Uni-color-2D graphics can't be spectaculair, let alone breathtaking. Unless you are in the know. Let me explain. I am now able to - almost - print tilings automatically from just a string of numbers as I wrote about in a previous post.
Not spectaculair at all, it's just a collection of 8 tilings containing 72 polygons in total. But if I change ( 4,2 ) by (10,5) and add a colorfunction the program creates immediately the following graphic.
Both are, in fact, representations of the same sequence.{6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4} but the variations from this sequence alone are endless. The program creates a graphic for every sequence input. Not all graphics will be tilings of the plane by definition. The following ( major ) step is to let the program reject sequences that do not lead to tilings of the plane.
Tne next part will include an implentation of a 'Point in Polygon' predicate. Many of those exist and have been implemented for many languages. That could be a topic for a more in-depth post next time.
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Sunday, December 15, 2013
Sunday, December 8, 2013
Challenging the eternal truth of mathematics
I learned that 0,999... = 1. I believe it was in M381 that I learned to prove it. The proof is quite simple actually.
Let
x = 0,999..., multiply both sides with 10
10x = 9,999..., now subtract x from 10x
9x = 9, and we have our result
x = 1.
What I like about mathematics is that it is timeless, i.e. we can still read math books of hundreds of years old and still learn things from them. That's not an advisable strategy hfor any other science than mathematics. Of course, new branches of mathematics appear, new discoveries are made, but they don't invalidate the truths of the past.
Today however I found someone who actually is challenging some of the established truths in mathematics, like for example that 0,999... = 1. I don't think that he is a crackpot, although I have no doubt that the mathematical establishment, professors who are 'safe' by all means, will call him like that.
The man is arrogant though, he calls the proof above, 'juvenile' for example.
Who is he, what are his ideas and how did he disproof that 1 = 0,999? The links below will help you abshereing these questions.
- The New Calculus - The first rigorous formulation of calculus in history.
- Proof that 0.999 not equal 1.pdf
Let
x = 0,999..., multiply both sides with 10
10x = 9,999..., now subtract x from 10x
9x = 9, and we have our result
x = 1.
What I like about mathematics is that it is timeless, i.e. we can still read math books of hundreds of years old and still learn things from them. That's not an advisable strategy hfor any other science than mathematics. Of course, new branches of mathematics appear, new discoveries are made, but they don't invalidate the truths of the past.
Today however I found someone who actually is challenging some of the established truths in mathematics, like for example that 0,999... = 1. I don't think that he is a crackpot, although I have no doubt that the mathematical establishment, professors who are 'safe' by all means, will call him like that.
The man is arrogant though, he calls the proof above, 'juvenile' for example.
Who is he, what are his ideas and how did he disproof that 1 = 0,999? The links below will help you abshereing these questions.
- The New Calculus - The first rigorous formulation of calculus in history.
- Proof that 0.999 not equal 1.pdf
Some progress...
If you read the previous posts on my tiling printing algorithms you'll understand the reason for this post: I made some progress that unraveled some serious knots in my stomach.
{ 4, 2, 4 } ->
{ 6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4 } ->
The second white polygon is made from the following points :
\begin{array}{cc}
-\frac{\sqrt{3}}{2} & \frac{1}{2} \\
-\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
0 & -1 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\
\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\
\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\
0 & 1+\sqrt{3} \\
-\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\
\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\
\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\
\end{array}
Ready to enter the next level of the problem. ;-)
{ 4, 2, 4 } ->
{ 6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4 } ->
The second white polygon is made from the following points :
\begin{array}{cc}
-\frac{\sqrt{3}}{2} & \frac{1}{2} \\
-\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
0 & -1 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\
\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\
\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\
0 & 1+\sqrt{3} \\
-\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\
\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\
\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\
\end{array}
Ready to enter the next level of the problem. ;-)