Graph sim (#29)

Jonathan Frech, 17 May 2015

Thinking of graphs — especially bar graphs — as a pretty piece of art I wanted to create something related. But due to me not having any data I could graph, I decided to just use randomly generated data. It has no real purpose rather than look good.

Some calculations

All of these calculations are based on random.randint() being perfectly random. Due to it being pseudo-random this will just be an approximation. To get further information, click on the equation to view them in Wolfram Alpha.
The chance of two adjacent bars having the same color:
$\Big(\big(\frac{1}{256}\big)^3\Big)^{2-1}=5.96\cdot 10^{-8}=\text{basically }0$ $\Big(\big(\frac{1}{256}\big)^3\Big)^{2-1}=5.96\cdot 10^{-8}=\text{basically }0$
The chance of two adjacent bars having the same color and size, where
$\text{max. size}=\text{window height}=720$ $\text{max. size}=\text{window height}=720$ :
$\Big(\big(\frac{1}{256}\big)^3\cdot\frac{1}{720}\Big)^{2-1}=8.28\cdot 10^{-11}=\text{basically }0$ $\Big(\big(\frac{1}{256}\big)^3\cdot\frac{1}{720}\Big)^{2-1}=8.28\cdot 10^{-11}=\text{basically }0$
The chance of the screen not showing anything, which means every bar has the background color and
$\text{max. quantity of bars on screen}=\frac{\text{window height}}{\text{bar width}}=\frac{720}{4}=180$ $\text{max. quantity of bars on screen}=\frac{\text{window height}}{\text{bar width}}=\frac{720}{4}=180$ :
$\Big(\big(\frac{1}{256}\big)^3\Big)^{180}=3.55\cdot 10^{-1301}=\text{basically }0.0$ $\Big(\big(\frac{1}{256}\big)^3\Big)^{180}=3.55\cdot 10^{-1301}=\text{basically }0.0$
Although these probabilities are very unlikely, they — theoretically — could occur.

Source code: graph-sim.py