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Computational Physics
Week 5 course work

Comments and questions to John Rowe.

This forms the first part of a two-week project.

The game of life

The game consists of a rectangular array of ones or zeros. A square is said to be alive if it has value one or dead if it has value zero. Each cell has eight neighbours, ie the cells that touch it including diagonally.

Initially the grid is populated at random. The game then iterates like this: for each iteration if a square is initially alive and it has exactly two or three live neighbours it stays alive. Otherwise it dies. On the other hand, if the cell is initially dead and it has exactly three occupied neighbours it becomes alive. Otherwise it stays dead.

Thus you will have a loop a bit like this:

  for (;;) {
      for (y = 0; y < NY; ++y)
          for (x = 0; x < NX; ++x)
              [Update the point (x,y)
      [Now draw the new grid out to the screem]
  }

Note how whilst logically all the cells are meant to be updated simultaneaously, in practice we have to do one at a time with a double for loop.

There are a few wrinkles we need to work out here:

How about the edges? We get round this by saying that the line x=0 is next to the line x=NX-1 (which makes it cylindrical). Similarly we say that y=0 is next to y=NY-1 (which makes it a torus or donut shape).
An extra complication is that we need not one but two arrays: one for the old values and one for the new values which we are updating. Thus our loop looks a bit like:
```
  for(;;) {
      [Update all the points in the new grid as above]
      [Draw the new grid on the screen]
      [Copy the new grid to the old one ready for next time]
    }
      
```
(This loop is a little inefficient as we shall see next week.)

The modulus operator

The % operator takes two integer arguments and returns the remainder of the second when divided by the first (eg 7 % 3 is 1, 13 % 8 is 5, 21 % 7 is zero). Thus i % j returns an integer in the range 0 to j-1. It can occasionally misbehave when i is negative so use (i+j-1) % j instead of (i-1) % j

Data structure

Assuming NX NY and NY are constants defined either at the top of a file or in a header file, what will we need in a structure to tell us everything we need to know about a game of life? If instead they are variables whose value is read in when the program is run, how would that change things? (You may assume whichever you prefer.)

Function structure

What are the functions we will need?

A function to allocate a life structure.
A function to initialise the grid at random.
A function to calculate the new grid from the old one.
A function to draw the new grid.
A function to copy the new grid to the old one.

Looking at these, 3 looks a bit complicated. What do we do when we need to write a complicated function? Answer: we split it up into two or more sub-functions! What are the steps involved in calculating the new grid? Well, for each cell we first calculate the number of live neighbours and then we use that number to calculate whether the cell is alive or dead in the next generation.

So now our structure looks like:

A function to allocate a life structure.
A function to initialise the grid at random.
A function to calculate the new grid from the old one:
1. Calculate the number of neighbours of a particular cell.
2. Use that to work out whether the cell lives or dies
A function to draw the new grid.
A function to copy the new grid to the old one.

3a should certainly be a separate function, it's up to you whether 3b is a separate function or just part of 3.

Writing the program

Do not write the program all at once. Think of a few functions you will need and write them individually and test each one at a time. Include a separate printout for each function and its test program. Try it at first with a very small grid and print it out as ones and zeros. Check that your update functions work properly and prove to me that you have checked this.

Play your game a couple of times. Does it always do the same thing? If so, try using srand. By the end of this week's work you should have been able to initialise a grid of cells and be able to print them out to the screen as ones and zeros. Ideally you will have debugged and tested your update routine.

There is no need to hand in this week's work as it will form the first half of next week's.

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