Pascal's Triangle (in 2.6)
""" Rows of Pascal's Triangle See: http://en.wikipedia.org/wiki/Pascal%27s_triangle "Calculating an individual row" Consider a row starting as follows: 1, 12... Initialize to [1] and multiply by (row_num/1) to get the next term, row[1]. Then decrement and increment again, getting (11 / 2), (10 / 3), (9 / 4) and so forth, and multiply by the last term so far. Stop when the numerator is 0. 1 * (12/1) = 12 12 * (11 / 2) = 66 66 * (10 / 3) = 220 220 * (9 / 4) = 495 etc. This is another way of computing successive values of C(n, k) without using a factorial function and dividing. Independently discovered by David Koski, implemented in Python by Kirby Urner """ def pascal_row(row_num): numer = row_num denom = 1 # initialize row of Pascal's Triangle row = [1] while numer > 0: row.append((row[-1] * numer/denom)) numer -= 1 # decrement numerator denom += 1 # increment denominator return row def pascal_mod2(row_num = 0): """ row integers mod 2, give a binary string which corresponds to Rule 60 in the Wolfram categorization scheme for cellular automata http://www.research.att.com/~njas/sequences/A001317 """ while True: therow = pascal_row(row_num) binary = "".join(str(i % 2) for i in therow) yield [int(binary,2), binary] row_num += 1 """ traditional generator for successive rows, included for completeness """ def pascal_gen(): row = [1] while True: yield row row = [i + j for i,j in zip(row + [0], [0] + row)]
Pascal's Triangle mod 2 is also a Sierpinski gasket fractal. This is one of the Python examples in Pippy in the Sugar education software. # Sierpinski triangles import sys size = 3 modulus = 2 lines = modulus**size vector = [1] for i in range(1,lines+1): vector.insert(0,0) vector.append(0) for i in range(0,lines): newvector = vector[:] for j in range(0,len(vector)-1): if (newvector[j] == 0): print " ", else: remainder = newvector[j] % modulus if (remainder == 0): print "O", else: print ".", newvector[j] = vector[j-1] + vector[j+1] print vector = newvector[:] On Sat, Jan 30, 2010 at 12:23, kirby urner <kirby.urner@gmail.com> wrote: This process below is how I learned Pascal's Triangle from my mother when I was 11.
-- Edward Mokurai (默雷/धर्ममेघशब्दगर्ज/دھرممیگھشبدگر ج) Cherlin Silent Thunder is my name, and Children are my nation. The Cosmos is my dwelling place, the Truth my destination. http://www.earthtreasury.org/
On Sat, Jan 30, 2010 at 11:21 AM, Edward Cherlin <echerlin@gmail.com> wrote:
Pascal's Triangle mod 2 is also a Sierpinski gasket fractal. This is one of the Python examples in Pippy in the Sugar education software.
Glad to see OLPC is getting the right stuff here. Your technique of backing everything with a rectangular array of 'O's, to be replaced by " ", "." or left alone, is a great way of formatting. You use "print comma" (print ,) to keep going on the line, no newline character before we're done in that loop. The same matting technique could be used for any rule-based row-by-row generating scheme, ala the Wolfram numbering using a bit pattern to encode the next permutation. I've done some work in this area in collaboration with other contributors to this archive (John Zelle, Gregor Lingl). 4dsolutions.net/ocn/python/nks.py http://4dsolutions.net/ocn/python/canvas3.py (note our back ending into PIL, Zelle's graphics.py or Gregor's wrapper for TkCanvas). Your use of a matte also mirrors my recent suggestion regarding building a "color sniffing" turtle that stores a shared "canvas" object in some array, perhaps of just integers, or RGB 3-tuples, if that's the only data we care about (in addition to geographic position). Turtles all read and write to the same matte through a handle (easy in Python as the default is to not copy). I did run it with gusto. Might've been a dot missing lower right?
Great work. Kirby PS: <geometry type="esoteric"> Pascal's is coming up in my research because I'm trying to do some technical writing about one David Koski's studies. He builds hexahedra (zonohedral rhomb-faced) from great circle networks, packs them out to the first spherical polyhedron, is finding some pattern in Pascal's that predicts how many. Example, when you use 10 mid-face axials of the icosahedron as a set of spokes from the origin, illuminate any three not at 180 degrees, you get successive corners of hexahedra, I think he said only 10 possible? You get a rhombic dodecahedron inside the enneacontahedron that way (as one of the zonohedra). You get cubes inside the great rhombicosadodecahedron. Hope I got that right, will double check with DAve. The resulting spherical polyhedra may not have all rhombic faces however as "zero volume" flattened hexahedra, predicted by Pascal's, come together in the form of other polygons. Here are a couple examples: http://www.flickr.com/photos/17157315@N00/4279038891/in/set-7215762279711854... (132 sides) http://www.flickr.com/photos/17157315@N00/4311083958/in/set-7215762279711854... (600 sides) http://mybizmo.blogspot.com/2008/11/enneacontahedron.html (older study) </geometry>
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kirby urner