Sunday

October 23, 2016
Posted by **astrid** on Friday, May 2, 2008 at 11:57pm.

Also, what is a conflict resoultion dealing with discrete math?

- algebra -
**Reiny**, Saturday, May 3, 2008 at 12:04amI googled your topic and got over 37000 hits

try this one, it shows how it grows.

try using the different colours for different stages in the iteration.

http://ejad.best.vwh.net/java/fractals/sierpinski.shtml - algebra -
**tchrwill**, Saturday, May 3, 2008 at 10:35amSierpinski's Triangle

Construction

1--Draw an all black equilateral triangle.

2--Define the midpoints of the three sides.

3--Connect the three midpoints forming a new inverted triangle within the triangle and 4 smaller equilateral triangles overall.

4--Let the inverted inner triangle be white.

5--Define the midpoints of the three sides on each of the other 3 black triangles.

6--Connect the midpoints of the sides again making the inverted inner triangles white.

7--Repeat this process to infinity resulting in Sierpinski's triangle.

With each stage of disection, the inverted white triangles are ignored and the remaining black triangles further disected into smaller triangles. The number of black triangles remaining after each disection is 3^(n-1), n being the stage number. Therefore, the remaining black triangles increase according to

n....1....2....3....4....5.....6.......7....etc.

B...1....3....9....27..81..243...729...etc. a geometric progression with common factor 3.

The number of black triangles remaining after each stage derives from this geometric progression, the nth term of which is L = ar^(n-1) = 3^(n-1).

If, by chance, you are interested in the sum total of black and white triangles existing after each stage (assuming the white triangles remain), the answer is different. The sum of black and white triangles increase according to

n....1....2....3....4.....5.....6.......7......etc.

B...1....4....13..40..121..364..1093...etc.

The total number of both black and white triangles existing at the end of each stage is the sum of the terms of the same geometric progression. This sum derives from S(b/w) = a(r^n - 1)/(r-1) = (3^n - 1)/2.

If you are interested in the sum total of all identifiable triangles within the starting triangle (the 4 smaller triangles form 1 larger triangle), their total increases according to

n....1....2....3....4.....5.....6.......7.....etc.

B...1....5...17...53..161..485..1457...etc.

The total number of identifiable triangles at the nth stage can be defined by adding the number of black and white triangles from the nth stage to the number of black and white triangles from the (n-1)th stage. This is expressed by summing the two individual sums as S(id) = a[r^n + r^(n-1) - 2]/(r-1) = [3^n + 3^(n-1)]/2.

The triangle numbers are summed as follows:

Stage n....................1....2....3.....4....5......6......7.....etc.

Black.B....................1....3....9....27..81...243...726...etc

Black & White b/w....1....4...13...40..121..364..1093..etc.

Identifiable................1....5...17...53..161..485..1457..etc.

The resulting black area is zero.