# How is a system for parametrizing a triangular mesh obtained?

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 $$$$a_{vw} = \begin{cases} 1, w = v,\\ −λ_{vw}, w ∈ Nv,\\ 0, otherwise \end{cases}\,.$$$$

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

• 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 $\lambda_{vw}$". Commented Jan 4 at 7:39
• @lightxbulb And another condition is that their sum must be equal to 1 Commented Jan 9 at 6:52
• @lightxbulb My question was rather how to get these 3 cases: 1, $- \lambda_{vw}$ and 0? Commented Jan 9 at 6:55
• 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 $Ax = b$. Commented Jan 9 at 9:28