图片相似度(汉明距离)

Google、Baidu 等搜索引擎相继推出了以图搜图的功能,测试了下效果还不错~ 那这种技术的原理是什么呢?计算机怎么知道两张图片相似呢?

根据Neal Krawetz博士的解释,原理非常简单易懂。我们可以用一个快速算法,就达到基本的效果。

这里的关键技术叫做”感知哈希算法”(Perceptual hash algorithm),它的作用是对每张图片生成一个”指纹”(fingerprint)字符串,然后比较不同图片的指纹。结果越接近,就说明图片越相似。

下面是一个最简单的实现:

第一步,缩小尺寸。

将图片缩小到8x8的尺寸,总共64个像素。这一步的作用是去除图片的细节,只保留结构、明暗等基本信息,摒弃不同尺寸、比例带来的图片差异。

第二步,简化色彩。

将缩小后的图片,转为64级灰度。也就是说,所有像素点总共只有64种颜色。

第三步,计算平均值。

计算所有64个像素的灰度平均值。

第四步,比较像素的灰度。

将每个像素的灰度,与平均值进行比较。大于或等于平均值,记为1;小于平均值,记为0。

第五步,计算哈希值。

将上一步的比较结果,组合在一起,就构成了一个64位的整数,这就是这张图片的指纹。组合的次序并不重要,只要保证所有图片都采用同样次序就行了。

= = 8f373714acfcf4d0

得到指纹以后,就可以对比不同的图片,看看64位中有多少位是不一样的。在理论上,这等同于计算”汉明距离”(Hamming distance)。如果不相同的数据位不超过5,就说明两张图片很相似;如果大于10,就说明这是两张不同的图片。

具体的代码实现,可以参见Wote用python语言写的imgHash.py。代码很短,只有53行。使用的时候,第一个参数是基准图片,第二个参数是用来比较的其他图片所在的目录,返回结果是两张图片之间不相同的数据位数量(汉明距离)。

这种算法的优点是简单快速,不受图片大小缩放的影响,缺点是图片的内容不能变更。如果在图片上加几个文字,它就认不出来了。所以,它的最佳用途是根据缩略图,找出原图。

实际应用中,往往采用更强大的pHash算法和SIFT算法,它们能够识别图片的变形。只要变形程度不超过25%,它们就能匹配原图。这些算法虽然更复杂,但是原理与上面的简便算法是一样的,就是先将图片转化成Hash字符串,然后再进行比较。

下面我们来看下上述理论用java来做一个DEMO版的具体实现:

import java.awt.Graphics2D;
import java.awt.color.ColorSpace;
import java.awt.image.BufferedImage;
import java.awt.image.ColorConvertOp;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.InputStream;

import javax.imageio.ImageIO;
/*
* pHash-like image hash.
* Author: Elliot Shepherd ([email protected]
* Based On: http://www.hackerfactor.com/blog/index.php?/archives/432-Looks-Like-It.html
*/
public class ImagePHash {

private int size = 32;
private int smallerSize = 8;

public ImagePHash() {
initCoefficients();
}

public ImagePHash(int size, int smallerSize) {
this.size = size;
this.smallerSize = smallerSize;

   initCoefficients();

}

public int distance(String s1, String s2) {
int counter = 0;
for (int k = 0; k < s1.length();k++) {
if(s1.charAt(k) != s2.charAt(k)) {
counter++;
}
}
return counter;
}

// Returns a ‘binary string’ (like. 001010111011100010) which is easy to do a hamming distance on.
public String getHash(InputStream is) throws Exception {
BufferedImage img = ImageIO.read(is);

   /* 1. Reduce size. 
    * Like Average Hash, pHash starts with a small image. 
    * However, the image is larger than 8x8; 32x32 is a good size. 
    * This is really done to simplify the DCT computation and not 
    * because it is needed to reduce the high frequencies.
    */
   img = resize(img, size, size);

   /* 2. Reduce color. 
    * The image is reduced to a grayscale just to further simplify 
    * the number of computations.
    */
   img = grayscale(img);

   double[][] vals = new double[size][size];

   for (int x = 0; x < img.getWidth(); x++) {
       for (int y = 0; y < img.getHeight(); y++) {
           vals[x][y] = getBlue(img, x, y);
       }
   }

