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麼呢？計算機怎麼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。
這裏的關鍵技術叫做”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，然後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。
將縮小後的圖片，轉爲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。大於或等於平均值，記爲1；小於平均值，記爲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。組合的次序並不重要，只要保證所有圖片都採用同樣次序就行了。
= = 8f373714acfcf4d0
得到指紋以後，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，這等同於計算”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；如果大於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，往往採用更強大的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，然後再進行比較。
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 ([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();
}
}
}
運行結果爲：
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；如果大於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