# Construção da curva que preenche o tapete Sierpinski.

it <- function(z){ # Iteração base
p=matrix(0,2,12);u=(z[,2]-z[,1])/3;t=pi/2
v=matrix(c(cos(t),sin(t),-sin(t),cos(t)),2,2)%*%u
p[,1]=z[,1];
p[,2]=p[,1]+u; 
p[,3]=p[,2]+v;
p[,4]=p[,3]+u;
p[,5]=p[,4]-v;
p[,6]=p[,5]-v;
p[,7]=p[,6]-u;
p[,8]=p[,7]+v;
p[,9]=p[,8]+v;
p[,10]=p[,9]+u;
p[,11]=p[,10]-v;
p[,12]=z[,2];p}

Sierpinski<-function(x,y,m){ # Replicação da iteração base.
z=matrix(c(x,y),2,2)
for ( i in 1:m){
q=matrix(0,2,12*length(z[1,])-12)
for (j in 1:(length(z[1,])-1)){
q[,(12*j-11):(12*j)]=it(z[,j:(j+1)]);}
z=q};z
}
 

# --- Teste -----------------------------------

sinal=1
x=c(0,0)
y=c(1,1)
z=matrix(c(x,y),2,2)

z=it(z);z
plot(z[1,],z[2,],'l')

z=Sierpinski(x,y,4);
plot(z[1,],z[2,],'l',col="red",axes=FALSE)
w=matrix(c(0,0,1,0,1,1,0,1),2,8);points(w[1,],w[2,],'l')