PHash.java

package haven.geoloc;

import java.awt.Graphics2D;
import java.awt.color.ColorSpace;
import java.awt.image.BufferedImage;
import java.awt.image.ColorConvertOp;

/*
 * Based on Elliot Shepherd's implementation of Neal Krawetz perceptual hashing algorithm.
 * http://www.hackerfactor.com/blog/index.php?/archives/432-Looks-Like-It.html
 */
public class PHash {
	private int size = 32;
	private int smallerSize = 8;
	private ColorConvertOp colorConvert = new ColorConvertOp(ColorSpace.getInstance(ColorSpace.CS_GRAY), null);
	
	public PHash(int size, int smallerSize) {
		this.size = size;
		this.smallerSize = smallerSize;

		initCoefficients();
	}

	public long getHash(BufferedImage img) {
		/*
		 * 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);

		/*
		 * 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] = ((img.getRGB(x, y)) & 0x0000ff00) >> 0x08;
			}
		}

		/*
		 * 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.
		 */
		double[][] dctVals = dct(vals);

		/*
		 * Reduce the DCT. While the DCT is 32x32, just keep the top-left 8x8. Those represent
		 * the lowest frequencies in the picture.
		 * 
		 * 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);

		/*
		 * 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.
		 */
		long hash = 0L;
		int i = 0;
		for (int x = 0; x < smallerSize; x++) {
			for (int y = 0; y < smallerSize; y++) {
				 if (x != 0 && y != 0) {
					 if (dctVals[x][y] > avg)
						 hash ^= 1L << (63 - i);
					 
					 i++;
				 }
			}
		}
		
		return hash;
	}

	private BufferedImage resize(BufferedImage img, int width, int height) {
		BufferedImage resizedImage = new BufferedImage(width, height, BufferedImage.TYPE_INT_ARGB);
		Graphics2D g = resizedImage.createGraphics();
		g.drawImage(img, 0, 0, width, height, null);
		g.dispose();
		return resizedImage;
	}

	private BufferedImage grayscale(BufferedImage img) {
		colorConvert.filter(img, img);
		return img;
	}

	// 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[][] dct(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;
	}
}