Imagine a meandering river. Imagine, furthermore, that every meandering has a meandering within it. This is one way that you might visualize the Hilbert Curve, an example of a space-filling curve. The complexity of this “river” may have implications for ecological study design. How, for example, would you measure the distance between the source of the river (lower left) and the ocean (lower right)? For a mathematical description of the Hibert Curve, see Sagan (1994).

Figure 1. Iterations in the construction of the Hilbert Curve.

The graphics in Figure 1 can be produced by cutting and pasting the R script given below into your R console. For more, see Fractals with R, Part 1 and Part 2.

**References**

Sagan, Hans. 1994. “Space-filling curves”. Springer-Verlag.

**R Script**

#Written by Allan Roberts, 2012.

hilbert.curve <- function(n){

Double <- function(A){

#The matrix for “H(n)” is equal to “Double(H(n-1))”.

m <- dim(A)[1];

n <- dim(A)[2];

N <- m*n;

B <- A+N;

C <- B+N;

D <- C+N;

E <- cbind(rbind(B,skew.transpose(A)),rbind(C,t(D)));

return(E);

}

Rotate <- function(A){

#Rotates the matrix A clockwise.

m <- dim(A)[1];

n <- dim(A)[2];

N <- m*n;

B <- matrix(0,m,n);

for (i in 1:m) for (j in 1:n) B[j,n+1-i] <- A[i,j]

return(B);

}

skew.transpose <- function(A){

return(Rotate(Rotate(t(A))));

}

rowofx <- function(A,x){

#Returns the row index of the matrix A for entry equal to x.

m <- dim(A)[1];

n <- dim(A)[2];

for (i in 1:m) for (j in 1:n) if (A[i,j]==x) return(i);

}

colofx <- function(A,x){

#Returns the column index of the matrix A for entry equal to x.

m <- dim(A)[1];

n <- dim(A)[2];

for (i in 1:m) for (j in 1:n) if (A[i,j]==x) return(j);

}

Draw <- function(A){

#Draws a graphical representation of the matrix A.

A <- Rotate(A);

m <- dim(A)[1];

n <- dim(A)[2];

N <- m*n;

plot( (rowofx(A,1)-1)/n, (colofx(A,1)-1)/n, pch=19,cex=0.5,ylim = c(0,1), xlim =c(0,1), ylab=character(1),xlab=character(1),axes=FALSE);

d <- 1/n;

for (i in 1:(N-1)) lines(c((rowofx(A,i)-1)/n,((rowofx(A,i+1)-1)/n)), c((colofx(A,i)-1)/n,((colofx(A,i+1)-1)/n)),lwd=1);

points((rowofx(A,N)-1)/n, (colofx(A,N)-1)/n, pch=19,cex=0.5);

}

H <- function(n){

#H(1) is shown in Figure 2.

if (n==0) return(matrix(c(2,1,3,4),2,2));

return(Double(H(n-1)));

}

Draw(H(n));

}

hilbert.curve(n=3);

** **

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How would you measure the distance from a river’s source and the ocean then? Would it be to fit the Hilbert curve to the shape of a river, or to include each of its branchings as a new “meander” to add to the curve? That’s an interesting idea, especially since I study sponges, which have complex, often-branching canals.

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