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Landscape generation using midpoint displacement

Today I will present how to implement in Python a simple yet effective algorithm for proceduraly generating 2D landscapes. It is called Midpoint Displacement (or Diamond-square algorithm, which seems less intuitive to me) and, with some tweaking it can also be used for creating rivers, lighting strikes or (fake) graphs. The final output may look like the following image.


Example terrain generated with the presented algorithm.

Algorithm overview

The main idea of the algorithm is as follows: Begin with a straight line segment, compute its midpoint and displace it by a bounded random value. This displacement can be done either by:

  1. Displacing the midpoint in the direction perpendicular to the line segment.
  2. Displacing only the y coordinate value of the midpoint.

Different methods for displacing the midpoint.

This first iteration will result in two straight line segments obtained from the displacement of the midpoint of the original segment.  The same process of computing and displacing the midpoint can be then applied to each of this new two segments and it will result in four straight line segments. Then we can repeat the process for each of this four segments to obtain eight and so on. The process can be repeated iteratively or recursively as many times as desired or until the segments cannot be reduced more (for graphical applications this limit would be two pixel’s width segments). The following image may help to clarify what I just said.


From top to bottom, successive iterations of the algorithm.

And that’s it! This is the core idea of the midpoint displacement algorithm. In pseudocode it looks like:

Initialize segment
While iterations < num_of_iterations and segments_length > min_length:
	For each segment:
		Compute midpoint
		Displace midpoint
		Update segments
        Reduce displacement bounds

However, before implementing the algorithm we should dig deeper in some of the concepts that have arisen so far. These are mainly:

  • How much should we displace the midpoint?
  • How much should the displacement bounds be reduced after each iteration?

How much should we displace the midpoint?

Sadly, there is no general answer for this question because it greatly depends on two aspects:

  1. The application the algorithm is being used for
  2. The desired effect

Since in this post our scope is terrain generation I will limit the explanation to the effects that this parameter has in this area. However, the ideas that I will explain now can be extrapolated to any other application where the Midpoint Displacement algorithm may be used. As I see it, there are two key considerations that should be taken into account when deciding the initial displacement value.

First of all, we should consider which is the desired type of terrain. Hopefully it makes sense to say that the bigger the mountains we want to generate the bigger the initial displacement value should be and viceversa. With a bit of trial and error it is easy to get an idea of the average profiles generated by different values and how do they look. The point here is that bigger mountains need bigger initial displacement values.

Secondly,  the overall dimensions (width and height) of the generated terrain. The initial displacement should be regarded as a value which depends on the generated terrain dimensions.  What I want to say is that an initial displacement of 5 may be huge when dealing with a 5×5 image but will hardly be noticed in a 1500×1500 image.


The same initial displacement value may be to big for a certain image size but may suit well another image size.

How much should the bounds be reduced after each iteration?

Well, the answer again depends on which is the desired output. It should be intuitive that the smaller the displacement reduction the more jagged the obtained profile will be and viceversa. The two extremes are no displacement reduction at all and setting the displacement to 0 after the first iteration. This two cases can be observed in the figure below.


On the left: no displacement reduction. On the right: Displacement reduction to 0 after first iteration.

Somewhere in between is the displacement reduction that will yield the desired output. There are plenty of ways to reduce the displacement bounds each iteration (linearly, exponentially, logarithmically, etc.) and I encourage you to try different ones and see how the results vary.

What I did was define a standard displacement reduction of 1/2, which means that the displacement is reduced by half each new iteration,  and a displacement decay power such that the displacement reduction is

displacement_reduction = 1/(2^i)


displacement_bounds(k+1) = displacement_bounds(k)*displacement_reduction

were is the current iteration and k+1 the next iteration. We can then talk about the obtained terrain profiles in terms of this decay power i. Below you can see how the algorithm performs for different decay powers.


Obtained profiles for different values of  the displacement decay powers.

Bear in mind that the two factors we just saw, the bounds reduction and initial displacement are related one to the other and that they do not affect the output independently. Smaller initial displacements may look good with smaller decay powers and viceversa. Here we have talked about some guidelines that may help when deciding which values to use but there will be some trial and error until the right parametres for the desired output are found. Finally, the number of iterations is another factor that also affects the output in relation with the initial displacement and the bounds reduction.

