function [result]=race_end(xt,yt,xt1,yt1)global Map_Data Finish_Data Pos_Vector_Indexes Speed_Vector_Indexes%     race_end : checks whether the transition from (xt,yt) to (xt1,yt1) crosses the finish line.% Arguments --------------------------------------------------------------%     (xt,yt) = The first position vector.%     (xt1,yt1) = The successor vector. These arguments are optional: if not present,%                 then xt and yt are states, not positions.% Evaluation -------------------------------------------------------------%     result = boolean vector; returns 1 if the car crosses the finish line.%--------------------------------------------------------------------------% In verbose mode, there is no difference to silent mode.%--------------------------------------------------------------------------% MDP Toolbox, INRA, BIA Toulouse, France%--------------------------------------------------------------------------% Initialisationst_len=size(Pos_Vector_Indexes,1)*size(Speed_Vector_Indexes,1);result=zeros(size(xt));% important racetrack data: (x1,y1) and (x2,y2) are the boundaries of the finish line, % (xc,yc) are the coordinates of the center of the finish line, % v2=VMAX squared (so as not to compute it every time) and alpha,beta are% the coefficients of the y=alpha*x+beta equation that the finish line representsx1=Finish_Data(1,2);y1=Finish_Data(1,1);x2=Finish_Data(2,2);y2=Finish_Data(3,1);xc=Finish_Data(3,2);yc=Finish_Data(3,1);v2=Finish_Data(4,1);alpha=Finish_Data(5,1);beta=Finish_Data(5,2);% % We check whether we have states or coordinates as argumentsif nargin==2    t=(yt==st_len+2 | xt==st_len+2);    result(t)=1;    t=(yt<=st_len | xt<=st_len);    xt1=zeros(size(xt));    yt1=zeros(size(xt));    [xt1(t),yt1(t),unused1,unused2]=get_values_from_state(yt(t));     [xt(t),yt(t),unused1,unused2]=get_values_from_state(xt(t));         xt(~t)=Inf;    yt(~t)=Inf;end% we only perform the check if the car is near enough% the center of the finish line to make it in 1 step.test_dist=((xc-xt).*(xc-xt)+(yc-yt).*(yc-yt)<3*v2);x_diff=(xt~=xt1);xx=zeros(size(xt));yy=zeros(size(xt));% We compute the coordinates of the intersection of the car's path% with the finish line t=find(test_dist&x_diff);a=zeros(size(xt));b=zeros(size(xt));a(t)=(yt(t)-yt1(t))./(xt(t)-xt1(t));b(t)=yt(t)-a(t).*xt(t);a_diff=(a~=alpha);t=find(test_dist&x_diff&a_diff);xx(t)=(-b(t)+beta)./(a(t)-alpha);        yy(t)=a(t).*xx(t)+b(t);t=find(test_dist&x_diff&(~a_diff)&(b==beta));result(t)=1;t=find(test_dist&(~x_diff));xx(t)=xt(t);yy(t)=alpha*xx(t)+beta;% We check whether the intersection is within the rectangle formed by the two closest points;% We add a small security margin to avois rounding errors.e=0.001;t=((xx>=min(x1,x2)-e)&(xx<=max(x1,x2)+e)&(yy>=min(y1,y2)-e)&(yy<=max(y1,y2)+e) & (xx>=min(xt,xt1)-e)&(xx<=max(xt,xt1)+e)&(yy>=min(yt,yt1)-e)&(yy<=max(yt,yt1)+e));result(t)=1;