Topology v. A topology

https://youtu.be/tdOaMOcxY7U

Equations an Solution for 2 robots moving on circle

Regions: Areas of Regions 1 & 2

Regions 1 & 2:

These regions were described previously in the post "Two robots moving in a circle". In summary, call region 1 to the flat drawing of our configuration of space without any alteration;  Region 2 would be the flat drawing of the configuration of space for the two robots moving in a circle but already transformed to a cylindrical flat drawing. It is important to have this transformation in mind since regions 1 and 2 are topologically equal, geometrically they are different. Dealing with the geometry of this flat drawing will be essential to present our solution to the two robots moving in a circle problem.

Areas $A_U$, $A_D$, sub-areas$A_2, A_4, A_1 and A_3$ and tranformations.-

On Region 2:

$A_D=A_2+A_4$

$A_{UT}=A_{1T}+A_{3T}$

$A_2=A_4 - A_D$

$A_{1T}=A_{3T}+A_{DT}$

Special cases.-   

$A_{3T}$

$x<y<-x+\ \frac{1}{2}$

$A_4$

$x>y>-x+\ \frac{8}{5}$

Antipodal and Non-Antipodal Cases:

Identification of an Antipodal and Non-Antipodal case.- 

Antipodal if $(x_0+y_0)-(x'_0+y'_0)= \mathbb{Z}$

Non Antipodal if $(x_0+y_0)-(x'_0+y'_0) \neq \mathbb{Z}$

Antipodal Cases:

Antipodal case in $A_2$ and $A_{1T}$.- 

$A_2$, then $P(x_0,y_0)$

$A_{1T}$, then $P(x_0,y_0 - 1)$

Deformation line for point $P(x_0,y_0)$.- 

$y=-x+(x_0+y_0)$

Interception between deformation line and Antipodal circle for point $P(x_0,y_0)$.- 

$( \frac {x_0+y_0}{2} + \frac {1}{4} , \frac {x_0+y_0}{2} - \frac {1}{4})$

Deformation line for point $P(x_0,y_0 - 1)$.- 

$y=-x+(x_0+y_0 - 1)$

Interception between deformation line and Antipodal circle for point $P(x_0,y_0 - 1)$.- 

$( \frac {x_0+y_0}{2} - \frac {1}{4} , \frac {x_0+y_0}{2} - \frac {3}{4})$

Antipodal Case in $A_{3T}$ and $A_4$.-

For $A_{3T}$:

Deformation line.- $y=-x+(x_0+y_0 - 1)$

Interception between deformation line and Antipodal circle.- 

$( \frac {x_0+y_0}{2} + \frac {3}{4} , \frac {x_0+y_0}{2} + \frac {1}{4})$

For $A_4$.-

Deformation line.- $y=-x+(x_0+y_0)$

Interception between deformation line and Antipodal circle,- 

$( \frac {x_0+y_0}{2} - \frac {3}{4} , \frac {x_0+y_0}{2} - \frac {5}{4})$

Algorithm for antipodal case.- 

Since the distance is constant when the pairs a strictly antipodal, our algorithm for this case will be to go clockwise.

  

Non-Antipodal Cases:

When $(x_0+y_0)-(x'_0+y'_0) \neq \mathbb{Z}$, then we have a non-antipodal case.

Non-Antipodal case in $A_2$ and $A_{1T}.- 

The generic equations for antipodal cases will also apply for non-antipodal cases, but the difference will be in the algorithm that we will create for this specific case. 

Algorithm for non-antipodal case.-  

In this case, the algorithm will be for the two robots, when laying on the Strictly Antipodal Circle (SAC) while having a distance $d \neq \frac {\sqrt {2}}{2}$, go the shortest path from initial intersection to final intersection.

Shortest path.- 

Depending on the on the distance between two points laying on SAC, the shortest distance will always vary. Therefore, we will have to set some conditions in order to classify different cases: 

$$d > \frac{\sqrt {2}}{2} \rightarrow d_{move}= d - \frac{\sqrt {2}}{2}$$ $$d \frac{\sqrt {2}}{2}\rightarrow d_{move}=d $$