Skip to content

Fix docs/components #1288

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
furqan463 wants to merge 4 commits into
base: main
Choose a base branch
from
+176 −221
Open

Fix docs/components #1288

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
4 commits
Select commit Hold shift + click to select a range
c9c3e91
Fix components
furqan463 Jan 28, 2026
24b6cc9
Merge branch 'main' into fix_docs
furqan463 Feb 4, 2026
8fdb8e7
Fix calculations
furqan463 Feb 4, 2026
087cd59
Fix LU Solver
furqan463 Feb 4, 2026
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
72 changes: 23 additions & 49 deletions docs/algorithms/lu-solver.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -207,13 +207,11 @@ $\boldsymbol{l}_k$ and $\boldsymbol{u}_k$ ($k < p$) denote sections of the matri
[Gaussian elimination](https://en.wikipedia.org/wiki/Gaussian_elimination) using
[LU decomposition](https://en.wikipedia.org/wiki/LU_decomposition) constructs the matrices

$$
\begin{align*}
\mathbf{L}_p &= \begin{bmatrix} 1 && \boldsymbol{0}^T \\ m_p^{-1} \boldsymbol{q}_p && \mathbf{1}_p\end{bmatrix} \\
\mathbf{U}_p &= \begin{bmatrix} m_p && \boldsymbol{r}_p^T \\\boldsymbol{0} && \mathbf{1}_p\end{bmatrix} \\
\mathbf{M}_{p+1} &= \hat{\mathbf{M}}_p - m_p^{-1} \boldsymbol{q}_p \boldsymbol{r}_p^T
\end{align*}
$$

where $\mathbf{1}$ is the matrix with ones on the diagonal and zeros off-diagonal, and $\mathbf{M}_{p+1}$ is the start
of the next iteration.
Expand Down Expand Up @@ -253,7 +251,6 @@ dimensions $\left(N-p-1\right) \times \left(N-p-1\right)$.

Iterating this process yields the matrices

$$
\begin{align*}
\mathbf{L} = \begin{bmatrix}
1 && 0 && \cdots && 0 \\
Expand All @@ -278,7 +275,6 @@ m_0 && \left(\boldsymbol{r}_0^T\right)_0 && \cdots && \left(\boldsymbol{r}_0^T\r
0 && \cdots && 0 && m_{N-1}
\end{bmatrix}
\end{align*}
$$

in which $\boldsymbol{l}_p$ is the first column of the lower triangle of $\mathbf{L}_p$ and $\boldsymbol{u}_p^T$ is the
first row of the upper triangle of $\mathbf{U}_p$.
Expand Down Expand Up @@ -376,7 +372,6 @@ confusion with the block-sparse matrix components.
The matrices constructed with [LU decomposition](https://en.wikipedia.org/wiki/LU_decomposition) for
[Gaussian elimination](https://en.wikipedia.org/wiki/Gaussian_elimination) are constructed accordingly.

$$
\begin{align*}
\mathbf{L}_p &= \begin{bmatrix}
\mathbf{1} && \mathbf{0}^T \\
Expand All @@ -388,15 +383,13 @@ $$
\end{bmatrix} \\
\mathbf{M}_{p+1} &= \hat{\mathbf{M}}_p - \widehat{\mathbf{m}_p^{-1}\mathbf{q}_p \mathbf{r}_p^T}
\end{align*}
$$

