LU decomposition: Difference between revisions

Content added Content deleted

Inline

Revision as of 21:54, 24 March 2011

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:

A={\begin{pmatrix}a_{11}&a_{21}&a_{31}\\a_{21}&a_{22}&a_{32}\\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

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.

A={\begin{pmatrix}a_{11}&a_{21}&a_{31}\\a_{21}&a_{22}&a_{32}\\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

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) $u_{ij}$ , so we get problems when $u_{ij}$ 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 columns means to multiply $A$ by a permutation matrix $P$ :

PA\Rightarrow A'

Example:

{\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}}

The decomposition algorithm is then applied on the rearranged matrix so that

PA=LU

Task description

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. You should then test it on the following two examples and include your output.

Example 1:

A

1   3   5
2   4   7
1   1   0

L

1.00000   0.00000   0.00000
0.50000   1.00000   0.00000
0.50000  -1.00000   1.00000

U

2.00000   4.00000   7.00000
0.00000   1.00000   1.50000
0.00000   0.00000  -2.00000

P

0   1   0
1   0   0
0   0   1

Example 2:

A

11    9   24    2
 1    5    2    6
 3   17   18    1
 2    5    7    1

L

1.00000   0.00000   0.00000   0.00000
0.27273   1.00000   0.00000   0.00000
0.09091   0.28750   1.00000   0.00000
0.18182   0.23125   0.00360   1.00000

U

11.00000    9.00000   24.00000    2.00000
 0.00000   14.54545   11.45455    0.45455
 0.00000    0.00000   -3.47500    5.68750
 0.00000    0.00000    0.00000    0.51079

P

1   0   0   0
0   0   1   0
0   1   0   0
0   0   0   1

Common Lisp

Uses the routine (mmul A B) from Matrix multiplication.

<lang lisp>;; Creates a nxn identity matrix. (defun eye (n)

 (let ((I (make-array `(,n ,n) :initial-element 0)))
   (loop for j from 0 to (- n 1) do
         (setf (aref I j j) 1))
   I))

Swap two rows l and k of a mxn matrix A, which is a 2D array.

(defun swap-rows (A l k)

 (let* ((n (cadr (array-dimensions A)))
        (row (make-array n :initial-element 0)))
   (loop for j from 0 to (- n 1) do
         (setf (aref row j) (aref A l j))
         (setf (aref A l j) (aref A k j))
         (setf (aref A k j) (aref row j)))))

Creates the pivoting matrix for A.

(defun pivotize (A)

 (let* ((n (car (array-dimensions A)))
        (P (eye n)))
   (loop for j from 0 to (- n 1) do
         (let ((max (aref A j j))
               (row j))
           (loop for i from j to (- n 1) do
                 (if (> (aref A i j) max)
                     (setq max (aref A i j)
                           row i)))
           (if (not (= j row))
               (swap-rows P j row))))

 ;; Return P.
 P))

Decomposes a square matrix A by PA=LU and returns L, U and P.

(defun lu (A)

 (let* ((n (car (array-dimensions A)))
        (L (make-array `(,n ,n) :initial-element 0))
        (U (make-array `(,n ,n) :initial-element 0))
        (P (pivotize A))
        (A (mmul P A)))

   (loop for j from 0 to (- n 1) do
         (setf (aref L j j) 1)
         (loop for i from 0 to j do
               (setf (aref U i j)
                     (- (aref A i j)
                        (loop for k from 0 to (- i 1)
                              sum (* (aref U k j)
                                     (aref L i k))))))
         (loop for i from j to (- n 1) do
               (setf (aref L i j)
                     (/ (- (aref A i j)
                           (loop for k from 0 to (- j 1)
                                 sum (* (aref U k j)
                                        (aref L i k))))
                        (aref U j j)))))

 ;; Return L, U and P.
 (values L U P)))</lang>

Example 1:

<lang lisp>(setf g (make-array '(3 3) :initial-contents '((1 3 5) (2 4 7)(1 1 0))))

