Create a program to display some approximation of a bell curve. You are free to use whatever method you wish for generating the data, but the data must be displayed vertically, as in oriented like so:
Other restrictions:
This is a popularity-contest, so be creative.
The method of generation is based off of the game Sugar, Sugar. In that game, sugar particles fall, traveling randomly left and right. In my program, Dust objects fall until they hit the ground. I collect data based on where they land; this is the bell curve:
import java.awt.image.BufferedImage; import java.io.File; import java.io.IOException; import java.util.Random; import java.util.Stack; import javax.imageio.ImageIO; /** * * @author Quincunx */ public class ApproxBellCurve { public static final int NUM_DUST = 10_000_000; public static final int NUM_THREADS = Runtime.getRuntime().availableProcessors() - 1; public static final int LOOP_LIMIT = NUM_DUST / (NUM_THREADS + 1); public static void main(String[] args) { BufferedImage output = new BufferedImage(2049, 2049, BufferedImage.TYPE_INT_RGB); Stack<Thread> threads = new Stack(); for (int i = 0; i < NUM_THREADS; i++) { threads.push(new Thread(new Duster())); threads.peek().start(); } Dust d; Random r = new Random(); for (int i = 0; i < LOOP_LIMIT + NUM_DUST - (LOOP_LIMIT * (NUM_THREADS + 1)); i++) { d = new Dust(r); while (!d.end) { d.step(); } if ((i & 1024) == 0) { r = new Random(); } } while (threads.size() > 0) { try { threads.pop().join(); } catch (InterruptedException ex) { } } int maxy = 0; for (int x = 0; x < Dust.data.length; x++) { maxy = Dust.data[x] > maxy ? Dust.data[x] : maxy; } for (int x = 0; x < Dust.data.length; x++) { for (int y = 0; y < output.getHeight(); y++) { output.setRGB(x, y, (y >= output.getHeight() - (Dust.data[x] * output.getHeight() / maxy)) ? 6591981 : 16777215); } } try { ImageIO.write(output, "png", new File("Filepath")); } catch (IOException ex) { ex.printStackTrace(); } } private static class Duster implements Runnable { @Override public void run() { Dust d; Random r = new Random(); for (int i = 0; i < LOOP_LIMIT; i++) { d = new Dust(r); while (!d.end) { d.step(); } if ((i & 1024) == 0) { r = new Random(); } } } } private static class Dust { public static int[] data = new int[2049]; static { for (int i = 0; i < data.length; i++) { data[i] = 0; } } private int[] location; private Random r; public boolean end; public Dust(Random rand) { location = new int[]{1024, 0}; r = rand; end = false; } public void step() { if (end) { return; } if (location[1] >= 1024) { end = true; synchronized (data) { data[location[0]] += 1; } return; } location[1] += 1; location[0] += r.nextInt(21) - 10; } } } Sample output (took 1 minute to create):
Because of the nature of the random number generation, it is possible to get an ArrayIndexOutOfBoundsException.
I used multithreading to speed up the process. The program determines the number of processors available and creates that many threads to run the simulations.
To obtain a finer image, NUM_DUST can be increased, but that would also lead to an increased risk of an ArrayIndexOutOfBoundsException.
Each thread creates a new Random after every 1024 Dust objects are simulated. When that code was removed, the image was more coarse.
rand.nextInt(21) - 10 is to widen the distribution. A change to rand.nextInt(3) - 1 would remove all chance of an ArrayIndexOutOfBoundsException.
I decided it should not only look like a bell curve, it should look like a bell. So hey, let's spin it around the Z-axis and make it 3D.
And hey, while we're at it, it should sound like a bell as well.
So, each time you evaluate the code, it makes a nice pleasing church-bell BWONNGGG!
Manipulate[ RevolutionPlot3D[PDF[NormalDistribution[mu, sigma], x], {x, -4, 4}, PlotRange -> {Automatic, Automatic, Automatic}, BoxRatios -> {1, 1, 1}, PlotStyle -> {Orange, Yellow}, ImageSize -> {500, 500}], {{mu, 0, "mean"}, -3, 3, Appearance -> "Labeled"}, {{sigma, 1.47, "standard deviation"}, .5, 2, Appearance -> "Labeled"}] EmitSound[Sound[SoundNote[0, 3, "TubularBells", SoundVolume -> 1]]] Here's what it looks like: 
And here's a brief video of the bell ringing.
Here is a practical application of the Central Limit Theorem. I even golfed it!
