## Phone Vision 15 – Median Filters

1 03 2011

Filters, filters, filters… Tired of them yet?  Me too.  After this we’ll leave them alone for a while.  I promise.  Until then, we have work to do.

What is a median filter?  If you understand the average filter (or mean filter) then the median filter should give you no troubles at all – assuming you know what the median is.  We’ll start there.

What is the Median?

The median is the middle number of a sample.  To find it, you sort your list and pick the middle item.  Yes, it really is that simple.

Here is a random set of 9 numbers:

Sort them:

Pick the middle value:

Like I said, it’s simple.

Why use median instead of mean?

The median is particularly effective against certain types of noise – specifically “salt and pepper” noise (black and white specks).  Let’s assume this is the neighborhood we’re dealing with:

Performing the mean gives:

While the median returns:

If we expect that the 0 values are noise then the median is probably a much closer estimate.

Code

private WriteableBitmap MedianFilter(WriteableBitmap grayscale, int radius)

{

// we are still going to create a new image

// because we don’t want to modify the

// old image as we are processing it

WriteableBitmap filtered =

new WriteableBitmap(

grayscale.PixelWidth,

grayscale.PixelHeight);

// boiler plate code for our

// histogram stuff

int[] histogram = new int[256];

int maxIntensity = 0;

// the math is still easier if we create two loops

for (int y = 0; y < grayscale.PixelHeight; y++)

{

for (int x = 0; x < grayscale.PixelWidth; x++)

{

//here’s the pixel we’re centered on

int pixel = x + y * grayscale.PixelWidth;

byte intensity = (byte)grayscale.Pixels[pixel];

// if we are on an edge we are going to leave it

// as the original intensity.  you will see the

// edges increasingly unsmoothed as the window

// size increases.  here we are using the radius

// to determine our bounds

if (y <= radius – 1 ||

x <= radius – 1 ||

y >= grayscale.PixelHeight – radius ||

{

histogram[intensity]++;

if (histogram[intensity] > maxIntensity)

{

maxIntensity = histogram[intensity];

}

continue;

}

/////////////////////////////////////////////////////////

// IMPORTANT PART ///////////////////////////////////////

/////////////////////////////////////////////////////////

// this list is the key

// it contains all of the neighboring pixels

List<byte> localIntensities = new List<byte>();

{

xoffset++)

{

(byte)grayscale.Pixels[(x + xoffset)

+ (y + yoffset) * grayscale.PixelWidth]));

}

}

//sort the intensities

localIntensities.Sort();

//pick the middle value

int medianLocalIntensity =

localIntensities[(int)(localIntensities.Count/2.0+.5)];

/////////////////////////////////////////////////////////

// END IMPORTANT PART ///////////////////////////////////

/////////////////////////////////////////////////////////

// and now just set the color

filtered.Pixels[pixel] = (255 << 24)

| (byte)medianLocalIntensity << 16

| (byte)medianLocalIntensity << 8

| (byte)medianLocalIntensity;

histogram[(byte)medianLocalIntensity]++;

if (histogram[(byte)medianLocalIntensity] > maxIntensity)

{

maxIntensity = histogram[(byte)medianLocalIntensity];

}

}

}

PlotHistogram(histogram, maxIntensity);

return filtered;

}

Results

Taking a simple gray image I added salt and pepper noise to the image then performed a median filter and an average filter to it.  The results are stunning.

flat gray image                              10% salt and pepper noise

after median filtering                      after mean filtering

There are a few specks after applying the median filter, but the noise is removed pretty well.  The average filter performed dismally to say the list.

Summary

If you expect salt and pepper noise then the median filter is great tool to have in your toolbox.  If you want you can explore max, min, and mode filters.  I don’t think we’ll cover them here unless we have a specific application for it.

http://cid-88e82fb27d609ced.office.live.com/embedicon.aspx/Blog%20Files/PhoneVision/PhoneVision%2015%20-%20Median%20Filter.zip

(code includes salt and pepper noise generation)

Up Next: Binary Images

## Project Euler 005 – C#

1 03 2011

Problem 5

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Solution

// 2520 is the smallest number that can be

// divided by each of the numbers from 1 to 10

// without any remainder.

// What is the smallest positive number

// that is evenly divisible by all of the

// numbers from 1 to 20?

namespace ProjectEulerCSharp_005

{

class Program

{

static void Main(string[] args)

{

//have we found the answer yet?

//let’s start off by saying "no"

bool notFound = true;

//20 seems like as good of a

//starting spot as any

long candidate = 20;

while(notFound)

{

//I am incrementing up front

//so that my loop exits elegantly

candidate++;

//here I am assuming that this candidate

//is the answer.  I set it back when I

//find a divisor that it doesn’t work for

notFound = false;

//this is not too important, but it’s a good

//they are more likely to fail

for (int divisor = 2; divisor <= 20; divisor++)

{

if (candidate % divisor != 0)

{

notFound = true;

break;

}

}

}

System.Console.WriteLine(candidate);

}

}

}

Discussion

The brute force approach is pretty silly, but it does work and within the 1 minute rule.  It takes about 7.5 seconds on my 6 month old laptop.  When this problem was announced (November 2001) the same code would have probably taken over 10 minutes to complete – at least according to my back of the napkin estimation of Moore’s law.  🙂