Mini exercise: a choice of two

Overview

Teaching: 10 min
Exercises: 50 min

Questions

• What are some real problems I can try out?

Objectives

• Work on one of two larger exercises, putting your new skills to use.

A choice…

We have two exercises you can choose to work on now. Both will allow you to make use of the Fortran that you have learnt today. The former exercise is the solution of a tri-diagonal system of equations, while the latter will allow you program an implementation of Conway’s Game of Life.

Whichever you choose (or if you choose both), we have time this afternoon and tomorrow for you to work on your solutions.

Solve a tri-diagonal system (30 minutes)

This exercise will solve a tri-diagonal system of equations. A linear system of equations can be represented by the problem Ax = b where A is a known matrix, b is a known vector, and x is an unknown vector we wish to solve for. A is tri-diagonal, meaning that it is 0 everywhere except for a stripe of non-zeroes three-elements wide moving down the diagonal.

This means that we can represent an n-by-n tri-diagonal matrix using three rank-1 arrays, one of size n representing the diagonal and two of size n-1 representing the upper and lower diagonals. Through two operations across the arrays, the first moving forwards and the second backwards, as described at Wikipedia, the solution x can be determined.

The accompanying template tri-diagonal.f90 has some further instructions, and some suggestions on how to test the result.

Solution

A solution to the problem appears as a template to the exercise in the later episode on dummy arguments.

Game of Life (60 minutes)

In this exercise you will create a simple implementation of Conway’s Game of Life. The Game of Life is a cellular automata model where an array of cells is updated based on some simple rules which give rise to complex patterns.

This model uses a ‘board’ of cells, each cell having a value of 0 (“dead”) or 1 (“alive”). The board’s initial state is chosen and then stepped forwards. In a new step, each cell calculates the sum of the surrounding eight cells. Depending on the value of that sum, the cell’s state may change or stay the same. The rules governing the update are summarised in the following table:

Cell state	Sum of neighbouring 8 cells	New Cell state
0	0,1,2,4,5,6,7,8	0
0	3	1
1	0,1	0
1	4,5,6,7,8	0
1	2,3	1

A template life.f90 is provided with some hints to get you started.

Reference

Your program should produce the following output for the initial board and the first update:

$ ./a.out
 Initial Board Set-up:
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 ........#.#.#........
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 First iteration:
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 .........#.#.........
 .......##...##.......
 .......###.###.......
 .......##...##.......
 .........#.#.........
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Solution

Two reference answers are given in the solutions directory. The first only computes the first update, while the second includes the time stepping loop.

Key Points

• With a fairly small amount of the Fortran language, you can already solve some real problems.