Here, $\overrightarrow{\mathbf{m}_p^{-1}\mathbf{q}_p}$ is symbolic notation for the block-vector of solutions to the
equation $\mathbf{m}_p x_{p;k} = \mathbf{q}_{p;k}$, where $k = 0..(p-1)$.
Similarly, $\widehat{\mathbf{m}_p^{-1}\mathbf{q}_p \mathbf{r}_p^T}$ is symbolic notation for the block-matrix of
solutions to the equation $\mathbf{m}_p x_{p;k,l} = \mathbf{q}_{p;k} \mathbf{r}_{p;l}^T$, where $k,l = 0..(p-1)$.
That is:

$$
\begin{align*}
\overrightarrow{\mathbf{m}_p^{-1}\mathbf{q}_p}
&= \begin{bmatrix}\mathbf{m}_p^{-1}\mathbf{q}_{p;0} \\
Expand All @@ -409,7 +402,6 @@ $$
\mathbf{m}_p^{-1} \mathbf{q}_{p;0} \mathbf{r}_{p;0}^T && \cdots &&
\mathbf{m}_p^{-1} \mathbf{q}_{p;N-1} \mathbf{r}_{p;N-1}^T \end{bmatrix}
\end{align*}
$$

Iteratively applying above factorization process yields $\mathbf{L}$ and $\mathbf{U}$, as well as $\mathbf{P}$ and
$\mathbf{Q}$.
Expand All @@ -435,7 +427,6 @@ $\mathbf{m}_p$ is a dense block that can be [LU factorized](#dense-lu-factorizat
$\mathbf{m}_p = \mathbf{p}_p^{-1} \mathbf{l}_p \mathbf{u}_p \mathbf{q}_p^{-1}$.
Partial Gaussian elimination constructs the following matrices.

$$
\begin{align*}
\mathbf{L}_p &= \begin{bmatrix}
\mathbf{l}_p && \mathbf{0}^T \\
Expand All @@ -448,7 +439,6 @@ $$
\mathbf{M}_{p+1} &=
\hat{\mathbf{M}}_p - \widehat{\mathbf{q}_p\mathbf{q}_p\mathbf{u}_p^{-1}\mathbf{l}_p^{-1}\mathbf{p}_p\mathbf{r}_p^T}
\end{align*}
$$

Note that the first column of $\mathbf{L}_p$ can be obtained by applying a right-solve procedure, instead of the regular
left-solve procedure, as is the case for $\mathbf{U}_p$.
Expand All @@ -457,7 +447,6 @@ left-solve procedure, as is the case for $\mathbf{U}_p$.

To illustrate the rationale, let's fully solve a matrix equation (without using blocks):

$$
\begin{align*}
\begin{bmatrix}
\mathbf{a} && \mathbf{b} \\
Expand Down Expand Up @@ -511,12 +500,10 @@ a_{11} && a_{12} && b_{11} && b_{12} \\
\frac{c_{22} - a_{12} \frac{c_{21}}{a_{11}}}{a_{22} - a_{12} \frac{a_{21}}{a_{11}}}
\end{bmatrix}
\end{align*}
$$

Using the following denotations, we can simplify the above as
$\begin{bmatrix} \mathbf{l}_a \mathbf{u}_a && \mathbf{u}_b \\ \mathbf{l}_c && \mathbf{l}_d \mathbf{u}_d \end{bmatrix}$.

$$
\begin{align*}
\mathbf{l}_a &= \begin{bmatrix}
1 && 0 \\
Expand Down Expand Up @@ -549,7 +536,6 @@ $$
\frac{c_{22} - a_{12} \frac{c_{21}}{a_{11}}}{a_{22} - a_{12} \frac{a_{21}}{a_{11}}}
\end{bmatrix}
\end{align*}
$$

Interestingly, the matrices $\mathbf{l}_c$, $\mathbf{u}_b$ and $\mathbf{l}_d\mathbf{u}_d$ can be obtained without doing
full pivoting on the sub-block level:
Expand Down Expand Up @@ -582,7 +568,6 @@ The following proves the equations $\mathbf{l}_c \mathbf{u}_a = \mathbf{c}$, $\m
and $\mathbf{l}_d\mathbf{u}_d = \mathbf{d} - \mathbf{l}_c \mathbf{u}_b$ mentioned in the
[previous section](#rationale-of-the-block-sparse-lu-factorization-process).

