source: CIVL/examples/experimental/gaussElim.cvl@ 53828dd

#include<civlc.h>

$input int N_BOUND; // bound on the number of rows
$input int M_BOUND; // bound on the number of columns
$input int N;  // number of rows
$input int M;  // number of columns
$assume N>0 && N<N_BOUND;
$assume M>0 && M<M_BOUND;
$assume N*M <= N_BOUND*M_BOUND;
$input double inputMatrix[N*M];
$output double outputMatrix[N*M];

// Perform Guass-Jordon elimination on the matrix.  Upon return, the
// matrix will be in reduced row echelon form.
void gaussElim(double* matrix) {
  int top = 0;           // index of current top row
  int col = 0;           // index of current left column
  int pivotRow = 0;      // index of row containing the pivot
  double pivot = 0.0;    // the value of the pivot
  int i = 0;             // loop variable over rows of matrix
  int j = 0;             // loop variable over columns of matrix
  double tmp = 0.0;      // temporary double variable

  col = 0;
  for (top=0; top<N && col< M; top++) {
    /* At this point we know that the submatarix consisting of the
     * first top rows of A is in reduced row-echelon form.  We will
     * now consider the submatrix B consisting of the remaining rows.
     * We know, additionally, that the first col columns of B are all
     * zero.
     */
        
    /* Step 1: Locate the leftmost column of B that does not consist
     * entirely of zeros, if one exists.  The top nonzero entry of
     * this column is the pivot. */

    pivot = 0.0;
    for (; col < M; col++) {
      for (pivotRow = top; pivotRow < N; pivotRow++) {
        pivot = matrix[pivotRow*M + col];
        if (pivot) break;
      }
      if (pivot) break;
    }

    if (col >= M) {
      break;
    }

    /* At this point we are guaranteed that pivot = A[pivotRow,col] is
     * nonzero.  We also know that all the columns of B to the left of
     * col consist entirely of zeros. */

            
    /* Step 2: Interchange the top row with the pivot row, if
     * necessary, so that the entry at the top of the column found in
     * Step 1 is nonzero. */

    if (pivotRow != top) {
      for (j = 0; j < M; j++) {
        tmp = matrix[top*M + j];
        matrix[top*M + j] = matrix[pivotRow*M + j];
        matrix[pivotRow*M + j] = tmp;
      }
    }

    /* At this point we are guaranteed that A[top,col] = pivot is
     * nonzero. Also, we know that (i>=top and j<col) implies
     * A[i,j] = 0. */

    /* Step 3: Divide the top row by pivot in order to introduce a
     * leading 1. */

    for (j = col; j < M; j++) {
      matrix[top*M + j] /= pivot;
    }

    /* At this point we are guaranteed that A[top,col] is 1.0,
     * assuming that floating point arithmetic guarantees that a/a
     * equals 1.0 for any nonzero double a. */

    /* Step 4: Add suitable multiples of the top row to all other rows
     * so that all entries above and below the leading 1 become
     * zero. */
    
    for (i = 0; i < N; i++) {
      if (i != top){
        tmp = matrix[i*M + col];
        for (j = col; j < M; j++) {
          matrix[i*M + j] -= matrix[top*M + j]*tmp;
        }
      }
    }
    col++;
  }

}

// Check that the output matrix is indeed in reduced row echelon form.
void checkOutput() {

}

void main() {
  $heap h;
  double* matrix = (double *) $malloc(&h, N*M*sizeof(double));
  
  for (int i = 0; i < N*M; i++) {
    matrix[i] = inputMatrix[i];
  }
  gaussElim(matrix);
  for (int i = 0; i < N*M; i++) {
    outputMatrix[i] = matrix[i];
  }
  checkOutput();
}