freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/gaussian-elimination.md

title	id	challengeType	forumTopicId
Gaussian elimination	5a23c84252665b21eecc7e77	5	302272

Description

Write a function to solve \(Ax = b\) 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.

Instructions

Tests

tests:
  - text: <code>gaussianElimination</code> should be a function.
    testString: assert(typeof gaussianElimination=='function');
  - text: <code>gaussianElimination([[1,1],[1,-1]], [5,1])</code> should return an array.
    testString: assert(Array.isArray(gaussianElimination([[1,1],[1,-1]], [5,1])));
  - text: <code>gaussianElimination([[1,1],[1,-1]], [5,1])</code> should return <code>[ 3, 2 ]</code>.
    testString: assert.deepEqual(gaussianElimination([[1,1],[1,-1]], [5,1]), [ 3, 2 ]);
  - text: <code>gaussianElimination([[2,3],[2,1]] , [8,4])</code> should return <code>[ 1, 2 ]</code>.
    testString: assert.deepEqual(gaussianElimination([[2,3],[2,1]] , [8,4]), [ 1, 2 ]);
  - text: <code>gaussianElimination([[1,3],[5,-2]], [14,19])</code> should return <code>[ 5, 3 ]</code>.
    testString: assert.deepEqual(gaussianElimination([[1,3],[5,-2]], [14,19]), [ 5, 3 ]);
  - text: <code>gaussianElimination([[1,1],[5,-1]] , [10,14])</code> should return <code>[ 4, 6 ]</code>.
    testString: assert.deepEqual(gaussianElimination([[1,1],[5,-1]] , [10,14]), [ 4, 6 ]);
  - text: <code>gaussianElimination([[1,2,3],[4,5,6],[7,8,8]] , [6,15,23])</code> should return <code>[ 1, 1, 1 ]</code>.
    testString: assert.deepEqual(gaussianElimination([[1,2,3],[4,5,6],[7,8,8]] , [6,15,23]), [ 1, 1, 1 ]);

Challenge Seed

function gaussianElimination(A,b) {

}

Solution

function gaussianElimination(A, b) {
  // Lower Upper Decomposition
  function ludcmp(A) {
  	// A is a matrix that we want to decompose into Lower and Upper matrices.
  	var d = true
  	var n = A.length
  	var idx = new Array(n) // Output vector with row permutations from partial pivoting
  	var vv = new Array(n) // Scaling information

  	for (var i=0; i<n; i++) {
  		var max = 0
  		for (var j=0; j<n; j++) {
  			var temp = Math.abs(A[i][j])
  			if (temp > max) max = temp
  		}
  		if (max == 0) return // Singular Matrix!
  		vv[i] = 1 / max // Scaling
  	}

		var Acpy = new Array(n)
		for (var i=0; i<n; i++) {
			var Ai = A[i]
			let Acpyi = new Array(Ai.length)
			for (j=0; j<Ai.length; j+=1) Acpyi[j] = Ai[j]
			Acpy[i] = Acpyi
		}
		A = Acpy

  	var tiny = 1e-20 // in case pivot element is zero
  	for (var i=0; ; i++) {
  		for (var j=0; j<i; j++) {
  			var sum = A[j][i]
  			for (var k=0; k<j; k++) sum -= A[j][k] * A[k][i];
  			A[j][i] = sum
  		}
  		var jmax = 0
  		var max = 0;
  		for (var j=i; j<n; j++) {
  			var sum = A[j][i]
  			for (var k=0; k<i; k++) sum -= A[j][k] * A[k][i];
  			A[j][i] = sum
  			var temp = vv[j] * Math.abs(sum)
  			if (temp >= max) {
  				max = temp
  				jmax = j
  			}
  		}
  		if (i <= jmax) {
  			for (var j=0; j<n; j++) {
  				var temp = A[jmax][j]
  				A[jmax][j] = A[i][j]
  				A[i][j] = temp
  			}
  			d = !d;
  			vv[jmax] = vv[i]
  		}
  		idx[i] = jmax;
  		if (i == n-1) break;
  		var temp = A[i][i]
  		if (temp == 0) A[i][i] = temp = tiny
  		temp = 1 / temp
  		for (var j=i+1; j<n; j++) A[j][i] *= temp
  	}
  	return {A:A, idx:idx, d:d}
  }

  // Lower Upper Back Substitution
  function lubksb(lu, b) {
  	// solves the set of n linear equations A*x = b.
  	// lu is the object containing A, idx and d as determined by the routine ludcmp.
  	var A = lu.A
  	var idx = lu.idx
  	var n = idx.length

		var bcpy = new Array(n)
		for (var i=0; i<b.length; i+=1) bcpy[i] = b[i]
		b = bcpy

  	for (var ii=-1, i=0; i<n; i++) {
  		var ix = idx[i]
  		var sum = b[ix]
  		b[ix] = b[i]
  		if (ii > -1)
  			for (var j=ii; j<i; j++) sum -= A[i][j] * b[j]
  		else if (sum)
  			ii = i
  		b[i] = sum
  	}
  	for (var i=n-1; i>=0; i--) {
  		var sum = b[i]
  		for (var j=i+1; j<n; j++) sum -= A[i][j] * b[j]
  		b[i] = sum / A[i][i]
  	}
  	return b // solution vector x
  }

	var lu = ludcmp(A)
	if (lu === undefined) return // Singular Matrix!
	return lubksb(lu, b)
}