Ernado - 10 months ago 49

C# Question

I want to automatically divide an image of ancient handwritten text by lines (and by words in future).

I'm just using a simple digitization (based on brightness of pixel). After that I store data into two-dimensional array.

- My first algorithm was pretty simple - if there are more black pixels in a row of the array than the root-mean-square of
*Maximum*and*Minimum*value, then this row is part of line.

After forming the list of lines I cut off lines with*height*that is less than average.

Finally it turned out into some kind of linear regression, trying to minimize the difference between the blank rows and text rows. (I assumed that fact)

- My second attempt - I tried to use GA with several fitness functions.

The chromosome contained 3 values -*xo, x1, x2. xo [-1;0] x1 [0;0.5] x2 [0;0.5]*

Function, that determines identity the row to line is

Another functions, that i tried is

The last function is the most efficient.

The fitness function is

Where range is difference between maximum and minimum. It represents the homogeneity of text. The global optimum of this function - the most smooth way to divide the image into lines.

I am using C# with my self-coded GA (classical, with 2-point crossover, gray-code chromosomes, maximum population is 40, mutation rate is 0.05)

Now I ran out of ideas how to divide this image into lines with ~100% accuracy.

What is the efficient algorithm to do this?

Original image

Original BMP (1.3 MB)

Improved results on this text to 100%

- fixed minor bug in range count
- changed fitness function to 1/(distancesRange+1)*(heightsRange+1))
- minimized classifying function to (1/xo + x2/range) > 0 (points in row now don't affect classification)

(i.e. optimized input data and made fitness function optimizations more explicit)

Problem:

GA surprisingly failed to recognize this line. I looked at debug data of 'find rages' function and found, that there is too much noise in 'unrecognized' place.

The function code is below:

`public double[] Ranges()`

{

var ranges = new double[_original.Height];

for (int y = 0; y < _original.Height; y++ )

{

ranges[y] = 0;

var dx = new List<int>();

int last = 0;

int x = 0;

while (last == 0 && x<_original.Width)

{

if (_bit[x, y])

last = x;

x++;

}

if (last == 0)

{

ranges[y] = 0;

continue;

}

for (x = last; x<_original.Width; x++)

{

if (!_bit[x, y]) continue;

if (last != x - 1)

{

dx.Add((x-last)+1);

}

last = x;

}

if (dx.Count > 2)

{

dx.Sort();

ranges[y] = dx[dx.Count / 2];

//ranges[y] = dx.Average();

}

else

ranges[y] = 0;

}

var maximum = ranges.Max();

for (int i = 0; i < ranges.Length; i++)

{

if (Math.Abs(ranges[i] - 0) < 0.9)

ranges[i] = maximum;

}

return ranges;

}

I'm using some hacks in this code. The main reason - I want to minimize the range between nearest black pixels, but if there are no pixels, the value becomes '0', and it becomes impossible to solve this problem with finding optimas. The second reason - this code is changing too frequently.

I'll try to fully change this code, but I have no idea how to do it.

- If there is more efficient fitness function?
- How to find more versatile determination function?

Answer Source

Although I'm not sure how to translate the following algorithm into GA (and I'm not sure why you need to use GA for this problem), and I could be off base in proposing it, here goes.

The simple technique I would propose is to count the number of black pixels per row. (Actually it's the dark pixel density per row.) This requires very few operations, and with a few additional calculations it's not difficult to find peaks in the pixel-sum histogram.

A raw histogram will look something like this, where the profile along the left side shows the number of dark pixels in a row. For visibility, the actual count is normalized to stretch out to x = 200.

After some additional, simple processing is added (described below), we can generate a histogram like this that can be clipped at some threshold value. What remains are peaks indicating the center of lines of text.

From there it's a simple matter to find the lines: just clip (threshold) the histogram at some value such as 1/2 or 2/3 the maximum, and optionally check that the width of the peak at your clipping threshold is some minimum value w.

One implementation of the full (yet still simple!) algorithm to find the nicer histogram is as follows:

- Binarize the image using a "moving average" threshold or similar local thresholding technique in case a standard Otsu threshold operating on pixels near edges isn't satisfactory. Or, if you have a nice black-on-white image, just use 128 as your binarization threshold.
- Create an array to store your histogram. This array's length will be the height of the image.
- For each pixel (x,y) in the binarized image, find the number of dark pixels above and below (x,y) at some radius R. That is, count the number of dark pixels from (x, y - R) to x (y + R), inclusive.
- If the number of dark pixels within a vertical radius R is equal or greater to R--that is, at least half the pixels are dark--then pixel (x,y) has sufficient vertical dark neighbors. Increment your bin count for row y.
- As you march along each row, track the leftmost and rightmost x-values for pixels with sufficient neighbors. As long as the width (right - left + 1) exceeds some minimum value, divide the total count of dark pixels by this width. This normalizes the count to ensure the short lines like the very last line of text are included.
- (Optional) Smooth the resulting histogram. I just used the mean over 3 rows.

