Two volunteer problem in a circular campus

Octave/Matlab Code for the solution :

Blue and Red pair stands for the positions of the Two volunteers in a given circular campus.

For various  random pair of seeds the optimization finally resulted in a set of four nonlinear integral equations.
 Solved them using iterative procedure.
The locus of the volunteer positions have come to be on a particular radius(=0.42676) circle with each one on antipodal positions.
  Later, based on the antipodal property observed from the computations the value of the radius was found to be exactly 4/(3pi)=0.42441 by evaluating the integral equations in polar form for the unknown radius.


Code starts here :


%%% Two volunteer selection problem

T=[0:0.01:2*pi];
R=[0:0.01:1];
[GridR, GridT]=meshgrid(R, T);
iX=GridR.*cos(GridT);
iY=GridR.*sin(GridT);
dA=GridR*0.01*0.01;

for ini=1:100
%%initialization
P=[rand*cos(rand*2*pi), rand*sin(rand*2*pi)];
Q=[rand*cos(rand*2*pi), rand*sin(rand*2*pi)];

 for it=1:40
   dD=sum(P.*P-Q.*Q ) ;
   sgn=sign(2*(Q(1)-P(1))*iX+2*(Q(2)-P(2))*iY+dD);
   %% x0, y0
   isgn=1-sgn;
   Nume=sum((iX.*isgn.*dA)(:));
   Deno=sum((isgn.*dA)(:));
   P(1)=Nume/Deno;
   Nume=sum((iY.*isgn.*dA)(:));
   Deno=sum((isgn.*dA)(:));
   P(2)=Nume/Deno;
 
   %% x1, y1
   isgn=1+sgn;
   Nume=sum((iX.*isgn.*dA)(:));
   Deno=sum((isgn.*dA)(:));
   Q(1)=Nume/Deno;
   Nume=sum((iY.*isgn.*dA)(:));
   Deno=sum((isgn.*dA)(:));
   Q(2)=Nume/Deno;
 
  endfor
   polar( atan2(P(2), P(1)), norm(P),'r*');
   hold on;
   polar( atan2(Q(2), Q(1)), norm(Q), 'b*');
   pause(1)
   endfor