<p>I am analyzing the article Parameterization of triangular meshes (Michael S. Floater November 10, 2009) and have reached the point where the uniqueness of the solution to a system of linear equations.</p>
<blockquote>
<p>We now describe a general method for constructing a parameterization of
triangular mesh in <span class="math-container">$R^3$</span>. We denote by <span class="math-container">$S$</span> the set of triangles in the mesh and
<span class="math-container">$V$</span> its vertices and <span class="math-container">$E$</span> its edges. We let <span class="math-container">$Ω_S ⊂ R^3$</span> be the union of the triangles
in <span class="math-container">$S$</span>. Then we define a parameterization of <span class="math-container">$S$</span> as a continuous piecewise linear mapping <span class="math-container">$ψ : Ω_S → R^2$</span>. Then <span class="math-container">$ψ$</span> maps each vertex, edge, and triangle of <span class="math-container">$S$</span> to a corresponding vertex, edge, and triangle in <span class="math-container">$R^2$</span>. Such a mapping is completely determined by the points <span class="math-container">$ψ(v), v ∈ V$</span>. Let <span class="math-container">$V_I$</span> denote the interior vertices of <span class="math-container">$S$</span> and <span class="math-container">$V_B$</span> the boundary ones. The boundary vertices of <span class="math-container">$S$</span> form a polygon <span class="math-container">$∂S$</span> in <span class="math-container">$R^3$</span> which we call the boundary polygon of <span class="math-container">$S$</span>. Two distinct vertices <span class="math-container">$v$</span> and <span class="math-container">$w$</span> in <span class="math-container">$S$</span> are neighbours if they are the end points of some edge
in <span class="math-container">$S$</span>. For each <span class="math-container">$v ∈ V$</span> , let <span class="math-container">$$N_v = {w ∈ V : [w, v] ∈ E},$$</span> the set of neighbours of <span class="math-container">$v$</span>, where <span class="math-container">$E = E(S)$</span> is the set of edges in <span class="math-container">$S$</span>.
The first step of the method is to choose any points <span class="math-container">$ψ(v) ∈ R^2$</span>, for <span class="math-container">$v ∈ V_B$</span>, such that the boundary polygon <span class="math-container">$∂S$</span> of <span class="math-container">$S$</span> is mapped into a simple polygon <span class="math-container">$ψ(∂S)$</span> in the plane. In the second step, for <span class="math-container">$v ∈ V_I$</span> , we choose a set of strictly positive values <span class="math-container">$λ_{vw}$</span>, for <span class="math-container">$w ∈ N_v$</span>, such that <span class="math-container">$$ \sum_{w∈N_v}λ_{vw} = 1. (3)$$</span>
Then we let the points <span class="math-container">$ψ(v)$</span> in <span class="math-container">$R^2$</span>, for <span class="math-container">$v ∈ V_I$</span> , be the unique solutions of the
linear system of equations <span class="math-container">$$ ψ(v) = \sum_{w∈N_v}λ_{vw}ψ(w), v ∈ V_I . (4)$$</span>
Since these equations force each point ψ(v) to be a convex combination of its neighbouring points <span class="math-container">$ψ(w)$</span>, we call <span class="math-container">$ψ$</span> a convex combination mapping.
Let us take a closer look at the linear system. We must show that it has a unique solution. To this end, note that it can be rewritten in the form
<span class="math-container">$$ψ(v) − \sum_{w∈N_v∩V_I}λ_{vw}ψ(w) = \sum_{w∈N_v∩V_B}λ_{vw}ψ(w), v ∈ VI . (5)$$</span>
This can be written as the matrix equation
<span class="math-container">$$Ax = b,$$</span>
where <span class="math-container">$x = (ψ(w))_{w∈V_I}$</span> is the column vector of unknowns in some arbitrary ordering, <span class="math-container">$b$</span> is the column vector whose elements are the right hand sides of (5), and the matrix <span class="math-container">$A = (a_{vw})_{v,w∈V_I}$</span> has dimension <span class="math-container">$n × n$</span>, with <span class="math-container">$n = |V_I|$</span>,
and elements
<span class="math-container">\begin{equation}
a_{vw} =
    \begin{cases}
      1, w = v,\\
      −λ_{vw}, w ∈ Nv,\\
      0, otherwise
    \end{cases}\,.
\end{equation}</span></p>
</blockquote>
<p>I don't understand where the <span class="math-container">$a_{vw}$</span> values come from. How can this be calculated, please tell me?</p>

