How to resolve the algorithm Galton box animation step by step in the XPL0 programming language
How to resolve the algorithm Galton box animation step by step in the XPL0 programming language
Table of Contents
Problem Statement
A Galton device Sir Francis Galton's device is also known as a bean machine, a Galton Board, or a quincunx.
In a Galton box, there are a set of pins arranged in a triangular pattern. A number of balls are dropped so that they fall in line with the top pin, deflecting to the left or the right of the pin. The ball continues to fall to the left or right of lower pins before arriving at one of the collection points between and to the sides of the bottom row of pins. Eventually the balls are collected into bins at the bottom (as shown in the image), the ball column heights in the bins approximate a bell curve. Overlaying Pascal's triangle onto the pins shows the number of different paths that can be taken to get to each bin.
Generate an animated simulation of a Galton device.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Galton box animation step by step in the XPL0 programming language
Source code in the xpl0 programming language
include c:\cxpl\codes; \intrinsic code declarations
define Balls = 80; \maximum number of balls
int Bx(Balls), By(Balls), \character cell coordinates of each ball
W, I, J, Peg, Dir;
[W:= Peek($40, $4A); \get screen width in characters
Clear; CrLf(6); CrLf(6);
for Peg:= 1 to 10 do \draw pegs
[for I:= 1 to 12-Peg do ChOut(6,^ ); \space over to first peg
for I:= 1 to Peg do [ChOut(6,^.); ChOut(6,^ )];
CrLf(6);
];
for J:= 0 to 12-1 do \draw slots
[for I:= 0 to 12-1 do [ChOut(6,^:); ChOut(6,^ )];
CrLf(6);
];
for I:= 0 to 23-1 do ChOut(6,^.); \draw bottom
for I:= 0 to Balls-1 do \make source of balls at top
[Bx(I):= 11; By(I):= 1];
Attrib($C); \make balls bright red
repeat \balls away! ...
for I:= 0 to Balls-1 do \for all the balls ...
[Cursor(Bx(I), By(I)); ChOut(6, ^ ); \erase ball's initial position
if Peek($B800, (Bx(I)+(By(I)+1)*W)*2) = ^ \is ball above empty location?
then By(I):= By(I)+1 \yes: fall straight down
else [Dir:= Ran(3)-1; \no: randomly fall right or left
if Peek($B800, (Bx(I)+Dir+(By(I)+1)*W)*2) = ^ then
[Bx(I):= Bx(I)+Dir; By(I):= By(I)+1];
];
Cursor(Bx(I), By(I)); ChOut(6, ^o); \draw ball at its new position
];
Sound(0, 3, 1); \delay about 1/6 second
until KeyHit; \continue until a key is struck
]
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