167 lines
5.0 KiB
Markdown
167 lines
5.0 KiB
Markdown
---
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title: Gaussian elimination
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id: 5a23c84252665b21eecc7e77
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challengeType: 5
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---
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## Description
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<section id='description'>
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Write a function to solve \(A.x = b\) using Gaussian elimination then backwards substitution. \(A\) being an \(n \times n\) matrix. Also, \(x\) and \(b\) are \(n\) by 1 vectors. To improve accuracy, please use partial pivoting and scaling.
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>gaussianElimination</code> should be a function.
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testString: assert(typeof gaussianElimination=='function','<code>gaussianElimination</code> should be a function.');
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- text: <code>gaussianElimination([[1,1],[1,-1]], [5,1])</code> should return an array.
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testString: assert(Array.isArray(gaussianElimination([[1,1],[1,-1]], [5,1])),'<code>gaussianElimination([[1,1],[1,-1]], [5,1])</code> should return an array.');
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- text: <code>gaussianElimination([[1,1],[1,-1]], [5,1])</code> should return <code>[ 3, 2 ]</code>.
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testString: assert.deepEqual(gaussianElimination([[1,1],[1,-1]], [5,1]), [ 3, 2 ],'<code>gaussianElimination([[1,1],[1,-1]], [5,1])</code> should return <code>[ 3, 2 ]</code>.');
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- text: <code>gaussianElimination([[2,3],[2,1]] , [8,4])</code> should return <code>[ 1, 2 ]</code>.
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testString: assert.deepEqual(gaussianElimination([[2,3],[2,1]] , [8,4]), [ 1, 2 ],'<code>gaussianElimination([[2,3],[2,1]] , [8,4])</code> should return <code>[ 1, 2 ]</code>.');
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- text: <code>gaussianElimination([[1,3],[5,-2]], [14,19])</code> should return <code>[ 5, 3 ]</code>.
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testString: assert.deepEqual(gaussianElimination([[1,3],[5,-2]], [14,19]), [ 5, 3 ],'<code>gaussianElimination([[1,3],[5,-2]], [14,19])</code> should return <code>[ 5, 3 ]</code>.');
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- text: <code>gaussianElimination([[1,1],[5,-1]] , [10,14])</code> should return <code>[ 4, 6 ]</code>.
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testString: assert.deepEqual(gaussianElimination([[1,1],[5,-1]] , [10,14]), [ 4, 6 ],'<code>gaussianElimination([[1,1],[5,-1]] , [10,14])</code> should return <code>[ 4, 6 ]</code>.');
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- text: <code>gaussianElimination([[1,2,3],[4,5,6],[7,8,8]] , [6,15,23])</code> should return <code>[ 1, 1, 1 ]</code>.
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testString: assert.deepEqual(gaussianElimination([[1,2,3],[4,5,6],[7,8,8]] , [6,15,23]), [ 1, 1, 1 ],'<code>gaussianElimination([[1,2,3],[4,5,6],[7,8,8]] , [6,15,23])</code> should return <code>[ 1, 1, 1 ]</code>.');
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function gaussianElimination (A,b) {
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// Good luck!
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}
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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function gaussianElimination(A, b) {
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// Lower Upper Decomposition
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function ludcmp(A) {
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// A is a matrix that we want to decompose into Lower and Upper matrices.
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var d = true
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var n = A.length
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var idx = new Array(n) // Output vector with row permutations from partial pivoting
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var vv = new Array(n) // Scaling information
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for (var i=0; i<n; i++) {
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var max = 0
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for (var j=0; j<n; j++) {
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var temp = Math.abs(A[i][j])
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if (temp > max) max = temp
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}
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if (max == 0) return // Singular Matrix!
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vv[i] = 1 / max // Scaling
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}
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var Acpy = new Array(n)
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for (var i=0; i<n; i++) {
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var Ai = A[i]
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let Acpyi = new Array(Ai.length)
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for (j=0; j<Ai.length; j+=1) Acpyi[j] = Ai[j]
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Acpy[i] = Acpyi
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}
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A = Acpy
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var tiny = 1e-20 // in case pivot element is zero
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for (var i=0; ; i++) {
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for (var j=0; j<i; j++) {
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var sum = A[j][i]
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for (var k=0; k<j; k++) sum -= A[j][k] * A[k][i];
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A[j][i] = sum
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}
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var jmax = 0
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var max = 0;
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for (var j=i; j<n; j++) {
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var sum = A[j][i]
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for (var k=0; k<i; k++) sum -= A[j][k] * A[k][i];
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A[j][i] = sum
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var temp = vv[j] * Math.abs(sum)
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if (temp >= max) {
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max = temp
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jmax = j
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}
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}
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if (i <= jmax) {
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for (var j=0; j<n; j++) {
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var temp = A[jmax][j]
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A[jmax][j] = A[i][j]
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A[i][j] = temp
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}
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d = !d;
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vv[jmax] = vv[i]
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}
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idx[i] = jmax;
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if (i == n-1) break;
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var temp = A[i][i]
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if (temp == 0) A[i][i] = temp = tiny
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temp = 1 / temp
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for (var j=i+1; j<n; j++) A[j][i] *= temp
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}
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return {A:A, idx:idx, d:d}
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}
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// Lower Upper Back Substitution
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function lubksb(lu, b) {
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// solves the set of n linear equations A*x = b.
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// lu is the object containing A, idx and d as determined by the routine ludcmp.
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var A = lu.A
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var idx = lu.idx
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var n = idx.length
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var bcpy = new Array(n)
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for (var i=0; i<b.length; i+=1) bcpy[i] = b[i]
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b = bcpy
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for (var ii=-1, i=0; i<n; i++) {
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var ix = idx[i]
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var sum = b[ix]
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b[ix] = b[i]
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if (ii > -1)
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for (var j=ii; j<i; j++) sum -= A[i][j] * b[j]
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else if (sum)
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ii = i
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b[i] = sum
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}
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for (var i=n-1; i>=0; i--) {
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var sum = b[i]
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for (var j=i+1; j<n; j++) sum -= A[i][j] * b[j]
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b[i] = sum / A[i][i]
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}
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return b // solution vector x
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}
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var lu = ludcmp(A)
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if (lu === undefined) return // Singular Matrix!
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return lubksb(lu, b)
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}
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```
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</section>
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