freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/rosetta-code/lu-decomposition.md

id, title, challengeType, isHidden

id	title	challengeType	isHidden
5e6decd8ec8d7db960950d1c	LU decomposition	5	false

Description

Every square matrix $A$ can be decomposed into a product of a lower triangular matrix $L$ and a upper triangular matrix $U$, as described in LU decomposition. $A = LU$ It is a modified form of Gaussian elimination. While the Cholesky decomposition only works for symmetric, positive definite matrices, the more general LU decomposition works for any square matrix. There are several algorithms for calculating $L$ and $U$. To derive Crout's algorithm for a 3x3 example, we have to solve the following system: \begin{align}A = \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{pmatrix}= \begin{pmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33}\\ \end{pmatrix} \begin{pmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{pmatrix} = LU\end{align} We now would have to solve 9 equations with 12 unknowns. To make the system uniquely solvable, usually the diagonal elements of $L$ are set to 1 $l_{11}=1$ $l_{22}=1$ $l_{33}=1$ so we get a solvable system of 9 unknowns and 9 equations. \begin{align}A = \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\\ \end{pmatrix} \begin{pmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{pmatrix} = \begin{pmatrix} u_{11} & u_{12} & u_{13} \\ u_{11}l_{21} & u_{12}l_{21}+u_{22} & u_{13}l_{21}+u_{23} \\ u_{11}l_{31} & u_{12}l_{31}+u_{22}l_{32} & u_{13}l_{31} + u_{23}l_{32}+u_{33} \end{pmatrix} = LU\end{align} Solving for the other $l$ and $u$, we get the following equations: $u_{11}=a_{11}$ $u_{12}=a_{12}$ $u_{13}=a_{13}$ $u_{22}=a_{22} - u_{12}l_{21}$ $u_{23}=a_{23} - u_{13}l_{21}$ $u_{33}=a_{33} - (u_{13}l_{31} + u_{23}l_{32})$ and for $l$: $l_{21}=\frac{1}{u_{11}} a_{21}$ $l_{31}=\frac{1}{u_{11}} a_{31}$ $l_{32}=\frac{1}{u_{22}} (a_{32} - u_{12}l_{31})$ We see that there is a calculation pattern, which can be expressed as the following formulas, first for $U$ $u_{ij} = a_{ij} - \sum_{k=1}^{i-1} u_{kj}l_{ik}$ and then for $L$ $l_{ij} = \frac{1}{u_{jj}} (a_{ij} - \sum_{k=1}^{j-1} u_{kj}l_{ik})$ We see in the second formula that to get the $l_{ij}$ below the diagonal, we have to divide by the diagonal element (pivot) $u_{jj}$, so we get problems when $u_{jj}$ is either 0 or very small, which leads to numerical instability. The solution to this problem is pivoting $A$, which means rearranging the rows of $A$, prior to the $LU$ decomposition, in a way that the largest element of each column gets onto the diagonal of $A$. Rearranging the rows means to multiply $A$ by a permutation matrix $P$: $PA \Rightarrow A'$ Example: \begin{align} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \end{align} The decomposition algorithm is then applied on the rearranged matrix so that $PA = LU$

Instructions

The task is to implement a routine which will take a square nxn matrix $A$ and return a lower triangular matrix $L$, a upper triangular matrix $U$ and a permutation matrix $P$, so that the above equation is fullfilled. The returned value should be in the form [L, U, P].