Python implementation

Finally it is time to, with all the ideas explained above, code our 2D terrain generator. For this particular implementation I have decided to:

  • Displace only the y coordinate of the midpoints (Second of the two displacement methods explained at the begining).
  • Use symmetric bounds with respect to zero for the displacement (if b is the upper bound then –b will be the lower bound.)
  • Choose the displacement value to be either the upper bound or the lower bound, but never allow values in between.
  • Reduce the bounds after each iteration by multiplying the current bounds by 1/(2^i)

We will have three functions: one that will generate the terrain, one that will draw the generated terrain and one that will handle the above processes.

Before implementing the functions we should first import the modules that we will use:

import os                             # path resolving and image saving
import random                         # midpoint displacement
from PIL import Image, ImageDraw      # image creation and drawing
import bisect                         # working with the sorted list of points

Terrain generation

For the terrain generation we need a function that, given a straight line segment returns the profile of the terrain. I have decided to provide as inputs the initial segment and displacement, the rate of decay or roughness of the displacement and the number of iterations:

# Iterative midpoint vertical displacement
def midpoint_displacement(start, end, roughness, vertical_displacement=None,
    Given a straight line segment specified by a starting point and an endpoint
    in the form of [starting_point_x, starting_point_y] and [endpoint_x, endpoint_y],
    a roughness value > 0, an initial vertical displacement and a number of
    iterations > 0 applies the  midpoint algorithm to the specified segment and
    returns the obtained list of points in the form
    points = [[x_0, y_0],[x_1, y_1],...,[x_n, y_n]]
    # Final number of points = (2^iterations)+1
    if vertical_displacement is None:
        # if no initial displacement is specified set displacement to:
        #  (y_start+y_end)/2
        vertical_displacement = (start[1]+end[1])/2
    # Data structure that stores the points is a list of lists where
    # each sublist represents a point and holds its x and y coordinates:
    # points=[[x_0, y_0],[x_1, y_1],...,[x_n, y_n]]
    #              |          |              |
    #           point 0    point 1        point n
    # The points list is always kept sorted from smallest to biggest x-value
    points = [start, end]
    iteration = 1
    while iteration <= num_of_iterations:
        # Since the list of points will be dynamically updated with the new computed
        # points after each midpoint displacement it is necessary to create a copy
        # of the state at the beginning of the iteration so we can iterate over
        # the original sequence.
        # Tuple type is used for security reasons since they are immutable in Python.
        points_tup = tuple(points)
        for i in range(len(points_tup)-1):
            # Calculate x and y midpoint coordinates:
            # [(x_i+x_(i+1))/2, (y_i+y_(i+1))/2]
            midpoint = list(map(lambda x: (points_tup[i][x]+points_tup[i+1][x])/2,
                                [0, 1]))
            # Displace midpoint y-coordinate
            midpoint[1] += random.choice([-vertical_displacement,
            # Insert the displaced midpoint in the current list of points         
            bisect.insort(points, midpoint)
            # bisect allows to insert an element in a list so that its order
            # is preserved.
            # By default the maintained order is from smallest to biggest list first
            # element which is what we want.
        # Reduce displacement range
        vertical_displacement *= 2 ** (-roughness)
        # update number of iterations
        iteration += 1
    return points

The initial line segment is specified by the coordinates of the points where it begins and ends. Both are a list in the form:

point = [x_coordinate, y_coordinate]

And the output is a list of lists containing all the points that should be connected to obtain the terrain profile in the form:

points = [[x_0, y_0], [x_1, y_1], ..., [x_n, y_n]]

Terrain drawing

For the graphical output we need a function that returns an image of the drawn terrain and that takes as inputs at least the profile generated by the midpoint displacement algorithm. I have also included as inputs the desired width and height of the image and the colors it should use for painting. What I did for drawing several layers of terrain was start with the layer in the background and draw each new layer on top of the previous one.

For drawing each layer I first infer the value of every value in the range (0, image width) based on the assumption that the known points, the ones obtained from the midpoint displacement, are connected with straight lines. Once knowing the value of each value in the range (0, image width) I traverse all the values iteratively and for each x value draw a line from its value to the bottom of the image.

def draw_layers(layers, width, height, color_dict=None):
    # Default color palette
    if color_dict is None:
        color_dict = {'0': (195, 157, 224), '1': (158, 98, 204),
                      '2': (130, 79, 138), '3': (68, 28, 99), '4': (49, 7, 82),
                      '5': (23, 3, 38), '6': (240, 203, 163)}
        # len(color_dict) should be at least: # of layers +1 (background color)
        if len(color_dict) < len(layers)+1:
            raise ValueError("Num of colors should be bigger than the num of layers")