   /* 3. Compute the DCT. 
    * The DCT separates the image into a collection of frequencies 
    * and scalars. While JPEG uses an 8x8 DCT, this algorithm uses 
    * a 32x32 DCT.
    */
   long start = System.currentTimeMillis();
   double[][] dctVals = applyDCT(vals);
   System.out.println("DCT: " + (System.currentTimeMillis() - start));

   /* 4. Reduce the DCT. 
    * This is the magic step. While the DCT is 32x32, just keep the 
    * top-left 8x8. Those represent the lowest frequencies in the 
    * picture.
    */
   /* 5. Compute the average value. 
    * Like the Average Hash, compute the mean DCT value (using only 
    * the 8x8 DCT low-frequency values and excluding the first term 
    * since the DC coefficient can be significantly different from 
    * the other values and will throw off the average).
    */
   double total = 0;

   for (int x = 0; x < smallerSize; x++) {
       for (int y = 0; y < smallerSize; y++) {
           total += dctVals[x][y];
       }
   }
   total -= dctVals[0][0];

   double avg = total / (double) ((smallerSize * smallerSize) - 1);

   /* 6. Further reduce the DCT. 
    * This is the magic step. Set the 64 hash bits to 0 or 1 
    * depending on whether each of the 64 DCT values is above or 
    * below the average value. The result doesn't tell us the 
    * actual low frequencies; it just tells us the very-rough 
    * relative scale of the frequencies to the mean. The result 
    * will not vary as long as the overall structure of the image 
    * remains the same; this can survive gamma and color histogram 
    * adjustments without a problem.
    */
   String hash = "";

   for (int x = 0; x < smallerSize; x++) {
       for (int y = 0; y < smallerSize; y++) {
           if (x != 0 && y != 0) {
               hash += (dctVals[x][y] > avg?"1":"0");
           }
       }
   }

   return hash;

}

private BufferedImage resize(BufferedImage image, int width, int height) {
BufferedImage resizedImage = new BufferedImage(width, height, BufferedImage.TYPE_INT_ARGB);
Graphics2D g = resizedImage.createGraphics();
g.drawImage(image, 0, 0, width, height, null);
g.dispose();
return resizedImage;
}

private ColorConvertOp colorConvert = new ColorConvertOp(ColorSpace.getInstance(ColorSpace.CS_GRAY), null);

private BufferedImage grayscale(BufferedImage img) {
colorConvert.filter(img, img);
return img;
}

private static int getBlue(BufferedImage img, int x, int y) {
return (img.getRGB(x, y)) & 0xff;
}

// DCT function stolen from http://stackoverflow.com/questions/4240490/problems-with-dct-and-idct-algorithm-in-java

private double[] c;
private void initCoefficients() {
c = new double[size];

   for (int i=1;i

}

private double[][] applyDCT(double[][] f) {
int N = size;

   double[][] F = new double[N][N];
   for (int u=0;u

}

public static void main(String[] args) {

   ImagePHash p = new ImagePHash();
   String image1;
   String image2;
   try {
       image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
       image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
       System.out.println("1:1 Score is " + p.distance(image1, image2));
       image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
       image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/2.jpg")));
       System.out.println("1:2 Score is " + p.distance(image1, image2));
       image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
       image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/3.jpg")));
       System.out.println("1:3 Score is " + p.distance(image1, image2));
       image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/2.jpg")));
       image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/3.jpg")));
       System.out.println("2:3 Score is " + p.distance(image1, image2));

       image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/4.jpg")));
       image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/5.jpg")));
       System.out.println("4:5 Score is " + p.distance(image1, image2));

   } catch (FileNotFoundException e) {
       e.printStackTrace();
   } catch (Exception e) {
       e.printStackTrace();
   }

}
}

运行结果为:

DCT: 163
DCT: 158
1:1 Score is 0
DCT: 168
DCT: 164
1:2 Score is 4
DCT: 156
DCT: 156
1:3 Score is 3
DCT: 157
DCT: 157
2:3 Score is 1
DCT: 157
DCT: 156
4:5 Score is 21
说明:其中1,2,3是3张非常相似的图片,图片分别加了不同的文字水印,肉眼分辨的不是太清楚,下面会有附图,4、5是两张差异很大的图,图你可以随便找来测试,这两张我就不上传了。

结果说明:汉明距离越大表明图片差异越大,如果不相同的数据位不超过5,就说明两张图片很相似;如果大于10,就说明这是两张不同的图片。从结果可以看到1、2、3是相似图片,4、5差异太大,是两张不同的图片。

附:图1、2、3

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