88 lines
2.8 KiB
JavaScript
88 lines
2.8 KiB
JavaScript
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import { csReach } from './csReach.js';
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import { factory } from '../../../utils/factory.js';
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var name = 'csSpsolve';
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var dependencies = ['divideScalar', 'multiply', 'subtract'];
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export var createCsSpsolve = /* #__PURE__ */factory(name, dependencies, _ref => {
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var {
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divideScalar,
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multiply,
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subtract
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} = _ref;
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/**
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* The function csSpsolve() computes the solution to G * x = bk, where bk is the
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* kth column of B. When lo is true, the function assumes G = L is lower triangular with the
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* diagonal entry as the first entry in each column. When lo is true, the function assumes G = U
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* is upper triangular with the diagonal entry as the last entry in each column.
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*
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* @param {Matrix} g The G matrix
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* @param {Matrix} b The B matrix
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* @param {Number} k The kth column in B
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* @param {Array} xi The nonzero pattern xi[top] .. xi[n - 1], an array of size = 2 * n
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* The first n entries is the nonzero pattern, the last n entries is the stack
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* @param {Array} x The soluton to the linear system G * x = b
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* @param {Array} pinv The inverse row permutation vector, must be null for L * x = b
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* @param {boolean} lo The lower (true) upper triangular (false) flag
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*
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* @return {Number} The index for the nonzero pattern
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*
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* Reference: http://faculty.cse.tamu.edu/davis/publications.html
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*/
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return function csSpsolve(g, b, k, xi, x, pinv, lo) {
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// g arrays
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var gvalues = g._values;
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var gindex = g._index;
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var gptr = g._ptr;
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var gsize = g._size; // columns
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var n = gsize[1]; // b arrays
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var bvalues = b._values;
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var bindex = b._index;
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var bptr = b._ptr; // vars
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var p, p0, p1, q; // xi[top..n-1] = csReach(B(:,k))
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var top = csReach(g, b, k, xi, pinv); // clear x
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for (p = top; p < n; p++) {
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x[xi[p]] = 0;
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} // scatter b
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for (p0 = bptr[k], p1 = bptr[k + 1], p = p0; p < p1; p++) {
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x[bindex[p]] = bvalues[p];
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} // loop columns
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for (var px = top; px < n; px++) {
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// x array index for px
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var j = xi[px]; // apply permutation vector (U x = b), j maps to column J of G
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var J = pinv ? pinv[j] : j; // check column J is empty
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if (J < 0) {
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continue;
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} // column value indeces in G, p0 <= p < p1
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p0 = gptr[J];
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p1 = gptr[J + 1]; // x(j) /= G(j,j)
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x[j] = divideScalar(x[j], gvalues[lo ? p0 : p1 - 1]); // first entry L(j,j)
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p = lo ? p0 + 1 : p0;
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q = lo ? p1 : p1 - 1; // loop
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for (; p < q; p++) {
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// row
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var i = gindex[p]; // x(i) -= G(i,j) * x(j)
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x[i] = subtract(x[i], multiply(gvalues[p], x[j]));
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}
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} // return top of stack
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return top;
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};
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});
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