Lu decomposition
This presentation is the property of its rightful owner.
Sponsored Links
1 / 16

LU Decomposition PowerPoint PPT Presentation


  • 63 Views
  • Uploaded on
  • Presentation posted in: General

LU Decomposition. Greg Beckham, Michael Sedivy. Overview. Step 1. This is handled implicitly in the code by only calculating the diagonal for β. Step 2. Calculating β ij. for(j = 0; j < n; j++) // This is the loop over columns of Crout's method {

Download Presentation

LU Decomposition

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Lu decomposition

LU Decomposition

Greg Beckham, Michael Sedivy


Overview

Overview


Step 1

Step 1

This is handled implicitly in the code by only calculating the diagonal for β


Step 2

Step 2


Calculating ij

Calculating βij

for(j = 0; j < n; j++) // This is the loop over columns of Crout's method

{

for(i = 0; i < j; i++) // Equation (2.3.12) except for i = j

{

sum = a[i][j];

for(k = 0; k < i; k++) sum -= a[i][k] * a[k][j];

a[i][j] = sum;

}

}


Calculating ij1

Calculating αij


Lu decomposition

for(j = 0; j < n; j++) // This is the loop over columns of Crout's method

{

for(i = j; i < n; i++)// This is i=j of equation (2.3.12) and i=j+1

{ // N-1 of equation (2.3.13)

sum = a[i][j];

for(k = 0; k < j; k++) sum -= a[i][k] * a[k][j];

a[i][j] = sum;

}

if(j != n - 1) // Divide by the pivot element

{

dum = 1.0/(a[j][j]);

for(i = j + 1; i < n; i++) a[i][j] *= dum;

}

}


Pivoting

Pivoting

  • Initially finds largest element in each row

  • Used as a “scaling factor”, not sure of use other than to rollover

for(i = 0; i < n; i++) // Loop over the rows to get implicit scaling

{ // information

big = 0.0;

for(j = 0; j < n; j++)

{

if((temp = fabs(a[i][j])) > big) big = temp;

}

if (big == 0.0)

{

printf("ERROR: Singular matrix\n");

}

// non-zero largest element.

vv[i] = 1.0/big; // Save the scaling

}


Pivoting1

Pivoting

  • Finds maximum

if((dum = vv[i] * fabs(sum)) >= big)

{

// Is the figure of merit for the pivot better than the best so far?

big = dum;

imax = i;

}


Pivoting2

Pivoting

  • Performs row interchanges

if(j != imax) // Do we need to interchange rows?

{

for(k = 0; k < n; k++) // Interchange rows

{

dum = a[imax][k];

a[imax][k] = a[j][k];

a[j][k] = dum;

}

d = -d; // change the parity of d

vv[imax] = vv[j]; // interchange scale factor

}

indx[j] = imax;


Related questions

Related Questions

  • What is the advantage of LU(P) solver over GJ(P) solver? (Complexity)

    • Both are O(N3)

    • After LU(P) is solved, more solutions supposed to be found in O (N2)

  • Are you keeping L and U in the same matrix, or separate? Advantage/disadvantage?

    • LU are being created in place in the same matrix.

    • The advantage to this strategy is lower memory usage

    • The disadvantage is that the original matrix is lost

  • I am somewhat confused with extraction of P in decomposition, and how it is then used in eq solving. Can you elaborate more?


Related questions1

Related Questions

  • Cormen et al., p 824, used a single array instead of P. Needs careful explanation.

    • From Cormen et al. 825. “we dynamically maintain the permutation matrix P as an array π, where π[i] = j means that the ith row of P contains a 1 in column j

      |2| | 0 1 0 |

      π = |3| => P | 0 0 1 |

      |1| | 1 0 0 |


Related questions2

Related Questions

  • How complex equations are solved? (in Text)

    • If only the right hand side vector is complex then the operation can be performed by solving for the real part, then the imaginary

    • If the matrix itself is complex then

      • Rewrite the algorithm for complex values

      • Split the real and imaginary parts into separate real number and solve using existing algorithm

        • A * x – C * y = b

        • C * x + A * y = b


Gj vs lup we found lup faster than gj but

GJ vs. LUP: we found lup faster than gj, but…


Gj vs lup lup not faster amortized

GJ vs. LUP: lup not faster amortized


Gj vs lup

GJ vs. LUP

Average Difference is 2.765471

Average Difference is 1.255924


  • Login