# A Fractal Height Map Generator in Ruby

[I'm migrating my old articles to the blog in order to switch to it entirely]

Date: 25th March 2010

## Introduction

This article describes the theory behind, and how to implement, a basic fractal height map generator. It is based upon the algorithm defined at http://gameprogrammer.com/fractal.html. Height maps are used in many things: positioning software, graphical rendering, and procedural map generation. Given the right conditions, height maps can be used to create visually stunning images, and are frequently used with an alpha channel to simulate clouds.

All source code used in this article shall be made freely available, under a Creative Commons GPL Licence

The implementation which will be defined here outputs an array of numbers, which on the surface seems fairly mundane. However, with a bit of tweaking with VT100 colour codes, or passing to an image generation program, the algorithm can produce outputs such as this:

The ASCII renderer assigns a colour code to a range of numbers and then depending on your ranges, you can create rainbow-like maps like the one above. The grey scale and transparent images had their number arrays passed to an image generation program called RMagick.

## The Theory

I will go through the basic idea again as it was defined in the gameprogrammer link, to keep a consistency with terms used further in the article. So before we get on to the DiamondSquare algorithm, I shall implement the simpler 1 dimensional algorithm, which will describe the height of a line, similar to a horizon. The patterns created show some fractal behaviour, in that it is said to be self-similar. An object is said to be self-similar when magnified subsets of the object look like, or are identical to, the whole and to each other[1].

In context of this algorithm, it means that as we add more detail, the more rougher and major features will still hold true, but more detail will become apparent as we look closer. This makes the algorithm ideal for recursion or iterative methods. This implementation uses an iterative method.

### Midpoint Displacement in One Dimension

For the creation of the 1 dimensional height map, similar to a horizon, a very simple algorithm is used:

    def generateLine(length)

# Due to this algorithm being simple for
# the articles sake, length should be
# constrained to the powers of 2.
#
# As we need a midpoint, however, length
# should be defined as (2^n)+1.

# Create an array which describes a line.
# Set the default values to 0American Standard Code for Information Interchange
line = Array.new(length, 0)

# Define the range of the terrain values.
# For this example we shall take the range
# of values to be -63 to 63, with the midpoint
# at 0, out default value.
# Range is then 128.
range = 128;

# Work out the number of line segments
# levels there are in the line. As each line
# segment level is defined by deviding by two
# until length is 1, the number of segments
# is the log_2 of the length.
noSegments = Math.log(length-1, 2)

#Iterate through the levels
(1 .. noSegments).each{ |level|

# Work out the line segment length so you
# can properly address the offset.
segLength = (length/2**level)

# Work out the number of line segments
# within this level
noSegL = (2**level)-1

# Iterate through the line segments and
(1 .. noSegL).each{ |segOffset|

# If value is not zero, skip over it, done on a previous
# level
if( line[segLength*segOffset] == 0 )

# Make sure the current value of the line
# is directly midway between its two parents.
line[segLength*segOffset] = (line[segLength*(segOffset-1)] + line[segLength*(segOffset+1)])/2

# Apply random function to value
line[segLength*segOffset] = randomFunction(line[segLength*segOffset], level, range)

end
}
}

return line
end

Now as you can see the most important part of that algorithm is that which I have purposely missed, randomFunction. This is where you, quite obviously, define how you want your heights defined. A good way is to simply use a bounded random function, where the bounds are defined by the line segment level you are currently within. For example, a function like: 

    def randomFunction(value, level, range)

# Roughness constant 0 &lt; h &lt; 1
# As h approachs 0, the level dependent
# bounds grow and tend to 1 for all values
# of level
h = 0.8