$$
\begin{align*}
\mathbf{l}_c \mathbf{u}_a
&=
Expand Down Expand Up @@ -656,7 +641,6 @@ $$
\end{bmatrix} \\
&= \mathbf{l}_d\mathbf{u}_d
\end{align*}
$$

We can see that $\mathbf{l}_c$ and $\mathbf{u}_b$ are affected by the in-block LU decomposition of the pivot block
$\left(\mathbf{l}_a,\mathbf{u}_a\right)$, as well as the data in the respective blocks ($\mathbf{c}$ and $\mathbf{b}$)
Expand All @@ -674,7 +658,6 @@ The structure of the block-sparse matrices is as follows.

This can be graphically represented as

$$
\begin{align*}
\mathbf{M} &\equiv \begin{bmatrix}
\mathbf{M}_{0,0} && \cdots && \mathbf{M}_{0,N-1} \\
Expand All @@ -701,7 +684,6 @@ $$
\mathbf{M}_{N-1,N-1}\left[N_i-1,0\right] && \cdots && \mathbf{M}_{N-1,N-1}\left[N_i-1,N_j-1\right] \end{bmatrix}
\end{bmatrix}
\end{align*}
$$

Because of the sparse structure and the fact that all $\mathbf{M}_{i,j}$ have the same shape, it is much more efficient
to store the blocks $\mathbf{M}_{i,j}$ in a vector $\mathbf{M}_{\tilde{k}}$ where $\tilde{k}$ is a reordered index from
Expand All @@ -711,7 +693,6 @@ by the row-index $i$, and the inner vector containing the values of $j$ for whic
All topologically relevant matrix elements, as well as [fill-ins](#pivot-operations), are included in this mapping.
The following illustrates this mapping.

$$
\begin{align*}
\begin{bmatrix}
\mathbf{M}_{0,0} && && && \mathbf{M}_{0,3} \\
Expand Down Expand Up @@ -749,7 +730,6 @@ $$
[0 && && 2 && && 4 && && && 7 && && && && 10]
\end{bmatrix}
\end{align*}
$$

In the first equation, the upper row contains the present block entries and the bottom row their column indices per row
to obtain a flattened representation of the matrix.
Expand Down Expand Up @@ -875,17 +855,17 @@ as well as the well-known
Let $\mathbf{M}$ be the matrix, $\left\|\mathbf{M}\right\|_{\infty ,\text{bwod}}$ the
[block-wise off-diagonal infinite norm](#block-wise-off-diagonal-infinite-matrix-norm) of the matrix.

1. $\epsilon \gets \text{perturbation_threshold} * \left\|\mathbf{M}\right\|_{\text{bwod}}$.
2. If $|\text{pivot_element}| \lt \epsilon$, then:
1. If $|\text{pivot_element}| = 0$, then:
1. $\text{phase_shift} \gets 1$.
1. $\epsilon \gets \text{perturbation\_threshold} * \left\|\mathbf{M}\right\|_{\text{bwod}}$.
2. If $|\text{pivot\_element}| \lt \epsilon$, then:
1. If $|\text{pivot\_element}| = 0$, then:
1. $\text{phase\_shift} \gets 1$.
2. Proceed.
2. Else:
1. $\text{phase_shift} \gets \text{pivot_element} / |\text{pivot_element}|$.
1. $\text{phase\_shift} \gets \text{pivot\_element} / |\text{pivot\_element}|$.
2. Proceed.
3. $\text{pivot_element} \gets \epsilon * \text{phase_shift}$.
3. $\text{pivot\_element} \gets \epsilon * \text{phase\_shift}$.

$\text{phase_shift}$ ensures that the sign (if $\mathbf{M}$ is a real matrix) or complex phase (if $\mathbf{M}$ is a
$\text{phase\_shift}$ ensures that the sign (if $\mathbf{M}$ is a real matrix) or complex phase (if $\mathbf{M}$ is a
complex matrix) of the pivot element is preserved.
The positive real axis is used as a fallback when the pivot element is identically zero.