    # Create image into which the terrain will be drawn
    landscape = Image.new('RGBA', (width, height), color_dict[str(len(color_dict)-1)])
    landscape_draw = ImageDraw.Draw(landscape)
    # Draw the sun
    landscape_draw.ellipse((50, 25, 100, 75), fill=(255, 255, 255, 255))
    # Sample the y values of all x in image
    final_layers = []
    for layer in layers:
        sampled_layer = []
        for i in range(len(layer)-1):
            sampled_layer += [layer[i]]
            # If x difference is greater than 1
            if layer[i+1][0]-layer[i][0] > 1:
                # Linearly sample the y values in the range x_[i+1]-x_[i]
                # This is done by obtaining the equation of the straight
                # line (in the form of y=m*x+n) that connects two consecutive
                # points
                m = float(layer[i+1][1]-layer[i][1])/(layer[i+1][0]-layer[i][0])
                n = layer[i][1]-m*layer[i][0]
                r = lambda x: m*x+n  # straight line
                for j in range(layer[i][0]+1, layer[i+1][0]):  # for all missing x
                    sampled_layer += [[j, r(j)]]  # Sample points
        final_layers += [sampled_layer]

    final_layers_enum = enumerate(final_layers)
    for final_layer in final_layers_enum:
        # traverse all x values in the layer
        for x in range(len(final_layer[1])-1):
            # for each x value draw a line from its y value to the bottom
            landscape_draw.line((final_layer[1][x][0], height-final_layer[1][x][1],
                                 final_layer[1][x][0], height),

    return landscape


The PIL module (and mostly all the modules that allow working with images) sets the origin of coordinates on the top left corner of the image. Also, the x value increases when moving right and the value when moving down. The values that have to be passed to the function that draws the lines have to be expressed in this system of reference and that is why for drawing the desired line its y values have to be transformed from our reference system (origin lower left) to PIL’s reference system.


Differences between PIL’s reference system and the one we have been using so far.

With these two functions we are now able to actually compute and draw our 2D proceduraly generated terrain.

Our main function

The final step is to define our main function. This function will compute the profiles of the desired number of layers, draw them and save the obtained terrain as a .png image:

def main():
    width = 1000  # Terrain width
    height = 500  # Terrain height
    # Compute different layers of the landscape
    layer_1 = midpoint_displacement([250, 0], [width, 200], 1.4, 20, 12)
    layer_2 = midpoint_displacement([0, 180], [width, 80], 1.2, 30, 12)
    layer_3 = midpoint_displacement([0, 270], [width, 190], 1, 120, 9)
    layer_4 = midpoint_displacement([0, 350], [width, 320], 0.9, 250, 8)

    landscape = draw_layers([layer_4, layer_3, layer_2, layer_1], width, height)



To call the main() function when we run the program we finally add the lines:

if __name__ == "__main__":


And we’re done! Now it’s your turn to code your own terrain generator and play with its different parametres for modifying the results (you can also change the colors). If you have any doubts do not hesitate to contact me. You can find the whole code at github.

As a bonus, some more terrain images obtained with the previous code:


Midpoint Displacement 2D generated landscapes.



Converting images to ASCII art (Part 1)

In the last post we discussed a simple method for procedurally generating 2D terrain. In today’s post, we will also generate an image that will be the final output of our program, but with a totally different purpose and method. Today we will try to generate ASCII art from images.  For those of you who are not familiar with ASCII art, what we will achieve will look (more or less) like the image below.


Original image (left) and output image (right).

Not as impressive as you thought? If still not yet convinced (I promise this point will be revisited later on) take a look at the image at full resolution and stare at it after moving a few feet away from the screen. And yes, it is only made of ASCII characters! This can be quickly checked by zooming in any section of the image.


Zooming a region of the image until characters are visible.

It should be clear from the previous example that the output image has two main characteristics: it is composed solely of ASCII characters and it looks like the original image. This is our main goal. However, we will implement and discuss some extra features or alternative implementations along the way in order to improve the results. Finally, before we embark I would like to point out three things:

  1. The ideas that will be explained can be extended to any kind of process were the resulting image is a composition of other images and were the aim is to resemble a target or input image.
  1. The post is about the algorithms and ideas that make the program work and not about a detailed explanation of the whole program.
  2. The example code presented is written in C#. However, the main ideas can be easily implemented in other programming languages. (Disclaimer: I am currently learning C# and this was my first project – done with learning purposes -. Sorry if the code isn’t as clean or efficient as it could be or if the project architecture isn’t the ideal one. Any suggestions on how to improve it are welcome. I am not teaching how to code in C#.)