# Define bounds in terms of level and a roughness
# function
multiplier = 2 ** (-h * level-1)

# Perform random function, bound it, and then apply
# multiplier
value += (rand() * (range) - (range / 2)) * multiplier

# Return
return value
end

  Would offset the value of the line a random amount which is bounded by a function dependent on the line segment level and a roughness coefficient. As this function is the heart of this algorithm and the 2 dimensional algorithm, I will go into some detail on the use of the roughness coefficient. If the roughness coefficient is set to 1, then the multiplier acts the same as : $2^{(-l-1)}; 0 < l < \infty ; l \in \mathbb{Z}^+$. Decreasing the value of h flattens out the function meaning the bounds are less constrictive and allowing for much rougher terrain. Here is a plot of the multiplier when h=0.8 and h=0.2, and a plot of the generated lines when those constraints are used. X axis is equal to line segment level.

As you can see, the roughness coefficient makes a massive difference to the outputted numbers. For those interested in recreating the plot, I piped the output of the above code into a text file, and then used GNUPlot to make the images.

### Extending into 2 dimensions - The diamond square algorithm

To extend this algorithm into the second dimension, we have to imagine the terrain data as a two dimensional array of numbers. Each number represents a height in this two dimensional field, and so each column and row can be treated similarly to above.

To split up a two dimensional array in a self-similar way we must use squares, or diamonds, which are analogous to the line segments of the 1 dimensional algorithm. Then, rather than using the ends of a line segment to work out a base height the corners of the square, or diamond, are used. For example:

Like the line segment algorithm, the mathematical 'meat' is the same random function as before. The complexity comes in managing which indexes are part of a diamond or a square. So, for example, here is a code segment which works out indexes of a square, depending on level and location, and applies the random function:

    # Get the corners of an arbitrary square and perform operation
# on center.
#
# @param  array           Array to use
# @param  topleftIndexX   X index of top left of square
# @param  topleftIndexY   Y index of top left of square
# @param  length          Length of the square
# @param  level           Level into the calculation
def processSquare(array, topleftIndexX, topleftIndexY, length, level, range, h)

# Get coordinates of the corners of the square
toprightIndexX    = topleftIndexX
toprightIndexY    = topleftIndexY + length + ((level == 0) ? -1 : 0)

bottomleftIndexX  = topleftIndexX + length + ((level == 0) ? -1 : 0)
bottomleftIndexY  = topleftIndexY

bottomrightIndexX = topleftIndexX + length + ((level == 0) ? -1 : 0)
bottomrightIndexY = topleftIndexY + length + ((level == 0) ? -1 : 0)

middleX           = topleftIndexX + (length)/2
middleY           = topleftIndexY + (length)/2

# Get values
topleftValue      = array[topleftIndexX][topleftIndexY]
toprightValue     = array[toprightIndexX][toprightIndexY]
bottomleftValue   = array[bottomleftIndexX][bottomleftIndexY]
bottomrightValue  = array[bottomrightIndexX][bottomrightIndexY]

# Get average
average = (topleftValue + toprightValue + bottomleftValue + bottomrightValue)/4

# Set new value
array[middleX][middleY] = average + calculateOffset(level, range, h)
end

Where calculateOffset is the random function in this application. The diamond calculation algorithm is very similar and looks like this:

    # Get the edges of an arbitrary diamond and perform operation
# on center
#
# @param  array           Array to use
# @param  topIndexX       X index of top of diamond
# @param  topIndexY       Y index of top of diamond
# @param  length          Length of diamond
# @param  level           Level into the calculation
def processDiamond(array, topIndexX, topIndexY, arraylength, level, range, h)

arraylength -= 1
length = arraylength/(2 ** level)

#Get coordinates of the diamond
rightIndexX   = topIndexX + length/2
rightIndexY   = (topIndexY == length) ? length/2 : topIndexY + length/2

leftIndexX    = topIndexX + length/2
leftIndexY    = (topIndexY == 0) ? arraylength - length/2 : topIndexY - length/2

bottomIndexX  = (topIndexX + length/2 == arraylength) ? length/2 : topIndexX + length
bottomIndexY  = topIndexY

middleX       = topIndexX + length/2
middleY       = topIndexY

# Get values
topValue      = array[topIndexX][topIndexY]
rightValue    = array[rightIndexX][rightIndexY]
bottomValue   = array[bottomIndexX][bottomIndexY]
leftValue     = array[leftIndexX][leftIndexY]