Expand Down Expand Up @@ -915,7 +895,7 @@ Solving for $\boldsymbol{\Delta x}$ and substituting back into
$\boldsymbol{x}_{i+1} = \boldsymbol{x}_i + \boldsymbol{\Delta x}$ provides the next best approximation
$\boldsymbol{x}_{i+1}$ for $\boldsymbol{x}$.

A measure for the quality of the approximation is given by the $\text{backward_error}$ (see also
A measure for the quality of the approximation is given by the $\text{backward\_error}$ (see also
[backward error formula](#backward-error-calculation)).

Since the matrix $\mathbf{M}$ remains static during this process, the LU decomposition is valid throughout the process.
Expand All @@ -928,11 +908,11 @@ algorithm is as follows:
1. Initialize:
1. Set the initial estimate: $\boldsymbol{x}_{\text{est}} = \boldsymbol{0}$.
2. Set the initial residual: $\boldsymbol{r} \gets \boldsymbol{b}$.
3. Set the initial backward error: $\text{backward_error} = \infty$.
3. Set the initial backward error: $\text{backward\_error} = \infty$.
4. Set the number of iterations to 0.
2. Iteratively refine; loop:
1. Check stop criteria:
1. If $\text{backward_error} \leq \epsilon$, then:
1. If $\text{backward\_error} \leq \epsilon$, then:
1. Convergence reached: stop the refinement process.
2. Else, if the number of iterations > maximum allowed amount of iterations, then:
1. Convergence not reached; iterative refinement not possible: raise a sparse matrix error.
Expand Down Expand Up @@ -967,21 +947,19 @@ We use the following backward error calculation, inspired by
[Li99](https://www.semanticscholar.org/paper/A-Scalable-Sparse-Direct-Solver-Using-Static-Li-Demmel/7ea1c3360826ad3996f387eeb6d70815e1eb3761),
with a few modifications described [below](#improved-backward-error-calculation).

$$
\begin{align*}
D_{\text{max}} &= \max_i\left\{
\left(\left|\mathbf{M}\right|\cdot\left|\boldsymbol{x}\right| + \left|\boldsymbol{b}\right|\right)_i
\right\} \\
\text{backward_error} &= \max_i \left\{
\text{backward\_error} &= \max_i \left\{
\frac{\left|\boldsymbol{r}\right|_i}{
\max\left\{
\left(\left|\mathbf{M}\right|\cdot\left|\boldsymbol{x}\right| + \left|\boldsymbol{b}\right|\right)_i,
\epsilon_{\text{backward_error}} D_{\text{max}}
\epsilon_{\text{backward\_error}} D_{\text{max}}
\right\}
}
\right\}
\end{align*}
$$

$\epsilon \in \left[0, 1\right]$ is a value that introduces a
[cut-off value to improve stability](#improved-backward-error-calculation) of the algorithm and should ideally be small.
Expand All @@ -1006,7 +984,7 @@ contains an early-out criterion for the [iterative refinement](#iterative-refine
for diminishing backward error in consecutive iterations.
It amounts to (in reverse order):

1. If $\text{backward_error} \gt \frac{1}{2}\text{last_backward_error}$, then:
1. If $\text{backward\_error} \gt \frac{1}{2}\text{last\_backward\_error}$, then:
1. Stop iterative refinement.
2. Else:
1. Go to next refinement iteration.
Expand All @@ -1033,9 +1011,8 @@ their sum is prone to rounding errors, which may be several orders larger than m
uses the following backward error in the
[iterative refinement algorithm](#iterative-refinement-of-lu-solver-solutions):

$$
\begin{align*}
\text{backward_error}_{\text{Li}}
\text{backward\_error}_{\text{Li}}
&= \max_i \frac{
\left|\boldsymbol{r}_i\right|
}{
Expand All @@ -1052,7 +1029,6 @@ $$
\left(\left|\mathbf{M}\right| \cdot \left|\boldsymbol{x}\right| + \left|\boldsymbol{b}\right|\right)_i
}
\end{align*}
$$