2A((1 3 5) (2 4 7) (1 1 0))

(lu g)

2A((1 0 0) (1/2 1 0) (1/2 -1 1))
2A((2 4 7) (0 1 3/2) (0 0 -2))
2A((0 1 0) (1 0 0) (0 0 1))</lang>

Example 2:

<lang lisp>(setf h (make-array '(4 4) :initial-contents '((11 9 24 2)(1 5 2 6)(3 17 18 1)(2 5 7 1))))

2A((11 9 24 2) (1 5 2 6) (3 17 18 1) (2 5 7 1))

(lup h)

2A((1 0 0 0) (3/11 1 0 0) (1/11 23/80 1 0) (2/11 37/160 1/278 1))
2A((11 9 24 2) (0 160/11 126/11 5/11) (0 0 -139/40 91/16) (0 0 0 71/139))
2A((1 0 0 0) (0 0 1 0) (0 1 0 0) (0 0 0 1))</lang>

D

Using some functions from Matrix multiplication task.

Translation of: Common Lisp

<lang d>import std.stdio, std.algorithm, std.typecons;

/// Creates a nxn identity matrix. pure T[][] identityMatrix(T)(const int n)

   in {
       assert(n >= 0);
   } body {
       auto m = new typeof(return)(n, n);
       foreach (i, row; m) {
           row[] = 0;
           row[i] = 1;
       }
       return m;
   }

/// Creates the pivoting matrix for m. pure T[][] pivotize(T)(const T[][] m)

   in {
       // is square
       assert(isRectangular(m) && m[0].length == m.length);
   } body {
       int n = m.length;
       auto P = identityMatrix!T(n);

       foreach (j; 0 .. n) {
           T max = m[j][j];
           int row = j;
           foreach (i; j .. n)
               if (m[i][j] > max) {
                   max = m[i][j];
                   row = i;
               }
           if (j != row)
               swap(P[j], P[row]);
       }

       return P;
   }

/// Decomposes a square matrix A by PA=LU and returns L, U and P. /*pure*/ Tuple!(T[][], T[][], T[][]) lu(T)(const T[][] A)

   in {
       // is square
       assert(isRectangular(A) && A[0].length == A.length);
   } body {
       int n = A.length;
       auto L = new T[][](n, n);
       auto U = new T[][](n, n);
       foreach (i; 0 .. n) {
           L[i][i .. $] = 0;
           U[i][0 .. i] = 0;
       }
       auto P = pivotize!T(A);
       auto A2 = matrixMul!T(P, A);

       foreach (j; 0 .. n) {
           L[j][j] = 1;
           foreach (i; 0 .. j+1) {
               T s1 = 0;
               foreach (k; 0 .. i)
                   s1 += U[k][j] * L[i][k];
               U[i][j] = A2[i][j] - s1;
           }
           foreach (i; j .. n) {
               T s2 = 0;
               foreach (k; 0 .. j)
                   s2 += U[k][j] * L[i][k];
               L[i][j] = (A2[i][j] - s2) / U[j][j];
           }
       }

       return tuple(L, U, P);
   }

void main() {

   enum double[][] a = [[1, 3, 5], [2, 4, 7], [1, 1, 0]];
   foreach (part; lu(a).tupleof)
       writeln(prettyPrint(part), "\n");
   writeln();

   enum double[][] b = [[11, 9, 24, 2], [1, 5, 2, 6],
                        [3, 17, 18, 1], [2, 5, 7, 1]];
   foreach (part; lu(b).tupleof)
       writeln(prettyPrint(part), "\n");

}</lang> Output:

[[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]]


[[1, 0, 0, 0],
 [0.272727, 1, 0, 0],
 [0.090909, 0.2875, 1, 0],
 [0.181818, 0.23125, 0.00359712, 1]]

[[11, 9, 24, 2],
 [0, 14.5455, 11.4545, 0.454545],
 [0, 0, -3.475, 5.6875],
 [0, 0, 0, 0.510791]]

