用汉明距离进行图片相似度检测
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<size;i++) {
c[i]=1;
}
c[0]=1/Math.sqrt(2.0);
}
private double[][] applyDCT(double[][] f) {
int N = size;
double[][] F = new double[N][N];
for (int u=0;u<N;u++) {
for (int v=0;v<N;v++) {
double sum = 0.0;
for (int i=0;i<N;i++) {
for (int j=0;j<N;j++) {
sum+=Math.cos(((2*i+1)/(2.0*N))*u*Math.PI)*Math.cos(((2*j+1)/(2.0*N))*v*Math.PI)*(f[i][j]);
}
}
sum*=((c[u]*c[v])/4.0);
F[u][v] = sum;
}
}
return F;
}
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
图1
图2
图3
参考地址：
代码参考：http://pastebin.com/Pj9d8jt5
原理参考：http://www.ruanyifeng.com/blog/2011/07/principle_of_similar_image_search.html
汉明距离：http://baike.baidu.com/view/725269.htm
相似度计算常用方法综述：http://stblog.baidu-tech.com/?p=1846
漫谈机器学习中的距离和相似性度量方法：http://blog.jobbole.com/84876/