Archive for the ‘Code Samples’ Category

How To Repair Images with ImageMagick

Monday, June 15th, 2009

I have been doing a lot of importing of photos on my photo site, and have noticed some images do not display properly. SinceI love ImageMagick so much,  I set up  my gallery to use it. However, there appears to be a check in the Coppermine software that aborts reading the file upon errors in the image. So I typed the command “display imagename.jpg” for some of the files that did not import properly, and everything appeared to be fine. But I did notice something peculiar about the images: they were taken with my phone’s camera and all had errors printed out when I tried to open them, but seem to display correctly.

Since these files could be opened by ImageMagick, I figured if I saved them again, I would be able to fix the errors. I ran “convert image1.jpg image1.jpg”, got the familiar errors printed and opened the file again. This time, no errors! But since I have a  lot of files I want to import, I needed a way to convert a bunch at once.

I created a neat little Perl script to do just that. All you need is ImageMagick and perl installed. This script does not rely on PerlMagick, as it runs the convert command. So here it is:

  1. #!/usr/bin/perl
  2.  
  3. processFile("$_") foreach (@ARGV);
  4.  
  5. #recursively fix images
  6. sub processFile {
  7.    my $file=shift;
  8.    return (-d "$file")?( do{ processFile("$_") foreach (<"$file"/*>) }) : `convert "$file" "$file"`;
  9. }

I saved this file as $HOME/bin/fiximage. All you need to do to run it is type “fiximage file1 file2 directory1 directory2 …”. One other thing to note is that this script has been purposefully shortened. I like to use the ternary operator whenever possible in my scripts for just this reason. As such, it does not print anything to the command line itself, but if the convert command encounters an error, that will be printed.

I hope you have found this article useful!

Posted in Code Samples, Linux, Perl, Programming | 5 Comments »

Fun With Numbers!

Friday, May 29th, 2009

I have always had a certain love for math and the neat things you can do with it. Here is a bit of information and shortcuts I have picked up in a few of my math classes.

Pascal’s Triangle

Pascal’s Triangle is a pretty neat thing. It is very simple to construct and can be used to understand a lot of different ideas. It follows a very simple form: start with a ’1′ and add the two digits above  to get the next number. The first few line look like this:

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
…

The line numbers start at 0, and continue on ad infinitum. In order to generate this triangle, programmatically, you would use something like this:

  1. vector<int> pascal(
  2.       vector<int> prev, //the current (old) row data
  3.       int *len,           //the length of the data
  4.       int end,                    //the row to retrieve
  5.       int cur=0           //the current row we are on
  6.       );
  7.  
  8. vector<int> pascal(vector<int> prev,int *len,int end,int cur){
  9.  
  10.        //return immediately if we are at the last row
  11.       if (cur==end) return prev;
  12.  
  13.            //if the current vector length is 0, then set it to 1
  14.           if (*len==0) *len=1;
  15.  
  16.           //create a temp vector (all 1′s) to store the new row data
  17.           vector<int> t((*len)+1,1);
  18.  
  19.           //sum the two rows
  20.           for(int i=1;i<(*len);i++)
  21.             t[i]=prev[i-1]+prev[i];
  22.  
  23.           //increase the length by 1
  24.           *len=(*len)+1;
  25.  
  26.           //return the new row data
  27.           return pascal(t,len,end,cur+1);
  28.  
  29. }

Binomial Expansion

Remember binomials from algebra? They were the pair of numbers used to create or simplify polynomial expressions, something like:

( x + 3 )^3 = x^3 + 9x^2 + 27x + 27

You can use Pascal’s triangle to find the coefficients of the polynomials. Let’s begin by solving for the generic case:

(a+b)^n =
(a+b)^{n-1}(a+b) =
(a+b)^{n-2}(a+b)^2 =
(a+b)^{n-2}((a+b)(a+b)) =
( (a*a) + (a*b) + (b*a) + (b*b) )(a+b)^{n-1} =
(a^2+2ab+b^2)(a+b)^{n-1} =

See the coefficients so far, with n=2 ? They are [1 2 1], which corresponds to the second row in Pascal’s triangle. But this could be a fluke, right, so let’s jump ahead to n=5 to see if that works as well.

(a+b)^n =
(a+b)^{n-6}(a+b)^5 =
(a+b)^{n-6}( (a+b) (a+b) (a+b) (a+b) (a+b) ) =
(a+b)^{n-6}( ( (a+b)(a+b) )( (a+b)(a+b) ) (a+b) ) =
[we know what (a+b)^2 is, so: ]
(( a^2+2ab+b^2 )( a^2+2ab+b^2 )(a+b) )(a+b)^{n-6} =
( ( (a^2*a^2)+(a^2*2ab) + (a^2*b^2) + ( 2ab*a^2) + (2ab*2ab) +
    (2ab*b^2) + (b^2*a^2)+(b^2*2ab) + (b^2*b^2) ) (a+b) )(a+b)^{n-6} =
( ( a^4 + 2a^3b+ a^2b^2 + 2a^3b + 4a^2b^2 + 2ab^3 + b^2a^2 + 2ab^3+ b^4) (a+b) )(a+b)^{n-6} =
( ( a^4 + 4a^3b + 6a2b^2 + 4ab^3+b^4) (a+b) )(a+b)^{n-6} =
[note: notice that the coefficients of (a+b)^4 are (1 4 6 4 1) ! ]
( ( a^4*a + a^4*b + 4a^3b*a + 4a^3b*b + 6a^2b^2*a + 6a^2b^2*b + 4ab^3*a + 4ab^3*b + b^4*a+b^4*b))(a+b)^{n-6} =
( ( a^5 + a^4b + 4a^4b + 4a^3b^2 + 6a^3b^2 + 6a^2b^3 + 4a^2b^3 + 4ab^4 + ab^4+b^5))(a+b)^{n-6}=
( ( a^5 +5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5)(a+b)^{n-6}

There it is! The coefficients correspond to the rows on Pascal’s Triangle!

Features

Now, to make things a little simpler, I will note some interesting “features” about what we just did.

General Formula for Binomial Expansion

The general formula for binomial expansion is:

(a+b)^n = \sum_{i=0}^{n}(P_{ni}a^{n-i}b^i)

Where P_{ni} is the coefficient at row n (starting from 0) and column i in Pascal’s Triangle. The formula means to add from i=0 to n all the terms (P_{ni}a^{n-i}b^i) , replacing i with the number you are at. For example, supposing i=3, you would get:

\sum_{i=0}^{3}(P_{3i}a^{3-i}b^i) =
(P_{(3,0)}a^{3-0}b^0) + (P_{(3,1)}a^{3-1}b^1) + (P_{(3,2)}a^{3-2}b^2) + (P_{(3,3)}a^{3-3}b^3) =
since P_3 = [1 3 3 1] , we finally get:
((1)a^{3-0}b^0) + ((3)a^{3-1}b^1) + ((3)a^{3-2}b^2) + ((1)a^{3-3}b^3)
cleaning up a bit :
a^3 + 3a^2b + 3ab^2 + b^3

Exponents

Note in all the expansions, the first variable counts down from n to 0, while the second variable counts up from 0 to n.

Does your binomial already have coefficients?

If your binomial already has coefficients, simply add the coefficient in with the variable it is with, so it merely multiplies the term by that amount. For example:

(3a+b)^3 =

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Posted in Code Samples, Math | 1 Comment »