Type this into the input box on Wolfram Alpha:
3d20 Here is a link to the results. It interprets this input as 3 icosahedral dice, and decides you want a graph of the distribution function of the pip sums. It prints the following graph:
We can make the graph more and more bell shaped by adding more dice. I don't see this as taking my data from the internet since, if I could run the query engine for it on my home computer I would. Unfortunately, Wolfram Research doesn't provide it for download, though.
Everyone's code is so ... huge, and their approximations are ... less than deterministic. Just to remind everyone of the simplicity of the actual task, here's a nine-liner in Haskell:
import Data.List; import Data.Ratio main = do putStrLn "please choose the width and the maximum height:" [w, mh] <- (map read.words) `fmap` getLine let dist = map length $ group $ sort $ map length $ subsequences [2..w] scale = ceiling $ maximum dist % mh maxD = scale * ceiling (maximum dist % scale) putStrLn $ unlines $ map (row dist) [maxD, maxD - scale .. 0] where row dist y = map (\x -> if x >= y then '*' else ' ') dist OK... that wasn't exactly fast. Here's another go. Does it still count as not hard-coded?
import Data.List; import Data.Ratio main = do putStrLn "please choose the width and the maximum height:" [w, mh] <- (map read.words) `fmap` getLine let dist = map ((w-1) `choose`) [0..(w-1)] scale = ceiling $ maximum dist % mh maxD = scale * ceiling (maximum dist % scale) putStrLn $ unlines $ map (row dist) [maxD, maxD - scale .. 0] where row dist y = map (\x -> if x >= y then '*' else ' ') dist factorial x = product [1..x] n `choose` k = factorial n `div` factorial k `div` factorial (n-k) 
from pygame import* t,u=640,400;init();d=display;s=d.set_mode((t,u)) s.fill(0xFFFFFF) for x in range(t): d.flip();event.get() draw.line(s,u,(x,int(u-2**(-((x-t*.5)/120)**2)*u)),(x,u)) while time.wait(50): for e in event.get(): if e.type==QUIT:exit() This is actually a normal distribution, using 2 as the base of the exponent instead of e, and not dividing by ⎷2π, because without axes, it doesn't really matter if the area under the curve isn't exactly 1.
The last 3 lines just keep the window open until the user closes it manually.
load 'plot' plot (3+i.58)([:+/=)"0 1]3+([:+/?,?,?)"0]1e6#20 Not very creative, but still fun!
Throws three twenty sided dice 1000000 times, summing the result each time. Then to get it in a plot-ready format, we take the array [3..60] and replace each element by the number of times it occures in the random sums.
plot then creates a PDF:
Simulates the classic example of bell curves, dropping a ball down a grid full of pole thingies.
a = [0]*30 1000.times do x = 15 49.times do x += (rand(3) - 1) end a[[[x.to_i, 29].min, 0].max] += 1 end until a.all?{|x| x == 0} a.each_with_index{|x,i| if x == 0 print ' ' else print 'x' a[i] -= 1 end } puts end Sample output:
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxx x xxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxxx xxxxxxxxxxxxxx xxxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxx xx xxxxxxx xx xxxxxxx xx xxxxxx xx xxxxxx xx xxxxxx x xxxxx x xxxxx x xxx x xxx x xxx x xxx x xxx x xxx xxx xxx xxx xxx xxx xxx xxx xxx x x x x x x x x x x x Oh, did I forget to mention it's upside down?
Generating the sums of 5 random integers, each between 0 and 100, 20000 times.
Histogram[Plus @@@ RandomInteger[100, {20000, 5}]] 
Here's the result for sums of 1,2, 3 numbers. As the number of integers to be summed increases, the curve approaches the bell shape.
When single random integers are generated, but not summed, the distribution should be fairly uniform, not favoring any value. The apparent anomaly for 100 may be the result of how Matheamtica bins data. (see https://mathematica.stackexchange.com/questions/43300/does-randominteger0-n-disfavor-n/43302#43302 )
Table[Histogram[Plus @@@ RandomInteger[100, {20000, k}]], {k, 1, 3}] 
Rough approximation with a hand-crafted Bézier curve. To make it more interesting: No use of Line, Path or similar, only Ellipse.
