Q | there are binary expansions of x,y with ai+bi 
1, i = 1,2,…}SIERPINSKI'S TRIANGLE BY IFS
We begin to see the Sierpinski gasket when coloring the odd entries in Pascal's triangle because of iterated function systems, IFS. Let us begin with the mod-2 pattern and binary addresses. (See figure 9) With this binary address, we can provide a number theoretical description of the Sierpinski gasket as follows:
S = { (x,y) Q | there are binary expansions of x,y with ai+bi 1, i = 1,2,…}
Figure 9
Example 6:
|
(x,y) |
(0,0) |
(1,0) |
(1/2, 1/2) |
(3/4,3/4) |
|
x in binary |
0.000… |
0.111… |
0.1000… |
0.11000… |
|
y in binary |
0.000… |
0.000… |
0.0111… |
0.10111… |
|
|
|
|
|
|
In the first 3 points we see that it is contained in S because the sum of each digit is less than or equal to 1. (i.e. ai+bi 1) However in the last point, from the first digit of x and y, we see 1+1=2 which in not
less than 1, so it does not satisfy the definition of S, and therefore not
contained in S. This method is known as the Kummer's carry-condition.
Using IFS, we can show that by the definition of S, it will generate the Sierpinski gasket. We start by looking at the IFS given by four similarity transformations transforming the unit square Q into its four congruent subsquares, see figure 10, and the next step in IFS in figure 11.
|
Figure 10 |
Figure 11 |
Then the four contractions W00, W01, W10, W11 will be:
W00 (x,y) = (
x , y )
W01 (x,y) = (x , y + )
W10 (x,y) = (x + , y )
W11 (x,y) = (x + , y + )
Continuing with the transformation, we can figure out which of the Wij
are in the Sierpinski gasket and which are not by the definition of S. Thus,
omitting all the squares not satisfying the definition of S will generate
the Sierpinski gasket from an unit square. And so, with three of these transformations,
S = W00(Q) 
W01(Q) W10(Q) called the Hutchinson equation for the Sierpinski gasket S is formed.
Thus, we can conclude that from figure 9, these 3 grids are the first 3 rescaled versions of the Pascal's triangle by 1/2, 1/4, or simply 1/2n in general. In other words, the mod-2 pattern we see in Pascal's triangle is exactly the pattern we generate when iterating the IFS which encodes the Sierpinski gasket.
S