How to resolve the algorithm Galton box animation step by step in the Wren programming language
How to resolve the algorithm Galton box animation step by step in the Wren 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 Wren programming language
Source code in the wren programming language
import "random" for Random
import "./iterate" for Reversed
var boxW = 41 // Galton box width.
var boxH = 37 // Galton box height.
var pinsBaseW = 19 // Pins triangle base.
var nMaxBalls = 55 // Number of balls.
var centerH = pinsBaseW + (boxW - pinsBaseW * 2 + 1) / 2 - 1
var Rand = Random.new()
class Cell {
static EMPTY { " " }
static BALL { "o" }
static WALL { "|" }
static CORNER { "+" }
static FLOOR { "-" }
static PIN { "." }
}
/* Galton box. Will be printed upside down. */
var Box = List.filled(boxH, null)
for (i in 0...boxH) Box[i] = List.filled(boxW, Cell.EMPTY)
class Ball {
construct new(x, y) {
if (Box[x][y] != Cell.EMPTY) Fiber.abort("The cell at (x, y) is not empty.")
Box[y][x] = Cell.BALL
_x = x
_y = y
}
doStep() {
if (_y <= 0) return // Reached the bottom of the box.
var cell = Box[_y - 1][_x]
if (cell == Cell.EMPTY) {
Box[_y][_x] = Cell.EMPTY
_y = _y - 1
Box[_y][_x] = Cell.BALL
} else if (cell == Cell.PIN) {
Box[_y][_x] = Cell.EMPTY
_y = _y - 1
if (Box[_y][_x - 1] == Cell.EMPTY && Box[_y][_x + 1] == Cell.EMPTY) {
_x = _x + Rand.int(2) * 2 - 1
Box[_y][_x] = Cell.BALL
return
} else if (Box[_y][_x - 1] == Cell.EMPTY){
_x = _x + 1
} else _x = _x - 1
Box[_y][_x] = Cell.BALL
} else {
// It's frozen - it always piles on other balls.
}
}
}
var initializeBox = Fn.new {
// Set ceiling and floor:
Box[0][0] = Cell.CORNER
Box[0][boxW - 1] = Cell.CORNER
for (i in 1...boxW - 1) Box[0][i] = Cell.FLOOR
for (i in 0...boxW) Box[boxH - 1][i] = Box[0][i]
// Set walls:
for (r in 1...boxH - 1) {
Box[r][0] = Cell.WALL
Box[r][boxW - 1] = Cell.WALL
}
// Set pins:
for (nPins in 1..pinsBaseW) {
for (pin in 0...nPins) {
Box[boxH - 2 - nPins][centerH + 1 - nPins + pin * 2] = Cell.PIN
}
}
}
var drawBox = Fn.new() {
for (row in Reversed.new(Box, 1)) {
for (c in row) System.write(c)
System.print()
}
}
initializeBox.call()
var balls = []
for (i in 0...nMaxBalls + boxH) {
System.print("\nStep %(i):")
if (i < nMaxBalls) balls.add(Ball.new(centerH, boxH - 2)) // Add ball.
drawBox.call()
// Next step for the simulation.
// Frozen balls are kept in balls list for simplicity
for (b in balls) b.doStep()
}
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