One of the (first) things I tend to do when developing software is to devise which are the steps the program should take in order to achieve the desired outcome. It is not the only method to develop a general scheme of what the program has to do, but in this case we will work our way backwards. Starting from the ASCII art image we will try to guess from where did it come and how it was obtained.

Clearly there has been a point in the process of generating the ASCII art image where each pixel or region of pixels from the original image has been mapped to a certain character. Since the final output is the ASCII image we won’t be too wrong if we assume that this is the last step of our process.


Last step of the process.

Now we have to deduce if the image used as input for the mapping is the original image or if there has been some pre-processing applied to it before this mapping step. Since the ASCII art image is made only of black characters over a white background, which from a certain distance looks like a grayscale image, a good guess is that a grayscale version of the original image is the image used as a reference for the mapping.


Getting closer to know the full the process.

Since a grayscale version of the original image can be easily obtained from the original image the final scheme of the process is:


Overview of the process.

A more general way of defining the “Convert to grayscale” step would be “Image pre-processing”. This would include, in the case we wanted to perform more operations than just a grayscale conversion, all the different operations and transformations applied to the image before the final mapping of the image to ASCII characters.

Image pre-processing

Grayscale conversion

There are many methods that can be applied to obtain a grayscale version of an RGB image and I will only focus on the one I decided to use. What I applied is a conversion that preserves the luminance (or luminous intensity) of the original image in the grayscale image. Each pixel’s intensity or grey value is simply the weighted sum of the values of its RGB components computed as follows:

I = 0.299·R+0.587·G+0.114·B ≈ 0.3·R+0.59·G+0.11·B

It is not the most efficient way, but in C# this can be easily achieved with:

public static Bitmap Grayscale(Image image)
Bitmap btm = new Bitmap(image);
      for (int i = 0; i < btm.Width; i++)
      	for (int j = 0; j < btm.Height; j++)
            	int ser = (int)(btm.GetPixel(i, j).R*0.3 + btm.GetPixel(i, j).G*0.59 + btm.GetPixel(i, j).B*0.11);

                    btm.SetPixel(i, j, Color.FromArgb(ser, ser, ser));
       return btm;


Were we are simply traversing all the pixels in the original image, computing the weighted sum and then assigning the result as the pixel’s new intensity. Now, we could keep going from here, but since the final results depend largely on the grayscale version of the original image, wouldn’t it be nice if we allowed the user to have some control over the grayscale conversion? One way to do this is to let the user adjust the contrast of the original image and when done, convert it to grayscale. We could let the user adjust directly the contrast of the grayscale image but I preferred to develop a simple interface that only displays two images: the original and the one made of ASCII characters which is why I decided to implement the contrast adjustment on the original image.


Updated program execution flow adding the contrast adjustment step.

Contrast adjustment

So, what is the contrast of an image and how can we modify it? Basically, the contrast of an image is the difference between the maximum and minimum pixel intensity, the “amount” of separation between its darkest and brightest areas. It is then determined by the colour and brightness of the different objects or shapes that appear in the image. This means that the higher the contrast, the bigger the difference between pixel intensities, and the easier it will be to recognize the different objects in the image (due to this bigger range of intensities). This can be appreciated in the image below.


Contrast variation. The contrast increases from bottom to top on the left column and from top to bottom on the right column. Source: Wikipedia.

Once knowing this it’s time to tackle the big question: How can we modify (increase or decrease) the contrast of an image?

We will achieve this by applying to the image a pixel by pixel operation (as we did when grayscaling) that modifies each pixel’s intensity with the objective of expanding or contracting the range of intensity values or luminance an image has. The value of the output image at a particular pixel will depend only on the value of the pixel located at that same position on the original image. This kind of operations are called in Computer Vision point operations.


Schematic of a point operation.

Note that this implies that two pixels that have the same intensity in the original image will be mapped to the same, and usually different from the original, intensity value in the output image. The transformation is then a single variable function

g(x,y) = T( f(x,y) )

Where g(x,y) is the intensity of the pixel (x,y) in the output image, f(x,y) is the intensity of the original image at the same position and T is the applied transformation. The pixel position in the image doesn’t play any role at all and its intensity is the only interesting property when varying the contrast. Point operations can be plotted in the same way single variable functions are. On the x axis, we will have the range of values the input property can take (I) and, on the y axis, the mapped values that result from the transformation (I’). We will make use of this to derive intuitively our contrast adjustment transformation. As an example, how would a transformation that didn’t modify the image at all look like?