CC BY-SA 4.0
2024-01-04T07:39:02.440Z
They have written the definition of $a_{vw}$ they have not written the definition of $\lambda_{vw}$ though, at least in the excerpt you posted, i.e. "we choose a set of strictly positive values $\lambda_{vw}$".
lightxbulb
CC BY-SA 4.0
2024-01-09T06:52:20.613Z
@lightxbulb And another condition is that their sum must be equal to 1
e7min
CC BY-SA 4.0
2024-01-09T06:55:32.250Z
@lightxbulb My question was rather how to get these 3 cases: 1, $- \lambda_{vw}$ and 0?
e7min
CC BY-SA 4.0
2024-01-09T09:28:03.847Z
It's just equation 5 rewritten in matrix-vector form: $\psi(v)- \sum_{w \in N_v\cap V_I} \lambda_{vw}\psi(w) = \sum_{w \in N_v \cap V_B} \lambda_{vw}\psi(w)$, now with the definition for $a_{vw}$ and $x= (\psi(w))_{w\in V_I}$ you may rewrite this as $\sum_{w} a_{vw} x_w = b_v$, and since you have $|V_I|$ such equations you can write it as $Ax = b$.
lightxbulb
<linear-algebra>
How is a system for parametrizing a triangular mesh obtained?
e7min
# How is a system for parametrizing a triangular mesh obtained?

(Asked by: username_0 on 2023-12-28)

I am analyzing the article Parameterization of triangular meshes (Michael S. Floater November 10, 2009) and have reached the point where the uniqueness of the solution to a system of linear equations.

> We now describe a general method for constructing a parameterization of
> triangular mesh in $R^3$. We denote by $S$ the set of triangles in the mesh and
> $V$ its vertices and $E$ its edges. We let $Ω\_S ⊂ R^3$ be the union of the triangles
> in $S$. Then we define a parameterization of $S$ as a continuous piecewise linear mapping $ψ : Ω\_S → R^2$. Then $ψ$ maps each vertex, edge, and triangle of $S$ to a corresponding vertex, edge, and triangle in $R^2$. Such a mapping is completely determined by the points $ψ(v), v ∈ V$. Let $V\_I$ denote the interior vertices of $S$ and $V\_B$ the boundary ones. The boundary vertices of $S$ form a polygon $∂S$ in $R^3$ which we call the boundary polygon of $S$. Two distinct vertices $v$ and $w$ in $S$ are neighbours if they are the end points of some edge
> in $S$. For each $v ∈ V$ , let $$N\_v = {w ∈ V : [w, v] ∈ E},$$ the set of neighbours of $v$, where $E = E(S)$ is the set of edges in $S$.
> The first step of the method is to choose any points $ψ(v) ∈ R^2$, for $v ∈ V\_B$, such that the boundary polygon $∂S$ of $S$ is mapped into a simple polygon $ψ(∂S)$ in the plane. In the second step, for $v ∈ V\_I$ , we choose a set of strictly positive values $λ\_{vw}$, for $w ∈ N\_v$, such that $$ \sum\_{w∈N\_v}λ\_{vw} = 1. (3)$$
> Then we let the points $ψ(v)$ in $R^2$, for $v ∈ V\_I$ , be the unique solutions of the
> linear system of equations $$ ψ(v) = \sum\_{w∈N\_v}λ\_{vw}ψ(w), v ∈ V\_I . (4)$$
> Since these equations force each point ψ(v) to be a convex combination of its neighbouring points $ψ(w)$, we call $ψ$ a convex combination mapping.
> Let us take a closer look at the linear system. We must show that it has a unique solution. To this end, note that it can be rewritten in the form
> $$ψ(v) − \sum\_{w∈N\_v∩V\_I}λ\_{vw}ψ(w) = \sum\_{w∈N\_v∩V\_B}λ\_{vw}ψ(w), v ∈ VI . (5)$$
> This can be written as the matrix equation
> $$Ax = b,$$
> where $x = (ψ(w))\_{w∈V\_I}$ is the column vector of unknowns in some arbitrary ordering, $b$ is the column vector whose elements are the right hand sides of (5), and the matrix $A = (a\_{vw})\_{v,w∈V\_I}$ has dimension $n × n$, with $n = |V\_I|$,
> and elements
> \begin{equation}
> a\_{vw} =
> \begin{cases}
> 1, w = v,\\
> −λ\_{vw}, w ∈ Nv,\\
> 0, otherwise
> \end{cases}\,.
> \end{equation}

I don't understand where the $a\_{vw}$ values come from. How can this be calculated, please tell me?

**Tags:** linear-algebra

> **Comments:**
> **username_1**: They have written the definition of $a_{vw}$ they have not written the definition of $\lambda_{vw}$ though, at least in the excerpt you posted, i.e. "we choose a set of strictly positive values $\lambda_{vw}$".  
> **username_0**: @username_1 And another condition is that their sum must be equal to 1  
> **username_0**: @username_1 My question was rather how to get these 3 cases: 1, $- \lambda_{vw}$ and 0?  
> **username_1**: It's just equation 5 rewritten in matrix-vector form: $\psi(v)- \sum_{w \in N_v\cap V_I} \lambda_{vw}\psi(w) = \sum_{w \in N_v \cap V_B} \lambda_{vw}\psi(w)$, now with the definition for $a_{vw}$ and $x= (\psi(w))_{w\in V_I}$ you may rewrite this as $\sum_{w} a_{vw} x_w = b_v$, and since you have $|V_I|$ such equations you can write it as $Ax = b$.  

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## 0 answers