# Get average
average = (topValue + rightValue + bottomValue + leftValue)/4

# Set new value
array[middleX][middleY] = average + calculateOffset(level, range, h)

# Wraps
if(middleX == arraylength)
array[0][middleY] = array[middleX][middleY]
end
if(middleY == 0)
array[middleX][arraylength] = array[middleX][middleY]
end
end

The only difference with the above snippet is the different indices it retrieves, and that it must handle wrap around for some of the edges.

So currently, we can create arbitrary diamonds and squares within a 2-dimensional array and assign a fuzzy average of the edges according the h value. Now all we need is some code to manage traversing through the levels of iterations and through the diamonds and squares themselves. Here is my solution:

    # The main control loop for the algorithm.
#
# @param  lengthExp       Length exponent
# @param  range           Value range
# @param  h               Roughness constant
def generateTerrain(lengthExp, range, h)

length = (2 ** lengthExp) + 1

array = Array.new
array = createArray(array, length)

#Go through Levels (irerative recursion)
(0 .. lengthExp - 1).each{ |level|

# Iterator for the Square part of the algorithm
# Will go through the x-axis coords
(0 .. (2 ** level) -1 ).each { |sqx|

# Y axis coords
(0 .. (2 ** level) -1).each { |sqy|

gap = length/2 ** level
x = (0 + (gap*sqx))
y = (0 + (gap*sqy))

processSquare(array, x, y, gap, level, range, h)
}
}

# Iterator for the diamond part of the algorithm
(0 ... (2 ** (level+1))).each { |dix|

# Offset in the number of points on the y-axis. Dependant
# on if x iteration is even or odd.
offset = (dix.even?) ? 1 : 2
(0 ... (2 ** (level+1)/2)).each { |diy|

gap = (length/2 ** (level+1))
ygap = 2 * gap

x = (0 + (gap*dix))
if (dix.even?)

y = 0 + (ygap*diy)
else

y = gap + (ygap*diy)
end
processDiamond(array, x, y, length, level, range, h)
}
}
}
return array
end

And this gives us our array with its height map hidden inside. Using a library like RMagick we can output images like the ones shown above. To create the gray scale image, the following code was used:

  image = Image.new(array.length, array.length)

(0 ... array.length).each { |x|
(0 ... array[x].length).each { |y|
val = array[x][y] * (2**9)
# Create greyscale image
image.pixel_color(x, y, Pixel.new(val, val, val, val))
}
}
image.display

Which just takes the value in the array, and multiplies it by 512 which gives the values a range of $0 \ge v \ge 2^{15}, \; \frac{v}{512} \in \mathbb{Z}$ . This gives us the gaseous image that has been generated above.

## Code Listings

A library version of the ruby code found in this tutorial can be found at GitHub.

## References

1. Voss, Richard D., FRACTALS in NATURE: characterization, measurement, and simulation. SIGGRAPH 1987

Based on a work at gameprogrammer.com.

# Code Golf

CSLU did code golf today. I did 1 and a half tasks, which were:

1. Output the first 100 prime numbers
2. Output e to 100 decimal places

## Prime numbers

I’m quite proud of this, I managed to do this in 55 characters initially but then after some collaboration with the rest of the club shrunk it down to 49 characters

2.upto(541){|a|i=2;i+=1 while a%i&gt;0;p a if i==a}

## e

I never got his fully working as I ended up getting caught up in list comprehensions. Ended up with:

1 + sum [1 / (product [m | m&lt;- [1..n] ]) | n &lt;- [1..300] ]

Which is the same as $e = \sum_{n=0}^{\infty } \frac{1}{n!}$ and shows how pretty Haskell is.