[[1, 0, 0, 0],
 [0, 0, 1, 0],
 [0, 1, 0, 0],
 [0, 0, 0, 1]]

Python

Translation of: D

<lang python>from pprint import pprint

def matrixMul(A, B):

   TB = zip(*B)
   return [[sum(ea*eb for ea,eb in zip(a,b)) for b in TB] for a in A]

def pivotize(m):

   """Creates the pivoting matrix for m."""
   n = len(m)
   ID = [[float(i == j) for i in xrange(n)] for j in xrange(n)]
   for j in xrange(n):
       row = max(xrange(j, n), key=lambda i: m[i][j])
       if j != row:
           ID[j], ID[row] = ID[row], ID[j]
   return ID

def lu(A):

   """Decomposes a nxn matrix A by PA=LU and returns L, U and P."""
   n = len(A)
   L = [[0.0] * n for i in xrange(n)]
   U = [[0.0] * n for i in xrange(n)]
   P = pivotize(A)
   A2 = matrixMul(P, A)
   for j in xrange(n):
       L[j][j] = 1.0
       for i in xrange(j+1):
           s1 = sum(U[k][j] * L[i][k] for k in xrange(i))
           U[i][j] = A2[i][j] - s1
       for i in xrange(j, n):
           s2 = sum(U[k][j] * L[i][k] for k in xrange(j))
           L[i][j] = (A2[i][j] - s2) / U[j][j]
   return (L, U, P)

a = [[1, 3, 5], [2, 4, 7], [1, 1, 0]] for part in lu(a):

   pprint(part, width=19)
   print

print b = [[11,9,24,2],[1,5,2,6],[3,17,18,1],[2,5,7,1]] for part in lu(b):

   pprint(part)
   print</lang>

Output:

[[1.0, 0.0, 0.0],
 [0.5, 1.0, 0.0],
 [0.5, -1.0, 1.0]]

[[2.0, 4.0, 7.0],
 [0.0, 1.0, 1.5],
 [0.0, 0.0, -2.0]]

[[0.0, 1.0, 0.0],
 [1.0, 0.0, 0.0],
 [0.0, 0.0, 1.0]]


[[1.0, 0.0, 0.0, 0.0],
 [0.27272727272727271, 1.0, 0.0, 0.0],
 [0.090909090909090912, 0.28749999999999998, 1.0, 0.0],
 [0.18181818181818182, 0.23124999999999996, 0.0035971223021580693, 1.0]]

[[11.0, 9.0, 24.0, 2.0],
 [0.0, 14.545454545454547, 11.454545454545455, 0.45454545454545459],
 [0.0, 0.0, -3.4749999999999996, 5.6875],
 [0.0, 0.0, 0.0, 0.51079136690647597]]

[[1.0, 0.0, 0.0, 0.0],
 [0.0, 0.0, 1.0, 0.0],
 [0.0, 1.0, 0.0, 0.0],
 [0.0, 0.0, 0.0, 1.0]]

Tcl

<lang tcl>package require Tcl 8.5 namespace eval matrix {

   namespace path {::tcl::mathfunc ::tcl::mathop}

   # Construct an identity matrix of the given size
   proc identity {order} {

set m [lrepeat $order [lrepeat $order 0]] for {set i 0} {$i < $order} {incr i} { lset m $i $i 1 } return $m

   # Produce the pivot matrix for a given matrix
   proc pivotize {matrix} {

set n [llength $matrix] set p [identity $n] for {set j 0} {$j < $n} {incr j} { set max [lindex $matrix $j $j] set row $j for {set i $j} {$i < $n} {incr i} { if {[lindex $matrix $i $j] > $max} { set max [lindex $matrix $i $j] set row $i } } if {$j != $row} { # Row swap inlined; too trivial to have separate procedure set tmp [lindex $p $j] lset p $j [lindex $p $row] lset p $row $tmp } } return $p

