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Start with a line segment (depth 0)
*Divide the line into thirds
*In the middle third produce
an equilateral triangle
* Then remove the base
Koch Snowflake
How to draw a Koch snowflake.
The Koch Snowflake
 The Koch Snowflake was studied by Niels
Fabian Helge von Koch (25 Jan 1870 - 11
March 1924), a Swedish mathematician and
nobility.
The Koch Curve
This gives Depth 1
Repeat the process
for each of the line segments
•The Koch Snowflake, which starts
with an equilateral triangle, consists
of three Koch curves.
•The snowflake is self-similar. This
means that it is roughly the same
on any scale.
•If you were to magnify any part
of the snowflake it would look
the same.
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This gives Depth 2
Repeating the process again
for each line segments gives
Depth 3
Starting with an equilateral
triangle produces the
Koch Snowflake
How to draw a Koch snowflake by
Computer?
P2
P1
P3
DrawKochSnowflake(Integer: depth, Float: length)
P1 = (0, 0)
P2 = (length/2, length*sqrt(3)/2)
P3 = (length, 0)
DrawKoch(depth, P1, 60, length)
DrawKoch(depth, P2, -60, length)
DrawKoch(depth, P3, 180, length)
End DrawKochSnowflake
DrawKoch(Integer: depth, Point: pt1, Float: angle, Float: length)
This iterative process can be repeated
again and again and again…..
Below is Depth 4
How to draw Koch snowflake by
Computer?
pt3
pt1
pt2
pt4
DrawKoch(Integer: depth, Point: pt1, Float: angle, Float: length)
If (depth == 0) drawLine(pt1, angle, length)
Else
<Find the points pt2, pt3, pt4>
DrawKoch(depth-1, pt1, angle, length/3)
DrawKoch(depth-1, pt2, angle+60, length/3)
DrawKoch(depth-1, pt3, angle-60, length/3)
DrawKoch(depth-1, pt4, angle, length/3)
End If
End DrawKoch
Run Time: T(n) = 4T(n-1)+O(1)  T(n) = O(n4)
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Questions: Given a Koch
Snowflake of depth n and initial
length S
Number of Sides
We end up with a series:
Nn ={3, 12, 48…}
• What is the length of each line segment?
• How many line segments?
• What is the total area inside the snowflake?
The common ratio then is four, and since we
started with three sides, the series begins
with three.
The formula is thenNn=(3)(4)n
n= depth, the iteration number
N0 =3
Length of Sides
N1 =12
N2=48
Common Sense
Observation: each iteration is (1/3) of the length
before it.
Let’s say that the equilateral triangle starts with a
length of S=3 and n=depth.
L0 =3→this is the side of original triangle
L1 =1
L2 =1/3
Ln ={3, 1, 1/3, 1/9, 1/27…}
The ratio of this series is (1/3)
If we know the length of each side, and the total number of
sides, we can also determine the formula for the
perimeter of the Koch snowflake!
We get the formula:
Ln= (1/3n)S
Number of Sides
 Like the length of the sides, the number of sides
for the snowflake is also a geometric series.
 The snowflake always starts with three sides.
When we make the first iteration, as shown in
the picture, we have 12 sides now.
 The second iteration shows 48 sides if we count
it.
Perimeter of a Koch Snowflake
We already knew these two formulas:
Nn=(3)(4)n
Ln= S(1/3)n
The perimeter of an object that has equal
sides, is always the total number of sides
times the length. This means:
Pn= Nn Ln
Pn=3*S(4/3)n
Again, if the side length is equal to 3, for P0 :
P0= 3*3 (4/3)0=9
P1= 3*3 (4/3)1=12
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Infinity
Area with iterations:
The first iteration:
Because we can continuously add more and
more sides to the snowflake, the
perimeter continues to get larger and
larger. This means that the snowflake has
a perimeter of infinite length!
BUT……
We get the area of the initial triangle, plus
the area of three more triangles with side
lengths of (1/3) of S.
What about area??
2nd Iteration
Formula for Area
It never exceeds the box area!
The area of the snowflake is finite…the
series converges!
How does this happen?
If we start with the area of a regular
equilateral triangle we get this formula….
Each time you iterate it, we add (3)(4)n triangles
of :
and the following formula is the total area.
The area converges to:
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Other variations
Hilbert Curve
eg: use different length segments
and add an isosceles triangle
 Each part of the larger curve is a copy of the smaller
curve suitably scaled, rotated and connected.
OR divide the initial line segment
into thirds and add another
shape in the gap
The 2D Hilbert Curve (1891)
A plane-filling Peano curve
Construction of 3D Hilbert Curve
You could start with a polygon
rather than a straight line
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Construction of 3D Hilbert Curve
 Use this element with proper orientation, mirroring.
Typical Early Solutions
Metal Sculpture at SIGGRAPH 2006
Design:
 closed loop
 maximal symmetry
 at most 3 coplanar segments
Sierpiński Curve
Design Flaws:
 2 collinear segments
 less than maximal
symmetry
 Each part of the larger curve is a copy of the smaller
curve suitably scaled and rotated
 4 coplanar segments
D. Garcia, and T. Eladi (1994)
Jane Yen: “Hilbert Radiator Pipe” (2000)
Sierpiński Curve Levels 0 - 3
Flaws
( from a sculptor’s
point of view ):
 4 coplanar
segments
 Not a closed loop
 Broken symmetry
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Sierpiński Gasket
(or sieve or triangle)
 Divide an area into shapes
 Recursively color some of the shapes
Sierpiński Carpet
All these figures can be drawn by simple recursive programs.
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