Java实现用汉明距离进行图片相似度检测的

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：

package reyo.sdk.utils.ai.pic;

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;

/* 
 * 
 * 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。
 * 
 * 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 {

    // 项目根目录路径
    public static final String path = System.getProperty("user.dir");
    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) {

        // 项目根目录路径
        String filename = ImagePHash.path + "\\images\\";
        ImagePHash p = new ImagePHash();
        String image1;
        String image2;
        try {
            for (int i = 0; i < 10; i++) {
                image1 = p.getHash(new FileInputStream(new File(filename + "example" + (i + 1) + ".jpg")));
                image2 = p.getHash(new FileInputStream(new File(filename + "source.jpg")));
                System.out.println("example" + (i + 1) + ".jpg:source.jpg Score is " + p.distance(image1, image2));
            }

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

    }
}

运行结果为：

DCT: 249
DCT: 237
example1.jpg:source.jpg Score is 25
DCT: 102
DCT: 103
example2.jpg:source.jpg Score is 16
DCT: 103
DCT: 104
example3.jpg:source.jpg Score is 17
DCT: 104
DCT: 103
example4.jpg:source.jpg Score is 2
DCT: 103
DCT: 103
example5.jpg:source.jpg Score is 0
DCT: 104
DCT: 104
example6.jpg:source.jpg Score is 10
DCT: 105
DCT: 104
example7.jpg:source.jpg Score is 25
DCT: 103
DCT: 103
example8.jpg:source.jpg Score is 28
DCT: 102
DCT: 103
example9.jpg:source.jpg Score is 25
DCT: 102
DCT: 103
example10.jpg:source.jpg Score is 31

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。

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

来自：http://stackoverflow.com/questions/6971966/how-to-measure-percentage-similarity-between-two-images