{all}_i\left(\text{jordan}\left(A_i\right) in B\right) \ \text{and}\]

For Simple in Simple, follows as bellow.

Simple in simple

This part describes how to compare cases when \(A\) and \(B\) are simple shapes. The following statements must be satisfied to \(A \subset B\).

A unbounded shape is not is not contained in a bounded shape

Translated as: If A.area < 0 and B.area > 0, then A not in B
The boundary of \(A\) must be inside of \(B\)

Translated as: If the A.jordan is not inside B, then A not in B
The area from \(A\) must not be greater than the area of \(B\)
Translated as: If A.area > B.area, then A not in B
- This consider the cases such, if A.area > 0 and B.area > 0, then it’s

Table simple shapes

List of geometric cases
Case 1	Case 2	Case 3	Case 4	Case 5

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(J_B\) in \(A\)	\(a_A ? a_B\)
1	\(+\)	\(+\)	F	F	F	?
	\(+\)	\(-\)	T	T	F	>
	\(-\)	\(+\)	F	F	T	<
	\(-\)	\(-\)	F	T	T	?
2	\(+\)	\(+\)	T	T	F	<
	\(+\)	\(-\)	F	F	F	>
	\(-\)	\(+\)		T	T	<
	\(-\)	\(-\)		F	T	>
3	\(+\)	\(+\)	F	F	T	<
	\(+\)	\(-\)		T	T	<
	\(-\)	\(+\)		F	F	>
	\(-\)	\(-\)	T	T	F	>
4	\(+\)	\(+\)	T	T	T	=
	\(+\)	\(-\)	F			>
	\(-\)	\(+\)	F			<
	\(-\)	\(-\)	T			=
5	\(+\)	\(+\)	F	F	F	?
	\(+\)	\(-\)				<
	\(-\)	\(+\)				>
	\(-\)	\(-\)				?

This table is translated to an algorithm. Unfortunatelly we don’t know which case the simples shapes are, so we will test by using some caracteristics.

For example, the first good information from the table is given by:

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(J_B\) in \(A\)	\(a_A ? a_B\)
1	\(-\)	\(+\)	F	F	T	<
2	\(-\)	\(+\)	F	T	T	<
3	\(-\)	\(+\)	F	F	F	<
4	\(-\)	\(+\)	F	T	T	<
5	\(-\)	\(+\)	F	F	F	<

# ...
shapea = SimpleShape(jordana)
shapeb = SimpleShape(jordanb)
# Decide if shapea in shapeb
if shapea.area < 0 and shapeb.area > 0:
    # For any presented cases it happens
    return False
# continue ...

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(J_B\) in \(A\)	\(a_A ? a_B\)
1	\(+\)	\(-\)	T	T	F	>
2	\(+\)	\(-\)	F	F	F	>
3	\(+\)	\(-\)	F	T	T	>
4	\(+\)	\(-\)	F	T	T	>
5	\(+\)	\(-\)	F	F	F	>

# ... continue
if shapea.area > 0 and shapeb.area < 0:
    # Only for case 1
    return (jordana in shapeb) and (jordanb not in shapea)
# continue ...

Taking out the already extracted values, and separating by when areaA > areaB:

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(a_A ? a_B\)
1	\(+\)	\(+\)	F	F	>
1	\(-\)	\(-\)	F	T	>
2	\(-\)	\(-\)	F	F	>
3	\(+\)	\(+\)	F	F	>
5	\(+\)	\(+\)	F	F	>
5	\(-\)	\(-\)	F	F	>

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(a_A ? a_B\)
1	\(+\)	\(+\)	F	F	<=
1	\(-\)	\(-\)	F	T	<=
2	\(+\)	\(+\)	T	T	<
3	\(-\)	\(-\)	T	T	<
4	\(+\)	\(+\)	T	T	=
4	\(-\)	\(-\)	T	T	=
5	\(+\)	\(+\)	F	F	<=
5	\(-\)	\(-\)	F	F	<=

# ... continue
if shapea.area > shapeb.area:
    return False
# continue ...

We see that when \(J_A \ \text{in} \ B\) gives \(F\), the \(A \ \text{in} \ B\) is also \(F\)

# ... continue
if jordana not in shapeb:
    return False
# continue ...

Rewriting the table we get

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(a_A ? a_B\)
1	\(-\)	\(-\)	F	T	<=
2	\(+\)	\(+\)	T	T	<
3	\(-\)	\(-\)	T	T	<
4	\(+\)	\(+\)	T	T	=
4	\(-\)	\(-\)	T	T	=

Taking out when areaA > 0 we get

Case	\(A\)	\(B\)	\(A\) in \(B\)	\(J_A\) in \(B\)	\(a_A ? a_B\)
1	\(-\)	\(-\)	F	T	<=
3	\(-\)	\(-\)	T	T	<
4	\(-\)	\(-\)	T	T	=

Boolean operations

The boolean operations are the main objective of this package. The following operations are available:

Name	Logic	Math	Python
Inversion	NOT	\(\overline{A}\)	~A
Union	OR	\(A \cup B\)	A \| B
Intersection	AND	\(A \cap B\)	A & B
Subtraction	SUB	\(A - B\)	A - B
Exclusive or	XOR	\(A \otimes B\)	A ^ B

From these operations above, only three of them are basic: NOT, OR and AND.

