# An Effective Algorithm for Finding Shortest Paths in Tubular Spaces
## Appendix A
### Proof of Remark 1  
We will prove the correctness of the proposed direction of the VP at each partition. 1. Case 1: \(\mathbf{X}\in\mathbf{P1}\) (\(\mathbf{X}\) _can see_ \(\mathbf{\Omega}\))  
As the line segment joining \(\mathbf{X}\) and \(\mathbf{\Omega}\) is the shortest path between them, the direction of the ESP \(\mathbf{p}(s)\) at \(\mathbf{X}\) must be towards \(\mathbf{\Omega}\):
2. Case 2: \(\mathbf{X}\in\mathbf{P2}\) (\(\mathbf{X}\) _can see_ \(S_{end}\), _but_ \(\mathbf{\Omega}\))  
Let \(\mathbf{Y}\in\sigma_{X}(S_{end})\) be the set of visible points on the ending cross-section such that the angle between \(\mathbf{X}\mathbf{Y}\) and \(\mathbf{X}\mathbf{\Omega}\) is the smallest one. We define a cone surface \((C_{0})\) with the apex \(\mathbf{X}\) and the generatrix makes an angle \(\widehat{\mathbf{Y}\mathbf{X}\mathbf{\Omega}}\) to the axis \(\mathbf{X}\mathbf{\Omega}\), then \(\mathbf{Y}\in(\sigma_{X}(S_{end})\cap C_{0})\) (see Figure 11). As \(\sigma_{X}(S_{end})\) is convex, we can easily prove that the existence of \(\mathbf{Y}\) is unique, moreover \(\mathbf{Y}\in(\partial\sigma_{X}(S_{end})\setminus\partial\Omega)\). Thus, \(\mathbf{X}\mathbf{Y}\) must be tangent to \(\partial\Omega\) at \(\mathbf{T}\). Let \((\alpha)\) be the corresponding tangent plane, then we obtain that \((\alpha)\) is also the tangent plane of \(\sigma_{X}(S_{end})\) (see the proof of Corollary 2 for a similar case). As \(\mathbf{Y}\) is the tangent point between \(\sigma_{X}(S_{end})\) and \((C_{0}\cap S_{end})\) (these two convex sets have only one common point \(\mathbf{Y}\)), \((\alpha)\) is also the tangent plane of \((C_{0})\). Let \(\mathbf{I}\) be the center of the cross-section at \(\mathbf{T}\). As \(\mathbf{I}\mathbf{T}\perp(\alpha)\), \(\mathbf{I}\mathbf{T}\) must intersect the axis \(\mathbf{X}\mathbf{\Omega}\) of \((C_{0})\). Thus, \(\mathbf{X}\mathbf{T}\mathbf{I}\mathbf{I}\mathbf{Y}\), and \(\mathbf{\Omega}\) are coplanar. We denote this coplanar plane by \((P_{c})\). Let \(\mathbf{W}\) be the intersection between the ending cross-section plane \(\beta(S_{end})\) and \(\mathbf{p}(s_{X})\). Now, we have to prove that \(\mathbf{W}\equiv\mathbf{Y}\). Using Corollary 2, we obtain: \(\mathbf{W}\in\sigma_{X}(S_{end})\). Let\((C_{1})\) be the closed surface enclosed by \(\sigma_{X}(S_{end})\) and the set of line segments from \(\mathbf{X}\) to every point of \(\partial\sigma_{X}(S_{end})\). Thus, \(\mathbf{X}\) can see every point in \((C_{1})\). If \(\mathbf{W}\
ot\in\partial\sigma_{X}(S_{end})\) (that is, \(\mathbf{W}\) belongs to the inner zone of \(\sigma_{X}(S_{end})\)), then the ESP goes into the inner space of \((C_{1})\) with the direction \(\mathbf{\dot{p}}(s_{X})\). As \(\mathbf{\Omega}\) is outside \((C_{1})\), the ESP must pass the boundary of \((C_{1})\). We denote \(\mathbf{H}\) as the passing point. Since \(\mathbf{X}\) can see \(\mathbf{H}\), the part of the ESP connecting \(\mathbf{X}\) and \(\mathbf{H}\) is not the shortest path (as it is longer than \(\overline{\mathbf{X}\mathbf{H}}\)). This leads to a contradiction with Lemma 1. Hence:  
\[\mathbf{W}\in\partial\sigma_{X}(S_{end})\] (A5)  
In addition, if \(\mathbf{W}\
ot\equiv\mathbf{Y}\), then \(\mathbf{W}\
ot\in(P_{c})\). By using Corollary 1, we can confirm that the particle will move far away from \((P_{c})\) so it cannot reach \(\mathbf{\Omega}\) on \((P_{c})\). Thus, \(\mathbf{W}\equiv\mathbf{Y}\). (Q.E.D.) Case 3: \(\mathbf{X}\in\mathbf{P}3\) (\(\mathbf{X}\)_cannot see \(S_{end}\)_)  
As \(\mathbf{X}\) cannot see \(S_{end}\), the farthest cross-section of the tube is \(S_{f}\) which can be seen by \(\mathbf{X}\). We will prove that \(\mathbf{X}\) can see only one point in this cross-section. In \(S_{f}\), if there exist two different visible points \(\mathbf{Y_{1}}\) and \(\mathbf{Y_{2}}\) by \(\mathbf{X}\), then \(\mathbf{X}\) can see the midpoint \(\mathbf{Y_{m}}\) of \(\overline{\mathbf{Y_{1}Y_{2}}}\) (using Lemma 2).