skibulk - 1 year ago 95

PHP Question

What I'd like here is a working, optimized version of my current code. While my function does return an array with actual results, I don't know if they are correct (I'm not a mathematics guru and I don't know Java code to compare my results against known implementations). Secondly, I'd like the function to be able to accept custom table sizes, but I don't know how to do that. Is table size equivalent to resampling the image? Am I applying the coefficients correctly?

`// a lot of processing is required for large images`

$image = imagecreatetruecolor(21, 21);

$black = imagecolorallocate($image, 0, 0, 0);

$white = imagecolorallocate($image, 255, 255, 255);

imagefilledellipse($image, 10, 10, 15, 15, $white);

print_r(imgDTC($image));

function imgDTC($img, $tableSize){

// m1 = Matrix1, an associative array with pixel data from the image

// m2 = Matrix2, an associative array with DCT Frequencies

// x1, y1 = coordinates in matrix1

// x2, y2 = coordinates in matrix2

$m1 = array();

$m2 = array();

// iw = image width

// ih = image height

$iw = imagesx($img);

$ih = imagesy($img);

// populate matrix1

for ($x1=0; $x1<$iw; $x1++) {

for ($y1=0; $y1<$ih; $y1++) {

$m1[$x1][$y1] = imagecolorat($img, $x1, $y1) & 0xff;

}

}

// populate matrix2

// for each coordinate in matrix2

for ($x2=0;$x2<$iw;$x2++) {

for ($y2=0;$y2<$ih;$y2++) {

// for each coordinate in matrix1

$sum = 1;

for ($x1=0;$x1<$iw;$x1++) {

for ($y1=0;$y1<$ih;$y1++) {

$sum +=

cos(((2*$x1+1)/(2*$iw))*$x2*pi()) *

cos(((2*$y1+1)/(2*$ih))*$y2*pi()) *

$m1[$x1][$y1]

;

}

}

// apply coefficients

$sum *= .25;

if ($x2 == 0 || $y2 == 0) {

$sum *= 1/sqrt(2);

}

$m2[$x2][$y2] = $sum;

}

}

return $m2;

}

My PHP function is a derivitive from this post in Java: Problems with DCT and IDCT algorithm in java. I have rewritten the code for php and readability. Ultimately, I am working on a script which will enable me to compare images and find similarities. The technique is outlined here: http://www.hackerfactor.com/blog/index.php?/archives/432-Looks-Like-It.html.

Thanks!

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Answer Source

This is how i performed my DCT what i'm doing here is to perform a 1 dimention DCT on each row. Then I took the result an perform the DTC on each column it's faster.

```
function dct1D($in){
$results = array();
$N = count($in);
for($k = 0; $k < $N; $k++){
$sum = 0;
for($n = 0; $n < $N; $n++){
$sum += $in[$n] * cos($k * pi() * ($n + 0.5) / ($N));
}
$sum *= sqrt(2 / $N);
if($k == 0){
$sum *= 1 / sqrt(2);
}
$results[$k] = $sum;
}
return $results;
}
function optimizedImgDTC($img){
$results = array();
$N1 = imagesx($img);
$N2 = imagesy($img);
$rows = array();
$row = array();
for($j = 0; $j < $N2; $j++){
for($i = 0; $i < $N1; $i++)
$row[$i] = imagecolorat($img, $i, $j);
$rows[$j] = dct1D($row);
}
for($i = 0; $i < $N1; $i++){
for($j = 0; $j < $N2; $j++)
$col[$j] = $rows[$j][$i];
$results[$i] = dct1D($col);
}
return $results;
}
```

Most algorithm i found on internet assume that the input matrix is 8x8. That's why you multiplyed by 0.25. In general you should multiply by sqrt(2 / N) a 1D matrix and here we are in 2D so sqrt(2/N1) * sqrt(2/N2). If you do this for N1 = 8 and N2 = 8: sqrt(2/8)^2 = 2/8 = 1/4 = 0.25

The other thing was to multiply by 1/sqrt(2) X0 it's for 1D matrix here we are in 2D so you multiply when k1 = 0 or k2 = 0. When k1 = 0 and k2 = 0 you have to do it twice.

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