Bidimensional Hilbert Curve -iterations 1 to 5- [*Courbe de Hilbert bidimensionnelle -itérations 1 à 5-*].

See the used color set to display the parameter T.

See the used color set to display the pi digits.

See the used color set to display the parameter T.

See the used color set to display the pi digits.

[0,1] --> [0,1]x[0,1]Let's

T = 0.ALet's_{1}A_{2}A_{3}... ∈ [0,1] with A_{i}∈ {0,1,2}

X(T) = 0.B_{1}B_{2}B_{3}... ∈ [0,1] with B_{i}∈ {0,1,2}

Y(T) = 0.Cwith :_{1}C_{2}C_{3}... ∈ [0,1] with C_{i}∈ {0,1,2}

B_{n}= A_{2n-1}if A_{2}+A_{4}+...+A_{2n-0}is even

B_{n}= 2-A_{2n-1}otherwise

C_{n}= A_{2n}if A_{1}+A_{3}+...+A_{2n-1}is even

C_{n}= 2-A_{2n}otherwise

These two functions X(T) and Y(T) are the coordinates of a point P(T) inside the [0,1]x[0,1] square. The displayed "curve" -as little spheres- is the trajectory of P(T) when T varies from 0 (lower left corner) to 1-epsilon (upper right corner).

Here are the four first bidimensional Peano curves with an increasing number of digits {2,4,6,8}:

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

[0,1] --> [0,1]x[0,1]x[0,1]as a generalization of the bidimensional one.

Let's

T = 0.ALet's_{1}A_{2}A_{3}... ∈ [0,1] with A_{i}∈ {0,1,2}

X(T) = 0.B_{1}B_{2}B_{3}... ∈ [0,1] with B_{i}∈ {0,1,2}

Y(T) = 0.C_{1}C_{2}C_{3}... ∈ [0,1] with C_{i}∈ {0,1,2}

Y(T) = 0.Dwith :_{1}D_{2}D_{3}... ∈ [0,1] with D_{i}∈ {0,1,2}

B_{n}= A_{3n-2}if A_{3}+A_{6}+...+A_{3n-0}is even

B_{n}= 2-A_{3n-2}otherwise

C_{n}= A_{3n-1}if A_{2}+A_{5}+...+A_{3n-1}is even

C_{n}= 2-A_{3n-1}otherwise

D_{n}= A_{3n}if A_{1}+A_{4}+...+A_{3n-2}is even

D_{n}= 2-A_{3n}otherwise

These three functions X(T), Y(T) and Z(T) are the coordinates of a point P(T) inside the [0,1]x[0,1]x[0,1] cube. The displayed "curve" is the trajectory of P(T) -displayed as little spheres- when T varies from 0 (lower left corner) to 1-epsilon (upper right corner).

Here are the three first tridimensional Peano curves with an increasing number of digits {3,6,9}:

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

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-]

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