public MainWindow() { InitializeComponent(); // Requires a Canvas named Canvas in XAML file DrawBellCurve(Canvas, 300, 300); } public static void DrawBellCurve(Canvas canvas, int width, int height) { Point p1 = new Point(0, height); Point p2 = new Point(0.3 * width, 1 * height); Point p3 = new Point(0.4 * width, 0.1 * height); Point peak = new Point(width / 2, 0); Point p3r = new Point(width - p3.X, p3.Y); Point p2r = new Point(width - p2.X, p2.Y); Point p1r = new Point(width, height); List<Point> controlPoints = new List<Point> { p1, p2, p3, peak, p3r, p2r, p1r }; DrawBezierCurve(canvas, controlPoints, Color.FromRgb(50, 200, 25)); DrawPoints(canvas, controlPoints, 5, Color.FromRgb(80, 80, 220)); } public static void DrawPoints(Canvas canvas, List<Point> points, double radius, Color color) { foreach (var p in points) { Ellipse ellipse = new Ellipse(); ellipse.Width = radius; ellipse.Height = radius; ellipse.SetValue(Canvas.LeftProperty, p.X - radius / 2); ellipse.SetValue(Canvas.TopProperty, p.Y - radius / 2); ellipse.Fill = new SolidColorBrush(color); canvas.Children.Add(ellipse); } } public static void DrawBezierCurve(Canvas canvas, List<Point> controlPoints, Color color) { var curvePoints = new List<Point>(); double tStep = 0.01; for (double t = 0; t <= 1; t += tStep) { curvePoints.Add(GetBezierPoint(controlPoints, t)); } DrawPoints(canvas, curvePoints, 3, color); } public static Point GetBezierPoint(List<Point> controlPoints, double t) { // De Casteljau's algorithm if (controlPoints.Count == 1) { return controlPoints[0]; } var newControlPoints = new List<Point>(); for (int i = 0; i < controlPoints.Count - 1; i++) { Point p1 = controlPoints[i]; Point p2 = controlPoints[i + 1]; Point p = new Point( t * p1.X + (1 - t) * p2.X, t * p1.Y + (1 - t) * p2.Y); newControlPoints.Add(p); } return GetBezierPoint(newControlPoints, t); } 
c=79; x=ifft(fft([ones(1,10) zeros(1,c-10)]).^8); y=(1:c)*0+' '; for l=-24:-1, for k=1:c, if -real(x(k))<l*2e5, y(k)='*'; end end printf([char(y) '\n']) end Instead of directly simulating coin flips, this just uses the observation that the pdf of a sum of random variables is the convolution of the pdfs of each of the random variables. An efficient way to do circular convolution is to take the inverse fourier transform of the element-wise product of the fourier transforms of the pdfs of each of the random variables. For the purpose of this problem (not true in general) circular convolution can stand in for linear convolution.
As an example, given the following pdf of a uniform random variable.
********** ********** ********** ********** ********** ********** ********** ********** ********** ********** ********** ********** The above random variable summed with itself
* *** *** ***** ******* ********* *********** *********** ************* *************** ***************** ******************* The uniform random variable summed with itself 7 times.
* ******* *********** ************* *************** ***************** ******************* *********************** ************************* *************************** ******************************* ************************************* It gets the height and the width from the user, chooses width random booleans and count how many are true, registering the counted frequency. Repeat this until some of the generated frequencies is enough to fill the height:
import java.awt.Dimension; import java.awt.EventQueue; import java.awt.Graphics; import java.awt.image.BufferedImage; import java.util.Random; import javax.swing.JComponent; import javax.swing.JFrame; import javax.swing.JOptionPane; public class Bell { public static int[] bell(int width, int height) { // I could have used Random.nextGaussian() method, but this way is more fun. Random r = new Random(); int max = 0; int[] freqs = new int[width]; for (int j = 0; max < height - 1; j++) { int count = 0; for (int i = 0; i < width; i++) { if (r.nextBoolean()) count++; } freqs[count]++; if (freqs[count] > max) max++; } return freqs; } public static BufferedImage