FastHadamardTransform.java
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* this work for additional information regarding copyright ownership.
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*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
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package org.apache.commons.math4.transform;
import java.util.function.UnaryOperator;
import org.apache.commons.numbers.core.ArithmeticUtils;
/**
* <a href="http://www.archive.chipcenter.com/dsp/DSP000517F1.html">Fast Hadamard Transform</a> (FHT).
* <p>
* The FHT can also transform integer vectors into integer vectors.
* However, this transform cannot be inverted directly, due to a scaling
* factor that may lead to rational results (for example, the inverse
* transform of integer vector (0, 1, 0, 1) is vector (1/2, -1/2, 0, 0).
*/
public class FastHadamardTransform implements RealTransform {
/** Operation to be performed. */
private final UnaryOperator<double[]> op;
/**
* Default constructor.
*/
public FastHadamardTransform() {
this(false);
}
/**
* @param inverse Whether to perform the inverse transform.
*/
public FastHadamardTransform(final boolean inverse) {
op = create(inverse);
}
/**
* {@inheritDoc}
*
* @throws IllegalArgumentException if the length of the data array is
* not a power of two.
*/
@Override
public double[] apply(final double[] f) {
return op.apply(f);
}
/**
* Returns the forward transform of the given data set.
* The integer transform cannot be inverted directly, due to a scaling
* factor which may lead to double results.
*
* @param f Data array to be transformed (signal).
* @return the transformed array (spectrum).
* @throws IllegalArgumentException if the length of the data array is
* not a power of two.
*/
public int[] apply(final int[] f) {
return fht(f);
}
/**
* FHT uses only subtraction and addition.
* It requires {@code N * log2(N) = n * 2^n} additions.
*
* <h3>Short Table of manual calculation for N=8</h3>
* <ol>
* <li><b>x</b> is the input vector to be transformed,</li>
* <li><b>y</b> is the output vector (Fast Hadamard transform of <b>x</b>),</li>
* <li>a and b are helper rows.</li>
* </ol>
* <table style="text-align: center" border="">
* <caption>manual calculation for N=8</caption>
* <tbody style="text-align: center">
* <tr>
* <th>x</th>
* <th>a</th>
* <th>b</th>
* <th>y</th>
* </tr>
* <tr>
* <th>x<sub>0</sub></th>
* <td>a<sub>0</sub> = x<sub>0</sub> + x<sub>1</sub></td>
* <td>b<sub>0</sub> = a<sub>0</sub> + a<sub>1</sub></td>
* <td>y<sub>0</sub> = b<sub>0</sub >+ b<sub>1</sub></td>
* </tr>
* <tr>
* <th>x<sub>1</sub></th>
* <td>a<sub>1</sub> = x<sub>2</sub> + x<sub>3</sub></td>
* <td>b<sub>0</sub> = a<sub>2</sub> + a<sub>3</sub></td>
* <td>y<sub>0</sub> = b<sub>2</sub> + b<sub>3</sub></td>
* </tr>
* <tr>
* <th>x<sub>2</sub></th>
* <td>a<sub>2</sub> = x<sub>4</sub> + x<sub>5</sub></td>
* <td>b<sub>0</sub> = a<sub>4</sub> + a<sub>5</sub></td>
* <td>y<sub>0</sub> = b<sub>4</sub> + b<sub>5</sub></td>
* </tr>
* <tr>
* <th>x<sub>3</sub></th>
* <td>a<sub>3</sub> = x<sub>6</sub> + x<sub>7</sub></td>
* <td>b<sub>0</sub> = a<sub>6</sub> + a<sub>7</sub></td>
* <td>y<sub>0</sub> = b<sub>6</sub> + b<sub>7</sub></td>
* </tr>
* <tr>
* <th>x<sub>4</sub></th>
* <td>a<sub>0</sub> = x<sub>0</sub> - x<sub>1</sub></td>
* <td>b<sub>0</sub> = a<sub>0</sub> - a<sub>1</sub></td>
* <td>y<sub>0</sub> = b<sub>0</sub> - b<sub>1</sub></td>
* </tr>
* <tr>
* <th>x<sub>5</sub></th>
* <td>a<sub>1</sub> = x<sub>2</sub> - x<sub>3</sub></td>
* <td>b<sub>0</sub> = a<sub>2</sub> - a<sub>3</sub></td>
* <td>y<sub>0</sub> = b<sub>2</sub> - b<sub>3</sub></td>
* </tr>
* <tr>
* <th>x<sub>6</sub></th>
* <td>a<sub>2</sub> = x<sub>4</sub> - x<sub>5</sub></td>
* <td>b<sub>0</sub> = a<sub>4</sub> - a<sub>5</sub></td>
* <td>y<sub>0</sub> = b<sub>4</sub> - b<sub>5</sub></td>
* </tr>
* <tr>
* <th>x<sub>7</sub></th>
* <td>a<sub>3</sub> = x<sub>6</sub> - x<sub>7</sub></td>
* <td>b<sub>0</sub> = a<sub>6</sub> - a<sub>7</sub></td>
* <td>y<sub>0</sub> = b<sub>6</sub> - b<sub>7</sub></td>
* </tr>
* </tbody>
* </table>
*
* <h3>How it works</h3>
* <ol>
* <li>Construct a matrix with {@code N} rows and {@code n + 1} columns,
* {@code hadm[n+1][N]}.<br>
* <em>(If I use [x][y] it always means [row-offset][column-offset] of a
* Matrix with n rows and m columns. Its entries go from M[0][0]
* to M[n][N])</em></li>
* <li>Place the input vector {@code x[N]} in the first column of the
* matrix {@code hadm}.</li>
* <li>The entries of the submatrix {@code D_top} are calculated as follows
* <ul>
* <li>{@code D_top} goes from entry {@code [0][1]} to
* {@code [N / 2 - 1][n + 1]},</li>
* <li>the columns of {@code D_top} are the pairwise mutually
* exclusive sums of the previous column.</li>
* </ul>
* </li>
* <li>The entries of the submatrix {@code D_bottom} are calculated as
* follows
* <ul>
* <li>{@code D_bottom} goes from entry {@code [N / 2][1]} to
* {@code [N][n + 1]},</li>
* <li>the columns of {@code D_bottom} are the pairwise differences
* of the previous column.</li>
* </ul>
* </li>
* <li>The computation of {@code D_top} and {@code D_bottom} are best
* understood with the above example (for {@code N = 8}).
