用漢明距離進行圖片相似度檢測的Java實現

jopen 11年前發布 | 157K 次閱讀算法圖形/圖像處理

Google、Baidu 等搜索引擎相繼推出了以圖搜圖的功能，測試了下效果還不錯~ 那這種技術的原理是什么呢？計算機怎么知道兩張圖片相似呢？

根據Neal Krawetz博士的解釋，原理非常簡單易懂。我們可以用一個快速算法，就達到基本的效果。

這里的關鍵技術叫做"感知哈希算法"（Perceptual hash algorithm），它的作用是對每張圖片生成一個"指紋"（fingerprint）字符串，然后比較不同圖片的指紋。結果越接近，就說明圖片越相似。

下面是一個最簡單的實現：

第一步，縮小尺寸。

將圖片縮小到8x8的尺寸，總共64個像素。這一步的作用是去除圖片的細節，只保留結構、明暗等基本信息，摒棄不同尺寸、比例帶來的圖片差異。

第二步，簡化色彩。

將縮小后的圖片，轉為64級灰度。也就是說，所有像素點總共只有64種顏色。

第三步，計算平均值。

計算所有64個像素的灰度平均值。

第四步，比較像素的灰度。

將每個像素的灰度，與平均值進行比較。大于或等于平均值，記為1；小于平均值，記為0。

第五步，計算哈希值。

將上一步的比較結果，組合在一起，就構成了一個64位的整數，這就是這張圖片的指紋。組合的次序并不重要，只要保證所有圖片都采用同樣次序就行了。

= = 8f373714acfcf4d0

得到指紋以后，就可以對比不同的圖片，看看64位中有多少位是不一樣的。在理論上，這等同于計算"漢明距離"（Hamming distance）。如果不相同的數據位不超過5，就說明兩張圖片很相似；如果大于10，就說明這是兩張不同的圖片。

具體的代碼實現，可以參見Wote用python語言寫的imgHash.py。代碼很短，只有53行。使用的時候，第一個參數是基準圖片，第二個參數是用來比較的其他圖片所在的目錄，返回結果是兩張圖片之間不相同的數據位數量（漢明距離）。

這種算法的優點是簡單快速，不受圖片大小縮放的影響，缺點是圖片的內容不能變更。如果在圖片上加幾個文字，它就認不出來了。所以，它的最佳用途是根據縮略圖，找出原圖。

實際應用中，往往采用更強大的pHash算法和SIFT算法，它們能夠識別圖片的變形。只要變形程度不超過25%，它們就能匹配原圖。這些算法雖然更復雜，但是原理與上面的簡便算法是一樣的，就是先將圖片轉化成Hash字符串，然后再進行比較。

下面我們來看下上述理論用java來做一個DEMO版的具體實現：

 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 (elliot@jarofworms.com
* 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();
       }

   }
}

運行結果為：

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

說明：其中1,2,3是3張非常相似的圖片，圖片分別加了不同的文字水印，肉眼分辨的不是太清楚，下面會有附圖，4、5是兩張差異很大的圖，圖你可以隨便找來測試，這兩張我就不上傳了。

結果說明：漢明距離越大表明圖片差異越大，如果不相同的數據位不超過5，就說明兩張圖片很相似；如果大于10，就說明這是兩張不同的圖片。從結果可以看到1、2、3是相似圖片，4、5差異太大，是兩張不同的圖片。

附：圖1、2、3