6.5.4 LU Decomposition

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/*

The LU decomposition of a matrix $a$ with row-partial pivoting is a factorization of $a$ (after some
rows are possibly permuted by a permutation matrix $p$) as a product of a lower triangular matrix
$l$ and an upper triangular matrix $u$. This factorization can be used to tackle many common
problems in linear algebra such as solving systems of linear equations and computing determinants.
An improvement on basic row reduction, LU decomposition by row-partial pivoting keeps the relative
magnitude of matrix values small, thus reducing the relative error due to rounding in computed
solutions.

- `lu_decompose(a, &p1col)` assigns the $m$ by $n$ matrix `a` to merged LU decomposition matrix
  `lu`, returning either 0 or 1 denoting the "sign" of the permutation parity (0 if the number of
  overall row swaps performed is even, or 1 if it is odd), or $-1$ denoting a degenerate matrix
  (i.e. singular for square matrices). The merged matrix `lu` has `lu[i][j]` = `l[i][j]` for `i` >
  `j` and `lu[i][j]` = `u[i][j]` for `i` $\leq$ `j`. Note that the algorithm always yields an atomic
  lower triangular matrix for which the diagonal entries `l[i][i]` are always equal to 1, so this is
  not explicitly stored in the resulting merged matrix. For general $i$ and $j$, the values of the
  lower and upper triangular matrices should be accessed via the `getl(lu, i, j)` and
  `getu(lu, i, j)` functions. Optionally, a `vector<int>` pointer `p1col` may be passed to return
  the permutation vector `p1col` where `p1col[i]` stores the only column that is equal to 1 in row
  $i$ of the permutation matrix $p$ (all other columns in row $i$ of $p$ are implicitly 0). The
  permutation matrix $p$ corresponding to `p1col` satisfies $pa = lu$.
- `solve_system(a, b, &x)` solves the system of linear equations $ax = b$ given an $m$ by $n$ matrix
  `a` of real values, and a length $m$ vector `b`, returning 0 if there is one solution or $-1$ if
  there are zero or infinite solutions. If there is exactly one solution, then the output pointer
  `x` is populated with the solution vector of length $n$.
- `det(a)` returns the determinant of an $n$ by $n$ matrix `a` using LU decomposition.
- `invert(a)` assigns the $n$ by $n$ matrix `a` to its inverse (if it exists), returning 0 if the
  inversion was successful or $-1$ if `a` has no inverse.

Time Complexity:
- O(m^2*n) per call to `lu_decompose(a)` and `solve_system(a, b)`, where $m$ and $n$ are the number
  of rows and columns of `a`, respectively.
- O(n^3) per call to `det(a)` and `invert(a)`, where $n$ is the length of square matrix `a`.

Space Complexity:
- O(1) auxiliary for `lu_decompose()`.
- O(n^2) for `det(a)` and `invert(a)`.
- O(m*n) auxiliary heap space for `solve_system(a, b)`.

*/

#include <algorithm>
#include <cmath>
#include <cstddef>
#include <limits>
#include <numeric>
#include <vector>

template<class Matrix>
int lu_decompose(Matrix &a, std::vector<int> *p1col = nullptr, const double EPS = 1e-10) {
  int rows = static_cast<int>(a.size());
  int cols = a.empty() ? 0 : static_cast<int>(a[0].size());
  int parity = 0;
  if (p1col != nullptr) {
    p1col->resize(rows);
    std::iota(p1col->begin(), p1col->end(), 0);
  }
  for (int i = 0; i < rows && i < cols; i++) {
    int pi = i;
    for (int k = i + 1; k < rows; k++) {
      if (fabs(a[k][i]) > fabs(a[pi][i])) {
        pi = k;
      }
    }
    if (fabs(a[pi][i]) < EPS) {
      return -1;
    }
    if (pi != i) {
      if (p1col != nullptr) {
        std::iter_swap(p1col->begin() + i, p1col->begin() + pi);
      }
      std::iter_swap(a.begin() + i, a.begin() + pi);
      parity = 1 - parity;
    }
    for (int j = i + 1; j < rows; j++) {
      a[j][i] /= a[i][i];
      for (int k = i + 1; k < cols; k++) {
        a[j][k] -= a[j][i] * a[i][k];
      }
    }
  }
  return parity;
}

