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Re: Pascal's Triangle
 To: mathgroup at smc.vnet.net
 Subject: [mg37356] Re: Pascal's Triangle
 From: "Borut L" <gollum at email.si>
 Date: Fri, 25 Oct 2002 02:46:40 0400 (EDT)
 References: <ap86hf$554$1@smc.vnet.net>
 Sender: ownerwrimathgroup at wolfram.com
Helo,
I should thank you for this pleasent mind excercise, though not tough, it
was just basic geometry.
$TextStyle={FontFamily>Verdana,FontSize>10};
(* first create a basic object, a colored triangle with text inscribed, note
that the height of the triangle is a unit *)
Triangle[j_,i_]:=
Module[
{
x0=(i.5 j)2/Sqrt[3],y0=j,a=2/Sqrt[3]
},
{
If[EvenQ[Binomial[j,i]],Hue[0,.5,1],Hue[.55,.5,1]],
Polygon[{{x0,y0},{x0.5 a,y01},{x0+.5 a,y01}}],
GrayLevel[0],Text[ToString[Binomial[j,i]],{x0,y0.5},{0,1}]
}
]
(* the following command will show the agglomerate, note that it doesn't
draw each of the triangles having no text, instead it draws a big triagle as
a background and puts the numbered ones onto it *)
ShowBigTriangle[jmax_]:=
Show[
Graphics[
Prepend[
Table[Triangle[j,i],{j,0,jmax},{i,0,j}],
{Hue[0,.5,1],
Polygon[{{0,0},{(jmax+1)/Sqrt[3],(jmax+1)},{(jmax+1)/Sqrt[3],(jmax+1)}}]}
]
]
,AspectRatio>Automatic
]
ShowBigTriangle[33]//Timing
(1.3 sec on my P2350, how's your timming?)
p.s. : According to S. Wolfram and his new kind of science, the solution to
your problem is just a simple program. He ilustrates this in 2nd Chapter. It
may be worthy to take a peak.
Bye,
Borut
"Al Mannon" <almannon at attbi.com> wrote in message
news:ap86hf$554$1 at smc.vnet.net...
 I want to write a program that will create an array of equilateral
triangles
 such that in the first row there is 1 triangle, in the second row there is
3
 triangles, in the third row there will be 5 triangles...in the nth row
there
 will be 2n  1 triangles. Putting all of these triangles together and
 calling the first row, row 0, we would have a large triangle with n + 1
rows
 along the side of the large triangle and 2n  1 columns along the base of
 the triangle.

 This would be phase one of the project.

 Phase 2: Fill in the binomial coefficients into the triangle.
 Phase 3: Color all odd numbered triangles blue.
 Phase 4: Color all even numbered triangles red.
 Phase 5: Any triangle that shares an edge with a red triangle, color red.

 The result is Zierpinski's Triangle! I have done this by hand for a
triangle
 with 16 rows. Needless to say the work was tedious. The result, however,
is
 quite satisfying and remarkable. I would like to be able to use this as a
 tool to teach some of the other derivations that are possible from
Pascal's
 triangle other than binomial coefficients and combinations. Therefore it
 would be beneficial to be able to reproduce this work at will.

 Since my Mathematica programming skills are practically nil, any help
would
 be appreciated.

 I can be reached directly by electronic mail by deleting "the" in the
 following:

 althemannon at attbi.com


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