How to resolve the algorithm Wireworld step by step in the C programming language

Published on 7 June 2024 03:52 AM
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How to resolve the algorithm Wireworld step by step in the C programming language

Table of Contents

Problem Statement

Wireworld is a cellular automaton with some similarities to Conway's Game of Life. It is capable of doing sophisticated computations with appropriate programs (it is actually Turing complete), and is much simpler to program for. A Wireworld arena consists of a Cartesian grid of cells, each of which can be in one of four states. All cell transitions happen simultaneously. The cell transition rules are this:

Create a program that reads a Wireworld program from a file and displays an animation of the processing. Here is a sample description file (using "H" for an electron head, "t" for a tail, "." for a conductor and a space for empty) you may wish to test with, which demonstrates two cycle-3 generators and an inhibit gate: While text-only implementations of this task are possible, mapping cells to pixels is advisable if you wish to be able to display large designs. The logic is not significantly more complex.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Wireworld step by step in the C programming language

This C code simulates the Game of Life, a cellular automaton popularized by John Conway in 1970, on a 14x7 grid and displays the result in an infinite loop. Let's break down the code step by step:

  1. Game of Life Simulation:

    • world_7x14[2][512]: A 2D array of characters representing the current and next game states.

    • next_world(const char *in, char *out, int w, int h): This function calculates the next generation of the Game of Life based on the current state (in) and stores it in the output buffer (out). It iterates through each cell and applies the following rules:

      • A space stays a space.
      • A 't' becomes a '.'.
      • An 'H' becomes a 't'.
      • A '.' becomes an 'H' if it has exactly 1 or 2 neighboring 'H' cells.
  2. Main Game Loop:

    • The main function enters an infinite loop, alternating between the two segments of the world_7x14 array.

    • In each iteration of the loop:

      • It prints the current world state from world_7x14[f].
      • It calculates the next world state using next_world.
      • For visualization purposes, it uses VT100 escape sequences (if ANIMATE_VT100_POSIX is defined) to move the cursor to the top left corner and refresh the display.
      • It pauses for 100 milliseconds using nanosleep to create an animation effect.
  3. VT100 Animation (Optional):

    • If ANIMATE_VT100_POSIX is defined, the code uses VT100 escape sequences to position the cursor and refresh the display, creating an animation effect.
    • The escape sequences used are:
      • \x1b[%dA: Move the cursor up n lines.
      • \x1b[%dD: Move the cursor right n columns.

Source code in the c programming language

/* 2009-09-27 <kaz@kylheku.com> */
#define ANIMATE_VT100_POSIX
#include <stdio.h>
#include <string.h>
#ifdef ANIMATE_VT100_POSIX
#include <time.h>
#endif

char world_7x14[2][512] = {
  {
    "+-----------+\n"
    "|tH.........|\n"
    "|.   .      |\n"
    "|   ...     |\n"
    "|.   .      |\n"
    "|Ht.. ......|\n"
    "+-----------+\n"
  }
};

void next_world(const char *in, char *out, int w, int h)
{
  int i;

  for (i = 0; i < w*h; i++) {
    switch (in[i]) {
    case ' ': out[i] = ' '; break;
    case 't': out[i] = '.'; break;
    case 'H': out[i] = 't'; break;
    case '.': {
      int hc = (in[i-w-1] == 'H') + (in[i-w] == 'H') + (in[i-w+1] == 'H') +
               (in[i-1]   == 'H')                    + (in[i+1]   == 'H') +
               (in[i+w-1] == 'H') + (in[i+w] == 'H') + (in[i+w+1] == 'H');
      out[i] = (hc == 1 || hc == 2) ? 'H' : '.';
      break;
    }
    default:
      out[i] = in[i];
    }
  }
  out[i] = in[i];
}

int main()
{
  int f;

  for (f = 0; ; f = 1 - f) {
    puts(world_7x14[f]);
    next_world(world_7x14[f], world_7x14[1-f], 14, 7);
#ifdef ANIMATE_VT100_POSIX
    printf("\x1b[%dA", 8);
    printf("\x1b[%dD", 14);
    {
      static const struct timespec ts = { 0, 100000000 };
      nanosleep(&ts, 0);
    }
#endif
  }

  return 0;
}


  

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