How to resolve the algorithm Conway's Game of Life step by step in the E programming language

Published on 12 May 2024 09:40 PM
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How to resolve the algorithm Conway's Game of Life step by step in the E programming language

Table of Contents

Problem Statement

The Game of Life is a   cellular automaton   devised by the British mathematician   John Horton Conway   in 1970.   It is the best-known example of a cellular automaton. Conway's game of life is described   here: A cell   C   is represented by a   1   when alive,   or   0   when dead,   in an   m-by-m   (or m×m)   square array of cells. We calculate   N   - the sum of live cells in C's   eight-location neighbourhood,   then cell   C   is alive or dead in the next generation based on the following table: Assume cells beyond the boundary are always dead. The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players.   One interacts with the Game of Life by creating an initial configuration and observing how it evolves.

Although you should test your implementation on more complex examples such as the   glider   in a larger universe,   show the action of the blinker   (three adjoining cells in a row all alive),   over three generations, in a 3 by 3 grid.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Conway's Game of Life step by step in the E programming language

Source code in the e programming language

def gridWidth := 3
def gridHeight := 3
def X := 0..!gridWidth
def Y := 0..!gridHeight

def makeFlexList := 
def makeGrid() {
  def storage := makeFlexList.fromType(, gridWidth * gridHeight)
  storage.setSize(gridWidth * gridHeight)

  def grid {
    to __printOn(out) {
      for y in Y {
        out.print("[")
        for x in X {
          out.print(grid[x, y].pick("#", " "))
        }
        out.println("]")
      }
    }
    to get(xb :int, yb :int) {
      return if (xb =~ x :X && yb =~ y :Y) {
        storage[y * gridWidth + x]
      } else {
        false
      }
    }
    to put(x :X, y :Y, c :boolean) { 
      storage[y * gridWidth + x] := c
    }
  }
  return grid
}

def mooreNeighborhood := [[-1,-1],[0,-1],[1,-1],[-1,0],[1,0],[-1,1],[0,1],[1,1]]
def computeNextLife(prevGrid, nextGrid) {
  for y in Y {
    for x in X {
      var neighbors := 0
      for [nx, ny] ? (prevGrid[x+nx, y+ny]) in mooreNeighborhood {
        neighbors += 1
      }
      def self := prevGrid[x, y]
      nextGrid[x, y] := (self && neighbors == 2 || neighbors == 3)
    }
  }
}

var currentFrame := makeGrid()
var nextFrame := makeGrid()
currentFrame[1, 0] := true
currentFrame[1, 1] := true
currentFrame[1, 2] := true

for _ in 1..3 {
  def frame := nextFrame
  computeNextLife(currentFrame, frame)
  nextFrame := currentFrame
  currentFrame := frame
  println(currentFrame)
}

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