1 // SPDX-FileCopyrightText: Copyright (c) 2017-2024, HONEE contributors.
2 // SPDX-License-Identifier: Apache-2.0 OR BSD-2-Clause
25 ///@brief Find the off-diagonal index in row i whose absolute value is largest
29 /// @returns   Index of absolute largest off-diagonal element in row i
39 ///        uses the max_idx_row[] array to find the answer in O(n) time.
48   for (CeedInt i = 1; i < N - 1; i++) {  in MaxEntry()
69   CeedScalar A_jj_ii = (A[j * N + j] - A[i * N + i]);  in CalcRot()
71     // kappa = (A[j][j] - A[i][i]) / (2*A[i][j])  in CalcRot()
77       // t satisfies: t^2 + 2*t*kappa - 1 = 0  in CalcRot()
80       if (kappa < 0.0) rotmat_cst[2] = -rotmat_cst[2];  in CalcRot()
108 ///      |          -s  ...   c          |
129 /// matrix elements in the upper-right triangle strictly above the diagonal.
157   A[i * N + i] -= rotmat_cst[2] * A[i * N + j];  in ApplyRot()
160   // A[i][i] = c*c*A[i][i] + s*s*A[j][j] - 2*s*c*A[i][j]  in ApplyRot()
163   // Update the off-diagonal elements of A which will change (above the diagonal)  in ApplyRot()
167   // compute A[w][i] and A[i][w] for all w!=i,considering above-diagonal elements  in ApplyRot()
170 …A[w * N + i] = rotmat_cst[0] * A[w * N + i] - rotmat_cst[1] * A[w * N + j];  // A[w][i], A[w][j] f…  in ApplyRot()
176 …A[i * N + w] = rotmat_cst[0] * A[i * N + w] - rotmat_cst[1] * A[w * N + j];  // A[i][w], A[w][j] f…  in ApplyRot()
180 …A[i * N + w] = rotmat_cst[0] * A[i * N + w] - rotmat_cst[1] * A[j * N + w];  // A[i][w], A[j][w] f…  in ApplyRot()
186   // compute A[w][j] and A[j][w] for all w!=j,considering above-diagonal elements  in ApplyRot()
218     A[i * N + v]   = rotmat_cst[0] * A[i * N + v] - rotmat_cst[1] * A[j * N + v];  in ApplyRotLeft()
232   for (CeedInt i = 0; i < N - 1; i++) {  in SortRows()
266 ///        on it, so divergence from normalized is due to finite-precision
276 … max_num_sweeps maximum number of iterations = max_num_sweeps * number of off-diagonals (N*(N-1)/2)
285   for (CeedInt i = 0; i < N - 1; i++) max_idx_row[i] = MaxEntryRow(A, N, i);  in Diagonalize()
287   // -- Iteration --  in Diagonalize()
289   CeedInt max_num_iters = max_num_sweeps * N * (N - 1) / 2;  in Diagonalize()