template<class Matrix>
double getl(const Matrix &lu, int i, int j) {
  return i > j ? lu[i][j] : (i < j ? 0 : 1);
}

template<class Matrix>
double getu(const Matrix &lu, int i, int j) {
  return i <= j ? lu[i][j] : 0;
}

template<class Matrix, class T>
int solve_system(
    const Matrix &a, const std::vector<T> &b, std::vector<T> *x, const double EPS = 1e-10
) {
  int rows = static_cast<int>(a.size());
  int cols = a.empty() ? 0 : static_cast<int>(a[0].size());
  if (x == nullptr || a.empty() || a.size() != b.size() || rows < cols) {
    return -1;
  }
  x->resize(cols);
  std::vector<int> p1col;
  Matrix lu;
  int status = lu_decompose(lu = a, &p1col, EPS);
  if (status < 0) {
    return status;
  }
  for (int i = 0; i < cols; i++) {
    (*x)[i] = b[p1col[i]];
    for (int k = 0; k < i; k++) {
      (*x)[i] -= getl(lu, i, k) * (*x)[k];
    }
  }
  for (int i = cols - 1; i >= 0; i--) {
    for (int k = i + 1; k < cols; k++) {
      (*x)[i] -= getu(lu, i, k) * (*x)[k];
    }
    (*x)[i] /= getu(lu, i, i);
  }
  for (int i = 0; i < rows; i++) {
    double val = 0;
    for (int j = 0; j < cols; j++) {
      val += a[i][j] * (*x)[j];
    }
    // Mixed absolute/relative tolerance: dividing by b[i] alone would skip the check for negative
    // b[i] (ratio goes negative) and divide by zero when b[i] == 0.
    if (fabs(val - b[i]) > EPS * (1.0 + fabs(b[i]))) {
      return -1;
    }
  }
  return 0;
}

template<class T>
double det(const T &a) {
  int n = static_cast<int>(a.size());
  T lu;
  int status = lu_decompose(lu = a);
  if (status < 0) {
    return 0;
  }
  double res = 1;
  for (int i = 0; i < n; i++) {
    res *= lu[i][i];
  }
  return status == 0 ? res : -res;
}

template<class T>
int invert(T &a) {
  int n = static_cast<int>(a.size());
  std::vector<int> p1col;
  int status = lu_decompose(a, &p1col);
  if (status < 0) {
    return status;
  }
  T ia(n, typename T::value_type(n, 0));
  for (int j = 0; j < n; j++) {
    for (int i = 0; i < n; i++) {
      if (p1col[i] == j) {
        ia[i][j] = 1.0;
      } else {
        ia[i][j] = 0.0;
      }
      for (int k = 0; k < i; k++) {
        ia[i][j] -= getl(a, i, k) * ia[k][j];
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int k = i + 1; k < n; k++) {
        ia[i][j] -= getu(a, i, k) * ia[k][j];
      }
      ia[i][j] /= getu(a, i, i);
    }
  }
  a.swap(ia);
  return 0;
}

/*** Example Usage ***/

#include <cassert>
using namespace std;

#define EQ(a, b) (fabs((a) - (b)) < 1e-10)

int main() {
  {  // Solve a system.
    vector<vector<double>> a{{-1, 2, 5}, {1, 0, -6}, {-4, 2, 2}};
    vector<double> b{3, 1, -2};
    vector<double> x;
    assert(solve_system(a, b, &x) == 0);
    for (int i = 0; i < static_cast<int>(a.size()); i++) {
      double sum = 0;
      for (int j = 0; j < static_cast<int>(a[0].size()); j++) {
        sum += a[i][j] * x[j];
      }
      assert(EQ(sum, b[i]));
    }
  }
  {  // Find the determinant.
    vector<vector<double>> a{{1, 3, 5}, {2, 4, 7}, {1, 1, 0}};
    assert(EQ(det(a), 4));
  }
  {  // Find the inverse.
    vector<vector<double>> a{{6, 1, 1}, {4, -2, 5}, {2, 8, 7}};
    vector<vector<double>> inv = a;
    int n = static_cast<int>(a.size());
    vector<vector<double>> res(n, vector<double>(n, 0));
    assert(invert(inv) == 0);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          res[i][j] += a[i][k] * inv[k][j];
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        assert(EQ(res[i][j], i == j ? 1 : 0));
      }
    }
  }
  return 0;
}