| /* |
| * Licensed to the Apache Software Foundation (ASF) under one or more |
| * contributor license agreements. See the NOTICE file distributed with |
| * this work for additional information regarding copyright ownership. |
| * The ASF licenses this file to You under the Apache License, Version 2.0 |
| * (the "License"); you may not use this file except in compliance with |
| * the License. You may obtain a copy of the License at |
| * |
| * http://www.apache.org/licenses/LICENSE-2.0 |
| * |
| * Unless required by applicable law or agreed to in writing, software |
| * distributed under the License is distributed on an "AS IS" BASIS, |
| * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. |
| * See the License for the specific language governing permissions and |
| * limitations under the License. |
| */ |
| 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="1" cellpadding="3" summary="manual calculation for N=8"> |
| * <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="1" cellpadding="3" style="text-align: center" summary="chipcenter algorithm"> |
| * <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); |
| } |
| } |
| } |