The others are decomposed as follows:

SUB: A - B = A & (~B)
XOR: A ^ B = (A - B) | (B - A)

To recall, De Morgan’s law

~(A & B) = (~A) | (~B)
~(A | B) = (~A) & (~B)

\[\overline{A \cap B} = \overline{A} \cup \overline{B}\]

\[\overline{A \cup B} = \overline{A} \cap \overline{B}\]

A general table with all the operations

True and False entities

For this package, to represent the quantities :

Empty: False, void set
Whole: True, whole plane

Inversion / logic NOT

Union / logic OR

The union between two python boolean objects

from shapepy import Primitive
# Create two simple shapes
circle = Primitive.circle()
square = Primitive.square()
# Union
newshape = circle | square

Schema of adding sets :math:`A` and :math:`B`

Table of union between two positive circles

Intersection / logic AND

The intersection between two python boolean objects

# Create two positive shapes
circle = section.shape.primitive.circle()
square = section.shape.primitive.square()
# Intersection
newshape = circle & square

Example of multiplication between two positive shapes

Table of intersection between two positive circles

Subtraction

The subtraction between two positive shapes means take out all part of \(A\) such is inside \(B\).

from shapepy import Primitive
# Create two positive shapes
circle = Primitive.circle()
square = Primitive.square()
# Subtract
newshape = circle - square

Schema of subtraction between sets :math:`A` and :math:`B`

Table of subtraction between two positive circles

Exclusive union / logic XOR

The xor between two shapes. For this operator, we use the symbol ^.

# Create two positive shapes
circle = section.shape.primitive.circle()
square = section.shape.primitive.square()
# Subtract
newshape = circle ^ square

Example of XOR between two positive shapes

Table of XOR between two positive circles

Generalities

One-dimensional integration

In the Integral section, the computation of the integral is needed:

\[I = \int_{a}^{b} f(t) \ dt\]

When analitic integration is not used, then numerical integration takes place

\[\tilde{I} = (b-a) \cdot \sum_{j=0}^{m-1} w_{j} \cdot f(t_j)\]

The values of \(t_j\) and \(w_j\) are the nodes and weights of the quadrature schema. There are available schemas are bellow, with some nodes/weights depending on \(m\)

Closed Newton-Cotes
Open Newton-Cotes
Chebyshev
Gauss-Legendre

Lebesgue Density function

The Density function is a function on the plane, based on the a shape \(S\), that

Is equal to \(1\) for interior points
Is equal to \(0\) for exterior points
Is between \(0\) and \(1\) at the boundary

At the boundary, this function measures how much ‘convex’ the boundary is.

For smooth boundaries, like a straight line or the edges of a polygon, it is equal to \(0.5\).
At the corners of a square, it’s equal to \(0.25\), cause only 25% of the neighborhood is inside the square.

The formal definition is given by

\[w_{S}(x, y) = \lim_{\varepsilon \to 0^{+}} \dfrac{\text{area}\left(S \cap D\left(x, y, \ \varepsilon\right)\right)}{\pi \varepsilon^2}\]

If \((x, \ y)\) lies on the boundary, that means there’s a \(t^{\star}\) for a curve \(p\), and therefore

\[w_{S}(x, \ y) = \dfrac{1}{2\pi} \arg\left(\langle v_0, \ v_1\rangle + i \cdot v_0 \times v_1\right)\]

\[v_0 = -\lim_{\delta \to 0^{+}} p'(t^{\star}-\delta)\]

\[v_1 = \lim_{\delta \to 0^{+}} p'(t^{\star}+\delta)\]

For smooth boundaries, \(p'\) is continuous at \(t^{\star}\). Meaning \(v_0 + v_1 = 0\) and then \(w_{S}(x, \ y) = 0.5\)

\(n\)	\(x_i\)	\(w_i\)
2	0	1/2
	1	1/2

3	0	1/6
	0.5	4/6
	1	1/6

\(n\)	\(x_i\)	\(w_i\)
1	1/2	1

2	0	1/2
	1	1/2

3	1/4	2/3
	2/4	-1/3
	3/4	2/3