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This problem has been solved by a variety of methods over the years. The first step is isolating the individual characters in the image. Many approaches then divide a character into a grid, and average the level in each grid cell. At this point, I've seen three approaches:
1. Input the cell values to a neural net to train it to recognize the characters,
2. Heuristically come up with rules that differentiate the characters, and
3. Use the averages and standard deviations of the rows and columns with discriminant analysis (http://statsoft.nl/uk/textbook/stdiscan.html[^]).
You can probably find source code somewhere on the web to do this.
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thanks a lot for ur help.....
i know theoretically some how to implement it but i cant the hang of how to start of with the code my characters are not getting segmented properly
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actually segmenting the characters worked only for one image , i applied thinning to individual character to get the skeleton but then i dunno how to proceed with recognition ...
plz help if u can
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Well, you can trace vertical lines in the bitmap, and see if they intersect characters. When you find a vertical line that doesn't intersect a character, it's in a boundry between two characters. You can then do a binary search to find the X coordinate where it changes from character to boundary.
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The basic steps will be something like:
1. Locate and extract the license plate within the image (not always easy, especially if the vehicle is moving)
2. Clean and enhance the license plate image (de-skew, de-noise, improve contrast, etc.)
3. Isolate individual glyphs (characters)
4. Extract informative features from characters
5. Recognize individual characters
Step 5 could be performed any number of ways: neural network, discriminant analysis, k-nearest neighbors, etc.
-Will Dwinnell
Data Mining in MATLAB
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Please help me to write a simulator program for Pre fetching global variables using the sieve concept for the Cell BE Processor.
Or give me the working code to simulate the same.
Arun
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Please deposit £10,000 in my company's bank account and we'll provide the code. Hell, we'll even throw some support in for you. Oh wait - you did realise this isn't rentacoder didn't you?
"WPF has many lovers. It's a veritable porn star!" - Josh Smith As Braveheart once said, "You can take our freedom but you'll never take our Hobnobs!" - Martin Hughes.
My blog | My articles | MoXAML PowerToys
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Hi,
I need to calculate the width and height of a child window in a mdi setup.
Now i have got the area of the window, say 40000 pixels, and i know that the width is size N, and that the height is (0.75 x N), the 0.75 is the ratio of the windows height, which will always be calculated.
I need to do this in C#, but any code or help could be translated into c# if you know how to calcualte this in any other language.
Please feel free to ask me to expalin it better.
Cheers
Cloughy
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If you know the area, width and height, I'm not sure I understand what the problem is? Are you trying to fix an aspect ratio?
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Cloughy wrote: ... in any other language
We aim to please:
Als de hoogte driekwart is van de breedte, vermenigvuldig dan de oppervlakte met 4/3 en trek de vierkantswortel, dit geeft de breedte.
Luc Pattyn [Forum Guidelines] [My Articles]
- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
- use the code block button (PRE tags) to preserve formatting when showing multi-line code snippets
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Sorry, I only speak English.
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Yusuf
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@Zep
I know that N is the width and the height is (0.75 x N). But i dont know the value of N.
I only know the area and the height ratio.
Cheers
Cloughy
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So what you are saying is solving 0.75*N*N = known_area for N is a problem?
Did you consider dividing both sides by 0.75?
Luc Pattyn [Forum Guidelines] [My Articles]
- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
- use the code block button (PRE tags) to preserve formatting when showing multi-line code snippets
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yes N is a problem. Just to clarify, N*(0.75*N)=Known_area.
Cheers
Cloughy
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Multiply out the brackets:
0.75*N^2 = A
0.75N^2 = A
Change subject:
N^2 = A/(3/4)
N^2 = (4/3)A
N^2 = 1.33A
N = (square root of)1.33A
Substitute:
A = 40000
1.33A = 53200
N = square root of 53200
N = 230.651252
Perhaps something like that?
modified on Sunday, March 8, 2009 7:40 AM
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Thanks for the input, but no.
Once i get the width(N) and the height(0.75*N). If i multiply H x W then i would still get 40000 for my area.
What i really need is a squared root function, but more like a Rectangle root.
The most important factor, is N.
So if my area is 480000, then width(N) = 800 and the height(0.75*N)= 600.
So my window would need to be set at 800x600. BTW area is never constant. I always get the ClientArea size of the mdi control.
Cheers
Cloughy
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Cloughy wrote: but more like a Rectangle root.
What is that 
See my post below for an equation
Yusuf
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Yusuf wrote: What is that
Oh didn't you notice, analogous to square roots, they recently introduced rectangular, circular, and diamond roots to determine the size of the corresponding shapes when given the area.
Luc Pattyn [Forum Guidelines] [My Articles]
- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
- use the code block button (PRE tags) to preserve formatting when showing multi-line code snippets
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oh, my bad.
Yusuf
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Welcome in the CP's Memorable Quotes list [^].
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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Don't forget - 25.8069758. That's the root of all evil.
"WPF has many lovers. It's a veritable porn star!" - Josh Smith As Braveheart once said, "You can take our freedom but you'll never take our Hobnobs!" - Martin Hughes.
My blog | My articles | MoXAML PowerToys
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Nah. I was trying to lift the scientific character of this forum, now you had to butt in.
Next you'll claim 6.480740698407860230965967436088 is the solution root to most questions here.
Luc Pattyn [Forum Guidelines] [My Articles]
- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
- use the code block button (PRE tags) to preserve formatting when showing multi-line code snippets
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Well it is.
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Computafreak wrote: Multiply out the brackets:
0.75*N*N^2 = A
No, you started out with wrong equation.
Shouldn't it be
A = h * w
A = N * .75N
A = .75 * N^2
A/.75 = N^2
N = sqrt(A/.75)
we know A = 480000
N = sqrt(480000/.75 )
N = sqrt(360000 640000)
N = 600 800
[edit] modified calculation error, thanks Computafreak[/edit]
Yusuf
modified on Sunday, March 8, 2009 7:53 AM
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