BeschreibungPeano-curve animated (140x120).gif	English: A peano curve. It continuously maps the interval [0,1] onto the square [0,1]×[0,1] as the parameter rises from 0 to 1.
Datum	23. März 2008
Quelle	own program, description below
Urheber	self
Genehmigung (Weiternutzung dieser Datei)	Public domainPublic domainfalsefalse Ich, der Urheberrechtsinhaber dieses Werkes, veröffentliche es als gemeinfrei. Dies gilt weltweit. In manchen Staaten könnte dies rechtlich nicht möglich sein. Sofern dies der Fall ist: Ich gewähre jedem das bedingungslose Recht, dieses Werk für jedweden Zweck zu nutzen, es sei denn, Bedingungen sind gesetzlich erforderlich.

Construction

The square on the right consists of two triangles, one at the lower left and one at the upper right. The description is for the lower left triangle

${\displaystyle C:\,[0,{\scriptstyle {\frac {1}{2}}}]\mapsto \{(x,y)\;/\;x\geqslant 0,\;y\geqslant 0,\;x+y\leqslant 1\}}$

For a point ${\displaystyle p=(x,y)}$in the image of ${\displaystyle C}$, i.e. in the lower left triangle, we construct its preimage ${\displaystyle \lambda }$ using the notation

${\displaystyle {\begin{aligned}r^{\bot }&=&(r_{x},r_{y})^{\bot }=(r_{y},-r_{x})\end{aligned}}}$

Starting with

${\displaystyle {\begin{aligned}p&=(x,y)\\r_{0}&=({\scriptstyle {\frac {1}{2}}},{\scriptstyle {\frac {1}{2}}})\\q_{0}&=(0,0)\end{aligned}}}$

we set

${\displaystyle {\begin{aligned}\alpha _{n}&=(p-q_{n})\cdot r_{n}^{\bot }\\\beta _{n}&={\begin{cases}-1,&{\text{if }}\alpha _{n}<0\\1,&{\text{if }}\alpha _{n}\geqslant 0\end{cases}}\\q_{n+1}&=q_{n}+r_{n}\\r_{n+1}&={\frac {1}{2}}{\begin{pmatrix}1&-\beta _{n}\\\beta _{n}&1\end{pmatrix}}r_{n}\\\lambda &={\frac {1}{2}}\sum _{n=0}^{\infty }{\frac {1+\beta _{n}(-1)^{n}}{2}}\cdot {\frac {1}{2^{n}}}\end{aligned}}}$

We then get ${\displaystyle C(\lambda )=p}$. Given ${\displaystyle \lambda _{0}}$ in [0,½], all pixels with ${\displaystyle \lambda (p)<\lambda _{0}}$ are colored in scarlet and all pixels with ${\displaystyle \lambda (p)>\lambda _{0}}$ are colored in white.