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curriculum/challenges/portuguese/10-coding-interview-prep/rosetta-code/lu-decomposition.md
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---
id: 5e6decd8ec8d7db960950d1c
title: LU decomposition
title: Decomposição de LU
challengeType: 5
forumTopicId: 385280
dashedName: lu-decomposition
@ -8,21 +8,21 @@ dashedName: lu-decomposition
# --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](https://en.wikipedia.org/wiki/LU decomposition).
Toda matriz quadrada $A$ pode ser decomposta em um produto de uma matriz triangular inferior $L$ e uma matriz triangular superior $U$, conforme descrito em [Decomposição de LU](https://en.wikipedia.org/wiki/LU decomposition).
$A = LU$
It is a modified form of Gaussian elimination.
Ela é uma forma modificada da eliminação de Gauss.
While the [Cholesky decomposition](http://rosettacode.org/wiki/Cholesky decomposition) only works for symmetric, positive definite matrices, the more general LU decomposition works for any square matrix.
Enquanto a [Decomposição do Cholesky](http://rosettacode.org/wiki/Cholesky decomposition) funciona somente para matrizes simétricas, definidas e positivas, a decomposição da LU é mais geral e funciona para qualquer matriz quadrada.
There are several algorithms for calculating $L$ and $U$.
Existem vários algoritmos para calcular $L$ e $U$.
To derive *Crout's algorithm* for a 3x3 example, we have to solve the following system:
Para derivar o *algoritmo de Crout* para um exemplo de 3x3, temos de resolver o seguinte sistema:
\\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
Agora, teríamos de resolver 9 equações com 12 incógnitas. Para tornar o sistema resolvível de forma única, geralmente os elementos diagonais de $L$ estão definidos como 1
$l\_{11}=1$
@ -30,11 +30,11 @@ $l\_{22}=1$
$l\_{33}=1$
so we get a solvable system of 9 unknowns and 9 equations.
portanto, obtemos um sistema resolvível de 9 incógnitas e 9 equações.
\\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:
Ao solucionar para os outros $l$ e $u$, recebemos as seguintes equações:
$u\_{11}=a\_{11}$
@ -48,7 +48,7 @@ $u\_{23}=a\_{23} - u\_{13}l\_{21}$
$u\_{33}=a\_{33} - (u\_{13}l\_{31} + u\_{23}l\_{32})$
and for $l$:
e para $l$:
$l\_{21}=\\frac{1}{u\_{11}} a\_{21}$
@ -56,41 +56,41 @@ $l\_{31}=\\frac{1}{u\_{11}} a\_{31}$
$l\_{32}=\\frac{1}{u\_{22}} (a\_{32} - u\_{12}l\_{31})$
We see that there is a calculation pattern, which can be expressed as the following formulas, first for $U$
Vemos que há um padrão de cálculo, que pode ser expresso com as seguintes fórmulas, primeiro para $U$
$u\_{ij} = a\_{ij} - \\sum\_{k=1}^{i-1} u\_{kj}l\_{ik}$
and then for $L$
e depois para $L$
$l\_{ij} = \\frac{1}{u\_{jj}} (a\_{ij} - \\sum\_{k=1}^{j-1} u\_{kj}l\_{ik})$
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.
Vemos na segunda fórmula que, para obter o $l\_{ij}$ abaixo da diagonal, temos de dividir pelo elemento da diagonal (pivô) $u\_{jj}$, de modo que temos problemas quando $u\_{jj}$ é 0 ou muito pequeno, o que gera uma instabilidade numérica.
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 $P$:
A solução para este problema é *rotar* $A$, o que significa reorganizar as linhas de $A$, antes da decomposição de $LU$, de modo que o maior elemento de cada coluna fique na diagonal de $A$. Reordenar as linhas significa multiplicar $A$ por uma matriz de permutação $P$:
$PA \\Rightarrow A'$
Example:
Exemplo:
\\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
O algoritmo de decomposição será então aplicado na matriz reorganizada para que
$PA = LU$
# --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]`.
A tarefa é implementar uma rotina que receba uma matriz quadrada $A$ nxn e retornar uma matriz triangular inferior $L$, uma matriz triangular superior $U$ e uma matriz de permutação $P$, para que a equação acima seja resolvida. O valor retornado deve estar na forma `[L, U, P]`.
# --hints--
`luDecomposition` should be a function.
`luDecomposition` deve ser uma função.
```js
assert(typeof luDecomposition == 'function');
```
`luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])` should return a array.
`luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])` deve retornar um array.
```js
assert(
@ -104,7 +104,7 @@ assert(
);
```
`luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])` should return `[[[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]]]`.
`luDecomposition([[1, 3, 5], [2, 4, 7], [1, 1, 0]])` deve retornar `[[[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]]]`.
```js
assert.deepEqual(
@ -133,7 +133,7 @@ assert.deepEqual(
);
```
`luDecomposition([[11, 9, 24, 2], [1, 5, 2, 6], [3, 17, 18, 1], [2, 5, 7, 1]])` should return `[[[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]]]`.
`luDecomposition([[11, 9, 24, 2], [1, 5, 2, 6], [3, 17, 18, 1], [2, 5, 7, 1]])` deve retornar `[[[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]]]`.
```js
assert.deepEqual(
@ -166,7 +166,7 @@ assert.deepEqual(
);
```
`luDecomposition([[1, 1, 1], [4, 3, -1], [3, 5, 3]])` should return `[[[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]]]`.
`luDecomposition([[1, 1, 1], [4, 3, -1], [3, 5, 3]])` deve retornar `[[[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]]]`.
```js
assert.deepEqual(
@ -195,7 +195,7 @@ assert.deepEqual(
);
```
`luDecomposition([[1, -2, 3], [2, -5, 12], [0, 2, -10]])` should return `[[[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]]]`.
`luDecomposition([[1, -2, 3], [2, -5, 12], [0, 2, -10]])` deve retornar `[[[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]]]`.
```js
assert.deepEqual(