Tests

tests:
  - text: <code>luDecomposition</code> should be a function.
    testString: assert(typeof luDecomposition == 'function');
  - text: <code>luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])</code> should return a array.
    testString: assert(Array.isArray(luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])));
  - text: <code>luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])</code> should return <code>[[[1, 0, 0], [0.5, 1, 0], [0.5, -1, 1]], [[2, 4, 7], [0, 1, 1.5], [0, 0, -2]], [[0, 1, 0], [1, 0, 0], [0, 0, 1]]]</code>.
    testString: assert.deepEqual(luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]]), [[[1, 0, 0], [0.5, 1, 0], [0.5, -1, 1]], [[2, 4, 7], [0, 1, 1.5], [0, 0, -2]], [[0, 1, 0], [1, 0, 0], [0, 0, 1]]]);
  - text: <code>luDecomposition([[11, 9, 24, 2], [1, 5, 2, 6], [3, 17, 18, 1], [2, 5, 7, 1]])</code> should return <code>[[[1, 0, 0, 0], [0.2727272727272727, 1, 0, 0], [0.09090909090909091, 0.2875, 1, 0], [0.18181818181818182, 0.23124999999999996, 0.0035971223021580693, 1]], [[11, 9, 24, 2], [0, 14.545454545454547, 11.454545454545455, 0.4545454545454546], [0, 0, -3.4749999999999996, 5.6875], [0, 0, 0, 0.510791366906476]], [[1, 0, 0, 0], [0, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 1]]]</code>.
    testString: assert.deepEqual(luDecomposition([[11, 9, 24, 2], [1, 5, 2, 6], [3, 17, 18, 1], [2, 5, 7, 1]]), [[[1, 0, 0, 0], [0.2727272727272727, 1, 0, 0], [0.09090909090909091, 0.2875, 1, 0], [0.18181818181818182, 0.23124999999999996, 0.0035971223021580693, 1]], [[11, 9, 24, 2], [0, 14.545454545454547, 11.454545454545455, 0.4545454545454546], [0, 0, -3.4749999999999996, 5.6875], [0, 0, 0, 0.510791366906476]], [[1, 0, 0, 0], [0, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 1]]]);
  - text: <code>luDecomposition([[1, 1, 1], [4, 3, -1], [3, 5, 3]])</code> should return <code>[[[1, 0, 0], [0.75, 1, 0], [0.25, 0.09090909090909091, 1]], [[4, 3, -1], [0, 2.75, 3.75], [0, 0, 0.9090909090909091]], [[0, 1, 0], [0, 0, 1], [1, 0, 0]]]</code>.
    testString: assert.deepEqual(luDecomposition([[1, 1, 1], [4, 3, -1], [3, 5, 3]]), [[[1, 0, 0], [0.75, 1, 0], [0.25, 0.09090909090909091, 1]], [[4, 3, -1], [0, 2.75, 3.75], [0, 0, 0.9090909090909091]], [[0, 1, 0], [0, 0, 1], [1, 0, 0]]]);
  - text: <code>luDecomposition([[1, -2, 3], [2, -5, 12], [0, 2, -10]])</code> should return <code>[[[1, 0, 0], [0, 1, 0], [0.5, 0.25, 1]], [[2, -5, 12], [0, 2, -10], [0, 0, -0.5]], [[0, 1, 0], [0, 0, 1], [1, 0, 0]]]</code>.
    testString: assert.deepEqual(luDecomposition([[1, -2, 3], [2, -5, 12], [0, 2, -10]]), [[[1, 0, 0], [0, 1, 0], [0.5, 0.25, 1]], [[2, -5, 12], [0, 2, -10], [0, 0, -0.5]], [[0, 1, 0], [0, 0, 1], [1, 0, 0]]]);

Challenge Seed

function luDecomposition(A) {

}

Solution

function luDecomposition(A) {

    function dotProduct(a, b) {
        var sum = 0;
        for (var i = 0; i < a.length; i++)
            sum += a[i] * b[i]
        return sum;
    }

    function matrixMul(A, B) {
        var result = new Array(A.length);
        for (var i = 0; i < A.length; i++)
            result[i] = new Array(B[0].length)
        var aux = new Array(B.length);

        for (var j = 0; j < B[0].length; j++) {

            for (var k = 0; k < B.length; k++)
                aux[k] = B[k][j];

            for (var i = 0; i < A.length; i++)
                result[i][j] = dotProduct(A[i], aux);
        }
        return result;
    }

    function pivotize(m) {
        var n = m.length;
        var id = new Array(n);
        for (var i = 0; i < n; i++) {
            id[i] = new Array(n);
            id[i].fill(0)
            id[i][i] = 1;
        }

        for (var i = 0; i < n; i++) {
            var maxm = m[i][i];
            var row = i;
            for (var j = i; j < n; j++)
                if (m[j][i] > maxm) {
                    maxm = m[j][i];
                    row = j;
                }

            if (i != row) {
                var tmp = id[i];
                id[i] = id[row];
                id[row] = tmp;
            }
        }
        return id;
    }

    var n = A.length;
    var L = new Array(n);
    for (var i = 0; i < n; i++) { L[i] = new Array(n); L[i].fill(0) }
    var U = new Array(n);
    for (var i = 0; i < n; i++) { U[i] = new Array(n); U[i].fill(0) }
    var P = pivotize(A);
    var A2 = matrixMul(P, A);

    for (var j = 0; j < n; j++) {
        L[j][j] = 1;
        for (var i = 0; i < j + 1; i++) {
            var s1 = 0;
            for (var k = 0; k < i; k++)
                s1 += U[k][j] * L[i][k];
            U[i][j] = A2[i][j] - s1;
        }
        for (var i = j; i < n; i++) {
            var s2 = 0;
            for (var k = 0; k < j; k++)
                s2 += U[k][j] * L[i][k];
            L[i][j] = (A2[i][j] - s2) / U[j][j];
        }
    }
    return [L, U, P];
}