drawBell(int[] ps, int width, int height) { BufferedImage bi = new BufferedImage(width, height, BufferedImage.TYPE_INT_RGB); for (int i = 0; i < width; i++) { for (int j = 0; j < ps[i]; j++) { bi.setRGB(i, height - j - 1, 0xFF00FFFF); // Cyan } for (int m = i - 1; m <= i + 1; m++) { if (m < 0 || m >= width) continue; for (int n = ps[i] - 1; n <= ps[i] + 1; n++) { if (n < 0 || n >= height) continue; bi.setRGB(m, height - n - 1, 0xFF0000FF); // Blue } } for (int j = ps[i] + 1; j < height; j++) { bi.setRGB(i, height - j - 1, 0xFFFFFFFF); // White } } return bi; } public static void main(String[] args) { String a = JOptionPane.showInputDialog("Give the width:"); String b = JOptionPane.showInputDialog("Give the height:"); int w, h; try { w = Integer.parseInt(a); } catch (NumberFormatException e) { JOptionPane.showMessageDialog(null, "Invalid width."); return; } try { h = Integer.parseInt(b); } catch (NumberFormatException e) { JOptionPane.showMessageDialog(null, "Invalid height."); return; } int[] freqs = bell(w, h); BufferedImage image = drawBell(freqs, w, h); EventQueue.invokeLater(() -> { JFrame j = new JFrame("Bell curve"); j.setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE); j.add(new SimpleSprite(image)); j.setResizable(false); j.pack(); j.setVisible(true); }); } public static class SimpleSprite extends JComponent { private final BufferedImage image; private final Dimension dim; public SimpleSprite(BufferedImage image) { this.image = image; this.dim = new Dimension(image.getWidth(), image.getHeight()); } @Override public Dimension getMinimumSize() { return dim; } @Override public Dimension getMaximumSize() { return dim; } @Override public Dimension getPreferredSize() { return dim; } @Override public void paintComponent(Graphics g) { g.drawImage(image, 0, 0, null); } } } Screenshot:
I don't know whether it is valid to use an application that is made to draw graphs, but if you enter this in the input, you get a Bell Curve:
f(x) = 3ℯ^(-(x - 1)²) Or, if you want a customizable bell curve:
a = 0 b = 0 c = 0 d = 0 f(x) = a ℯ^((-(x - b)²) / 2 c²) + d Then, make sure that you enable the sliders for the variables (if they are disabled, click on the circle next to the variable names). Then, you can set a value for them by sliding.
Image of the curve with the sliders (click to enlarge):
Throwing 6 dice in every possible combination approaches a 'normal distribution', which happens to be a bell curve.
public class Temp { public static void main(String[] args) throws Exception { int[] hist = new int[31]; for (int d1 = 0; d1 < 6; d1++) for (int d2 = 0; d2 < 6; d2++) for (int d3 = 0; d3 < 6; d3++) for (int d4 = 0; d4 < 6; d4++) for (int d5 = 0; d5 < 6; d5++) for (int d6 = 0; d6 < 6; d6++) hist[d1 + d2 + d3 + d4 + d5 + d6]++; for (int y = 4104; y >= 0; y -= 228) for (int x = 0; x < 32; x++) System.out.print(x == 31? '\n' : hist[x] <= y? '\u3000' : (char)('\u2580' + Math.min(Math.ceil((hist[x] - y) / 28.5), 8))); } } Some Unicode tricks make the output smoother:
▅█▅ ▂███▂ █████ ▁█████▁ ███████ ███████ ▅███████▅ █████████ █████████ ▇█████████▇ ███████████ ▃███████████▃ █████████████ ▁█████████████▁ ███████████████ ▃███████████████▃ █████████████████ ▁███████████████████▁ ▁▁▁▂▅█████████████████████▅▂▁▁▁ 
I liked Doorknob's generation method and Victor's output (his generation is actually equivalent), so I decided to combine the two with tkinter!
from tkinter import * import random bins = [0]*201 for i in range(20000): x = 0 for j in range(100): if random.randint(0, 1): x += 0.5 else: x -= 0.5 bins[int(x)+100] += 1 master = Tk() w = Canvas(master, width = 201, height=200) w.pack() for i in range(201): w.create_line(i, 200, i, 200-bins[i]/10, fill="light blue") w.create_line(i-1, 200-bins[i-1]/10, i, 200-bins[i]/10, fill="blue") 
Based on random walks
#include <stdio.h> #include <stdlib.h> #define X_SIZE 40 #define Y_SIZE 15 #define SMOOTH 100 int main(void) { int i,j; int T[X_SIZE*2]; int w; for (i=0; i<X_SIZE*2; i++) T[i]=0; // random walks do { w=X_SIZE; for (i=0;i<X_SIZE;i++) w+=rand()%2?1:-1; T[w]++; } while (T[w]<Y_SIZE*SMOOTH); // display for (j=Y_SIZE; j>=0; j--) { for (i=0; i<X_SIZE*2; i+=2) printf("%c", (T[i]>j*SMOOTH)?'#':' '); printf("\n"); } } Output :
