ndarray-lu-solve
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solve a linear system of equations from an LU decomposition
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# ndarray-lu-solve
solve a system of linear equations from an LU decomposition
[](https://ci.testling.com/substack/ndarray-lu-solve)
[](http://travis-ci.org/substack/ndarray-lu-solve)
# example
``` js
var solve = require('ndarray-lu-solve');
var show = require('ndarray-show');
var crout = require('ndarray-crout-decomposition');
var ndarray = require('ndarray');
var zeros = require('zeros');
var A = ndarray(
[ 2, 1, -1, 8, -3, -1, 2, -11, -2, 1, 2, -3 ],
[ 4, 3 ], [ 1, 4 ]
);
var L = zeros([ 3, 3 ]);
var U = zeros([ 3, 3 ]);
crout(A.hi(3,3), L, U);
var X = ndarray(new Float64Array(3));
var Y = ndarray(new Float64Array(3));
var solution = solve(L, U, A.lo(3,0).pick(0), X, Y);
console.log('input:\n' + show(A), '\n');
console.log('solution:\n' + show(solution));
```
output:
```
input:
2.000 1.000 -1.000 8.000
-3.000 -1.000 2.000 -11.000
-2.000 1.000 2.000 -3.000
solution:
2.000 3.000 -1.000
```
You can pass fewer of the parameters if you want and use a packed representation
for L and U in the same matrix:
``` js
var solve = require('ndarray-lu-solve');
var show = require('ndarray-show');
var crout = require('ndarray-crout-decomposition');
var ndarray = require('ndarray');
var zeros = require('zeros');
var A = ndarray(
[ 2, 1, -1, 8, -3, -1, 2, -11, -2, 1, 2, -3 ],
[ 4, 3 ], [ 1, 4 ]
);
var LU = zeros([ 3, 3 ]);
crout(A.hi(3,3), LU);
var solution = solve(LU, A.lo(3,0).pick(0));
console.log('input:\n' + show(A), '\n');
console.log('solution:\n' + show(solution));
```
output:
```
input:
2.000 1.000 -1.000 8.000
-3.000 -1.000 2.000 -11.000
-2.000 1.000 2.000 -3.000
solution:
2.000 3.000 -1.000
```
Just note that it's up to you to
[`ndscratch.free()`](https://npmjs.org/package/ndarray-scratch)
the `solution` you get back to prevent leaking memory.
# methods
``` js
var solve = require('ndarray-lu-solve')
```
## var solution = solve(L, U, B, X, Y)
Given an `L` and `U` ndarrays from a decomposition and a vector `B`, solve the
system `LY = B` for `Y` and `UX = Y` for `X`.
`L` and `U` can also be the same matrix.
The `solution` is written to `X` as the computation procedes but is also the
return value.
Optional `X` and `Y` parameters are modified in-place and contain elements of
the result. If not given, `X` and `Y` will be allocated.
## var solution = solve(LU, B, X, Y)
You can omit the `U` parameter if you have a `L` and `U` values packed into the
same matrix. The other parameters work the same.
# install
With [npm](https://npmjs.org) do:
```
npm install ndarray-lu-solve
```
# license
MIT