In this equation, the symbolic notation $\left|\mathbf{M}\right|$ and $\left|\boldsymbol{x}\right|$ are the matrix and
vector with absolute values of the elements of $\mathbf{M}$ and $\boldsymbol{x}$ as elements, i.e.,
Expand All @@ -1065,21 +1041,19 @@ iterative refinement to fail.
The power grid model therefore uses a modified version, in which the denominator is capped to a minimum value,
determined by the maximum across all denominators:

$$
\begin{align*}
D_{\text{max}} &= \max_i\left\{
\left(\left|\mathbf{M}\right|\cdot\left|\boldsymbol{x}\right| + \left|\boldsymbol{b}\right|\right)_i
\right\} \\
\text{backward_error} &= \max_i \left\{
\text{backward\_error} &= \max_i \left\{
\frac{\left|\boldsymbol{r}\right|_i}{
\max\left\{
\left(\left|\mathbf{M}\right|\cdot\left|\boldsymbol{x}\right| + \left|\boldsymbol{b}\right|\right)_i,
\epsilon_{\text{backward_error}} D_{\text{max}}
\epsilon_{\text{backward\_error}} D_{\text{max}}
\right\}
}
\right\}
\end{align*}
$$

$\epsilon$ may be chosen.
$\epsilon = 0$ means no cut-off, while $\epsilon = 1$ means that only the absolute values of the residuals are
Expand Down Expand Up @@ -1122,24 +1096,24 @@ with dimensions $N_i\times N_j$.

1. $\text{norm} \gets 0$.
2. Loop over all block-rows: $i = 0..(N-1)$:
1. $\text{row_norm} \gets 0$.
1. $\text{row\_norm} \gets 0$.
2. Loop over all block-columns: $j = 0..(N-1)$ (beware of sparse structure):
1. If $i = j$, then:
1. Skip this block: continue with the next block-column.
2. Else, calculate the $L_{\infty}$ norm of the current block and add to the current row norm:
1. the current block: $\mathbf{M}_{i,j} \gets \mathbf{M}\left[i,j\right]$.
2. $\text{block_norm} \gets 0$.
2. $\text{block\_norm} \gets 0$.
3. Loop over all rows of the current block: $k = 0..(N_{i,j} - 1)$:
1. $\text{block_row_norm} \gets 0$.
1. $\text{block\_row\_norm} \gets 0$.
2. Loop over all columns of the current block: $l = 0..(N_{i,j} - 1)$:
1. $\text{block_row_norm} \gets \text{block_row_norm} + \left\|\mathbf{M}_{i,j}\left[k,l\right]\right\|$.
1. $\text{block\_row\_norm} \gets \text{block\_row\_norm} + \left\|\mathbf{M}_{i,j}\left[k,l\right]\right\|$.
3. Calculate the new block norm: set
$\text{block_norm} \gets \max\left\{\text{block_norm}, \text{block_row_norm}\right\}$.
$\text{block\_norm} \gets \max\left\{\text{block\_norm}, \text{block\_row\_norm}\right\}$.
4. Continue with the next row of the current block.
4. $\text{row_norm} \gets \text{row_norm} + \text{block_norm}$.
4. $\text{row\_norm} \gets \text{row\_norm} + \text{block\_norm}$.
5. Continue with the next block-column.
3. Calculate the new norm: set
$\text{norm} \gets \max\left\{\text{norm}, \text{row_norm}\right\}$.
$\text{norm} \gets \max\left\{\text{norm}, \text{row\_norm}\right\}$.
4. Continue with the next block-row.

##### Illustration of the block-wise off-diagonal infinite matrix norm calculation
Expand Down
Loading