60 lines
2.0 KiB
JavaScript
60 lines
2.0 KiB
JavaScript
var vows = require('vows');
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var assert = require('assert');
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var suite = vows.describe('jStat');
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require('../env.js');
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suite.addBatch({
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'linearalgebra': {
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'topic': function() {
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return jStat;
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},
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// See "gauss elimination {{1, -3, 1, 4}, {2, -8, 8, -2}, {-6, 3, -15, 9}}"
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// on Wolfram Alpha
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'gauss elimination example 1': function(jStat) {
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var A = [[1, -3, 1], [2, -8, 8], [-6, 3, -15]];
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var B = [[4], [-2], [9]];
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var tol = 0.000001;
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var result = jStat.gauss_elimination(A, B);
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assert.epsilon(tol, result[0], 3);
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assert.epsilon(tol, result[1], -1);
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assert.epsilon(tol, result[2], -2);
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},
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// See "gauss elimination {{1, 1, 3}, {3, -2, 4}}" on Wolfram Alpha
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'gauss elimination example 2': function(jStat) {
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var A = [[1, 1], [3, -2]];
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var B = [[3], [4]];
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var tol = 0.000001;
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var result = jStat.gauss_elimination(A, B);
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assert.epsilon(tol, result[0], 2);
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assert.epsilon(tol, result[1], 1);
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},
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//Sanity check, should not do anything if already solved
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'gauss elimination example 3': function(jStat) {
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var A = jStat.identity(4);
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var B = [[1], [2], [3], [4]];
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var result = jStat.gauss_elimination(A, B);
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var tol = 0.000001;
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assert.epsilon(tol, result[0], 1);
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assert.epsilon(tol, result[1], 2);
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assert.epsilon(tol, result[2], 3);
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assert.epsilon(tol, result[3], 4);
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},
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// Use "{{1,1,1,1,10}, {0,1,1,1,9},{0,0,1,1,7},{0,0,0,1,4}}" on Wolfram
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// Since the matrix already has a lower left hand zero triangle, this just
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// tests the back-substitution portion of the algorithm
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'gauss elimination example 4': function(jStat) {
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var A = [[1,1,1,1],[0,1,1,1],[0,0,1,1],[0,0,0,1]];
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var B = [[10], [9], [7], [4]];
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var result = jStat.gauss_elimination(A, B);
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var tol = 0.000001;
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assert.epsilon(tol, result[3], 4);
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assert.epsilon(tol, result[2], 3);
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assert.epsilon(tol, result[1], 2);
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assert.epsilon(tol, result[0], 1);
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}
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}
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});
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suite.export(module);
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