   # Decompose a square matrix A by PA=LU and return L, U and P
   proc luDecompose {A} {

set n [llength $A] set L [lrepeat $n [lrepeat $n 0]] set U $L set P [pivotize $A] set A [multiply $P $A]

for {set j 0} {$j < $n} {incr j} { lset L $j $j 1 for {set i 0} {$i <= $j} {incr i} { lset U $i $j [- [lindex $A $i $j] [SumMul $L $U $i $j $i]] } for {set i $j} {$i < $n} {incr i} { set sum [SumMul $L $U $i $j $j] lset L $i $j [/ [- [lindex $A $i $j] $sum] [lindex $U $j $j]] } }

return [list $L $U $P]

   # Helper that makes inner loop nicer; multiplies column and row,
   # possibly partially...
   proc SumMul {A B i j kmax} {

set s 0.0 for {set k 0} {$k < $kmax} {incr k} { set s [+ $s [* [lindex $A $i $k] [lindex $B $k $j]]] } return $s

}</lang> Support code: <lang tcl># Code adapted from Matrix_multiplication and Matrix_transposition tasks namespace eval matrix {

   # Get the size of a matrix; assumes that all rows are the same length, which
   # is a basic well-formed-ness condition...
   proc size {m} {

set rows [llength $m] set cols [llength [lindex $m 0]] return [list $rows $cols]

   # Matrix multiplication implementation
   proc multiply {a b} {

lassign [size $a] a_rows a_cols lassign [size $b] b_rows b_cols if {$a_cols != $b_rows} { error "incompatible sizes: a($a_rows, $a_cols), b($b_rows, $b_cols)" } set temp [lrepeat $a_rows [lrepeat $b_cols 0]] for {set i 0} {$i < $a_rows} {incr i} { for {set j 0} {$j < $b_cols} {incr j} { lset temp $i $j [SumMul $a $b $i $j $a_cols] } } return $temp

   # Pretty printer for matrices
   proc print {matrix {fmt "%g"}} {

set max [Widest $matrix $fmt] lassign [size $matrix] rows cols foreach row $matrix { foreach val $row width $max { puts -nonewline [format "%*s " $width [format $fmt $val]] } puts "" }

   }
   proc Widest {m fmt} {

lassign [size $m] rows cols set max [lrepeat $cols 0] foreach row $m { for {set j 0} {$j < $cols} {incr j} { set s [format $fmt [lindex $row $j]] lset max $j [max [lindex $max $j] [string length $s]] } } return $max

}</lang> Demonstrating: <lang tcl># This does the decomposition and prints it out nicely proc demo {A} {

   lassign $A L U P
   foreach v {A L U P} {

upvar 0 $v matrix puts "${v}:" matrix::print $matrix %.5g if {$v ne "P"} {puts "---------------------------------"}

} demo {{1 3 5} {2 4 7} {1 1 0}} puts "=================================" demo {{11 9 24 2} {1 5 2 6} {3 17 18 1} {2 5 7 1}}</lang> Output:

A:
1 3 5 
2 4 7 
1 1 0 
---------------------------------
L:
  1  0 0 
0.5  1 0 
0.5 -1 1 
---------------------------------
U:
2 4   7 
0 1 1.5 
0 0  -2 
---------------------------------
P:
0 1 0 
1 0 0 
0 0 1 
=================================
A:
11  9 24 2 
 1  5  2 6 
 3 17 18 1 
 2  5  7 1 
---------------------------------
L:
       1       0         0 0 
 0.27273       1         0 0 
0.090909  0.2875         1 0 
 0.18182 0.23125 0.0035971 1 
---------------------------------
U:
11      9     24       2 
 0 14.545 11.455 0.45455 
 0      0 -3.475  5.6875 
 0      0      0 0.51079 
---------------------------------
P:
1 0 0 0 
0 0 1 0 
0 1 0 0 
0 0 0 1