### ### ##### ##### ##### ####### ####### ######### ######### ######### ########### ########### ############# ############### ####################### 
≔⟦Ｘφ±⁹⟧θＦ⁹⁹≔Ｅ⊞Ｏθ⁰⁺κ§θ⊖λθ↑θ Verbose version:
Assign([Power(1000, Negate(9))], q); for (99) Assign(Map(PushOperator(q, 0), Plus(k, AtIndex(q, Decremented(l)))), q); Print(:Up, q); Try it online! Output:
|| || |||| |||| |||| |||| |||||| |||||| |||||| |||||| |||||| |||||||| |||||||| |||||||| |||||||| |||||||| |||||||| |||||||||| |||||||||| |||||||||| |||||||||| |||||||||| |||||||||| |||||||||||| |||||||||||| |||||||||||| |||||||||||| |||||||||||| |||||||||||| |||||||||||||| |||||||||||||| |||||||||||||| |||||||||||||| |||||||||||||| |||||||||||||||| |||||||||||||||| |||||||||||||||| |||||||||||||||| |||||||||||||||| |||||||||||||||||| |||||||||||||||||| |||||||||||||||||| |||||||||||||||||||| |||||||||||||||||||| |||||||||||||||||||| |||||||||||||||||||||| |||||||||||||||||||||| |||||||||||||||||||||||| |||||||||||||||||||||||||| |||||||||||||||||||||||||||| 
spark for the shell:r=RANDOM%6;(for i ({0..99999})echo $[r+r+r])|sort -n|uniq -c|spark ▁▁▁▁▂▁▃▁▄▁▆▁▇▁▇▁█▁▇▁▆▁▄▁▃▁▂▁▁▁▁▁ Unfortunately Zsh $RANDOM is deterministic/pseudo-random so perhaps I should use jot -r. The following example uses rand48().
plotutils/graph:#!/bin/zsh zmodload zsh/mathfunc bm()printf %.1f\\n $[sqrt(-2*log(rand48()))*$1(8*atan(1)*rand48())] (repeat 99999 bm cos&&bm sin)|sed 's/-0.0/0.0/'|sort -n\ |uniq -c|graph -t -g0 -Tpng > $$.png # -t transpose axes # -g0 no labels Uses Box-Muller transform to convert rand48() values to a normal distribution. 
This short code uses the Box-Muller transform for simulating a normally distributed variable:
from math import cos, log, pi from random import random f = open("normDist.txt",'w') for i in xrange(10**6): f.write(str(cos(2*pi*random())*(-2*log(random()))**.5)+'\n') f.close() You can plug the values on GNUPLOT or use this script
f=open("normDist.txt",'r') v = [0 for i in xrange(51)] for i in f: v[int(float(i)*10) if float(i)<5 and float(i)>-5 else 0]+=1 m=0 for i in v: if i>m: m=i for i in v[3:]: for j in xrange(int(70.*i/m)): print '*', print to produce something like this:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 
#!/ust/bin/awk -f BEGIN { # # do WALKS random walks with OOPS times left or right per walk # WALKS=100000 OOPS=39 # # output will be H lines # the line width will be 2*OOPS+1 # H=23 # # shake awk's pseudorandom generator # srand() # # do the walks # for(i=0;i<WALKS;i++) { x=0 for(j=0;j<OOPS;j++) rand()<0.5?--x:++x X[x]++ } # # scan for maximum height # M=0 for(x=-OOPS;x<=OOPS;x++) { if(X[x]>M) M=X[x] } # # "draw" into S[x]: strings of H*X[x]/M "*"s # for(x=-OOPS;x<=OOPS;x++) { s=sprintf("%*s",H*X[x]/M," ") gsub(/./,"*",s) S[x]=s } # # print S[x] transposed # for(y=H;y>0;y--) { for(x=-OOPS;x<=OOPS;x++) { if(c=substr(S[x],y,1)) printf c else printf " " } print "" } } Output:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ******************************************************************************* ⌐∞¶J♠fr"▐cS╥aÆ~a╚∩┌I This isn't code-golf but I still want to give Stax a try. Calculates frequency of sum of dices rolled 6 times. Note there are no support for random numbers in Stax.
ASCII equivalent:
6R6|^{|+m{o:G{AJ/'**mMrm Output:
* *** *** *** ***** ***** ***** ***** ***** ******* ******* ******* ******* ******* ******* ********* ********* ********* ********* ********* ********* *********** *********** *********** *********** *********** *********** ************* ************* ************* ************* ************* *************** *************** *************** *************** ***************** ***************** ***************** ******************* ******************* ********************* *********************** 
getInternetBellCurve+0? What does "hard-coding" imply? This rule is currently highly opinion based. \$\endgroup\$