Point operations identity transformation.

Yep, that’s right. Each intensity value in the original image would be mapped to its same value in the output image. Recall now that what we want to achieve when increasing the contrast is to make wider the range of intensities in the image. Any ideas on the shape of the transformation that would achieve this? You probably guessed right. If we increase the slope of the line presented above, say double it, any considered range of intensities in the input image will be mapped to a range in the output image twice as big. And if we decrease the slope, say to a half, any considered range of intensities in the input image will half its size in the output image. This can be appreciated in the image below.


Deriving our contrast adjustment transformation.

We are getting close… Note that all the lines presented above have a point in common, the origin in this case. This is nice because if we apply the transformation directly to the original image, without creating a copy of it, having a fixed point will allow us to recover the original image. This wouldn’t be possible without having a fixed point. However, is the origin the best fixed point we can get? I don’t think so. Note that increasing the contrast in this way will result in many intensities being mapped to 255, the maximum allowed intensity in images that store each channel’s intensity with 8 bits. I hope that you agree in that it would be better to divide this region of “out of bounds” intensities between the max and min allowed intensities. This would make the loss of information symmetric. This results in choosing 127.5 as the fixed point of the transformation. Then our contrast transformation so far looks like

I’ = (I-127.5)·c + 127.5

where 127.5 can be rounded to 127 or 128.


Finally, our contrast adjustment transformation.

Not that difficult eh? See how in the previous figure half of the values “out of bounds” get mapped to 0 and the other half to 255?  The value of c will be the one that decides how much we increase or decrease the contrast. Its value can be decided in many ways. In my case I am using a quadratic function to determine the value of c. A scrollbar in the GUI of the program reads a value between -100 and 100. This value is then mapped to the range [0,2] and squared, so that c finally varies between 0 and 4. Finally, note that once knowing the value of c the mapping of intensities can be already computed so storing the transformations in a lookup table will improve the efficiency and speed of this step.


Example of the implemented contrast adjustment in action. For anyone who’s wondering, yeah, on the bottom right image you can see a nrf24L01+ module connected to a Pyboard.


The code below, which is in charge of varying the contrast should now be easy to follow. It is based on this BitBank’s  stackoverflow answer.

public unsafe static void AdjustContrast(Bitmap bmp, double contrast)
      	byte[] contrast_lookup_table = new byte[256];
            double newValue = 0;
            double c = (100.0 + contrast) / 100.0;

            c *= c;

            for (int i = 0; i < 256; i++)
            	newValue = (double)i;
                  newValue /= 255.0;
                  newValue -= 0.5;
                  newValue *= c;
                  newValue += 0.5;
                  newValue *= 255;

                  if (newValue < 0)
                  	newValue = 0;
                  if (newValue > 255)
                  	newValue = 255;
                  contrast_lookup_table[i] = (byte)newValue;

            var bitmapdata = bmp.LockBits(new Rectangle(0, 0, bmp.Width, bmp.Height),
            System.Drawing.Imaging.ImageLockMode.ReadWrite, System.Drawing.Imaging.PixelFormat.Format32bppArgb);

            int PixelSize = 4;

            for (int y = 0; y < bitmapdata.Height; y++)
            	byte* destPixels = (byte*)bitmapdata.Scan0 + (y * bitmapdata.Stride);
                  for (int x = 0; x < bitmapdata.Width; x++)
                  	destPixels[x * PixelSize] = contrast_lookup_table[destPixels[x * PixelSize]]; // B
                  	destPixels[x * PixelSize + 1] = contrast_lookup_table[destPixels[x * PixelSize + 1]]; // G
                  	destPixels[x * PixelSize + 2] = contrast_lookup_table[destPixels[x * PixelSize + 2]]; // R


I think that’s enough for today. We now have finished the pre-processing of the image (contrast adjustment + grayscaling) and on the next post we’ll see how to finally get the ASCII image. If you enjoyed it so far be sure to stay tuned for part 2!

PS: If you want to play with the code or take a look at it, it is available hereI wanted to upload the final program (.exe) somewhere so you can download it and play with it. However, I have no idea on the best way to achieve this so it is easy for you to get it. If you have any ideas please tell me. In the meantime, you can reach me and ask for it via e-mail at juangallostra@gmail.com

Update: You can download the compiled program now here. (Thanks to Campbell for pointing out that this could be easily done via Github releases.)

PS2: I know I haven’t revisited yet the point I made at the beginning but I will, for sure, revisit it on the next post.