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