The "vertical count" (step 3) eliminates horizontal strokes that happen to be located above or below the center line of text. A more sophisticated algorithm would just check directly above and below (x,y), but also to the upper left, upper right, lower left, and lower right.

With my rather crude implementation in C# I was able to process the image in less than 75 milliseconds. In C++, and with some basic optimization, I've little doubt the time could be cut down considerably.

This histogram method assumes the text is horizontal. Since the algorithm is reasonably fast, you may have enough time to calculate pixel count histograms at increments of every 5 degrees from the horizontal. The scan orientation with the greatest peak/valley differences would indicate the rotation.

I'm not familiar with GA terminology, but if what I've suggested is of some value I'm sure you can translate it into GA terms. In any case, I was interested in this problem anyway, so I might as well share.

EDIT: maybe for use GA, it's better to think in terms of "distance since previous dark pixel in X" (or along angle theta) and "distance since previous dark pixel in Y" (or along angle [theta - pi/2]). You might also check distance from white pixel to dark pixel in all radial directions (to find loops).

```
byte[,] arr = get2DArrayFromBitamp(); //source array from originalBitmap
int w = arr.GetLength(0); //width of 2D array
int h = arr.GetLength(1); //height of 2D array
//we can use a second 2D array of dark pixels that belong to vertical strokes
byte[,] bytes = new byte[w, h]; //dark pixels in vertical strokes
//initial morph
int r = 4; //radius to check for dark pixels
int count = 0; //number of dark pixels within radius
//fill the bytes[,] array only with pixels belonging to vertical strokes
for (int x = 0; x < w; x++)
{
//for the first r rows, just set pixels to white
for (int y = 0; y < r; y++)
{
bytes[x, y] = 255;
}
//assume pixels of value < 128 are dark pixels in text
for (int y = r; y < h - r - 1; y++)
{
count = 0;
//count the dark pixels above and below (x,y)
//total range of check is 2r, from -r to +r
for (int j = -r; j <= r; j++)
{
if (arr[x, y + j] < 128) count++;
}
//if half the pixels are dark, [x,y] is part of vertical stroke
bytes[x, y] = count >= r ? (byte)0 : (byte)255;
}
//for the last r rows, just set pixels to white
for (int y = h - r - 1; y < h; y++)
{
bytes[x, y] = 255;
}
}
//count the number of valid dark pixels in each row
float max = 0;
float[] bins = new float[h]; //normalized "dark pixel strength" for all h rows
int left, right, width; //leftmost and rightmost dark pixels in row
bool dark = false; //tracking variable
for (int y = 0; y < h; y++)
{
//initialize values at beginning of loop iteration
left = 0;
right = 0;
width = 100;
for (int x = 0; x < w; x++)
{
//use value of 128 as threshold between light and dark
dark = bytes[x, y] < 128;
//increment bin if pixel is dark
bins[y] += dark ? 1 : 0;
//update leftmost and rightmost dark pixels
if (dark)
{
if (left == 0) left = x;
if (x > right) right = x;
}
}
width = right - left + 1;
//for bins with few pixels, treat them as empty
if (bins[y] < 10) bins[y] = 0;
//normalize value according to width
//divide bin count by width (leftmost to rightmost)
bins[y] /= width;
//calculate the maximum bin value so that bins can be scaled when drawn
if (bins[y] > max) max = bins[y];
}
//calculated the smoothed value of each bin i by averaging bin i-1, i, and i+1
float[] smooth = new float[bins.Length];
smooth[0] = bins[0];
smooth[smooth.Length - 1] = bins[bins.Length - 1];
for (int i = 1; i < bins.Length - 1; i++)
{
smooth[i] = (bins[i - 1] + bins[i] + bins[i + 1])/3;
}
//create a new bitmap based on the original bitmap, then draw bins on top
Bitmap bmp = new Bitmap(originalBitmap);
using (Graphics gr = Graphics.FromImage(bmp))
{
for (int y = 0; y < bins.Length; y++)
{
//scale each bin so that it is drawn 200 pixels wide from the left edge
float value = 200 * (float)smooth[y] / max;
gr.DrawLine(Pens.Red, new PointF(0, y), new PointF(value, y));
}
}
pictureBox1.Image = bmp;
```