* <li>The output vector {@code y} is now in the last column of
* {@code hadm}.</li>
* <li><em>Algorithm from <a href="http://www.archive.chipcenter.com/dsp/DSP000517F1.html">chipcenter</a>.</em></li>
* </ol>
* <h3>Visually</h3>
* <table border="" style="text-align: center">
* <caption>chipcenter algorithm</caption>
* <tbody style="text-align: center">
* <tr>
* <td></td><th>0</th><th>1</th><th>2</th><th>3</th>
* <th>…</th>
* <th>n + 1</th>
* </tr>
* <tr>
* <th>0</th>
* <td>x<sub>0</sub></td>
* <td colspan="5" rowspan="5" align="center" valign="middle">
* ↑<br>
* ← D<sub>top</sub> →<br>
* ↓
* </td>
* </tr>
* <tr><th>1</th><td>x<sub>1</sub></td></tr>
* <tr><th>2</th><td>x<sub>2</sub></td></tr>
* <tr><th>…</th><td>…</td></tr>
* <tr><th>N / 2 - 1</th><td>x<sub>N/2-1</sub></td></tr>
* <tr>
* <th>N / 2</th>
* <td>x<sub>N/2</sub></td>
* <td colspan="5" rowspan="5" align="center" valign="middle">
* ↑<br>
* ← D<sub>bottom</sub> →<br>
* ↓
* </td>
* </tr>
* <tr><th>N / 2 + 1</th><td>x<sub>N/2+1</sub></td></tr>
* <tr><th>N / 2 + 2</th><td>x<sub>N/2+2</sub></td></tr>
* <tr><th>…</th><td>…</td></tr>
* <tr><th>N</th><td>x<sub>N</sub></td></tr>
* </tbody>
* </table>
*
* @param x Data to be transformed.
* @return the transformed array.
* @throws IllegalArgumentException if the length of the data array is
* not a power of two.
*/
private double[] fht(double[] x) {
final int n = x.length;
if (!ArithmeticUtils.isPowerOfTwo(n)) {
throw new TransformException(TransformException.NOT_POWER_OF_TWO,
n);
}
final int halfN = n / 2;
// Instead of creating a matrix with p+1 columns and n rows, we use two
// one dimension arrays which we are used in an alternating way.
double[] yPrevious = new double[n];
double[] yCurrent = x.clone();
// iterate from left to right (column)
for (int j = 1; j < n; j <<= 1) {
// switch columns
final double[] yTmp = yCurrent;
yCurrent = yPrevious;
yPrevious = yTmp;
// iterate from top to bottom (row)
for (int i = 0; i < halfN; ++i) {
// Dtop: the top part works with addition
final int twoI = 2 * i;
yCurrent[i] = yPrevious[twoI] + yPrevious[twoI + 1];
}
for (int i = halfN; i < n; ++i) {
// Dbottom: the bottom part works with subtraction
final int twoI = 2 * i;
yCurrent[i] = yPrevious[twoI - n] - yPrevious[twoI - n + 1];
}
}
return yCurrent;
}
/**
* Returns the forward transform of the specified integer data set.
* FHT only uses subtraction and addition.
*
* @param x Data to be transformed.
* @return the transformed array.
* @throws IllegalArgumentException if the length of the data array is
* not a power of two.
*/
private int[] fht(int[] x) {
final int n = x.length;
if (!ArithmeticUtils.isPowerOfTwo(n)) {
throw new TransformException(TransformException.NOT_POWER_OF_TWO,
n);
}
final int halfN = n / 2;
// Instead of creating a matrix with p+1 columns and n rows, we use two
// one dimension arrays which we are used in an alternating way.
int[] yPrevious = new int[n];
int[] yCurrent = x.clone();
// iterate from left to right (column)
for (int j = 1; j < n; j <<= 1) {
// switch columns
final int[] yTmp = yCurrent;
yCurrent = yPrevious;
yPrevious = yTmp;
// iterate from top to bottom (row)
for (int i = 0; i < halfN; ++i) {
// Dtop: the top part works with addition
final int twoI = 2 * i;
yCurrent[i] = yPrevious[twoI] + yPrevious[twoI + 1];
}
for (int i = halfN; i < n; ++i) {
// Dbottom: the bottom part works with subtraction
final int twoI = 2 * i;
yCurrent[i] = yPrevious[twoI - n] - yPrevious[twoI - n + 1];
}
}
// return the last computed output vector y
return yCurrent;
}
/**
* Factory method.
*
* @param inverse Whether to perform the inverse transform.
* @return the transform operator.
*/
private UnaryOperator<double[]> create(final boolean inverse) {
if (inverse) {
return f -> TransformUtils.scaleInPlace(fht(f), 1d / f.length);
} else {
return f -> fht(f);
}
}
}