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mathutils.h (4630B)


      1 /*
      2 * Copyright (c) 2017, Alliance for Open Media. All rights reserved.
      3 *
      4 * This source code is subject to the terms of the BSD 2 Clause License and
      5 * the Alliance for Open Media Patent License 1.0. If the BSD 2 Clause License
      6 * was not distributed with this source code in the LICENSE file, you can
      7 * obtain it at www.aomedia.org/license/software. If the Alliance for Open
      8 * Media Patent License 1.0 was not distributed with this source code in the
      9 * PATENTS file, you can obtain it at www.aomedia.org/license/patent.
     10 */
     11 
     12 #ifndef AOM_AOM_DSP_MATHUTILS_H_
     13 #define AOM_AOM_DSP_MATHUTILS_H_
     14 
     15 #include <assert.h>
     16 #include <math.h>
     17 #include <string.h>
     18 
     19 #include "aom_dsp/aom_dsp_common.h"
     20 
     21 static const double TINY_NEAR_ZERO = 1.0E-16;
     22 
     23 // Solves Ax = b, where x and b are column vectors of size nx1 and A is nxn
     24 static inline int linsolve(int n, double *A, int stride, double *b, double *x) {
     25  int i, j, k;
     26  double c;
     27  // Forward elimination
     28  for (k = 0; k < n - 1; k++) {
     29    // Bring the largest magnitude to the diagonal position
     30    for (i = n - 1; i > k; i--) {
     31      if (fabs(A[(i - 1) * stride + k]) < fabs(A[i * stride + k])) {
     32        for (j = 0; j < n; j++) {
     33          c = A[i * stride + j];
     34          A[i * stride + j] = A[(i - 1) * stride + j];
     35          A[(i - 1) * stride + j] = c;
     36        }
     37        c = b[i];
     38        b[i] = b[i - 1];
     39        b[i - 1] = c;
     40      }
     41    }
     42    for (i = k; i < n - 1; i++) {
     43      if (fabs(A[k * stride + k]) < TINY_NEAR_ZERO) return 0;
     44      c = A[(i + 1) * stride + k] / A[k * stride + k];
     45      for (j = 0; j < n; j++) A[(i + 1) * stride + j] -= c * A[k * stride + j];
     46      b[i + 1] -= c * b[k];
     47    }
     48  }
     49  // Backward substitution
     50  for (i = n - 1; i >= 0; i--) {
     51    if (fabs(A[i * stride + i]) < TINY_NEAR_ZERO) return 0;
     52    c = 0;
     53    for (j = i + 1; j <= n - 1; j++) c += A[i * stride + j] * x[j];
     54    x[i] = (b[i] - c) / A[i * stride + i];
     55  }
     56 
     57  return 1;
     58 }
     59 
     60 ////////////////////////////////////////////////////////////////////////////////
     61 // Least-squares
     62 // Solves for n-dim x in a least squares sense to minimize |Ax - b|^2
     63 // The solution is simply x = (A'A)^-1 A'b or simply the solution for
     64 // the system: A'A x = A'b
     65 //
     66 // This process is split into three steps in order to avoid needing to
     67 // explicitly allocate the A matrix, which may be very large if there
     68 // are many equations to solve.
     69 //
     70 // The process for using this is (in pseudocode):
     71 //
     72 // Allocate mat (size n*n), y (size n), a (size n), x (size n)
     73 // least_squares_init(mat, y, n)
     74 // for each equation a . x = b {
     75 //    least_squares_accumulate(mat, y, a, b, n)
     76 // }
     77 // least_squares_solve(mat, y, x, n)
     78 //
     79 // where:
     80 // * mat, y are accumulators for the values A'A and A'b respectively,
     81 // * a, b are the coefficients of each individual equation,
     82 // * x is the result vector
     83 // * and n is the problem size
     84 static inline void least_squares_init(double *mat, double *y, int n) {
     85  memset(mat, 0, n * n * sizeof(double));
     86  memset(y, 0, n * sizeof(double));
     87 }
     88 
     89 // Round the given positive value to nearest integer
     90 static AOM_FORCE_INLINE int iroundpf(float x) {
     91  assert(x >= 0.0);
     92  return (int)(x + 0.5f);
     93 }
     94 
     95 static inline void least_squares_accumulate(double *mat, double *y,
     96                                            const double *a, double b, int n) {
     97  for (int i = 0; i < n; i++) {
     98    for (int j = 0; j < n; j++) {
     99      mat[i * n + j] += a[i] * a[j];
    100    }
    101  }
    102  for (int i = 0; i < n; i++) {
    103    y[i] += a[i] * b;
    104  }
    105 }
    106 
    107 static inline int least_squares_solve(double *mat, double *y, double *x,
    108                                      int n) {
    109  return linsolve(n, mat, n, y, x);
    110 }
    111 
    112 // Matrix multiply
    113 static inline void multiply_mat(const double *m1, const double *m2, double *res,
    114                                const int m1_rows, const int inner_dim,
    115                                const int m2_cols) {
    116  double sum;
    117 
    118  int row, col, inner;
    119  for (row = 0; row < m1_rows; ++row) {
    120    for (col = 0; col < m2_cols; ++col) {
    121      sum = 0;
    122      for (inner = 0; inner < inner_dim; ++inner)
    123        sum += m1[row * inner_dim + inner] * m2[inner * m2_cols + col];
    124      *(res++) = sum;
    125    }
    126  }
    127 }
    128 
    129 static inline float approx_exp(float y) {
    130 #define A ((1 << 23) / 0.69314718056f)  // (1 << 23) / ln(2)
    131 #define B \
    132  127  // Offset for the exponent according to IEEE floating point standard.
    133 #define C 60801  // Magic number controls the accuracy of approximation
    134  union {
    135    float as_float;
    136    int32_t as_int32;
    137  } container;
    138  container.as_int32 = ((int32_t)(y * A)) + ((B << 23) - C);
    139  return container.as_float;
    140 #undef A
    141 #undef B
    142 #undef C
    143 }
    144 #endif  // AOM_AOM_DSP_MATHUTILS_H_