The sierpinski carpet 1 is a well known hierarchical decomposition of the square plane tiling associated with that is pairs of integers consider the sierpinski graph 2 which is the adjacency graph of the complement of in where is one of the hierarchical subsets of gray squares are used to depict the intersection of with a subset of.
Dimension of sierpinski carpet.
Here bright colors are used on a canvas size of 558x558px.
Remember it is a 2d fractal.
Whats people lookup in this blog.
1 the theorem is proved in section 2.
It was first described by waclaw sierpinski in 1916.
Between 1 and 2.
A curve that is homeomorphic to a subspace of plane.
The metric dimension of r is given by.
The sierpinski carpet is self similar with 8 non overlapping copies of itself each scaled by the factor r 1.
Possible sizes are powers of 3 squared.
A very challenging extension is to ask students to find the perimeter of each figure in the task.
The sierpinski carpet is a plane fractal curve i e.
The figures students are generating at each step are the figures whose limit is called sierpinski s carpet this is a fractal whose area is 0 and perimeter is infinite.
The hausdorff dimension of the carpet is log 8 log 3 1 8928.
Notion of metric dimension and discuss the following result.
In section 3 we recall the.
Let s see if this is true.
Since the sierpinski triangle fits in plane but doesn t fill it completely its dimension should be less than 2.
Solved now we can apply this formula for dimension to fra the sierpinski triangle area and perimeter of a you fractal explorer solved finding carpet see exer its decompositions scientific sierpiński sieve from wolfram mathworld oftenpaper net htm as constructed by removing center.
Sierpiήski carpetrform 2 n 3 andr 0 0 1 1 2 0.
Sierpiński demonstrated that his carpet is a universal plane curve.
First you have to decide which scale your sierpinski carpet should be.
Let s use the formula for scaling to determine the dimension of the sierpinski triangle fractal.
In these type of fractals a shape is divided into a smaller copy of itself removing some of the new copies and leaving the remaining copies in specific order to form new shapes of fractals.
First take a rough guess at what you might think the dimension will be.
Therefore the similarity dimension d of the unique attractor of the ifs is the solution to 8 k 1rd 1 d log 1 8 log r log 1 8 log 1 3 log 8 log 3 1 89279.
3x3 9x9 27x27 or 81x81.