Jean-François COLONNA

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(CMAP28 WWW site: this page was created on 05/16/2022 and last updated on 11/14/2023 17:54:58 -CET-)

In the year 1890 Giuseppe Peano discovered curves that are going through all the points of a square (and of a cube) showing then continuous surjections between [0,1] and [0,1]x[0,1] (and [0,1]x[0,1]x[0,1]). During the following years, David Hilbert defined more such curves that can be generalized.

X_{1}(T=0)=0 Y_{1}(T=0)=0 (lower left corner)

X_{1}(T=1)=1 Y_{1}(T=1)=0 (lower right corner)

Then one defines a sequence of curves

C_{i}(T) = {X_{i}(T),Y_{i}(T)} ∈ [0,1]x[0,1] --> C_{i+1}(T) = {X_{i+1}(T),Y_{i+1}(T)} ∈ [0,1]x[0,1]

if T ∈ [0,1/4[: X_{i+1}(T) = Y_{i}(4T-0) Y_{i+1}(T) = X_{i}(4T-0)Transformation 1

if T ∈ [1/4,2/4[: X_{i+1}(T) = X_{i}(4T-1) Y_{i+1}(T) = 1+Y_{i}(4T-1)Transformation 2

if T ∈ [2/4,3/4[: X_{i+1}(T) = 1+X_{i}(4T-2) Y_{i+1}(T) = 1+Y_{i}(4T-2)Transformation 3

if T ∈ [3/4,1]: X_{i+1}(T) = 2-Y_{i}(4T-3) Y_{i+1}(T) = 1-X_{i}(4T-3)Transformation 4

Please note that 4=2

See a special

Here are the five first bidimensional Hilbert curves with an increasing number of iterations :

[See the used color set to display the parameter T]

Here are some examples of Hilbert-like bidimensional curves using different generating curves :

X_{1}(T=0)=0 Y_{1}(T=0)=0 Z_{1}(T=0)=0 (lower left foreground corner)

X_{1}(T=1)=1 Y_{1}(T=1)=0 Z_{1}(T=1)=0 (lower right foreground corner)

Then one defines a sequence of curves

C_{i}(T) = {X_{i}(T),Y_{i}(T),Z_{i}(T)} ∈ [0,1]x[0,1]x[0,1] --> C_{i+1}(T) = {X_{i+1}(T),Y_{i+1}(T),Z_{i+1}(T)} ∈ [0,1]x[0,1]x[0,1]

if T ∈ [0,1/8[: X_{i+1}(T) = X_{i}(8T-0) Y_{i+1}(T) = Z_{i}(8T-0) Z_{i+1}(T) = Y_{i}(8T-0)Transformation 1

if T ∈ [1/8,2/8[: X_{i+1}(T) = Z_{i}(8T-1) Y_{i+1}(T) = 1+Y_{i}(8T-1) Z_{i+1}(T) = X_{i}(8T-1)Transformation 2

if T ∈ [2/8,3/8[: X_{i+1}(T) = 1+X_{i}(8T-2) Y_{i+1}(T) = 1+Y_{i}(8T-2) Z_{i+1}(T) = Z_{i}(8T-2)Transformation 3

if T ∈ [3/8,4/8[: X_{i+1}(T) = 1+Z_{i}(8T-3) Y_{i+1}(T) = 1-X_{i}(8T-3) Z_{i+1}(T) = 1-Y_{i}(8T-3)Transformation 4

if T ∈ [4/8,5/8[: X_{i+1}(T) = 2-Z_{i}(8T-4) Y_{i+1}(T) = 1-X_{i}(8T-4) Z_{i+1}(T) = 1+Y_{i}(8T-4)Transformation 5

if T ∈ [5/8,6/8[: X_{i+1}(T) = 1+X_{i}(8T-5) Y_{i+1}(T) = 1+Y_{i}(8T-5) Z_{i+1}(T) = 1+Z_{i}(8T-5)Transformation 6

if T ∈ [6/8,7/8[: X_{i+1}(T) = 1-Z_{i}(8T-6) Y_{i+1}(T) = 1+Y_{i}(8T-6) Z_{i+1}(T) = 2-X_{i}(8T-6)Transformation 7

if T ∈ [7/8,1]: X_{i+1}(T) = X_{i}(8T-7) Y_{i+1}(T) = 1-Z_{i}(8T-7) Z_{i+1}(T) = 2-Y_{i}(8T-7)Transformation 8

Please note that 8=2

See a special

Here are the four first tridimensional Hilbert curves with an increasing number of iterations :

[See the used color set to display the parameter T]

Here are some examples of Hilbert-like tridimensional curves using different generating curves and in particular "open" knots :

[More information about Peano Curves and Infinite Knots -in english/en anglais-]

[Plus d'informations à propos des Courbes de Peano et des Nœuds Infinis -en français/in french-]

Let's use a 3-foil torus knot as the parametric curve

Here are the five first curves

See the used color set to display the parameter T.

After an infinite number of iterations the curve fills the [0,1]x[0,1]x[0,1] cube: it is a first "

Now, let's use a 5-foil torus knot as the parametric curve

Here are the five first curves

See the used color set to display the parameter T.

Now, let's use a 7-foil torus knot as the parametric curve

Here are the five first curves

See the used color set to display the parameter T.

At last, it is obviously possible to fill more "complex" tridimensional manifolds than a cube and for example:

Tridimensional Manifold | ==> | C Curve | ==> | Tridimensional Manifold filling Node |

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